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Long time ago in a land far away system wrote: Hi all this is my riddle: "A bus rider is going down the street. He arrives at a crossing where he must turn right or continue forward, but he turns left. He does all this infront of a police car and they do nothing. Why?" Hint: + Show Spoiler +
Riddle:"A bus rider is going down the street. He arrives at a crossing where he must turn right or continue forward, but he turns left. He does all this infront of a police car and they do nothing. Why?"
HAHAH, awesome riddle. It was easy. Here is the answere: + Show Spoiler +
Hi all this is my riddle (decode this): "TLTLLTTLTLLTLLLLTLLLTLTLTTLTTTTTTLTLTTLLTLLTLLLLTLLTLLTTTLLLTLLTTLLTTLTLTLLTTLTTTTLTTTTTTLTLTLTTTLLTLTTTTLLTTLTLTTLTTTTTTLTLTTLTTLLTLTTLTLLTTLTTTLLTTLTTTLLTLLTTTLLTTLTLTTLTTTTL" Hint 1: + Show Spoiler +
Ok, here is how to solve this riddle: 1. You open 2 text editors. 2. Paste the "TLTLL..." into both of them, 3. In one text editor change all the "T"s to "1" and the "L"s to "0" and in the text editor the other way around. 4.Paste the two different combinations into a Binary to Text convertor like this one (http://www.roubaixinteractive.com/PlayGround/Binary_Conversion/Binary_To_Text.asp) and decode the message 5. "You Solved The Riddle!" Is what you are supposed to get as an answere with one of them.
Good Luck
Please tell me if you solved it, this is the hardes riddle I know.
You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?
Second Problem: You need exactly four gallons. How do you do it?
On April 18 2012 23:02 kochanfe wrote: You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?
Second Problem: You need exactly four gallons. How do you do it?
Fill up the three gallon with water, pour over to the five gallon jug. Now fill the three gallon jug again, and pour into the five gallon until you can't fit any more, then pour the five gallons out. Now you have one gallon of water in the three gallon one, and the other one is empty.
Pour the one gallon of water over to the five gallon jug, then fill the three gallon jug and pour over to the five jug again. Then fill the three gallon again. Now you have four gallons of water in the five gallon jug, and a full three gallon jug.
His nose would grow, as yours doesn't have the ability to grow (or so I suppose). It is ambiguous as to whether "my" refers to Pinoccios or yours. If it refer's to Pinoccio, then we can't actually know.
On April 18 2012 23:02 kochanfe wrote: You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?
Second Problem: You need exactly four gallons. How do you do it?
Seven Gallons: Fill 5 gallon jug. 5 Gallons / 0 Gallons Pour 5 gallon jug into 3 gallon jug. 2 gallons / 3 gallons Pour out 3 gallon jug and pour rest of 5 gallon jug into 3 gallons. 0 gallons / 2 gallons Fill 5 gallon jug. 5 gallons / 2 gallons for total of 7 between the 2.
Four gallons: Fill 3 gallon jug and pour into 5 gallon jug. 3 gallon / 0 gallon. Fill 3 gallon jug and pour into 5 gallon jug to fill. 5 gallon / 1 gallon. Empty 5 gallon jug and pour 3 gallon jug into 5 gallon jug. 1 gallon / 0 gallon. Fill 3 gallon jug. 1 gallon / 3 gallon for total of 4 between 2.
On April 18 2012 23:02 kochanfe wrote: You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?
Second Problem: You need exactly four gallons. How do you do it?
1)Fill the 3 gallon Jug 2) Pour the 3 gal into the 5 gal 3) Repeat until the 5gal is full 4) Empty the 5 gal and pour the 1gal left in the 3gal into the 5gal jug. 5) Fill 3 gal jug 6) 4 Gals
1)Fill the 3 gallon Jug 2) Pour the 3 gal into the 5 gal 3) Repeat until the 5gal is full 4) Empty the 5 gal and pour the 1gal left in the 3gal into the 5gal jug. 5) Repeat 3 more times. 6) Fill 3 gal jug 7) 7gals
You actually messed up the riddle. It's supposed to be *his* nose growing, not yours. That would be the paradox or impossibility or back and forth nose growth lol.
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
^ I'll let someone do the liar puzzle that hasn't heard it before.
Here's one from me.
A man is lying dead in the middle of a desert. He is entirely naked and shows no signs of violence, such as stabbing wounds, strangling, bullet wounds etc. The sands of the desert have not been disturbed by weather in a long time, and tracks would show for a long while. Yet, there are none that lead to or away from the body. The only object with him is a broken match stick, held tightly in his grip. How did he die?
He was traveling with a companion in a hot air balloon, when the balloon started losing altitude. They started throwing off everything with weigh, including their clothes, but it wasn't enough. They then drew short straws, and the one with the broken match stick had to jump off the balloon, to his death, saving the other companion.
On April 18 2012 23:26 mentallyafk wrote: you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
"Wich door would the other guard say hides the money?" Both guards will answer the same door, the one with the lion behind it, so you take the other one!
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
You ask "What would your brother say if I asked him behind which door the million dollars are?".
You will get the door where the dragon is.
you are presented with 3 doors and 3 gaurds. same story, 1 door has money, other 2 dont. 1 guard always tells the truth and the other 2 always lie. what do you ask?
You are presented with 3 doors and 3 gaurds. same story, 1 door has money, other 2 dont. 1 guard always tells the truth and the other 2 always lie. what do you ask?
On April 18 2012 23:26 mentallyafk wrote: you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
You are presented with 3 doors and 3 gaurds. same story, 1 door has money, other 2 dont. 1 guard always tells the truth and the other 2 always lie. what do you ask?
On April 18 2012 23:26 mentallyafk wrote: you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
On April 18 2012 23:26 mentallyafk wrote: you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
You count the places where there is a circular shape in the number and you add them.
I would probably have never figure it out if I did not write down the riddle on a paper, then I got bored and started coloring the circular shapes and that is when I saw it. I LOVE TO SCRIBLE
Lol GR final solution is 2.
EDIT: accidently wrote 45 minutes except 35minutes. Typo
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
On April 18 2012 23:26 mentallyafk wrote: you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
If its not yes/no questions & they're not colorblind: "What color is my shirt?..."
you can only ask one question
Sorry I forgot to add the "not" check now. It was a typo. Sorry
i'm saying that you can only ask one question. so after you ask that question you can't ask any more questions. so you still wouldn't know which door to open
A female friend of mine showed it to me. I figured out that each digit is replaced by a number and then summed to obtain the number on the right. I then computed that value for all the digits and barely managed to find the answer in under 10 minutes.
On April 18 2012 23:26 mentallyafk wrote: you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
If its not yes/no questions & they're not colorblind: "What color is my shirt?..."
you can only ask one question
Sorry I forgot to add the "not" check now. It was a typo. Sorry
i'm saying that you can only ask one question. so after you ask that question you can't ask any more questions. so you still wouldn't know which door to open
Sorry, fail would you give the answere in a spoiler please. Thanks Please dont say that it is accually impossible.
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
A female friend of mine showed it to me. I figured out that each digit is replaced by a number and then summed to obtain the number on the right. I then computed that value for all the digits and barely managed to find the answer in under 10 minutes.
Then she goes like "YOU COUNT THE CIRCLES OMG".
Stupid problem.
yea, it would be better if they left the number 5 out or something like that.
On April 18 2012 23:26 mentallyafk wrote: you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
If its not yes/no questions & they're not colorblind: "What color is my shirt?..."
you can only ask one question
Sorry I forgot to add the "not" check now. It was a typo. Sorry
i'm saying that you can only ask one question. so after you ask that question you can't ask any more questions. so you still wouldn't know which door to open
Sorry, fail would you give the answere in a spoiler please. Thanks Please dont say that it is accually impossible.
its not impossible. people have solved it earlier in the thread.
A female friend of mine showed it to me. I figured out that each digit is replaced by a number and then summed to obtain the number on the right. I then computed that value for all the digits and barely managed to find the answer in under 10 minutes.
Then she goes like "YOU COUNT THE CIRCLES OMG".
Stupid problem.
yea, it would be better if they left the number 5 out or something like that.
I was solving it and my sister knew the riddle allitle bit different. She knew it with the borthers and a road split she gave me hints, dont ask me how I got the dragon.
You *read off* the previous row, matching numbers that are alike and together in the preceding row, to come up with the next row. You say how many of each number there are, in order.
Row two: Row one is "one one". Row three: Row two is "two ones" Row four: Row three is "one two, one one". Row five: Row four is "one one, one two, two ones". Therefore, Row six: Row five is "three ones, two twos, one one", or 312211.
Riddle 2:
A farmer has three piles of straw in his front yard, five piles of straw in his backyard, and just returned home with two more piles of straw in his truck. How many piles of straw would he have if he put them all together?
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man say to them?
I saw this one in the blog section a few days ago. I read the solution but I still dont get it.
The other two I laughed.
Second one:
A farmer has three piles of straw in his front yard, five piles of straw in his backyard, and just returned home with two more piles of straw in his truck. How many piles of straw would he have if he put them all together?
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man say to them?
That one has many possibilites, but these two I thought off:
Don't know if you know this - but a teacher told it to me 7 years ago. So it's probably ancient. So here goes: (it was another currency but in order not to confuse too much, let's use dollars) The riddle: + Show Spoiler +
3 men enter a bar. They order 3 beers that all cost 10$, so the men give the waitress 30$. However when she goes up to the bartender and tells him to get 3 beers and hands over the 30$ he says: Oh no, we have a special offer with 3 beers for 25$. So now the waitress has to go back with 5$. But you can't split 5$ into 3. So she takes 2$ for herself and hands the gentlemen their beers and 1$ each.
So this adds up. Doesn't it? Or wait... Now they all paid 9$ - so 9$ * 3 = 27$... And the waitress took 2$ for herself. That's 29$. Where did the last dollar go?
There is no missing dollar. The men were to pay 25$ and the waitress took 2$ for herself. Meaning the men paying 27$ is the correct amount. It's a common deception to just add up all the money instead of realizing the actual numbers are from different points of view.
A family inherit 6 million dollars from a friend of the family who has passed away. The family consist of two fathers and two sons. The family decides to spilt the money so that they each get 2 millions. How is that possible?
3 men enter a bar. They order 3 beers that all cost 10$, so the men give the waitress 30$. However when she goes up to the bartender and tells him to get 3 beers and hands over the 30$ he says: Oh no, we have a special offer with 3 beers for 25$. So now the waitress has to go back with 5$. But you can't split 5$ into 3. So she takes 2$ for herself and hands the gentlemen their beers and 1$ each.
This seemd easy, but I might be wrong. Here is my solution: + Show Spoiler +
30=25+2+1+1+1
30 = The original ammount 25 = They cost 2 = She keept 1 = First guy 1 = Second guy 1 = Third guy
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
You ask "What would your brother say if I asked him behind which door the million dollars are?".
You will get the door where the dragon is.
Eh the first time I heard it I solved it in about 10-15 minutes or so. I didn't think it was a hard riddle but I had trouble thinking of the perfect question to ask. I'm not the brightest guy around it's not hard to believe someone else figured it out in 2 minutes.
For the question of counting I must confess it was rather fast. The 2 guards question is in pretty much in every fantasy/archaelogy children movie/series on the planet. The answer should be:
If I ask the truth guard and the door on the right was the money one, he would say its the left one, if the money door was on the left he would say the right one. If I asked the Lie guard and the money door was on the right he would say the left one, if the money door was on the right he would say the left one. In all these cases they all point towards the Lion Door, which means the other one is the good one.
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
You ask "What would your brother say if I asked him behind which door the million dollars are?".
You will get the door where the dragon is.
Eh the first time I heard it I solved it in about 10-15 minutes or so. I didn't think it was a hard riddle but I had trouble thinking of the perfect question to ask. I'm not the brightest guy around it's not hard to believe someone else figured it out in 2 minutes.
Check the riddle and then his answer, guards magically switched to brothers and lion turned into a dragon?
A family inherit 6 million dollars from a friend of the family who has passed away. The family consist of two fathers and two sons. The family decides to spilt the money so that they each get 2 millions. How is that possible?
Solution: inspiration: song named I am my own grand pa
Ok, here I go pay carefull atention.
I mary a widdow with a grown up daugher so I have a daugher in law. My grown up daughter marries my father so she is my mother in law. I am the father of my daugher in law so I am my fathers father in law. Becuase I am my fathers father I am my own grand pa. And my wife and daugher in law die.
So there are acctully only 2 of them so they split the 6 million dollars in two and they each get 3 million dollars.
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
You ask "What would your brother say if I asked him behind which door the million dollars are?".
You will get the door where the dragon is.
you are presented with 3 doors and 3 gaurds. same story, 1 door has money, other 2 dont. 1 guard always tells the truth and the other 2 always lie. what do you ask?
For reference sake the doors are labeled 1-3 and the million dollars is behind door 1. You ask them what door the person to their right would go through. The truthful person and 1 liar will say "They would go through either door 2 or 3". 1 liar would say door 1. Believe the person that lists only a single door.
3 men enter a bar. They order 3 beers that all cost 10$, so the men give the waitress 30$. However when she goes up to the bartender and tells him to get 3 beers and hands over the 30$ he says: Oh no, we have a special offer with 3 beers for 25$. So now the waitress has to go back with 5$. But you can't split 5$ into 3. So she takes 2$ for herself and hands the gentlemen their beers and 1$ each.
This seemd easy, but I might be wrong. Here is my solution: + Show Spoiler +
30=25+2+1+1+1
30 = The original ammount 25 = They cost 2 = She keept 1 = First guy 1 = Second guy 1 = Third guy
the point is you shouldn't actually calculate anything with 30. You should say: Actual price = 25 Their price = 27 Waitress = 2
Basicly 25+2 is all there is to say. I'm not saying you're wrong at all. You're obviously not. But you pretty much just summed it up
Alternatively you could say 27 (the 30 - the three they god back) - the 2 she took would be 25. However this is where alot of people get confused. That's the tricky part. When you have 27 but the waitress has 2 in her pockets. Doesn't that add up to 29? + Show Spoiler +
Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Ok you work in a gay bar, then 4 new gays come to the bar but you have only one chair left. How do you sit them on the chair makng sure that all of them are happy.
On April 19 2012 01:01 Rob28 wrote: Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
You switch on switch #1. You wait for 10 minutes. You switch it off and switch on switch #2. You never touch switch #3. You enter the room. The lit bulb is switch #2. You touch the remaining bulbs. The hot one is switch #1, the cold one is switch #3.
On April 19 2012 01:01 Rob28 wrote: Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Turn on switch 1 for 30 mins, then turn off. Turn on switch 2 Enter room The light bulb that is on is from switch 2, the off one is either switch 1 or 3, the one that is hot to the touch is from switch 1.
A family inherit 6 million dollars from a friend of the family who has passed away. The family consist of two fathers and two sons. The family decides to spilt the money so that they each get 2 millions. How is that possible?
Solution: Easier solution - the family consists of three people; a grandfather, a father, and a son. The grandfather and father are both fathers, while the son is not. The son and father are both sons, while the grandfather is not. 6 million divided by 3 is 2 million.
On April 19 2012 01:01 Rob28 wrote: Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Leave one switch off for the whole time, and turn two on. After a long time, turn one of the two "on" switches off. Then leave and check out the bulbs. One will be off and cool (corresponding to the switch you never touched), one will be off and warm (corresponding to the switch you left on for a while and recently turned off), and one will still be on
Ok you work in a gay bar, then 4 new gays come to the bar but you have only one chair left. How do you sit them on the chair makng sure that all of them are happy.
Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
What year are you in. If around 2000 this is my guess:
You take on switch and repidetly turn it on and off. In my experience this would cuase teh light builbs to explode. Then you turn on one other light and go into the other room. The one that is on is the one that you turned on, if you did well there should be on that is smelly and black thats the one you exploded, if you are lucky the glass might of have even shattered.
You *read off* the previous row, matching numbers that are alike and together in the preceding row, to come up with the next row. You say how many of each number there are, in order.
Row two: Row one is "one one". Row three: Row two is "two ones" Row four: Row three is "one two, one one". Row five: Row four is "one one, one two, two ones". Therefore, Row six: Row five is "three ones, two twos, one one", or 312211.
I'd say that riddle is ambiguous since the following solution also works: + Show Spoiler +
Arrived at by the following. Apply 1 of the following rules listed in order of precedence with tokens processed left to right: - for a pair of consecutive #s add 1 to the leftmost - otherwise add a 1 to the left of that #
Though of course the listed solution is more riddle-esque.
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
On April 19 2012 01:01 Rob28 wrote: Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Will post solution after seeing some guesses.
this is really old. but there are a lot of solutions to it:
As I was going to St Ives I met a man with seven wives Every wife had seven sacks Every sack had seven cats Every cat had seven kits Kits, cats, sacks, wives. How many were going to St Ives?
On April 19 2012 01:19 SomniGiggles wrote: As I was going to St Ives I met a man with seven wives Every wife had seven sacks Every sack had seven cats Every cat had seven kits Kits, cats, sacks, wives. How many were going to St Ives?
It seems that the other triangle is not as high as the other one so I guess that there the 1 blank sport comes form. Only a guess, I work on a very old small dark screen so I might see it wrong I go and check once more
On April 19 2012 01:19 SomniGiggles wrote: As I was going to St Ives I met a man with seven wives Every wife had seven sacks Every sack had seven cats Every cat had seven kits Kits, cats, sacks, wives. How many were going to St Ives?
On April 19 2012 01:01 Rob28 wrote: Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Will post solution after seeing some guesses.
this is really old. but there are a lot of solutions to it:
On April 19 2012 01:01 Rob28 wrote: Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Will post solution after seeing some guesses.
this is really old. but there are a lot of solutions to it:
Correct. The easiest way to see it is to look at the 2nd bottom row, where the 'hypotenuse' is (ie left most of 2nd row). One image has a complete square there, one has the top corner slightly clipped by white space. That's how I convinced myself anyway.
It seems that the other triangle is not as high as the other one so I guess that there the 1 blank sport comes form. Only a guess, I work on a very old small dark screen so I might see it wrong I go and check once more
thats basically right. The issue is that since the triangles have different ratios of height to length they don't form a true triangle. so the hypotenuse of the first is slightly 'bulged' outwards and the second is inwards. its just hard to discern that visually.
edit: nm I misread your guess, w/out relating the height and width of the 2 triangle pieces that isn't quite an acceptable solution.
On April 19 2012 01:01 Rob28 wrote: Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Will post solution after seeing some guesses.
this is really old. but there are a lot of solutions to it:
Correct. The easiest way to see it is to look at the 2nd bottom row, where the 'hypotenuse' is (ie left most of 2nd row). One image has a complete square there, one has the top corner slightly clipped by white space. That's how I convinced myself anyway.
Well thanks to the grid, you see quickly that triangles don't have the same angles. It would be much much harder if instead of the grid it was simply saying that both shapes are 13 cm wide and 5 cm high.
On April 19 2012 01:01 Rob28 wrote: Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Will post solution after seeing some guesses.
this is really old. but there are a lot of solutions to it:
Correct. The easiest way to see it is to look at the 2nd bottom row, where the 'hypotenuse' is (ie left most of 2nd row). One image has a complete square there, one has the top corner slightly clipped by white space. That's how I convinced myself anyway.
Well thanks to the grid, you see quickly that triangles don't have the same angles. It would be much much harder if instead of the grid it was simply saying that both shapes are 13 cm wide and 5 cm high.
2, but didnt notice it has anything to do with circles, i simply noticed the numbers have a "weight" most of them 0, so i went on and discovered 6, 9, 0 had a "weight" of one, and 8 has "2".
On April 19 2012 00:34 Mentalizor wrote: Don't know if you know this - but a teacher told it to me 7 years ago. So it's probably ancient. So here goes: (it was another currency but in order not to confuse too much, let's use dollars) The riddle: + Show Spoiler +
3 men enter a bar. They order 3 beers that all cost 10$, so the men give the waitress 30$. However when she goes up to the bartender and tells him to get 3 beers and hands over the 30$ he says: Oh no, we have a special offer with 3 beers for 25$. So now the waitress has to go back with 5$. But you can't split 5$ into 3. So she takes 2$ for herself and hands the gentlemen their beers and 1$ each.
So this adds up. Doesn't it? Or wait... Now they all paid 9$ - so 9$ * 3 = 27$... And the waitress took 2$ for herself. That's 29$. Where did the last dollar go?
There is no missing dollar. The men were to pay 25$ and the waitress took 2$ for herself. Meaning the men paying 27$ is the correct amount. It's a common deception to just add up all the money instead of realizing the actual numbers are from different points of view.
On April 19 2012 01:01 Rob28 wrote: Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Will post solution after seeing some guesses.
this is really old. but there are a lot of solutions to it:
You can solve this by looking closely. Instead of thinking of the triangles as replacing each other, think of them as being extended to fit the description.
The sizes are 5x2 and 8x3.
So to replace the smaller triangle by the bigger one, you need to extend it by one vertical box, and 2 horizontal boxes. However, we see that if we extend it, it doesnt fit in with the grid.
Basically, while the two triangles seem to form a straight line to the naked eye, they aren't actually straight. The smaller triangle has a slightly higher angle ( i.e, the hypotenuse is closer to the horz line ) and reverse for the bigger triangle. So, in the second case, the paritions are eating some space outside of what the first diagram did. Because the volume doesn't change, it is compensated by a "hole" inside.
On April 19 2012 01:01 Rob28 wrote: Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Will post solution after seeing some guesses.
this is really old. but there are a lot of solutions to it:
You can solve this by looking closely. Instead of thinking of the triangles as replacing each other, think of them as being extended to fit the description.
The sizes are 5x2 and 8x3.
So to replace the smaller triangle by the bigger one, you need to extend it by one vertical box, and 2 horizontal boxes. However, we see that if we extend it, it doesnt fit in with the grid.
Basically, while the two triangles seem to form a straight line to the naked eye, they aren't actually straight. The smaller triangle has a slightly higher angle ( i.e, the hypotenuse is closer to the horz line ) and reverse for the bigger triangle. So, in the second case, the paritions are eating some space outside of what the first diagram did. Because the volume doesn't change, it is compensated by a "hole" inside.
And this is why I think it's a horrible riddle. Since it's not about thinking of a smart solution to solve it. The maker of the riddle intentionally deceives you with something very small and you just need to be very observant to notice.
On April 19 2012 01:01 Rob28 wrote: Riddle: You are in a room with three light switches, each of which connects to a lightbulb in an adjacent room that you cannot see into. You do not know which switch goes with which bulb. You may only leave the switch room and can never re-enter it. How do you tell which switch controls which bulb?
Non-spoilered kinda hint: This puzzle involves no clever wordplay in it's description that is indicative of the solution. So don't try to solve it by reading too much into it.
Will post solution after seeing some guesses.
this is really old. but there are a lot of solutions to it:
You can solve this by looking closely. Instead of thinking of the triangles as replacing each other, think of them as being extended to fit the description.
The sizes are 5x2 and 8x3.
So to replace the smaller triangle by the bigger one, you need to extend it by one vertical box, and 2 horizontal boxes. However, we see that if we extend it, it doesnt fit in with the grid.
Basically, while the two triangles seem to form a straight line to the naked eye, they aren't actually straight. The smaller triangle has a slightly higher angle ( i.e, the hypotenuse is closer to the horz line ) and reverse for the bigger triangle. So, in the second case, the paritions are eating some space outside of what the first diagram did. Because the volume doesn't change, it is compensated by a "hole" inside.
And this is why I think it's a horrible riddle. Since it's not about thinking of a smart solution to solve it. The maker of the riddle intentionally deceives you with something very small and you just need to be very observant to notice.
Its not the best but as mentioned by another poster, the answer doesn't hinge entirely on visual observation. However, a bit of math background is required which is why this is usually found in geometry books or the like (at least thats where I recall seeing it o so long ago).
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man say to them?
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man say to them?
Is it just me or this riddle doesn't make sense. Something is up with the structure.
It's like saying:
A single mom has two kids, a boy and a girl. The mother always had a fancy for cake but she never wore blue coloured coveralls.
How old is the bus driver?
I don't see what the problem is, but maybe I can clarify: the sheikh has decided to give his fortune to exactly one of his two sons. Who he gives it to will be decided by a competition between the two boys. The competition is a race on camels, but it's not who can get to the city first... it's whoever's camel is slower will be the victor. The two sons must figure out a way to decide that out, as each one wants to obtain the fortune.
2, but didnt notice it has anything to do with circles, i simply noticed the numbers have a "weight" most of them 0, so i went on and discovered 6, 9, 0 had a "weight" of one, and 8 has "2".
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man say to them?
Is it just me or this riddle doesn't make sense. Something is up with the structure.
It's like saying:
A single mom has two kids, a boy and a girl. The mother always had a fancy for cake but she never wore blue coloured coveralls.
How old is the bus driver?
I don't see what the problem is, but maybe I can clarify: the sheikh has decided to give his fortune to exactly one of his two sons. Who he gives it to will be decided by a competition between the two boys. The competition is a race on camels, but it's not who can get to the city first... it's whoever's camel is slower will be the victor. The two sons must figure out a way to decide that out, as each one wants to obtain the fortune.
Ahh ok, I didn't get the part about the slowest camel.
2, but didnt notice it has anything to do with circles, i simply noticed the numbers have a "weight" most of them 0, so i went on and discovered 6, 9, 0 had a "weight" of one, and 8 has "2".
Wow... dude I like your way much better : )
That's cool Like a whole slew of linear equations, where each digit from 0-9 can be replaced with a variable ^^
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man say to them?
Is it just me or this riddle doesn't make sense. Something is up with the structure.
It's like saying:
A single mom has two kids, a boy and a girl. The mother always had a fancy for cake but she never wore blue coloured coveralls.
How old is the bus driver?
I don't see what the problem is, but maybe I can clarify: the sheikh has decided to give his fortune to exactly one of his two sons. Who he gives it to will be decided by a competition between the two boys. The competition is a race on camels, but it's not who can get to the city first... it's whoever's camel is slower will be the victor. The two sons must figure out a way to decide that out, as each one wants to obtain the fortune.
Ahh ok, I didn't get the part about the slowest camel.
If we switch camels, then I'm on your camel and you're on mine. If I force your camel to go its top speed and you force mine to go its top speed, and if I win the race, then my camel (the one you're riding on) is slower, meaning I win the fortune.
On April 18 2012 23:02 kochanfe wrote: You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?
Second Problem: You need exactly four gallons. How do you do it?
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
The 200 people don't know there are 200 people with green eyes and blue eyes divided equally?
edit1:Can people communicate with the captain other than stating the color of their eyes?
edit2: Are all the people near each other? Are they separated? Can every person meet the 199 other people?
On an island, there has to be some form of water they can see their reflection in and know their own eye color.
edit: also, if you say your eye color is blue, just guessing, and the captain doesnt let you leave, then you know you have green eyes and can leave the next time.
Pretty sure 100 people leave the island by just saying green. Because they have evidence that shows they could. And there is and will nvr be anymore certainty available to any of them. Poor blue eyes .
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important note: They are all perfect logicians.
The original, or at least from where it became somewhat famous can be found here, it's more fleshed out (might be helpful considering the difficulty).
The first day everyone knows that if there was only one person with green eyes then they would leave. Since nobody leaves they know that there are at least 2 people with green eyes.
Now the second day happens and everyone knows that if there are only 2 people with green eyes then they will leave since they will see only one other person with green eyes and will know that they can leave. Since they don't leave, now there must be three people with green eyes.
This process continues until the 100 green eyed people leave on the 100th night.
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
I think we're missing some information here, can everybody see every one elses eye color? if so, the message on day 1 is irrelevant, because everyone will see at least 99 people with green eyes. Also whats the punishment for answering wrong? Everyone could just guess state "blue" as their eye color on day 1, and then "green" on day 2
The first day everyone knows that if there was only one person with green eyes then they would leave. Since nobody leaves they know that there are at least 2 people with green eyes.
Now the second day happens and everyone knows that if there are only 2 people with green eyes then they will leave since they will see only one other person with green eyes and will know that they can leave. Since they don't leave, now there must be three people with green eyes.
This process continues until the 100 green eyed people leave on the 100th night.
I don't understand why it has to be that complicated. The blue-eyed people are screwed because they will nvr be able to know they have blue eyes so in a logical manner everyone would just say they have green eyes because they know its a possibility.
Now if there was a possibility of more information becoming available or something then sure no need to bother risking your chance to get off the island.
I think its possible that 100 people leave on 1 day and 100 are trapped forever since in this scenario they seem to be trapped regardless.
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important note: They are all perfect logicians.
I thought this one led to a paradox ? Or at least there are two solutions and both are disputable.
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important note: They are all perfect logicians.
The original, or at least from where it became somewhat famous can be found here, it's more fleshed out (might be helpful considering the difficulty).
Yeah, that's how I heard it once before.
My math friends and I went back and forth with brain teasers until I was given this one, and we worked it out using a much simpler case (like 1 blue and 1 green first, then moving upwards).
Also, I think it's important to note that the inhabitants don't know how many people of each eye color there are. Otherwise, you could just line everyone up and count, and figure out which color you should be (as one color would be one short).
The first day everyone knows that if there was only one person with green eyes then they would leave. Since nobody leaves they know that there are at least 2 people with green eyes.
Now the second day happens and everyone knows that if there are only 2 people with green eyes then they will leave since they will see only one other person with green eyes and will know that they can leave. Since they don't leave, now there must be three people with green eyes.
This process continues until the 100 green eyed people leave on the 100th night.
Also this doesn't make sense because everyone knows how many others have green/blue eyes. Why would it take 100 days?
Something seems just not right with this answer srry...
2, but didnt notice it has anything to do with circles, i simply noticed the numbers have a "weight" most of them 0, so i went on and discovered 6, 9, 0 had a "weight" of one, and 8 has "2".
Wow... dude I like your way much better : )
I feel smart since I figured it out in 5-10 minutes but then I realised that I my mind is comparable to a 5 year old. Should I be proud or not... hmmm
On April 18 2012 23:26 mentallyafk wrote: you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
What will the other guard say if I ask him which door has the lion ? Go through that door. The idea is you are removing the uncertainity as you are sure there is one lie and one truth, so you negate the lie and get the answer.
Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals. The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live. The chief agrees to give them one chance, but to live they have to solve a riddle. He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that: The man in front can't see any of the hats The man in the middle can see the man in front's hat And the man in the back can see the hats of the two others. None of them can see the color of their own hat. The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed. Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
On April 19 2012 03:52 Bubris wrote: Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals. The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live. The chief agrees to give them one chance, but to live they have to solve a riddle. He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that: The man in front can't see any of the hats The man in the middle can see the man in front's hat And the man in the back can see the hats of the two others. None of them can see the color of their own hat. The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed. Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
What color is his hat?
I actually had my geometry class reenact a similar problem to this while I was student teaching and I was demonstrating logic and reasoning. They totally loved it
I would look at pinochio's nose as a state based affect. When he says that his nose will now grow he is lying because nothing except him lying would have that affect. Because he is lying his nose would grow. At this point you would check to see if the parameters are still met 1- he lied and his nose grew and 2- he told the truth and his nose shrunk. If we assume a priority system then he would first increase in size and then decrease in size. Regardless I think this would only happen once. So there would be a growth and a decrease in size and then nothing would change.
On April 18 2012 23:02 kochanfe wrote: You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?
Second Problem: You need exactly four gallons. How do you do it?
Fill the 5 gallon jug. Use the 5 gallon jug to fill the 3 gallon jug so you have 2 gallons left over. dump the 3 gallon jug. fill the 3 gallon jug w/ the 2 gallons from the 5 gallon jug. fill the 5 gallon jug.
Part 2- fill the 5 gallon jug. dump 3 gallons from the 5 gallon jug into the 3 gallon jug. dump the 3 gallon jug, fill the 3 gallon jug w/ 2 gallons from the 5 gallon jug. fill the 5 gallon jug. dump 1 gallon from the 5 gallon jug into the 3 gallon jug and you have 4 gallons left in the 5 gallon jug
On April 18 2012 23:26 mentallyafk wrote: you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
"What would the other guard choose" and then go for the opposite. the liar would say that the truthful guard would always choose the lion door while the truthful guy would always say what the liar would say, ie choose the lion door. If you choose the opposite you will be rich
On April 19 2012 00:22 DarkPlasmaBall wrote:
Riddle 3:
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man say to them?
I don't get why the camels would make a difference for each son. I would think that the wise man would have said that it doesn't matter how long it takes to finish but the sons may die otw.
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain.
Day 1: There is at least one person that has green eyes.
Start from base case-- 1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes.
Though not sure if correct.
Edit: hmm I feel slightly better after reading other posts in this thread about this.
On April 19 2012 03:52 Bubris wrote: Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals. The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live. The chief agrees to give them one chance, but to live they have to solve a riddle. He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that: The man in front can't see any of the hats The man in the middle can see the man in front's hat And the man in the back can see the hats of the two others. None of them can see the color of their own hat. The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed. Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
If the first two were both wearing hats of colour black, the third would have known that he is wearing red. So this possibility is ruled out. This means at least one of the first 2 is wearing a hat of colour red.
Deducing this, if the first person were wearing a hat of colour black, the second person would have known for sure that his hat is of colour red. So, this possibility is also ruled out.
The first person can deduce this, meaning, he is not wearing a hat of colour black. So the first person is wearing a hat of colour red.
On April 19 2012 03:52 Bubris wrote: Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals. The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live. The chief agrees to give them one chance, but to live they have to solve a riddle. He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that: The man in front can't see any of the hats The man in the middle can see the man in front's hat And the man in the back can see the hats of the two others. None of them can see the color of their own hat. The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed. Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
If the 2 guys in the front both had a white hat, then the last guy would know he has a red hat. He doesn't know the colour of his hat, therefore, there is at least one red hat on the first 2 guys. If the middle guy would see a white hat in front of him, he would know that he has a red hat (because there was at least one red hat between the two of them. He doesn't know the colour of his hat, therefore the first guy has to be wearing a red hat.
Day 1: There is at least one person that has green eyes.
Start from base case-- 1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes.
Though not sure if correct.
Edit: hmm I feel slightly better after reading other posts in this thread about this.
Yes sir, very well done! If you came up with it yourself, very well done.
On April 19 2012 03:52 Bubris wrote: Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals. The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live. The chief agrees to give them one chance, but to live they have to solve a riddle. He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that: The man in front can't see any of the hats The man in the middle can see the man in front's hat And the man in the back can see the hats of the two others. None of them can see the color of their own hat. The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed. Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
Red The man in back doesn't see 2 black hats so says nothing.
The man in the middle doesn't see a black hat so he says nothing.
The man in the front can deduce that both men are seeing that he is wearing a red hat and can thus both also being wearing red hats so would be eaten if they said black.
On April 19 2012 03:52 Bubris wrote: Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals. The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live. The chief agrees to give them one chance, but to live they have to solve a riddle. He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that: The man in front can't see any of the hats The man in the middle can see the man in front's hat And the man in the back can see the hats of the two others. None of them can see the color of their own hat. The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed. Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
If the 2 guys in the front both had a white hat, then the last guy would know he has a red hat. He doesn't know the colour of his hat, therefore, there is at least one red hat on the first 2 guys. If the middle guy would see a white hat in front of him, he would know that he has a red hat (because there was at least one red hat between the two of them. He doesn't know the colour of his hat, therefore the first guy has to be wearing a red hat.
Thats exactly what I said :D Though u seem to have confused black with white
Day 1: There is at least one person that has green eyes.
Start from base case-- 1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes.
Though not sure if correct.
Edit: hmm I feel slightly better after reading other posts in this thread about this.
Yes sir, very well done! If you came up with it yourself, very well done.
On April 19 2012 03:52 Bubris wrote: Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals. The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live. The chief agrees to give them one chance, but to live they have to solve a riddle. He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that: The man in front can't see any of the hats The man in the middle can see the man in front's hat And the man in the back can see the hats of the two others. None of them can see the color of their own hat. The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed. Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
There are only two black hats. If we look at case 3 where the hat order is BBR if the guy in the back sees two black hats he knows he'll be wearing red. Thus everybody can be freed. Since in the problem the first person says he knows the answer (without anybody saying anything) this is not the case.
So this means that at least one of the front two people is wearing a red hat.
Since the third person hasn't said anything, the second person in line can deduce what I wrote above about somebody in the first two wearing a red hat.
Second person in line can logically deduce "if I see a black hat on person 1, then I am wearing red."
But since second person says nothing, this is not the case.
By thinking through what the second and third persons in line would think about, the first person in line can arrive at the conclusion that he is wearing a red hat.
Day 1: There is at least one person that has green eyes.
Start from base case-- 1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes.
Though not sure if correct.
Edit: hmm I feel slightly better after reading other posts in this thread about this.
Yes sir, very well done! If you came up with it yourself, very well done.
Day 1: There is at least one person that has green eyes.
Start from base case-- 1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes.
Though not sure if correct.
Edit: hmm I feel slightly better after reading other posts in this thread about this.
Yes sir, very well done! If you came up with it yourself, very well done.
Oh right, this problem doesn't state that "there are only blue and green eyes", and that's not part of the information given to the inhabitants. So I guess in the end all the blue-eyed people would still be stuck.
On April 19 2012 03:52 Bubris wrote: Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals. The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live. The chief agrees to give them one chance, but to live they have to solve a riddle. He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that: The man in front can't see any of the hats The man in the middle can see the man in front's hat And the man in the back can see the hats of the two others. None of them can see the color of their own hat. The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed. Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
So the back two do not know which hat they have. In order for the person in the back to know his color, the two in front of him would have to be black. Assuming the other two are logical people, they know that they don't have two black hats.
now that the man in the middle knows there is either 1 black and 1 red or 2 reds, in order for him not to know the person in front would have to be red.
Since the man in front can deduce these two occurances since none of them have spoken, he must be wearing a red hat.
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
The dog barks since he can hear the car coming? Or really if the driver is making a right turn already he wouldn't be crossing the middle of the intersection anyway.
Mathematically, as many times as you want since you can go negative. Yet literally, once you subtract 5 from 25 you can't do it again since you're subtracting 5 from 20, not subtracting 5 from 25.
On April 19 2012 03:52 Bubris wrote: Don't think this has been posted, the best one I know:
3 men travel to Africa for work, they live and work together for around 2 years. At some point they decide to go see the wilds, and unfortunately on the trip, they're caught by a tribe of cannibals. The Cannibals prepare a feast with the 3 men as the main course, but as they're about to be cooked one of the men asks the chief of the tribe to please let them live. The chief agrees to give them one chance, but to live they have to solve a riddle. He shows them a bag containing five hats, 2 black and 3 red. He then places the the three men on a line so that and places a hat on top of each of them so that: The man in front can't see any of the hats The man in the middle can see the man in front's hat And the man in the back can see the hats of the two others. None of them can see the color of their own hat. The chief tells them that if one of them can tell what color hat they're wearing, they will all be freed. Time passes and no one says anything for a looooooong time. When hours have passed the chief says "ok, last chance is now, otherwise we'll eat you". The guy standing in front then steps forward and says "I know the color of my hat"
easy! Ok so if the two in the front were both wearing black, the one in the black would have known he had red and would have said so. But he hasn't said anything so they are either wearing a red and a black or two reds. Now the one in the middle knows this too, and he sees the hat in front of him and knows that if it is black then since the guy in the back hasn't said anything his hat must be red. But he hasn't said anything because the color of the hat infront of him is red. So the guy at the front knows that neither of the two behind him are sure and therefor knows his hat is red.
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
Has anyone come up with the right answer for riddle number two yet? I've been thinking about it for a while now and I can only come up with lame explanations that don't seem "right".
On April 19 2012 00:34 Mentalizor wrote: Don't know if you know this - but a teacher told it to me 7 years ago. So it's probably ancient. So here goes: (it was another currency but in order not to confuse too much, let's use dollars) The riddle: + Show Spoiler +
3 men enter a bar. They order 3 beers that all cost 10$, so the men give the waitress 30$. However when she goes up to the bartender and tells him to get 3 beers and hands over the 30$ he says: Oh no, we have a special offer with 3 beers for 25$. So now the waitress has to go back with 5$. But you can't split 5$ into 3. So she takes 2$ for herself and hands the gentlemen their beers and 1$ each.
So this adds up. Doesn't it? Or wait... Now they all paid 9$ - so 9$ * 3 = 27$... And the waitress took 2$ for herself. That's 29$. Where did the last dollar go?
The waitress is handing them back 3 dollars and charging them 25 for drinks. that equals 28 dollars and a 2 dollar tip. This becomes confusing when you try to subtract 3 dollars from the 30 (which is also accurate, they paid 27 dollars, 25 for drinks and 2 for girl) and then divide that by 3. The numbers are close enough that people don't realize that 25/3 is 8.33 which when you add .66x3 (2) you get the 9 dollars that they paid.
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
Has anyone come up with the right answer for riddle number two yet? I've been thinking about it for a while now and I can only come up with lame explanations that don't seem "right".
Number 2 has to be that it was daytime, as mentioned above. Couldn't imagine it being anything else
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
Has anyone come up with the right answer for riddle number two yet? I've been thinking about it for a while now and I can only come up with lame explanations that don't seem "right".
Number 2 has to be that it was daytime, as mentioned above. Couldn't imagine it being anything else
On April 19 2012 05:14 Go1den wrote: Actually, all of the correct answers have been given so far, just not all in the same posts. See spoilers above if you want some hints/answers.
On April 19 2012 05:14 Go1den wrote: Actually, all of the correct answers have been given so far, just not all in the same posts. See spoilers above if you want some hints/answers.
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
They just have to stand in a line facing one direction. The last guy of the line comes to the front and walks along the line untill he sees two persons with different eyecolors standing next to eachother, in which case he steps between them. They have to repeat this till the one who was first is at the end of the line again. He and the one after him have to go a second time. This will seperate the first group of eycolors from the rest. This group can leave.
This process can be repeated untill someone walks along the whole line whithout stepping between two persons. If this is the case he either has the same eyecolor as everyone in his group or is the only one with a different one and will never know.
On April 19 2012 05:14 Go1den wrote: Actually, all of the correct answers have been given so far, just not all in the same posts. See spoilers above if you want some hints/answers.
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
On April 19 2012 05:14 Go1den wrote: Actually, all of the correct answers have been given so far, just not all in the same posts. See spoilers above if you want some hints/answers.
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
They just have to stand in a line facing one direction. The last guy of the line comes to the front and walks along the line untill he sees two persons with different eyecolors standing next to eachother, in which case he steps between them. They have to repeat this till the one who was first is at the end of the line again. He and the one after him have to go a second time. This will seperate the first group of eycolors from the rest. This group can leave.
This process can be repeated untill someone walks along the whole line whithout stepping between two persons. If this is the case he either has the same eyecolor as everyone in his group or is the only one with a different one and will never know.
If this counts as communicating then I dont know.
I like this solution, but I do think it's communication. I think the people are more supposed to be in a state of isolation.
Day 1: There is at least one person that has green eyes.
Start from base case-- 1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes.
Though not sure if correct.
Edit: hmm I feel slightly better after reading other posts in this thread about this.
Yes sir, very well done! If you came up with it yourself, very well done.
Everyone knew that there was at least one green eyed person at the start, because they could see everyone else's eye color. So why does it matter that the message was given?
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
They just have to stand in a line facing one direction. The last guy of the line comes to the front and walks along the line untill he sees two persons with different eyecolors standing next to eachother, in which case he steps between them. They have to repeat this till the one who was first is at the end of the line again. He and the one after him have to go a second time. This will seperate the first group of eycolors from the rest. This group can leave.
This process can be repeated untill someone walks along the whole line whithout stepping between two persons. If this is the case he either has the same eyecolor as everyone in his group or is the only one with a different one and will never know.
If this counts as communicating then I dont know.
I'm puzzled by this 'there is at least one person that has green eyes'...aren't there 100 persons with green eyes, like stated above in the text?
On the first day, first person looks into eyes of every other and counts how many green eyes he saw. If he saw 100, he has blue eyes. If he saw 99, he has green eyes. He leaves.
Next person knows that the person who left has green eyes ("Everyone knows the eyecolour of every OTHER inhabitant at all times"). He counts green eyes, if he sees 99, he knows he has blue eyes. If he sees 98, he knows he has green eyes. He leaves.
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
They just have to stand in a line facing one direction. The last guy of the line comes to the front and walks along the line untill he sees two persons with different eyecolors standing next to eachother, in which case he steps between them. They have to repeat this till the one who was first is at the end of the line again. He and the one after him have to go a second time. This will seperate the first group of eycolors from the rest. This group can leave.
This process can be repeated untill someone walks along the whole line whithout stepping between two persons. If this is the case he either has the same eyecolor as everyone in his group or is the only one with a different one and will never know.
If this counts as communicating then I dont know.
I'm puzzled by this 'there is at least one person that has green eyes'...aren't there 100 persons with green eyes, like stated above in the text?
On the first day, first person looks into eyes of every other and counts how many green eyes he saw. If he saw 100, he has blue eyes. If he saw 99, he has green eyes. He leaves.
Next person knows that the person who left has green eyes ("Everyone knows the eyecolour of every OTHER inhabitant at all times"). He counts green eyes, if he sees 99, he knows he has blue eyes. If he sees 98, he knows he has green eyes. He leaves.
And so on, everyone leaves on the first day.
The people don't know that there are 100 people with green eyes and 100 people with blue eyes at the start. They only know that there are 200 people total and all of them have either green eyes or blue eyes.
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
Solution: on day 100 everyone knows, half will leave, the rest the next day.
Proof by induction: If only one person has green eyes, he leaves the first night, as he knows the rest have brown eyes. As he knew he was the only one, the rest can deduce that they have brown eyes, and leave the next day.
Let's assume that with X people with green eyes, those X people will leave on the Xth day, the rest the next. Then, if there are X+1 people with green eyes, they would know for a fact that there are more than X people at day X+1, but they can only see X others, and can each deduce that they must have green eyes. The next day, the rest will leave, as they are perfect logicians and understand why the X+1 left.
Hence, this is true for any X that is a natural number, in this case, 100.
I have two good ones that i bugged my mind with for quite some time (credits to my math teacher Bo)
First problem, you have 9 identical small stones, an old fashioned weight and nothing else. One of the stones are heavier, you have to weightings to figure out which one. How? (easy one)
second problem (a lot harder) You have 12 identical stones, and the same setup as before. One stone is odd, but can be either heavier or lighter, you have three weightings to figure out which one, AND if it is in fact heavier or lighter. good luck! :D
The following puzzle was posed to me and a group of co-workers (we're software engineers):-
You and 20 other prisoners are in jail who are all gathered in the main hall and addressed by the warden who explains the following. "In 1 hour you will all be placed in solitary confinement and never see each other again. You will individually be escorted by the guards to the "Lever Room" in which the are a pair of levers that have no function except being able to be flipped either Up or Down. You will flip exactly one lever one time and then be escorted back to your cell. We make no guarantees about the order you will be called to the Lever Room in, and we may take any of you there more than once. If any of you prisoners tell the guard "We have all been to the lever room" and this is true you will all be released, otherwise you will all be shot."
There is one hour for you and the other prisoners to devise a strategy to surivive, but there is no way to communicate once you are escorted to your solitary confinement cells. How do you survive? (Solution will follow, I'll keep any eye on this thread and answer questions if needed though).
#1 A princess is as old as the prince will be, when the princess is twice as old as the prince was when the princess' age was half the sum of their present age. How old are they now?
There is more than one correct answer to this question, you don't need to guess which one, if you get any of the correct combinations you win ^^. + Show Spoiler [hint 1] +
Find all the data you can use. The riddle tells you one more constant to work with, that isn't mentioned
Prince age = x Princess age = y Age diffirence = a x=y-a y=x+a
princess was ½(x+y) when prince was ½(x+y)-a princess age will be 2[½(x+y)-a] so the princess age is now 2[½(x+y)-a]-a now having y=2[½(x+y)-a]-a we just solve the equation (x+y-2a)-a=y y=4a x=3a
so the possible solutions are all that keep the proportions right: Prince is 3 and princess is 4 years old, or they may be 30 and 40 years old etc.
Second one comes from a certain dragon in the High Forrest of Faerun if i remember correctly, it goes as fallows:
2# Old arab was near his death. He had two sons, but he decided that he won't split his land, but rather will give it to one of them. He gave them a challange. An unusual race, because the son whose camel would arrive last at the gates of neighboring city would get the land of his father. However, they weren't allowed to ride away from their destination, only towards it.
After long day, on the backs of their camels. Exhausted from moving at turtle's pace and being exposed to the buring sun of the dessert they decided to hold the challange and go to the inn. They told their story to innkeeper, and he listened carefully. Then he said only two words of advice to them.
The very morning of the next day, people could see both sons riding on their camels fast as if they were chased by the devil. Kicking and screaming at their mounts to hurry them up. They were heading towards the neighboring city.
On April 19 2012 06:44 TWIX_Heaven wrote: LOL diavlo how is that even possible! hahahaha, the odds of us both asking the same riddle at the same time is hilarious!
We didn't. I posted a shitty riddle one minute before you posted yours. Though yours was better, stole it and typed it back in my earlier message to take all the credit.
On April 19 2012 06:44 keplersfolly wrote: The following puzzle was posed to me and a group of co-workers (we're software engineers):-
You and 20 other prisoners are in jail who are all gathered in the main hall and addressed by the warden who explains the following. "In 1 hour you will all be placed in solitary confinement and never see each other again. You will individually be escorted by the guards to the "Lever Room" in which the are a pair of levers that have no function except being able to be flipped either Up or Down. You will flip exactly one lever one time and then be escorted back to your cell. We make no guarantees about the order you will be called to the Lever Room in, and we may take any of you there more than once. If any of you prisoners tell the guard "We have all been to the lever room" and this is true you will all be released, otherwise you will all be shot."
There is one hour for you and the other prisoners to devise a strategy to surivive, but there is no way to communicate once you are escorted to your solitary confinement cells. How do you survive? (Solution will follow, I'll keep any eye on this thread and answer questions if needed though).
Hmm, isn't this puzzle usually with one lever? Am I misinterpreting it or is there a reason you need two:
Leader algorithm: If lever is down, move lever up. Add one to a mental counter "x". Everyone else: You keep a mental variable "y" which starts at 2. If y > 0 and the lever is up, move the lever down and subtract one from y.
When x >= 40, the leader knows that everyone has been in the room. x counts the number of times the lever has been pulled up (It's off by at most one, which is why we need to start y at 2 instead of 1. It might be off by one because the lever might have started down, which would make x 1 larger than the number of times the lever has been pulled up.) Once x >= 40, we then know that the lever has been pulled up at least 39 times. Then, if not everyone else has been in the room, there are at most 19 non-leaders who have entered the room, each who could have pulled up the lever at most 2 times, so the lever could have been pulled up at most 38 times.
I've seen a version where everyone must have the same strategy, and here you need two levers; is this the version you want? (And if so, do you want the version where we know the starting position of the two levers or we don't?)
On April 19 2012 06:44 keplersfolly wrote: The following puzzle was posed to me and a group of co-workers (we're software engineers):-
You and 20 other prisoners are in jail who are all gathered in the main hall and addressed by the warden who explains the following. "In 1 hour you will all be placed in solitary confinement and never see each other again. You will individually be escorted by the guards to the "Lever Room" in which the are a pair of levers that have no function except being able to be flipped either Up or Down. You will flip exactly one lever one time and then be escorted back to your cell. We make no guarantees about the order you will be called to the Lever Room in, and we may take any of you there more than once. If any of you prisoners tell the guard "We have all been to the lever room" and this is true you will all be released, otherwise you will all be shot."
There is one hour for you and the other prisoners to devise a strategy to surivive, but there is no way to communicate once you are escorted to your solitary confinement cells. How do you survive? (Solution will follow, I'll keep any eye on this thread and answer questions if needed though).
The prisoners will assign one person as a counter. The FIRST time a prisoner (non-counter) enters the lever room they put it downward. Nobody will then change the lever until the counter enters the room and puts the lever back in the upward position. The next person who enters the lever room will put the lever downwards again (unless it is someone who already put the lever downwards)
This process will repeat itself in which every person is only allowed to put the lever downwards once. Otherwise they will just let the lever be. If the counter has seen the lever down 20 times (and he has put the lever back upwards 20 times), he will know that every other prisoner has been in the lever room. Therefore he can tell the guards that everyone has been in the room.
I don't know what is up with the 'Pair' of levers. If i'm missing something please tell me! But I guess that if you have had your turn earlier you can just switcheroo the one that is not being used, if you are FORCED to change direction of any of the levers (you decide which lever you're going to use beforehand for the counting process then.)
Hope this is clear enough. My wording is a bit complicated at times.
NINJA EDIT: I saw someone already posted a better result (with 40x instead of 20x because of the possible downward direction of the lever at the start. clever!
If a blue house is made of blue bricks, an orange house is made of orange bricks, and a red house is made of red bricks, what is a green house made of?
Everyone knew that there was at least one green eyed person at the start, because they could see everyone else's eye color. So why does it matter that the message was given?
Since everyone knows that there's either 99 g.e. persons and 100 b.e. persons or 100 g.e. persons and 99 b.e. persons, can't they (by the same method of induction) conclude that since no one leaves on the 99th day, their eye colour must be the same as the one of the 99 blue/green eyed persons that they see?
This would work only if the islanders know that there are only blue and green coloured people on the island. Otherwise it'd screw with the base case. Right?
Everyone knew that there was at least one green eyed person at the start, because they could see everyone else's eye color. So why does it matter that the message was given?
Since everyone knows that there's either 99 g.e. persons and 100 b.e. persons or 100 g.e. persons and 99 b.e. persons, can't they (by the same method of induction) conclude that since no one leaves on the 99th day, their eye colour must be the same as the one of the 99 blue/green eyed persons that they see?
This would work only if the islanders know that there are only blue and green coloured people on the island. Otherwise it'd screw with the base case. Right?
There actually is a key difference even if the villagers know that there are only blue/green eyes:
The idea is that, at the start, everyone knows that there is at least one person with green eyes. Everyone also knows that everyone knows that there is at least one person with green eyes. (Because if someone else didn't see anyone with green eyes, there could be at most 1 person with green eyes, so everyone knows that everyone knows that there is at least one person with green eyes.)
And we can take this even further: Everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that there is at least one person with green eyes. But the issue is, without the message, if you take this chain too far (around 100 "everyone knows", maybe 99 or 98 or 101), the statement no longer becomes true. But with the message, the statement is true no matter how many "everyone knows" there are.
An example with lower numbers: with 198 blue-eyed people, and 2 green-eyed people, everyone knows that there is at least one green-eyed person on the island. But if you are one of the green-eyed people, you don't know that you're green-eyed, so there's a possibility that the other green-eyed person sees only blue-eyed people. So everyone knows that there is a green-eyed person, but somebody doesn't realize that everybody knows that there is a green-eyed person. Once you add on more "everyone knows", the statement becomes harder and harder to process, but the same basic idea holds.
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
All will leave. On Night 1 (N1) No one will leave because everyone sees 99 other green eyes, this continues until N99 and on N100 No one will have left and everyone will figure out that there are only 100 green eyes and leave.
Anne, Bill and Cath each have a positive natural number written on their foreheads. They can only see the foreheads of others. One of the numbers is the sum of the other two. All the previous is common knowledge. The following truthful conversation takes place:
Anne: I don't know my number. Bill: I don't know my number. Cath: I don't know my number. Anne: I now know my number, and it is 50.
What are the numbers of Bill and Cath?
Just to give credit, reportedly it's from the November 2004 issue of Math Horizons (whatever that is).
Why? Let's see what happens: 1. A sees 20 and 30. She could have either 10 or 50. 2. B sees 50 and 30. Also he knows that his number is not equal to 30 (C's number), because then A would've known for sure that her number was 60 (0 is not allowed). He could have either 20 or 80. 3. C sees 50 and 20 and she knows that her number is neither of them. Still, she could have 30 or 70. 4. How does A now know that it's 50 and not 10?
Let's see what would've happened if it was 10 instead: 1. A sees 20 and 30. Same stuff, 10 or 50. 2. B sees 10 and 30. Again, he knows it's not 30. This time it could be 20 or 40. 3. C sees 10 and 20. It could be 10 or 30, but in this situation she knows that it's not 10 and she would've screamed "I know! I know!".
Since she didn't, A's number can't be 10 and the only other option is 50.
Bonus points if you can prove that it's the only solution.
Everyone knew that there was at least one green eyed person at the start, because they could see everyone else's eye color. So why does it matter that the message was given?
Since everyone knows that there's either 99 g.e. persons and 100 b.e. persons or 100 g.e. persons and 99 b.e. persons, can't they (by the same method of induction) conclude that since no one leaves on the 99th day, their eye colour must be the same as the one of the 99 blue/green eyed persons that they see?
This would work only if the islanders know that there are only blue and green coloured people on the island. Otherwise it'd screw with the base case. Right?
There actually is a key difference even if the villagers know that there are only blue/green eyes:
The idea is that, at the start, everyone knows that there is at least one person with green eyes. Everyone also knows that everyone knows that there is at least one person with green eyes. (Because if someone else didn't see anyone with green eyes, there could be at most 1 person with green eyes, so everyone knows that everyone knows that there is at least one person with green eyes.)
And we can take this even further: Everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that there is at least one person with green eyes. But the issue is, without the message, if you take this chain too far (around 100 "everyone knows", maybe 99 or 98 or 101), the statement no longer becomes true. But with the message, the statement is true no matter how many "everyone knows" there are.
An example with lower numbers: with 198 blue-eyed people, and 2 green-eyed people, everyone knows that there is at least one green-eyed person on the island. But if you are one of the green-eyed people, you don't know that you're green-eyed, so there's a possibility that the other green-eyed person sees only blue-eyed people. So everyone knows that there is a green-eyed person, but somebody doesn't realize that everybody knows that there is a green-eyed person. Once you add on more "everyone knows", the statement becomes harder and harder to process, but the same basic idea holds.
I see the difference now. This stuff is very mind-boggling (and also very interesting). Thanks for the explanation.
Ever play the game 24? Use the numbers 1, 5, 5, 5 and the four basic math operations ( + / - *) to get the answer 24. You can only use each number once. The operations can be used as many times as you want.
second problem (a lot harder) You have 12 identical stones, and the same setup as before. One stone is odd, but can be either heavier or lighter, you have three weightings to figure out which one, AND if it is in fact heavier or lighter. good luck! :D
- Rune
Just one question is there a method to figure this out with 100% accurancy or can you get occurences where you have bad "luck" and can't figure it out in three goes? + Show Spoiler +
So I think I've figured this out now or at least some of the scenarios. Had to find pen and paper to do it though:D
PS: I'm dividing the stones into 3 bulks of four :A B C, in A you have Aa Ab Ac Ad, the same with B and C. I started off writing out my thoughts but it just made the whole thing messy. X-Y is how I write my scale.
1: You divide your stones into A B C - 4 stones in each block. The benefit of this is if you weigh A - B and its equal you know that C is the where the light or heavy one is. Lets say that happens. You then know that the 8 stones on the scale are the same.
You then take 2 of the Cs on one side of the scale and on the other you take 1C and 1 from A. *Ca Cb - Cc A is equal, you now know the Cd is the odd stone. Cd - A is you last move. Cd goes up when its light and down when its heavy.
2 We continue on 1, this time though: *CaCb - Cc A is not equal. You also get to know that IF the odd one is on the left and the left side go up that the odd one is light, IF its on the right its heavy. (visaversa if right side go up.) Next move. Ca-Cb. If they are equal you know that Cc is the odd one and its either heavy or light depending on the result of the previous weighing if Cc went up its light, down its heavy.
Ca-Cb are unequal. You have to look on the 2nd result to know wich is the odd one. CaCb was lighter then CcA you know that the light one in this weighing will be the odd one. Similar story if its heavy in weighing number 2. Sick wall of text.
3:
A-B are unequal. A goes down B up. Either the odd one is in A and is heavy or its in B and is light. Edit: Gave up
Update: so I have figured out some of the scenarios.
On April 19 2012 07:33 zJayy962 wrote: Ever play the game 24? Use the numbers 1, 5, 5, 5 and the four basic math operations ( + / - *) to get the answer 24. You can only use each number once. The operations can be used as many times as you want.
I'm impressed :D Was that off the top of your heads or had you heard it before? It took me and my colleagues a while to work it out from a position of total ignorance. Nehsb is more correct though as he doesn't assume the starting state of the levers and he hints towards fully understanding the issue (but falls a little short). Basically there are two related facets:- You MUST flip a lever exactly once + There are 2 levers. What this effectively means is that the prisoners decide that the left lever is the "noop" lever. So the counter only cares about the right lever and you follow the logic you both described, so if a non-counter enters the room and the right lever is flipped to indicate someone needs counting then they throw away their flip on the left lever. You can think of this as changing the puzzle conditions to be having just one lever but flipping it is optional. Still, congrats if you worked this out from scratch!
Ninja Edit: Nehsb's solution accounts for the unknown starting state - good job! Miicro's solution accounts for the fact there are two levers - good job!
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
An easy way to do it would be to weigh coins one and two first. If they balance, then you know that neither of them is the coin you are looking for. In this case, remove coin 2 and replace it with coin 3 and repeat until you find a coin that doesn't balance with coin one. This is the coin you are looking for. The balance will tell you whether it is too light or too heavy.
If the initial weighing of coins 1 and 2 is not balanced, then you know that one of them is the coin you are looking for. You also know that the rest of the coins all weigh the same. Pick up coin two and set it aside. Put any other coin on the scale with coin 1. If it balances, you know that coin 2 is the coin you are looking for, and you can replace coin 1 with coin two to see if your coin is too light or too heavy. If it doesn't balance, then you know that coin 1 is your coin, and the scale will tell you if its too heavy or too light.
Those math forum guys made things way too complicated. (This is coming from an IDIOT)
EDIT: I am a moron. I did not read the whole question. I wear women's underpants. My face is stupid. I am a poop-nosed butt sniffer. The end.
An easy way to do it would be to weigh coins one and two first. If they balance, then you know that neither of them is the coin you are looking for. In this case, remove coin 2 and replace it with coin 3 and repeat until you find a coin that doesn't balance with coin one. This is the coin you are looking for. The balance will tell you whether it is too light or too heavy.
If the initial weighing of coins 1 and 2 is not balanced, then you know that one of them is the coin you are looking for. You also know that the rest of the coins all weigh the same. Pick up coin two and set it aside. Put any other coin on the scale with coin 1. If it balances, you know that coin 2 is the coin you are looking for, and you can replace coin 1 with coin two to see if your coin is too light or too heavy. If it doesn't balance, then you know that coin 1 is your coin, and the scale will tell you if its too heavy or too light.
Those math forum guys made things way too complicated. (This is coming from a mathematician)
Did you read the part where you only have three weighing?
The answer posted on the math forum is the correct one.
An easy way to do it would be to weigh coins one and two first. If they balance, then you know that neither of them is the coin you are looking for. In this case, remove coin 2 and replace it with coin 3 and repeat until you find a coin that doesn't balance with coin one. This is the coin you are looking for. The balance will tell you whether it is too light or too heavy.
If the initial weighing of coins 1 and 2 is not balanced, then you know that one of them is the coin you are looking for. You also know that the rest of the coins all weigh the same. Pick up coin two and set it aside. Put any other coin on the scale with coin 1. If it balances, you know that coin 2 is the coin you are looking for, and you can replace coin 1 with coin two to see if your coin is too light or too heavy. If it doesn't balance, then you know that coin 1 is your coin, and the scale will tell you if its too heavy or too light.
Those math forum guys made things way too complicated. (This is coming from a mathematician)
Did you read the part where you only have three weighing?
The answer posted on the math forum is the correct one.
No. I skipped that part. My mistake. I feel like such an IDIOT right now. Excuse: Haven't slept in 30 hours. Excusable? You decide.
I'm impressed :D Was that off the top of your heads or had you heard it before? It took me and my colleagues a while to work it out from a position of total ignorance. Nehsb is more correct though as he doesn't assume the starting state of the levers and he hints towards fully understanding the issue (but falls a little short). Basically there are two related facets:- You MUST flip a lever exactly once + There are 2 levers. What this effectively means is that the prisoners decide that the left lever is the "noop" lever. So the counter only cares about the right lever and you follow the logic you both described, so if a non-counter enters the room and the right lever is flipped to indicate someone needs counting then they throw away their flip on the left lever. You can think of this as changing the puzzle conditions to be having just one lever but flipping it is optional. Still, congrats if you worked this out from scratch!
Ninja Edit: Nehsb's solution accounts for the unknown starting state - good job! Miicro's solution accounts for the fact there are two levers - good job!
I'd heard it before. There's also a really hard version that I'm curious if anyone has an easier solution to:
20 prisoners are in jail who are all gathered in the main hall and addressed by the warden who explains the following. "In 1 hour you will all be placed in solitary confinement and never see each other again. You will individually be escorted by the guards to the "Lever Room" in which the are a pair of levers that have no function except being able to be flipped either Up or Down. You can flip any number of levers, including all of them or none of them. We make no guarantees about the order you will be called to the Lever Room in, and we may take any of you there more than once. If any of you prisoners tell the guard "We have all been to the lever room" and this is true you will all be released, otherwise you will all be shot."You do not know the starting position of the levers. You are allowed to write down a strategy and give it to all of the prisoners, but the strategy each prisoner gets must be the same; you are not allowed to "pick a leader" or anything like that.
The changes are in bold: - Still 2 levers - You can now flip any number of levers - Every prisoner must have the same strategy - You do not know the starting states of the levers
The problem gets much easier if you do know the starting state of the levers.
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
You ask "What would your brother say if I asked him behind which door the million dollars are?".
You will get the door where the dragon is.
Ask "Which one of you is lying?" ^that assumes i'm asking them both the question simultaneously, which i am, right?
I assumed you can only ask one. Even if you could ask the same question to both, they will both say the other is lying. In any case, you still don't know which door is which.
The solution is actually exactly the same, you just have to count through everyone twice instead of just once. Perhaps you can also speed up the process by using the second lever somehow but I don't care to figure out if that's possible.
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
The solution is actually exactly the same, you just have to count through everyone twice instead of just once. Perhaps you can also speed up the process by using the second lever somehow but I don't care to figure out if that's possible.
Are you taking in account the everyone must have the same strategy restriction?
The solution is actually exactly the same, you just have to count through everyone twice instead of just once. Perhaps you can also speed up the process by using the second lever somehow but I don't care to figure out if that's possible.
Are you taking in account the everyone must have the same strategy restriction?
On April 19 2012 07:33 zJayy962 wrote: Ever play the game 24? Use the numbers 1, 5, 5, 5 and the four basic math operations ( + / - *) to get the answer 24. You can only use each number once. The operations can be used as many times as you want.
On April 19 2012 07:33 zJayy962 wrote: Ever play the game 24? Use the numbers 1, 5, 5, 5 and the four basic math operations ( + / - *) to get the answer 24. You can only use each number once. The operations can be used as many times as you want.
Your answer is not 24, your answer is actually 21. Order of operations: (5 - 1/5) = 4.2, and 5*4.2 = 21. I believe this puzzle as presented has no solution.
On April 19 2012 07:33 zJayy962 wrote: Ever play the game 24? Use the numbers 1, 5, 5, 5 and the four basic math operations ( + / - *) to get the answer 24. You can only use each number once. The operations can be used as many times as you want.
Your answer is not 24, your answer is actually 21. Order of operations: (5 - 1/5) = 4.2, and 5*4.2 = 21. I believe this puzzle as presented has no solution.
BEDMAS 5 * (5 - 1 / 5) 5 * (5 - 0.2) 5 * 4.8 24
Excuse my dyslexia, I really need to spend more time double checking my own reading before I correct others T_T.
Think of words ending in "-gry". "Angry" and "hungry" are two of them. There are only three words in "the English language." What is the third word? The word is something that everyone uses every day. If you have listened carefully, I have already told you what it is.
Think of words ending in "-gry". "Angry" and "hungry" are two of them. There are only three words in "the English language." What is the third word? The word is something that everyone uses every day. If you have listened carefully, I have already told you what it is.
Think of words ending in "-gry". "Angry" and "hungry" are two of them. There are only three words in "the English language." What is the third word? The word is something that everyone uses every day. If you have listened carefully, I have already told you what it is.
You count the places where there is a circular shape in the number and you add them.
I would probably have never figure it out if I did not write down the riddle on a paper, then I got bored and started coloring the circular shapes and that is when I saw it. I LOVE TO SCRIBLE
Lol GR final solution is 2.
EDIT: accidently wrote 45 minutes except 35minutes. Typo
when I read that it would take a preschooler such a short time to solve, I immediately abandoned trying to find a mathematical pattern and just looked at the shapes haha.
Think of words ending in "-gry". "Angry" and "hungry" are two of them. There are only three words in "the English language." What is the third word? The word is something that everyone uses every day. If you have listened carefully, I have already told you what it is.
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man say to them?
If son A is riding son B's camel, he wants to make sure that he gets there before son B, because then son B's camel will have been the faster one and therefore son A wins. The unclear part of this riddle is that they don't switch ownership of the camels even if they ride each other's.
On April 19 2012 07:33 zJayy962 wrote: Ever play the game 24? Use the numbers 1, 5, 5, 5 and the four basic math operations ( + / - *) to get the answer 24. You can only use each number once. The operations can be used as many times as you want.
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
I would say this one is easily solved by simply scaling down the numbers. 200 people is a lot for the mind to wrap it's head around so lets just say there are 2 people on the island, one blue eyed one green eyed. If the green eyed guy sees the other guy has blue eyes then he will leave on the first day since he knows at least one of them has green eyes.
Now how about 2 green eyed people and 2 blue eyed people? The first day each green eyed person will see 2 blue eyed people and 1 green eyed person. They know that if they don't have green eyes then the other green eyed person will leave on the first night, because he would not see any other green eyes. So on the second night when it is confirmed that the other didn't leave they will both leave. Then you just work your way up until on the 100th day all 100 of the leave.
I think you don't even have to consider the blue eyed people, as the riddle will work without them.
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man say to them?
If son A is riding son B's camel, he wants to make sure that he gets there before son B, because then son B's camel will have been the faster one and therefore son A wins. The unclear part of this riddle is that they don't switch ownership of the camels even if they ride each other's.
On April 19 2012 07:33 zJayy962 wrote: Ever play the game 24? Use the numbers 1, 5, 5, 5 and the four basic math operations ( + / - *) to get the answer 24. You can only use each number once. The operations can be used as many times as you want.
5 - 1 / 5 is 25/5 - 1/5, by order of operations. 5*24/5 is 24, so he's correct. Also- and this wasn't explained- you're supposed to use every number exactly once.
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man say to them?
If son A is riding son B's camel, he wants to make sure that he gets there before son B, because then son B's camel will have been the faster one and therefore son A wins. The unclear part of this riddle is that they don't switch ownership of the camels even if they ride each other's.
On April 19 2012 07:37 TanGeng wrote:
On April 19 2012 07:33 zJayy962 wrote: Ever play the game 24? Use the numbers 1, 5, 5, 5 and the four basic math operations ( + / - *) to get the answer 24. You can only use each number once. The operations can be used as many times as you want.
5 - 1 / 5 is 25/5 - 1/5, by order of operations. 5*24/5 is 24, so he's correct. Also- and this wasn't explained- you're supposed to use every number exactly once.
Oh yeah, derp lol I was doing the subtraction first.
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
I would say this one is easily solved by simply scaling down the numbers. 200 people is a lot for the mind to wrap it's head around so lets just say there are 2 people on the island, one blue eyed one green eyed. If the green eyed guy sees the other guy has blue eyes then he will leave on the first day since he knows at least one of them has green eyes.
Now how about 2 green eyed people and 2 blue eyed people? The first day each green eyed person will see 2 blue eyed people and 1 green eyed person. They know that if they don't have green eyes then the other green eyed person will leave on the first night, because he would not see any other green eyes. So on the second night when it is confirmed that the other didn't leave they will both leave. Then you just work your way up until on the 100th day all 100 of the leave.
I think you don't even have to consider the blue eyed people, as the riddle will work without them.
Since logic tells us that you can only get off the island eventually if you're green eyed, shouldn't everyone just state their eye color is green on the first day? all of the green eyed will be saved, and the others are screwed either way
edit: does everyone know "not green" means blue? in that case i retract my statement
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
I would say this one is easily solved by simply scaling down the numbers. 200 people is a lot for the mind to wrap it's head around so lets just say there are 2 people on the island, one blue eyed one green eyed. If the green eyed guy sees the other guy has blue eyes then he will leave on the first day since he knows at least one of them has green eyes.
Now how about 2 green eyed people and 2 blue eyed people? The first day each green eyed person will see 2 blue eyed people and 1 green eyed person. They know that if they don't have green eyes then the other green eyed person will leave on the first night, because he would not see any other green eyes. So on the second night when it is confirmed that the other didn't leave they will both leave. Then you just work your way up until on the 100th day all 100 of the leave.
I think you don't even have to consider the blue eyed people, as the riddle will work without them.
Since logic tells us that you can only get off the island eventually if you're green eyed, shouldn't everyone just state their eye color is green on the first day? all of the green eyed will be saved, and the others are screwed either way
edit: does everyone know "not green" means blue? in that case i retract my statement
Once all of the blue eyes leave, the green eyes will know they have green eyes and will leave as well.
There is a missing part of the question I think. Why not just everyone say "i have green eyes" on the first day and "I have blue eyes the second day" and everyone is off after two days.
Also, do they KNOW that it's 100 and 100? it doesn't state that in the problem, which has massive implications.
On April 19 2012 14:17 BluePanther wrote: There is a missing part of the question I think. Why not just everyone say "i have green eyes" on the first day and "I have blue eyes the second day" and everyone is off after two days.
Also, do they KNOW that it's 100 and 100? it doesn't state that in the problem, which has massive implications.
I think its because you have to know for certain, if you guess green, even if you have green eyes you do not know for certain you have green eyes so they wont let you go. Just because you don't get off the first night doesn't mean you have the other color, it just means you weren't sure and can still have green or blue eyes.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
You just count how many green eyes there are and how many blue eyes there are, your eye color is the one that has 99 people in it.
or
You get 3 people, one person with blue eyes and one with green and a random one. You stare at the guy until he puts you with a partner.
or
You split them into two halves, blue and green eyes, if they dont trus you just wink with your eye, that is a sign of trust so they should let you split them, then you ask a friend and a person you trust to put you on one half.
or
You go to the captain and see if you can go if he says yes then you have green eyes if he says no you have blue eyes.
People will have to get a lot of babies and see their eye colours. EDIT: eye colour schematic pic But it looks like there's still some uncertainty, so it's theoretically possible to mate with blue+blue and get 10 children with green eyes :D not very likely, but possible
On April 19 2012 14:17 BluePanther wrote: There is a missing part of the question I think. Why not just everyone say "i have green eyes" on the first day and "I have blue eyes the second day" and everyone is off after two days.
Also, do they KNOW that it's 100 and 100? it doesn't state that in the problem, which has massive implications.
They would still not know their own eye colour with certainty on day 1.
This problem can be solved with any mix of blue and green eyed people.
If you only see blue eyed people, you would be the only one with green eyes and leave on day 1. If you see only 1 with green eyes, he would leave on day 1 if you have blue eyes. If he doesn't leave on day 1, it means he sees someone else with green eyes which can only be you and you'll both leave on day 2. If you see 2 green eyed people, they'll leave on day 2 if you have blue eyes. If they don't, it means you have green eyes and You'll leave on day 3. If you see 3 green eyed people, they'll leave on day 3 if you have blue eyes. If they don't, it means you have green eyes and You'll leave on day 4.
This pattern is repeating and end result is: If there's X number of green eyed people, they'll leave on day X, and the blue eyed people 1 day later.
A family inherit 6 million dollars from a friend of the family who has passed away. The family consist of two fathers and two sons. The family decides to spilt the money so that they each get 2 millions. How is that possible?
Make two people stand next to eachother, then another one (w/e the color of his eyes) will come stand between you if you have the same color of eyes. Otherwise he will stand next to them.You do this till no one is left. This way everyone will know what color their eyes are. Example:
1: gg 2: Bgg 3: bBgg 3: bbGgg 4: bbGggg 5: ... 199: B x 100 and G x 100
Make two people stand next to eachother, then another one (w/e the color of his eyes) will come stand between you if you have the same color of eyes. Otherwise he will stand next to them.You do this till no one is left. This way everyone will know what color their eyes are. Example:
1: gg 2: Bgg 3: bBgg 3: bbGgg 4: bbGggg 5: ... 199: B x 100 and G x 100
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
After 100 days, if no one leaves, you have green eyes, and so on. I believe the riddle is supposed to include an announcement about eye color daily, but I could be wrong.
On April 19 2012 03:24 Bahamuth wrote: green eyes problem
Haha, I have a vaguely related one, but it's not so much a brain teaser as a math problem (and I think I've posted it here before). I think it's way more counterintuitive ><
Infinitely many prisoners, labeled 1, 2, 3, 4, ... are standing in a line so that person N can see everyone with smaller label (and not himself). Each prisoner k is assigned a hat labeled with a [integer, rational number, or real number; doesn't actually matter (!)], c_k. Now, starting from prisoner 1 and going up, each prisoner will try to guess the number on their own hat; if he fails, he is shot, and if he succeeds, he lives.
Show that, if the prisoners are given (an infinite amount of) time beforehand to formulate a strategy, they can ensure that only finitely many prisoners are shot.
you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask?
You ask "What would your brother say if I asked him behind which door the million dollars are?".
You will get the door where the dragon is.
you are presented with 3 doors and 3 gaurds. same story, 1 door has money, other 2 dont. 1 guard always tells the truth and the other 2 always lie. what do you ask?
Heh this is interesting, I'll write what I think while I try to solve this.
The basic premise of the riddle is simple enough, especially when you know the answer to the version with 2 doors and 2 guards. You need to phrase the question in a way that will get you the same answer regardless of which guard you happen to ask. Phrasing the problem as a pseudo math equation makes it even simpler. 1 (truth) * -1 (lie) = -1 and of course -1 * 1 = -1 as well, so regardless which guard you get, you will always get -1, ie the wrong answer.
Now, if this problem was just 3 guards but still 2 doors, the answer would be similar and very simple. You would just ask guard 1 'what would guard 2 say that guard 3 would say if I asked him where the money was?'. The equation would turn into 1 * -1 * -1 = 1 and you would always get the correct answer.
The problem here is that we need to eliminate 2 doors, not one and that adds some ambiguity. If we asked the same question as above, the liars would always have 2 possible lies, not just 1. This would both make different outcomes possible and it would make it so that none of the guards would actually know which of the 2 lies the lying guards would pick, so they could not accurately tell the truth or lie about which door the lying guards would pick.
As an example: Guard 1 liar, guard 2 truth-teller, guard 3 liar. You ask guard 1: 'What would guard 2 say that guard 3 would say if I asked him where the money was? Guard 3 would say one of the two doors without money behind it. Guard 2 wouldn't know which one so his answer would have to be 'I don't know' or 'one of these two'. Even if we assume that guard 2 knows which lie guard 3 would pick, guard 1 would again have two lies available. He could either choose the right door or the wrong door not picked by guard 3.
The answer then has to be that you go to guard 1 and you ask: 'If I asked guard 2 what guard 3 would never say if I asked him where the money is, which door would never be the answer?'. (yay negatives) The point and difference being that with the never included, it removes the ambiguity of the liars being able to lie in two different ways.
Let's go through the 3 different options:
Option 1 (guard 1 truth-teller, 2 and 3 liars): Guard 3 would never pick the door with the money behind it, because he's a liar. Because guard 2 is also a liar, he would lie and say the answer is one of the two doors with no money behind it (there's still ambiguity, but it doesn't matter here). Because guard 1 knows this, he knows the door with the money behind it would never be the answer, so he will give that as the answer. So we pick that door.
Option 2 (guard 1 liar guard 2 truth-teller, guard 3 liar): Again, guard 3 would never pick the door with the money behind it. The truth-teller knows this so his answer would be the right door. Because guard 1 is a liar, he will also give us the right door as the answer (because we're asking him which door won't be the answer).
Option 3 (guards 1 and 2 liras, 3 truth-teller): Guard 3, the truth-teller, would never say either of the two doors without the money behind it. Because guard 2 is a liar, he will give us the door with the money behind it as the answer. Similarly to option 2, guard 1 will also give us this answer, as the question is what door wouldn't be the answer.
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. - The only possible eyecolours are green and blue.
Seems pretty straightforward. Let's imagine for a second that there are 199 with blue eyes and only 1 with green. Since the guy with the green eyes knows there's someone with green eyes and he knows noone else has green eyes, he has to be that one, so he could go home on the first night.
Similarly, if there were 2 people with green eyes, they would see one person with green eyes. If they themselves didn't have green eyes, that person would leave day 1, since he would see noone else with green eyes. Since that person doesn't leave day 1, they can deduce that they themselves must have green eyes too and they could leave day 2.
And on and on. If there were 3 people with green eyes they'd leave on day 3, if 10 they'd leave on day 10. Since there's no communication possible, this is the only way they can figure out what eye colour they have.
In this example, all the people with green eyes will leave on day 100. On day 99 they expect all the people with green eyes to go home if they themselves didn't have green eyes. On day 100 they know they in fact do.
You said that the only possible eye colours are green and blue. If the people on the island don't know that, the people with blue eyes are fucked. They can never know for sure what eye colour they have. If they do know, they will all leave on day 101 (or day 100 if they can immediately react to all the green eyed people leaving). All along, they saw 100 people with green eyes. Since all of them left on day 100, they can deduce that they themselves didn't have green eyes and therefore must have blue eyes.
I'm impressed :D Was that off the top of your heads or had you heard it before? It took me and my colleagues a while to work it out from a position of total ignorance. Nehsb is more correct though as he doesn't assume the starting state of the levers and he hints towards fully understanding the issue (but falls a little short). Basically there are two related facets:- You MUST flip a lever exactly once + There are 2 levers. What this effectively means is that the prisoners decide that the left lever is the "noop" lever. So the counter only cares about the right lever and you follow the logic you both described, so if a non-counter enters the room and the right lever is flipped to indicate someone needs counting then they throw away their flip on the left lever. You can think of this as changing the puzzle conditions to be having just one lever but flipping it is optional. Still, congrats if you worked this out from scratch!
Ninja Edit: Nehsb's solution accounts for the unknown starting state - good job! Miicro's solution accounts for the fact there are two levers - good job!
I'd heard it before. There's also a really hard version that I'm curious if anyone has an easier solution to:
20 prisoners are in jail who are all gathered in the main hall and addressed by the warden who explains the following. "In 1 hour you will all be placed in solitary confinement and never see each other again. You will individually be escorted by the guards to the "Lever Room" in which the are a pair of levers that have no function except being able to be flipped either Up or Down. You can flip any number of levers, including all of them or none of them. We make no guarantees about the order you will be called to the Lever Room in, and we may take any of you there more than once. If any of you prisoners tell the guard "We have all been to the lever room" and this is true you will all be released, otherwise you will all be shot."You do not know the starting position of the levers. You are allowed to write down a strategy and give it to all of the prisoners, but the strategy each prisoner gets must be the same; you are not allowed to "pick a leader" or anything like that.
The changes are in bold: - Still 2 levers - You can now flip any number of levers - Every prisoner must have the same strategy - You do not know the starting states of the levers
The problem gets much easier if you do know the starting state of the levers.
I have a question : do you know how often people go to the lever room, that is, do you have a way to keep track of the time ?
If not, then I don't know the answer yet. If so, then I have a much much harder version of your riddle, which can be stated as follows :
The basic set up is the same. Each day, a prisoner is brought at random to the lever room, where he has to switch exactly one lever. However, the prisoners don't know in advance how the lever room will look, that is, they can't agree on which lever is the left lever, and which one is the right lever, and even more, they can't agree on what is down position and what is up position. They just know that, when they will eventually get there, they will be able to switch exactly one of two devices which can have two distinct states. They don't know the initial position of the levers.
Then they will get freed not only if they are able to say that everyone went to the room at least one, but that this happened in the last 20 days too.
They are still allowed to make the strategy they want beforehand. They are not forced to all have the same strategy.
On April 19 2012 20:54 Tanukki wrote: The easier a riddle is, the more pleasant the feeling when you manage to beat someone with it. Guess I didn't get that today :p
On April 19 2012 18:04 ]343[ wrote: Haha, I have a vaguely related one, but it's not so much a brain teaser as a math problem (and I think I've posted it here before). I think it's way more counterintuitive ><
Infinitely many prisoners, labeled 1, 2, 3, 4, ... are standing in a line so that person N can see everyone with smaller label (and not himself). Each prisoner k is assigned a hat labeled with a [integer, rational number, or real number; doesn't actually matter (!)], c_k. Now, starting from prisoner 1 and going up, each prisoner will try to guess the number on their own hat; if he fails, he is shot, and if he succeeds, he lives.
Show that, if the prisoners are given (an infinite amount of) time beforehand to formulate a strategy, they can ensure that only finitely many prisoners are shot.
Nice -- but you don't mean an infinite amount of time, you mean an arbitrarily large finite amount of time. If they had an infinite amount of time to prepare, you could ensure that everyone lived by planning forever, and that's a stupid linguistic loophole. I'll have to think on that a bit more to get it down to cofinitely many survivors, but arbitrarily high proportions of survivors is easy (see spoiler as food for thought).
You can code lists of numbers into a single number in any number of ways -- Gödel codes, pairing functions, a bijection from N to N^k (or Q, or C, or whatever), etc. Suppose you want at most one person out of 10,000 to die. Then when you're planning, line everyone up, and have the first person memorize the number that encodes the numbers of the next 9,999 people, and have the 10,001st person do the same thing for the 9,999 people after him, and so on with the (10000k+1)th people for all natural numbers k. Then those people announce that they think their number is the relevant code, and are almost certainly killed, since that would be a ridiculous coincidence, and the next 9,999 people can immediately calculate their numbers with just arithmetic, and so on and so on.
Obviously moving from "arbitrarily close to 100% survival rate" to "all but finitely many survivors" is a huge gap that requires a whole new technique to bridge, but I think it's neat that it's so simple to get that much. I'll get my mental hamsters on their wheels and come back later with a full solution.
Teams of three players are invited to engage in a guessing game and told the rules of the game ahead of the time and allowed time to come up with a strategy.
On the set, there are three soundproof rooms. Above each room, there are two lights, one blue, one red. The rooms are windowed such that someone inside can only see the lights above other the rooms and nothing else. Inside the room, there are three buttons, a red button, a blue button, and a white button labeled pass. At the start of the game, each player will be led into a separate room, and one of the two lights will be turned on above each rooms. The light that is turned on is determined by a flip of a fair coin independently for all three rooms. The game ends when all players have pressed a button inside their respective rooms.
The team is adjudged to win if at least one of them pressed the button of the matching color of the light turn on above their room and none of them pressed the button of the non-matching color. A press of the white pass button counts as neither matching nor non-matching. In all other cases, the team loses.
You lead a team of three. Do you come up with a strategy to maximize your chances of winning? What is it?
On April 20 2012 01:22 TanGeng wrote: Teams of three players are invited to engage in a guessing game and told the rules of the game ahead of the time and allowed time to come up with a strategy.
On the set, there are three soundproof rooms. Above each room, there are two lights, one blue, one red. The rooms are windowed such that someone inside can only see the lights above other the rooms and nothing else. Inside the room, there are three buttons, a red button, a blue button, and a white button labeled pass. At the start of the game, each player will be led into a separate room, and one of the two lights will be turned on above each rooms. The light that is turned on is determined by a flip of a fair coin independently for all three rooms. The game ends when all players have pressed a button inside their respective rooms.
The team is adjudged to win if at least of them pressed the button of the matching color of the light turn on above their room and none of them pressed the button of the non-matching color. A press of the white pass button counts as neither matching nor non-matching. In all other cases, the team loses.
You lead a team of three. Do you come up with a strategy to maximize your chances of winning? What is it?
Just to clarify, I'm guessing players don't have to press their buttons simultaneously, and that other players can see what button the others pressed?
Am I allowed to use some notion of time? As in, the strategy is to do something under a certain circumstance. If no player does anything for a time x, proceed to the next step?
On April 20 2012 01:22 TanGeng wrote: Teams of three players are invited to engage in a guessing game and told the rules of the game ahead of the time and allowed time to come up with a strategy.
On the set, there are three soundproof rooms. Above each room, there are two lights, one blue, one red. The rooms are windowed such that someone inside can only see the lights above other the rooms and nothing else. Inside the room, there are three buttons, a red button, a blue button, and a white button labeled pass. At the start of the game, each player will be led into a separate room, and one of the two lights will be turned on above each rooms. The light that is turned on is determined by a flip of a fair coin independently for all three rooms. The game ends when all players have pressed a button inside their respective rooms.
The team is adjudged to win if at least of them pressed the button of the matching color of the light turn on above their room and none of them pressed the button of the non-matching color. A press of the white pass button counts as neither matching nor non-matching. In all other cases, the team loses.
You lead a team of three. Do you come up with a strategy to maximize your chances of winning? What is it?
Only two lights can ever be on, red or blue. As you know each person can see the other two peoples lights and not their own you set it up as followed. If you can see one of each colour you hit the white button. If you see two of the same colour you hit the opposite button. IE if you see two red lights above your fellows you will hit the blue button.
There is a chance obviously that all three people have the same colour, however to maximize your chances of winning I believe this is the best strategy.
Agreed. Wasn't sure if he messed up his ending (as a math educator, I've seen many instances of this in the past with people getting sloppy trying to generalize sequence numbers) or if this was the trick. If it was done on purpose, then it's definitely 3rd for the reason you cited. Otherwise, I have no idea.
I loved that Island puzzle so much! But the xkcd version is the one you need to read to have a chance of getting the 'right' answer; the version posted here is too ambiguous.
That said, I think even the xkcd version isn't strict enough to mandate the 'right' answer.
Here's what I think the intended correct solution is:
Upon being told that at least one of them has green eyes, anyone who can see only blue eyes can leave on the first ferry.
Thus if only one person had green eyes, they would leave on day 1.
If two people had green eyes, each would think 'I can see one person with green eyes, so it might be him and not me', and not leave the first night. However, upon waking the next day to find the other green-eyed person still around, and knowing that they themselves are the only candidate for having the second pair of green eyes, they can go down to the ferry and leave on the second night.
If three people had green eyes, each would look at the other two, follow the logic above, and think 'Since It takes two green-eyed people two days to figure out they can leave, if I wait two days and they're still here, there must be a third - and the only candidate is me." Thus the three can leave on the third day.
If four people had green eyes, each would see three greens, follow the logic above and conclude that the three greens would leave on the third day unless there was a fourth - them.
Thus, in general, if there are N green-eyed people, each green-eyed one can see N-1. He knows the N-1 would take N-1 days to figure out they could leave if he himself didn't have green eyes. The remaining blue-eyed ones, able to see N greens and wondering if there are N+1, need to wait an extra day and will be disappointed.
However, I think there's another answer that gets everyone off the island in just two days without the need for a note or message:
EDIT: Strike all that. I think there might be a way to do it without the note/message but the idea I had wasn't right.
I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard:
1. Forwards I'm heavy, backwards I'm not. What am I?
2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time?
3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep?
4. How can you rearrange the letters in "new door" to make one word?
5. How many times can you subtract the number 5 from the number 25?
I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor D ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
The group designates a single person the "counter". This person is the only one who will reset switch A to the bottom position. A starts in the down position. If it is your first visit to the room, you move switch A to up. If switch A is in the up position, you don't touch it, you just move B (also, if you have been there before you move B). If the counter finds the switch in the up position, he moves it down and adds +1 to his count. When the counter's count reaches 22, he frees himself and all his co-prisoners!
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor B ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
The group designates a single person the "counter". This person is the only one who will reset switch A to the bottom position. A starts in the down position. If it is your first visit to the room, you move switch A to up. If switch A is in the up position, you don't touch it, you just move B (also, if you have been there before you move B). If the counter finds the switch in the up position, he moves it down and adds +1 to his count. When the counter's count reaches 22, he frees himself and all his co-prisoners!
what if the warden picks the counter to go 22 times in a row?
I'm impressed :D Was that off the top of your heads or had you heard it before? It took me and my colleagues a while to work it out from a position of total ignorance. Nehsb is more correct though as he doesn't assume the starting state of the levers and he hints towards fully understanding the issue (but falls a little short). Basically there are two related facets:- You MUST flip a lever exactly once + There are 2 levers. What this effectively means is that the prisoners decide that the left lever is the "noop" lever. So the counter only cares about the right lever and you follow the logic you both described, so if a non-counter enters the room and the right lever is flipped to indicate someone needs counting then they throw away their flip on the left lever. You can think of this as changing the puzzle conditions to be having just one lever but flipping it is optional. Still, congrats if you worked this out from scratch!
Ninja Edit: Nehsb's solution accounts for the unknown starting state - good job! Miicro's solution accounts for the fact there are two levers - good job!
I'd heard it before. There's also a really hard version that I'm curious if anyone has an easier solution to:
20 prisoners are in jail who are all gathered in the main hall and addressed by the warden who explains the following. "In 1 hour you will all be placed in solitary confinement and never see each other again. You will individually be escorted by the guards to the "Lever Room" in which the are a pair of levers that have no function except being able to be flipped either Up or Down. You can flip any number of levers, including all of them or none of them. We make no guarantees about the order you will be called to the Lever Room in, and we may take any of you there more than once. If any of you prisoners tell the guard "We have all been to the lever room" and this is true you will all be released, otherwise you will all be shot."You do not know the starting position of the levers. You are allowed to write down a strategy and give it to all of the prisoners, but the strategy each prisoner gets must be the same; you are not allowed to "pick a leader" or anything like that.
The changes are in bold: - Still 2 levers - You can now flip any number of levers - Every prisoner must have the same strategy - You do not know the starting states of the levers
The problem gets much easier if you do know the starting state of the levers.
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor B ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
A can not be lying, if she did, B would also be lying C can not be lying, if he did D would also be lying D can not be lying, if she did, C would be lying B could be lying, if he was the one to eat it, A C and D would still be telling the truth.
or it could be the boy that says he had his cake eaten that's lying, seriously what kid goes around saving cake for later, he probably ate his cake and have it too.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
I didn't see it in the thread earlier, so here's my try:
Since there's no way to store the information of how many people have been in the room with the levers themselves, one of the prisoners needs to act as counter. The procedure is simple:
Simplified solution assuming switch 1 is down to begin with: If one of the prisoners (except the counter) enters the room for the first time, they flip switch 1 up. If switch 1 is already flipped up, they instead flip switch 2. If any of them are brought in again after the first time, they also flip switch 2.
Whenever the counter enters the room and finds switch 1 flipped up, he flips it back down. Otherwise he flips switch 2.
Switch 2 is completely useless here, it's simply used because the prisoners have to flip one switch. All the information is stored in switch 1.
If the prisoners knew that switch 1 was flipped down to begin with, this would be the end of the puzzle. The counter would simply announce that everyone had been in the room after he sees switch 1 flipped up for the 22nd time.
Because he doesn't know however, we need to account for the fact that the switch might have been up to begin with. I've been thinking about an elegant solution to this, but I don't think there is one. The prisoners simply have to take the longer way around.
The idea is that you can account for the possible mistake of 1 by doubling the amount of counts. If each of the non-counting prisoners flips the switch up the first two times they enter the room and the counter waits for switch 1 being up 44 times, then one of those times being a potential error doesn't matter. There would still be 43 real counts remaining, which could only have come from all the 22 non-counters (21 * 2 + 1 * 1).
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
The group designates a single person the "counter". This person is the only one who will reset switch A to the bottom position. A starts in the down position. If it is your first visit to the room, you move switch A to up. If switch A is in the up position, you don't touch it, you just move B (also, if you have been there before you move B). If the counter finds the switch in the up position, he moves it down and adds +1 to his count. When the counter's count reaches 22, he frees himself and all his co-prisoners!
your version doesn't account for the fact that the switch could be up to begin with^^
Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)
Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl
Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy
Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.
Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33
Edit + Update: This question turned out a lot more ambiguous than I had intended. The problem is in how one interprets "a male child answers." In case 1, a random child is chosen and happened to be male. In case 2, a male child is chosen if at all possible. Imagining a community of 100 houses with 2 children each, 25 BB, 50 BG, and 25 GG. In the first case, the 25 BB and 50% of the 50 BG houses are considered (since the chance of a boy answering the door in a BG house is 50%), leaving 25 BB and 25 BG with the probability of the unknown child being a boy equal to 25/50=50%. In the second case, a boy answers at all 50 of the BG houses which combined with the 25 BB houses makes 75 with 25/75=33% of the houses having the unknown child being a boy.
I have honestly always interpreted the question as the second case, but it seems like most people interpret it as the first case. The hint "The answer is not 50%" suggests that it is not case 1, but I should definitely have added some additional language to be more specific.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
The group designates a single person the "counter". This person is the only one who will reset switch A to the bottom position. A starts in the down position. If it is your first visit to the room, you move switch A to up. If switch A is in the up position, you don't touch it, you just move B (also, if you have been there before you move B). If the counter finds the switch in the up position, he moves it down and adds +1 to his count. When the counter's count reaches 22, he frees himself and all his co-prisoners!
your version doesn't account for the fact that the switch could be up to begin with^^
all he has to add to ensure that his solution is correct is add a couple conditions: 1) they decide who the counter is beforehand. 2) he doesn't start his count until after the first time he adjusts the switch
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
The group designates a single person the "counter". This person is the only one who will reset switch A to the bottom position. A starts in the down position. If it is your first visit to the room, you move switch A to up. If switch A is in the up position, you don't touch it, you just move B (also, if you have been there before you move B). If the counter finds the switch in the up position, he moves it down and adds +1 to his count. When the counter's count reaches 22, he frees himself and all his co-prisoners!
your version doesn't account for the fact that the switch could be up to begin with^^
all he has to add to ensure that his solution is correct is add a couple conditions: 1) they decide who the counter is beforehand. 2) he doesn't start his count until after the first time he adjusts the switch
How does he know that he's the first one to adjust the switch? What if one of the other prisoners originally saw it in the down position and flipped it up?
On April 20 2012 04:41 el_dawg wrote: Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)
Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl
Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy
Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.
I have to disagree with this one. Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33
You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.
The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.
In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.
In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.
This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!
On April 20 2012 04:41 el_dawg wrote: Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)
Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl
Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy
Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.
I have to disagree with this one. Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33
You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.
The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.
In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.
In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.
This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!
So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement.
On April 20 2012 04:41 el_dawg wrote: Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)
Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl
Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy
Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.
I have to disagree with this one. Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33
You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.
The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.
In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.
In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.
This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!
So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement.
Yeah the point is that you're supposed to introduce the idea that one is a male, what is the probability of the other being male?
Then you introduce the fact that you visited the house and opened the door - what is the probability now? (Since it changes to 50% upon opening the door).
On April 20 2012 04:41 el_dawg wrote: Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)
Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl
Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy
Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.
I have to disagree with this one. Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33
You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.
The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.
In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.
In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.
This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!
So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement.
Hmm, I didn't think the child answering the door made a difference. I will have to rethink this a bit.
If we play a game where I flip 2 coins (re-flipping both if both are tails) and show you that one is heads while hiding the other, you would guess that the hidden coin has a 66% chance of being tails, right?
I really like this question, but the hardest part is finding a way that asks it correctly.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers. for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers. for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.
On April 20 2012 04:41 el_dawg wrote: Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)
Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl
Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy
Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.
I have to disagree with this one. Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33
You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.
The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.
In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.
In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.
This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!
So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement.
Hmm, I didn't think the child answering the door made a difference. I will have to rethink this a bit.
If we play a game where I flip 2 coins (re-flipping both if both are tails) and show you that one is heads while hiding the other, you would guess that the hidden coin has a 66% chance of being tails, right?
I really like this question, but the hardest part is finding a way that asks it correctly.
I think what is missing from my question is whether or not the child opening the door was random. For example, in the Monty hall problem, it is understood that the host will never open the door with the car. So this question needs something to say that not only did a boy happen to answer the door, but a boy will always answer the door.
Edit: the monty hall problem still works if the reveal was random (assuming a goat was revealed), I'm just using it as an example where the revealed info is usually chosen by the asker who knows the status of all the doors/ coins/ children.
On April 20 2012 04:41 el_dawg wrote: Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)
Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl
Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy
Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.
Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33
What's funny is that even if your answer is mathematically incorrect. The answer is not 50% even if you consider the chance of having a girl or a boy to be 50/50.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers. for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.
The same solution works, you just count to 43 or whatever (I'm too lazy to figure out the exact number) and make sure everyone flips the counting lever twice
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers. for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.
The same solution works, you just count to 43 or whatever (I'm too lazy to figure out the exact number) and make sure everyone flips the counting lever twice
just make sure everybody flips the switch down twice, so at some point (too lazy to figure the number too) it still insures everybody has AT LEAST flipped the switch once, which is the only requirement
On April 20 2012 04:41 el_dawg wrote: Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)
Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl
Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy
Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.
I have to disagree with this one. Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33
You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.
The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.
In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.
In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.
This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!
So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement.
Hmm, I didn't think the child answering the door made a difference. I will have to rethink this a bit.
If we play a game where I flip 2 coins (re-flipping both if both are tails) and show you that one is heads while hiding the other, you would guess that the hidden coin has a 66% chance of being tails, right?
I really like this question, but the hardest part is finding a way that asks it correctly.
I think what is missing from my question is whether or not the child opening the door was random. For example, in the Monty hall problem, it is understood that the host will never open the door with the car. So this question needs something to say that not only did a boy happen to answer the door, but a boy will always answer the door.
Edit: the monty hall problem still works if the reveal was random (assuming a goat was revealed), I'm just using it as an example where the revealed info is usually chosen by the asker who knows the status of all the doors/ coins/ children.
First of all, the monty hall problem DOES NOT work if the reveal was random, if a goat is revealed.
Proof by decision tree:
Assume car behind 1, goats behind 2 and 3. P = player, H = host. All scenarios are known to be equally likely because both the player's choice and the host's subsequent choice are random:
P1H2 = stick to win P1H3 = stick to win P2H1 = either impossible to win or you win automatically depending upon unstated rule (host revealed car) P2H2 = switch to win P3H1 = impossible to win / automatic win (host revealed car) P3H2 = switch to win
As you can see, in this scenario seeing a goat revealed is just as likely when you've picked the other goat as it is when you picked the car, so it's a complete toss-up.
Secondly, the 2-child problem. Assuming the odds of a child being born male or female are in fact 50/50:
Here's the WRONG way to work it out:
Four equally likely possibilities: MM, MF, FM and FF Seeing a male child (or being told one exists) eliminates FF, leaving three equally likely possibilities. In only one of these three are there two male children. Therefore the odds are 33% male, 66% female.
Again, this answer is wrong.
Proof that it is wrong:
Let's say you have 1000 friends, each with 2 children you've never seen. Could you make a nice profit by betting even money the other child is female each time the first child you see or are told about is male?
Immediately you can see that the answer depends upon how often, out of the 1000 iterations, you are shown a male child first. Remember: the odds of a child being male or female are 50%. Thus out of your 1000 friends, 250ish will have MM, 250ish MF, 250ish FM and 250ish FF.
If the first child you see is random each time, then you will see boys first at all the MM houses, girls first at all the FF houses, and boys first roughly half the time at the MF/FM houses. So you'll see about 500 boys and 500 girls. And since we assume the genders of siblings are independent, we know that roughly half the boys we saw will have sisters and half of them brothers, and the same for the girls we saw. So no, we can't make money by betting 'F' when we see 'M'
Only if your chances of seeing a boy first are skewed somehow can we deduce different odds.
For instance, what if we the first child we see is a boy 750 times and a girl 250? The odds against that happening by chance if there are 250MM,250MF,250FM and 250FF are astronomical - we would have to assume people weren't showing us girls unless they had no choice. In which case yes, we could make money by betting, since in roughly 500 of the 750 times we are shown a boy, the other is a girl - and we would make even more money betting that the other is a girl if we are shown a girl, since it's true every time.
So if we see a boy AND know we would only have seen or been told about a girl as a last resort, the odds are 33 and 66. Ot herwise they're 50/50
On April 20 2012 01:22 TanGeng wrote: Teams of three players are invited to engage in a guessing game and told the rules of the game ahead of the time and allowed time to come up with a strategy.
On the set, there are three soundproof rooms. Above each room, there are two lights, one blue, one red. The rooms are windowed such that someone inside can only see the lights above other the rooms and nothing else. Inside the room, there are three buttons, a red button, a blue button, and a white button labeled pass. At the start of the game, each player will be led into a separate room, and one of the two lights will be turned on above each rooms. The light that is turned on is determined by a flip of a fair coin independently for all three rooms. The game ends when all players have pressed a button inside their respective rooms.
The team is adjudged to win if at least of them pressed the button of the matching color of the light turn on above their room and none of them pressed the button of the non-matching color. A press of the white pass button counts as neither matching nor non-matching. In all other cases, the team loses.
You lead a team of three. Do you come up with a strategy to maximize your chances of winning? What is it?
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor B ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
A can not be lying, if she did, B would also be lying C can not be lying, if he did D would also be lying D can not be lying, if she did, C would be lying B could be lying, if he was the one to eat it, A C and D would still be telling the truth.
or it could be the boy that says he had his cake eaten that's lying, seriously what kid goes around saving cake for later, he probably ate his cake and have it too.
Look at the original post, either he edited it or you changed his quote, because what B says differs between the 2 versions.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
Since the Warden says prisoners are chosen at random rather than at whim, it's best to just wait for a long time without a counting strategy. The first person to visit the room N time with N being sufficiently large (such as 40) should give a good probability of going free.
On April 20 2012 01:22 TanGeng wrote: Teams of three players are invited to engage in a guessing game and told the rules of the game ahead of the time and allowed time to come up with a strategy.
On the set, there are three soundproof rooms. Above each room, there are two lights, one blue, one red. The rooms are windowed such that someone inside can only see the lights above other the rooms and nothing else. Inside the room, there are three buttons, a red button, a blue button, and a white button labeled pass. At the start of the game, each player will be led into a separate room, and one of the two lights will be turned on above each rooms. The light that is turned on is determined by a flip of a fair coin independently for all three rooms. The game ends when all players have pressed a button inside their respective rooms.
The team is adjudged to win if at least one of them pressed the button of the matching color of the light turn on above their room and none of them pressed the button of the non-matching color. A press of the white pass button counts as neither matching nor non-matching. In all other cases, the team loses.
You lead a team of three. Do you come up with a strategy to maximize your chances of winning? What is it?
On April 20 2012 01:22 TanGeng wrote: Teams of three players are invited to engage in a guessing game and told the rules of the game ahead of the time and allowed time to come up with a strategy.
On the set, there are three soundproof rooms. Above each room, there are two lights, one blue, one red. The rooms are windowed such that someone inside can only see the lights above other the rooms and nothing else. Inside the room, there are three buttons, a red button, a blue button, and a white button labeled pass. At the start of the game, each player will be led into a separate room, and one of the two lights will be turned on above each rooms. The light that is turned on is determined by a flip of a fair coin independently for all three rooms. The game ends when all players have pressed a button inside their respective rooms.
The team is adjudged to win if at least of them pressed the button of the matching color of the light turn on above their room and none of them pressed the button of the non-matching color. A press of the white pass button counts as neither matching nor non-matching. In all other cases, the team loses.
You lead a team of three. Do you come up with a strategy to maximize your chances of winning? What is it?
Just to clarify, I'm guessing players don't have to press their buttons simultaneously, and that other players can see what button the others pressed?
Am I allowed to use some notion of time? As in, the strategy is to do something under a certain circumstance. If no player does anything for a time x, proceed to the next step?
Players don't know if other team members have pressed buttons unless they are being led out of the room in which case the game is already over. You may have a concept of time.
To the OP of the boy girl thing: It is 50%. BB,BG,GB,GG accounts for order (BG and GB are the same but in a different order). If you want to do it this way, then you must say that the boy you met at the door is either: Boy 1 being case of B_ or he is Boy 2 being case of _B. Either way, there is one case of second child being boy and one case of second child being a girl.
If he is boy 1, then: BB and BG are you only option. If he is boy 2, then: BB and GB are you only option. GB and BG can't be in the same case because the boy is in two different spots (think of it as the one you met vs the one you didn't meet).
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are unknown at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
Since the Warden says prisoners are chosen at random rather than at whim, it's best to just wait for a long time without a counting strategy. The first person to visit the room N time with N being sufficiently large (such as 40) should give a good probability of going free.
Would I be on the right track if I treated the 2 switches as a 2 bit memory cell, and allow only a certain amount of prisoners to change certain states?
On April 18 2012 23:02 kochanfe wrote: You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?
Second Problem: You need exactly four gallons. How do you do it?
There are two solutions for both problems, and they are each related.
Start by filling the 5 gallon jug. Pour 3 gallons into the 3 gallon jug (filling it) leaving 2 gallons in the 5 gallon jug. Dump the 3 gallon jug. Pour the remaining 2 gallons from the 5 gallon jug into the 3 gallon jug. Fill the 5 gallon jug. Grats, you now have 7 gallons. Pour 1 gallon from the 5 gallon jug into the 3 gallon jug. (Dump or ignore the 3 gallon jug.) Grats, you now have 4 gallons (in the 5 gallon jug).
#2 -- the 4 + 3 approach.
Start by filling the 3 gallon jug. Pour all 3 gallons into the 5 gallon jug. Again fill the 3 gallon jug. Pour 2 gallons into the 5 gallon jug (filling it) leaving 1 gallon in the 3 gallon jug. Dump the 5 gallon jug. Pour the remaining 1 gallon from the 3 gallon jug into the 5 gallon jug. Fill the 3 gallon jug. Pour the 3 gallons into the 5 gallon jug. Grats, you now have 4 gallons (in the 5 gallon jug). Fill the 3 gallon jug. Grats, you now have 7 gallons.
[QUOTE]On April 20 2012 10:51 drew-chan wrote: [QUOTE]On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
[QUOTE]
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory.
i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory.
i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years.
Solution is more or less the same, simply have one of the two switches be the one the counter uses, and have people who aren't going to modify it flip the other switch. Say there's one on the left and one on the right, have the counter only count when the left switch is up, and have him flip it down when he goes in. Everyone else only flips the right switch, unless the left switch is down, then they flip it up. They only ever flip the left switch up twice, never more than that, even if it is down.
Personally, I prefer the method of "wait until I've gone 50 times." Assuming the selection is truly random and that the probability of choosing any given prisoner is approximately the same as choosing any other prisoner, the odds of having been chosen 50 times before every other person has gone at least once is absurdly small.
On April 20 2012 04:41 el_dawg wrote: Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)
Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl
Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy
Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.
Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33
Like others have said, you're wrong and the correct answer is 50%.
Before answering the door there are 4 possible outcomes:
Male | Male Male | Female Female | Male Female | Female
Once you open the door to reveal a male child, you know that there is at least one male child. Therefore there are only two remaining outcomes left: Male | Male Male | Female
I have no idea why you thought Female | Male was an additional outcome, as it would only be an outcome if the one opening the door was female.
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor D ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
One of these children is lying. Who is it?
Alice said: "I didn't eat it!" Brad said: "Neither Alice nor Dena ate it." Cory said: "I swear on my momma's grave I didn't eat it!" Dena said: "Cory is telling the truth!"
[For reference, I changed their names so it would be easier to talk about]
Well let's work through it. For starters we know that one of these children are lying. That means, since no one mentions Brad at all, he can't be the one who ate it.
Alice Cory Dena
Now, Cory claims he didn't eat it, and Dena says that Cory is telling the truth. If Cory did eat it, both of them would be lying, and since we know only one person lied, that can't be true. Which means that Cory did not eat it either.
Alice Dena
Finally Alice claims she did not eat it, and Brad claims that neither Alice or Dena ate it. If Alice really did eat it, then both Brad and Alice would be lying, so Alice cannot be the one who ate it.
Dena
The only suspect left is Dena. Brad claims that neither Alice nor Dena ate the cake. However Dena was the one who ate the cake, and so Brad is lying.
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor D ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
One of these children is lying. Who is it?
Alice said: "I didn't eat it!" Brad said: "Neither Alice nor Dena ate it." Cory said: "I swear on my momma's grave I didn't eat it!" Dena said: "Cory is telling the truth!"
[For reference, I changed their names so it would be easier to talk about]
Well let's work through it. For starters we know that one of these children are lying. That means, since no one mentions Brad at all, he can't be the one who ate it.
Alice Cory Dena
Now, Cory claims he didn't eat it, and Dena says that Cory is telling the truth. If Cory did eat it, both of them would be lying, and since we know only one person lied, that can't be true. Which means that Cory did not eat it either.
Alice Dena
Finally Alice claims she did not eat it, and Brad claims that neither Alice or Dena ate it. If Alice really did eat it, then both Brad and Alice would be lying, so Alice cannot be the one who ate it.
Dena
The only suspect left is Dena. Brad claims that neither Alice nor Dena ate the cake. However Dena was the one who ate the cake, and so Brad is lying.
Easiest way: If D is lying, so is C. If D is telling the truth, so is C. Both C and D must be telling the truth. If A is lying, both A an B are lying. Not possible. A is telling truth. Only B is lying because D must have ate it.
Ya, I should have written it as "A boy answers the door if at all possible" since the way it is written now is ambiguous as to whether a boy just happened to open the door or whether the question is controlling that event. In the Monty hall example, "the host opens a door and reveals a goat" is always assumed (or explicitly stated) to be not random. As Umpteen mentioned, you get different answers if the information was revealed randomly or not.
I really like the 2 part version of the children question that was suggested.
On April 20 2012 13:32 el_dawg wrote: Re: all the 2 children replies.
Ya, I should have written it as "A boy answers the door if at all possible" since the way it is written now is ambiguous as to whether a boy just happened to open the door or whether the question is controlling that event. In the Monty hall example, "the host opens a door and reveals a goat" is always assumed (or explicitly stated) to be not random. As Umpteen mentioned, you get different answers if the information was revealed randomly or not.
I really like the 2 part version of the children question that was suggested.
If that's the case it's still 50% as soon as you find out a boy answers the door lol.
Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)
So at first there are four possiblities: Male | Male Male | Female Female | Male Female | Female
Where the left column is who answer the door, and the right column is the unseen child. However you change the rule to say if a boy can answer it, he will. So, if a female answers the door, it's because there is no chance of a male answering it .Which means there are only three possibilities to begin with now..
Male | Male Male | Female Female | Female
The thing is, in the question itself, you already determine that the person to answer the door is male. This eliminates the last outcome leaving only the two outcomes with males answering the door:
Male | Male Male | Female
Which means there's a 50/50 chance the other child is male.
The question would actually be a thinking question if you had instead asked:
What are the chances of a Male child answering the door?
Before telling us who answers the door, in which case the odds would be 66% since there are only 3 possible scenarios and in two of them a male answers it.
Alternatively, you could not tell us the gender of who answer the door at all, and instead say this:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, and you know that a male child will always answer the door first if possible, but you don't know the gender of either of them. Upon arriving, one of the children answers the door, but you can't tell if they are a boy or girl. What is the % chance that the other child is a male. (hint: it is not 50%)
By not telling them who answers the door, you have set it up so each of the three possibilities are still possible, so there's a 66% chance the second child is female, and a 33% chance they are male.
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor D ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
One of these children is lying. Who is it?
Alice said: "I didn't eat it!" Brad said: "Neither Alice nor Dena ate it." Cory said: "I swear on my momma's grave I didn't eat it!" Dena said: "Cory is telling the truth!"
[For reference, I changed their names so it would be easier to talk about]
Well let's work through it. For starters we know that one of these children are lying. That means, since no one mentions Brad at all, he can't be the one who ate it.
Alice Cory Dena
Now, Cory claims he didn't eat it, and Dena says that Cory is telling the truth. If Cory did eat it, both of them would be lying, and since we know only one person lied, that can't be true. Which means that Cory did not eat it either.
Alice Dena
Finally Alice claims she did not eat it, and Brad claims that neither Alice or Dena ate it. If Alice really did eat it, then both Brad and Alice would be lying, so Alice cannot be the one who ate it.
Dena
The only suspect left is Dena. Brad claims that neither Alice nor Dena ate the cake. However Dena was the one who ate the cake, and so Brad is lying.
Easiest way: If D is lying, so is C. If D is telling the truth, so is C. Both C and D must be telling the truth. If A is lying, both A an B are lying. Not possible. A is telling truth. Only B is lying because D must have ate it.
As per the rules of the thread, please spoiler that Release
On April 20 2012 13:32 el_dawg wrote: Re: all the 2 children replies.
Ya, I should have written it as "A boy answers the door if at all possible" since the way it is written now is ambiguous as to whether a boy just happened to open the door or whether the question is controlling that event. In the Monty hall example, "the host opens a door and reveals a goat" is always assumed (or explicitly stated) to be not random. As Umpteen mentioned, you get different answers if the information was revealed randomly or not.
I really like the 2 part version of the children question that was suggested.
If that's the case it's still 50% as soon as you find out a boy answers the door lol.
Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)
So at first there are four possiblities: Male | Male Male | Female Female | Male Female | Female
Where the left column is who answer the door, and the right column is the unseen child. However you change the rule to say if a boy can answer it, he will. So, if a female answers the door, it's because there is no chance of a male answering it .Which means there are only three possibilities to begin with now..
Male | Male Male | Female Female | Female
The thing is, in the question itself, you already determine that the person to answer the door is male. This eliminates the last outcome leaving only the two outcomes with males answering the door:
Male | Male Male | Female
Which means there's a 50/50 chance the other child is male.
The question would actually be a thinking question if you had instead asked:
What are the chances of a Male child answering the door?
Before telling us who answers the door, in which case the odds would be 66% since there are only 3 possible scenarios and in two of them a male answers it.
Alternatively, you could not tell us the gender of who answer the door at all, and instead say this:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, and you know that a male child will always answer the door first if possible, but you don't know the gender of either of them. Upon arriving, one of the children answers the door, but you can't tell if they are a boy or girl. What is the % chance that the other child is a male. (hint: it is not 50%)
By not telling them who answers the door, you have set it up so each of the three possibilities are still possible, so there's a 66% chance the second child is female, and a 33% chance they are male.
If we play a game where I flip 2 coins (re-flipping both if both are tails) and show you that one is heads while hiding the other, you would guess that the hidden coin has a 33% chance of being heads, right? Since the only way for me to show a heads while hiding a heads is if both were heads (33% chance), otherwise I will show a heads while hiding a tails (66% chance). That is sort of what I was trying to ask with the original question.
You have two lengths of rope. Each rope has the property that if you light it on fire at one end, it will take exactly 60 minutes to burn to the other end. Note that the ropes will not burn at a consistent speed the entire time (for example, it's possible that the first 90% of a rope will burn in 1 minute, and the last 10% will take the additional 59 minutes to burn).
Given these two ropes and a matchbook, can you find a way to measure out exactly 45 minutes?
You and a friend are standing in front of two houses. In each house lives a family with two children. "The family on the left has a boy who loves history, but their other child prefers math," your friend tells you. "The family on the right has a 7-year old boy, and they just had a new baby," he explains. "Does either family have a daughter?" you ask. "I'm not sure," your friend says. "But pick the family that you think is more likely to have a daughter. If they do have a daughter, I'll give you $100."
Which family should you pick, or does it not matter?
You and a friend are standing in front of two houses. In each house lives a family with two children. "The family on the left has a boy who loves history, but their other child prefers math," your friend tells you. "The family on the right has a 7-year old boy, and they just had a new baby," he explains. "Does either family have a girl?" you ask. "I'm not sure," your friend says. "But pick the family that you think is more likely to have a girl. If they do have a girl, I'll give you $100."
Which family should you pick, or does it not matter?
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
Your story has holes. Where do they arrive? You didn't state at the prison. Where can they meet and plan a strategy? Therefore it can be anywhere. You didn't specify if the switches are fastened to a wall/switchboard. Therefore I can move them.
A very simple answer is that they all go into the switch room on planning day because the warden didn't say that you couldn't. Then they say "We have all visited the switch room."
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory.
i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years.
Solution is more or less the same, simply have one of the two switches be the one the counter uses, and have people who aren't going to modify it flip the other switch. Say there's one on the left and one on the right, have the counter only count when the left switch is up, and have him flip it down when he goes in. Everyone else only flips the right switch, unless the left switch is down, then they flip it up. They only ever flip the left switch up twice, never more than that, even if it is down.
Personally, I prefer the method of "wait until I've gone 50 times." Assuming the selection is truly random and that the probability of choosing any given prisoner is approximately the same as choosing any other prisoner, the odds of having been chosen 50 times before every other person has gone at least once is absurdly small.
The problem is that the first time the counter goes in, he doesn't know if the switch is up because it started that way or because someone else entered before him.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory.
i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years.
Solution is more or less the same, simply have one of the two switches be the one the counter uses, and have people who aren't going to modify it flip the other switch. Say there's one on the left and one on the right, have the counter only count when the left switch is up, and have him flip it down when he goes in. Everyone else only flips the right switch, unless the left switch is down, then they flip it up. They only ever flip the left switch up twice, never more than that, even if it is down.
Personally, I prefer the method of "wait until I've gone 50 times." Assuming the selection is truly random and that the probability of choosing any given prisoner is approximately the same as choosing any other prisoner, the odds of having been chosen 50 times before every other person has gone at least once is absurdly small.
The problem is that the first time the counter goes in, he doesn't know if the switch is up because it started that way or because someone else entered before him.
On April 20 2012 14:42 Aelfric wrote: Riddle: You have two lengths of rope. Each rope has the property that if you light it on fire at one end, it will take exactly 60 minutes to burn to the other end. Note that the ropes will not burn at a consistent speed the entire time (for example, it's possible that the first 90% of a rope will burn in 1 minute, and the last 10% will take the additional 59 minutes to burn).
Given these two ropes and a matchbook, can you find a way to measure out exactly 45 minutes?
you light both ends of 1 rope and 1 end of the other rope at the same time. when the rope you light at both ends burns out, you light the other end of the other rope. the total time it takes to burn will be 45 mins.
Wow you guys are fast, then lets try something harder:
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
On April 20 2012 14:49 Aelfric wrote: More Riddle:
You and a friend are standing in front of two houses. In each house lives a family with two children. "The family on the left has a boy who loves history, but their other child prefers math," your friend tells you. "The family on the right has a 7-year old boy, and they just had a new baby," he explains. "Does either family have a girl?" you ask. "I'm not sure," your friend says. "But pick the family that you think is more likely to have a girl. If they do have a girl, I'll give you $100."
Which family should you pick, or does it not matter?
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
Is it 6? First take the horses into 5 different races and then race the 5 winners against each other in 1 race. Then the top 3 in your last race will be your 3 fastest horses.
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor D ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
On April 20 2012 15:06 Aelfric wrote: Wow you guys are fast, then lets try something harder:
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
Is it 6? First take the horses into 5 different races and then race the 5 winners against each other in 1 race. Then the top 3 in your last race will be your 3 fastest horses.
On April 20 2012 15:06 Aelfric wrote: Wow you guys are fast, then lets try something harder:
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
Is it 6? First take the horses into 5 different races and then race the 5 winners against each other in 1 race. Then the top 3 in your last race will be your 3 fastest horses.
On April 20 2012 15:06 Aelfric wrote: Wow you guys are fast, then lets try something harder:
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
Is it 6? First take the horses into 5 different races and then race the 5 winners against each other in 1 race. Then the top 3 in your last race will be your 3 fastest horses.
I see. The 2nd and 3rd place could be in the original race with the fastest horse. So from my method from before you split the horses into 5 different groups and race them all (thats 5 races) then you race the fastest one from each group (that will give you #1 spot). Now take the 2nd and 3rd from the 6th race and gather the horses from the race you got the fastest horse and have them race (thats 7). Answer is 7 races.
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
you have 5 5horse races. horses that place 4/5 are out. you write down what places the remaining horses place.
15 horses left in contention.
you have all the horses that placed 1st race against eachother. 4/5 are out. you have all the horses that placed 2nd race against eachother. 4/5 are out. you have all the horses that placed 3rd race against eachother. 4/5 are out.
you write down what places the remaining horses place.
The horse that placed 1st both races is your fastest horse.
8 horses left in contention.
you have 2 races of 4 horses each. 4th is out
6 horses left in contention.
you race 5 of your last 6 horses. 1 horse sits out. 4/5 are out.
4 horses left.
you race them. 1/2 are your 2nd/3rd fastest horse in that order.
On April 20 2012 15:06 Aelfric wrote: Wow you guys are fast, then lets try something harder:
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
i think the answer is 11. you have 5 races initially, and since you only want the top 3 fastest horses, you can say that the slowest 2 horses from each group couldn't possibly be the "top 3" so you can count them out. you are left with 15 horses. you then have 3 more races between the remaining horses. again repeat the process of negating the slowest 2 from each group. after this you are left with 9 horses. have 1 race with 5 of the horses and negate the bottom two again. after that you are only left with 7 horses. have another race with 5 of them, and negate the bottom two. there are now 5 horses left. have one more race and pick the top 3.
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor D ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
lol, since this is the third time B has been selected as the answer (including the same logic), I'll clarify something in regards to the matter. It's a simple logical oversight: if B is the liar, that would also mean that there are two cake eaters, which as indicated in the original riddle, is not the case. Remember to focus on what the riddle is asking.
On April 20 2012 15:06 Aelfric wrote: Wow you guys are fast, then lets try something harder:
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
i think the answer is 11. you have 5 races initially, and since you only want the top 3 fastest horses, you can say that the slowest 2 horses from each group couldn't possibly be the "top 3" so you can count them out. you are left with 15 horses. you then have 3 more races between the remaining horses. again repeat the process of negating the slowest 2 from each group. after this you are left with 9 horses. have 1 race with 5 of the horses and negate the bottom two again. after that you are only left with 7 horses. have another race with 5 of them, and negate the bottom two. there are now 5 horses left. have one more race and pick the top 3.
[QUOTE]On April 20 2012 11:30 whatwhatanut wrote: [QUOTE]On April 20 2012 10:51 drew-chan wrote: [QUOTE]On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
[QUOTE]
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory.
i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years.
1 person is the counter. the1st time the counter is in the room he moves switch no1 to off does not count anything but notes the position of the switches. Anyone else who comes in the room moves switch no1 to on if the switch is off. if its on then just moves switch no2. Everytime the counter goes in the room. He adds +1 if switch no1 is moved or adds 2 if both switches have changed position since he was last here. the minimum time it would take would be 11 visits or the maximum of 66. 66 being he visits 3 times ina row in between every unique visit..
btw i really want to know if there's an more efficient way to do this....
On April 20 2012 15:06 Aelfric wrote: Wow you guys are fast, then lets try something harder:
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
you have 5 5horse races. horses that place 4/5 are out. you write down what places the remaining horses place.
15 horses left in contention.
you have all the horses that placed 1st race against eachother. 4/5 are out. you have all the horses that placed 2nd race against eachother. 4/5 are out. you have all the horses that placed 3rd race against eachother. 4/5 are out.
you write down what places the remaining horses place.
The horse that placed 1st both races is your fastest horse.
8 horses left in contention.
you have 2 races of 4 horses each. 4th is out
6 horses left in contention.
you race 5 of your last 6 horses. 1 horse sits out. 4/5 are out.
4 horses left.
you race them. 1/2 are your 2nd/3rd fastest horse in that order.
On April 20 2012 15:06 Aelfric wrote: Wow you guys are fast, then lets try something harder:
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
you have 5 5horse races. horses that place 4/5 are out. you write down what places the remaining horses place.
15 horses left in contention.
you have all the horses that placed 1st race against eachother. 4/5 are out. you have all the horses that placed 2nd race against eachother. 4/5 are out. you have all the horses that placed 3rd race against eachother. 4/5 are out.
you write down what places the remaining horses place.
The horse that placed 1st both races is your fastest horse.
8 horses left in contention.
you have 2 races of 4 horses each. 4th is out
6 horses left in contention.
you race 5 of your last 6 horses. 1 horse sits out. 4/5 are out.
4 horses left.
you race them. 1/2 are your 2nd/3rd fastest horse in that order.
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor D ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
lol, since this is the third time B has been selected as the answer (including the same logic), I'll clarify something in regards to the matter. It's a simple logical oversight: if B is the liar, that would also mean that there are two cake eaters, which as indicated in the original riddle, is not the case. Remember to focus on what the riddle is asking.
Not true, A person can be lying without every single thing in their sentence being untrue. If half the sentence is false, it is still a lie. For example: I live in the US and own a computer. That is a true statement. I live in the US, and do not own a computer. That is a lie, even though part of the statement is true.
If you don't like that example, try this one: I am a man who attends college. That is true. I am a woman who attends college. A lie, even though i do attend college. The same applies to this scenario. B's statement is "A didn't eat it" and "D didn't eat it". For him to be lying, only one of those has to be false. A didn't eat the cake, that must be true. However, D did eat the cake. since part of his statement was false the statement was a lie.
On April 20 2012 15:06 Aelfric wrote: Wow you guys are fast, then lets try something harder:
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
i think the answer is 11. you have 5 races initially, and since you only want the top 3 fastest horses, you can say that the slowest 2 horses from each group couldn't possibly be the "top 3" so you can count them out. you are left with 15 horses. you then have 3 more races between the remaining horses. again repeat the process of negating the slowest 2 from each group. after this you are left with 9 horses. have 1 race with 5 of the horses and negate the bottom two again. after that you are only left with 7 horses. have another race with 5 of them, and negate the bottom two. there are now 5 horses left. have one more race and pick the top 3.
On April 20 2012 15:06 Aelfric wrote: Wow you guys are fast, then lets try something harder:
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
you have 5 5horse races. horses that place 4/5 are out. you write down what places the remaining horses place.
15 horses left in contention.
you have all the horses that placed 1st race against eachother. 4/5 are out. you have all the horses that placed 2nd race against eachother. 4/5 are out. you have all the horses that placed 3rd race against eachother. 4/5 are out.
you write down what places the remaining horses place.
The horse that placed 1st both races is your fastest horse.
8 horses left in contention.
you have 2 races of 4 horses each. 4th is out
6 horses left in contention.
you race 5 of your last 6 horses. 1 horse sits out. 4/5 are out.
4 horses left.
you race them. 1/2 are your 2nd/3rd fastest horse in that order.
the bottom 2 of each race are eliminated, 15/ 25 Left
#2)The winner of each group go into a second round
the winner of the race is the top horse
the 1st runner of this race stays (1st runner run of his group in #1 stays, 2nd runner up of his group in #2 eliminated as no chance to be in top 3) the 2nd runner of this race up stays (all other from his group in #1 got eliminated )
4th / 5th guy and all others belong to their group in #1 eliminated
6 lefts, with the top 1 determined
#3) the remaining five with uncertain seating run the last one
<--this sounds a lot like dual tournament format for some part
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory.
i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years.
Solution is more or less the same, simply have one of the two switches be the one the counter uses, and have people who aren't going to modify it flip the other switch. Say there's one on the left and one on the right, have the counter only count when the left switch is up, and have him flip it down when he goes in. Everyone else only flips the right switch, unless the left switch is down, then they flip it up. They only ever flip the left switch up twice, never more than that, even if it is down.
Personally, I prefer the method of "wait until I've gone 50 times." Assuming the selection is truly random and that the probability of choosing any given prisoner is approximately the same as choosing any other prisoner, the odds of having been chosen 50 times before every other person has gone at least once is absurdly small.
The problem is that the first time the counter goes in, he doesn't know if the switch is up because it started that way or because someone else entered before him.
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor D ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
lol, since this is the third time B has been selected as the answer (including the same logic), I'll clarify something in regards to the matter. It's a simple logical oversight: if B is the liar, that would also mean that there are two cake eaters, which as indicated in the original riddle, is not the case. Remember to focus on what the riddle is asking.
Not true, A person can be lying without every single thing in their sentence being untrue. If half the sentence is false, it is still a lie. For example: I live in the US and own a computer. That is a true statement. I live in the US, and do not own a computer. That is a lie, even though part of the statement is true.
If you don't like that example, try this one: I am a man who attends college. That is true. I am a woman who attends college. A lie, even though i do attend college. The same applies to this scenario. B's statement is "A didn't eat it" and "D didn't eat it". For him to be lying, only one of those has to be false. A didn't eat the cake, that must be true. However, D did eat the cake. since part of his statement was false the statement was a lie.
Why uh...thank you for the rather untoward presentation of lies but what you're disputing has honestly has no relevancy to the riddle at hand. At point blank, you're examples are also a facetious misrepresentation of girl B's statement. She sad neither A nor D ate the cake as one statement. If I were to take examples to your, I suppose flippant, level of abstract, it would be: Neither China nor the US nuked Russia. She made only a single claim that involves both parties, there is no "part" to it. If one of the two nuked Russia, that would have to mean the other nation did too. Or else the girl is lying; ergo there are more than two nukers. Juxtapose that with the cake, and you presto. Hence, your lecture didn't..establish much of anything.
P.S. No offense to Russians, nobody was nuked in the crafting of this explanation.
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor D ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
lol, since this is the third time B has been selected as the answer (including the same logic), I'll clarify something in regards to the matter. It's a simple logical oversight: if B is the liar, that would also mean that there are two cake eaters, which as indicated in the original riddle, is not the case. Remember to focus on what the riddle is asking.
Not true, A person can be lying without every single thing in their sentence being untrue. If half the sentence is false, it is still a lie. For example: I live in the US and own a computer. That is a true statement. I live in the US, and do not own a computer. That is a lie, even though part of the statement is true.
If you don't like that example, try this one: I am a man who attends college. That is true. I am a woman who attends college. A lie, even though i do attend college. The same applies to this scenario. B's statement is "A didn't eat it" and "D didn't eat it". For him to be lying, only one of those has to be false. A didn't eat the cake, that must be true. However, D did eat the cake. since part of his statement was false the statement was a lie.
Why uh...thank you for the rather untoward presentation of lies but what you're disputing has honestly has no relevancy to the riddle at hand. At point blank, you're examples are also a facetious misrepresentation of girl B's statement. She sad neither A nor D ate the cake as one statement. If I were to take examples to your, I suppose flippant, level of abstract, it would be: Neither China nor the US nuked Russia. She made only a single claim that involves both parties, there is no "part" to it. If one of the two nuked Russia, that would have to mean the other nation did too. Or else the girl is lying; ergo there are more than two nukers. Juxtapose that with the cake, and you presto. Hence, your lecture didn't..establish much of anything.
P.S. No offense to Russians, nobody was nuked in the crafting of this explanation.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory.
i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years.
Solution is more or less the same, simply have one of the two switches be the one the counter uses, and have people who aren't going to modify it flip the other switch. Say there's one on the left and one on the right, have the counter only count when the left switch is up, and have him flip it down when he goes in. Everyone else only flips the right switch, unless the left switch is down, then they flip it up. They only ever flip the left switch up twice, never more than that, even if it is down.
Personally, I prefer the method of "wait until I've gone 50 times." Assuming the selection is truly random and that the probability of choosing any given prisoner is approximately the same as choosing any other prisoner, the odds of having been chosen 50 times before every other person has gone at least once is absurdly small.
The problem is that the first time the counter goes in, he doesn't know if the switch is up because it started that way or because someone else entered before him.
1 person is the counter. the1st time the counter is in the room he moves switch no1 to off does not count anything but notes the position of the switches. Anyone else who comes in the room moves switch no1 to on if the switch is off. if its on then just moves switch no2. Everytime the counter goes in the room. He adds +1 if switch no1 is moved or adds 2 if both switches have changed position since he was last here. the minimum time it would take would be 11 visits or the maximum of 66. 66 being he visits 3 times ina row in between every unique visit..
btw i really want to know if there's an more efficient way to do this....
I don't think that works. In fact I can't see how this riddle is suppose to be solved with a counter. I think it needs to be more elaborate then that, and I don't have a clue as to how it's suppose to be solved. I would go ahead and just wait for a really really long time as someone else suggested and just bet on it
Lets say the counter comes in and sees ON ON-so he turns on 1 #1 (the counter) leaves the switches as OFF ON #2 changes it to ON ON #3 changes it to ON OFF #1 (the counter) sees that the switches are both different from when he was there, so he adds +2 and changes the switches back to OFF ON #2 changes it to ON ON #3 changes it to ON OFF #1 (the counter) sees that the switches are both different from when he was there, so he adds +2 and changes the switches back to OFF ON
The problem here is that theoretically #2 and #3, followed by #1, in an endless cycle. 1,2,3,1,2,3,1,2,3, and eventually #1 will reach any agreed upon number, (1000) and say "we've all been to the switch room" and be wrong, since in fact only 3 out of the 23 people have been there.
Five coworkers want to know what the average of all their salaries is, but refuse to reveal ANY information about their own salaries to their coworkers. How can they calculate the average?
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory.
i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years.
Solution is more or less the same, simply have one of the two switches be the one the counter uses, and have people who aren't going to modify it flip the other switch. Say there's one on the left and one on the right, have the counter only count when the left switch is up, and have him flip it down when he goes in. Everyone else only flips the right switch, unless the left switch is down, then they flip it up. They only ever flip the left switch up twice, never more than that, even if it is down.
Personally, I prefer the method of "wait until I've gone 50 times." Assuming the selection is truly random and that the probability of choosing any given prisoner is approximately the same as choosing any other prisoner, the odds of having been chosen 50 times before every other person has gone at least once is absurdly small.
The problem is that the first time the counter goes in, he doesn't know if the switch is up because it started that way or because someone else entered before him.
1 person is the counter. the1st time the counter is in the room he moves switch no1 to off does not count anything but notes the position of the switches. Anyone else who comes in the room moves switch no1 to on if the switch is off. if its on then just moves switch no2. Everytime the counter goes in the room. He adds +1 if switch no1 is moved or adds 2 if both switches have changed position since he was last here. the minimum time it would take would be 11 visits or the maximum of 66. 66 being he visits 3 times ina row in between every unique visit..
btw i really want to know if there's an more efficient way to do this....
I don't think that works. In fact I can't see how this riddle is suppose to be solved with a counter. I think it needs to be more elaborate then that, and I don't have a clue as to how it's suppose to be solved. I would go ahead and just wait for a really really long time as someone else suggested and just bet on it
Lets say the counter comes in and sees ON ON-so he turns on 1 #1 (the counter) leaves the switches as OFF ON #2 changes it to ON ON #3 changes it to ON OFF #1 (the counter) sees that the switches are both different from when he was there, so he adds +2 and changes the switches back to OFF ON #2 changes it to ON ON #3 changes it to ON OFF #1 (the counter) sees that the switches are both different from when he was there, so he adds +2 and changes the switches back to OFF ON
The problem here is that theoretically #2 and #3, followed by #1, in an endless cycle. 1,2,3,1,2,3,1,2,3, and eventually #1 will reach any agreed upon number, (1000) and say "we've all been to the switch room" and be wrong, since in fact only 3 out of the 23 people have been there.
I got stuck for a while with this...i could get the count to 22 but couldn't get the last one due to not knowing the starting positions of the switches.
The theory i used was pretty much the same as anyone elses, if the left switch was up, the prisoner would flick it down. they could only do it once, and had to be the first time they visited with the switch being up. IF they visited and it was down then they just flick the right switch. The counter always flicks the left switch up. The only circumstance in which the counter would see the switch was UP when he got there, was if he was the first prisoner, in which case he could easily count the 23 prisoners. Seeing as you dont know whether it started up or down, you have to assume it started down, and cant guarantee that the 23rd prisoner has flicked it down( the very first to enter the room, but #23 for your count).
BUT if each prisoner flicks the left switch down TWICE and the count goes to 45 he can guarantee that all 23 prisoners have been atleast once.
He would get a count of 44 for 22 prisoners(including himself, and the 45th would be the gaurantee that the last person had entered the room too. You can't say a count of 46 because again you wouldn't know if that very first time the counter entered and saw the switch as down whether it was due to a prisoner or starting position.
I hope that makes sense to someone else @@
so simply my solution is: If the counter enters the first time and sees the switch UP, he counts to 23(including himself twice) from his next visit. If the counter enters the first time and sees the switch DOWN, he counts to 45(including himself twice) from his next visit. The strategy being: The first 2 times that each prinsoner enters and sees the left switch UP, they flick it down. Every time after that, or if they visit and have not flicked it down twice, but the switch is already down, they flick the right switch instead. Only the counter flicks the left switch up, and only flicks it if it is down, otherwise flicks the right switch. The counter adds +1 for every visit after the first, in which he finds the left switch down, and flicks it up.
On April 20 2012 15:06 Aelfric wrote: Wow you guys are fast, then lets try something harder:
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
you have 5 5horse races. horses that place 4/5 are out. you write down what places the remaining horses place.
15 horses left in contention.
you have all the horses that placed 1st race against eachother. 4/5 are out. you have all the horses that placed 2nd race against eachother. 4/5 are out. you have all the horses that placed 3rd race against eachother. 4/5 are out.
you write down what places the remaining horses place.
The horse that placed 1st both races is your fastest horse.
8 horses left in contention.
you have 2 races of 4 horses each. 4th is out
6 horses left in contention.
you race 5 of your last 6 horses. 1 horse sits out. 4/5 are out.
4 horses left.
you race them. 1/2 are your 2nd/3rd fastest horse in that order.
the bottom 2 of each race are eliminated, 15/ 25 Left
#2)The winner of each group go into a second round
the winner of the race is the top horse
the 1st runner of this race stays (1st runner run of his group in #1 stays, 2nd runner up of his group in #2 eliminated as no chance to be in top 3) the 2nd runner of this race up stays (all other from his group in #1 got eliminated )
4th / 5th guy and all others belong to their group in #1 eliminated
6 lefts, with the top 1 determined
#3) the remaining five with uncertain seating run the last one
<--this sounds a lot like dual tournament format for some part
Race the winners of each group against each other. You got the fastest horse. Keep 2nd and 3rd (Lets call them #1, #2)
Race the 2nd place finishers from the group stages with each other. Keep 1st and 2nd (#3, #4) Race the 3rd place from the group stages with each other. Keep the winner here (#5)
Race #1-#5. Winner here is the second fastest. Runner up is third fastest.
So thats... 9?
actually im wrong lol. I just couldnt understand your wording.
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor D ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
lol, since this is the third time B has been selected as the answer (including the same logic), I'll clarify something in regards to the matter. It's a simple logical oversight: if B is the liar, that would also mean that there are two cake eaters, which as indicated in the original riddle, is not the case. Remember to focus on what the riddle is asking.
Not true, A person can be lying without every single thing in their sentence being untrue. If half the sentence is false, it is still a lie. For example: I live in the US and own a computer. That is a true statement. I live in the US, and do not own a computer. That is a lie, even though part of the statement is true.
If you don't like that example, try this one: I am a man who attends college. That is true. I am a woman who attends college. A lie, even though i do attend college. The same applies to this scenario. B's statement is "A didn't eat it" and "D didn't eat it". For him to be lying, only one of those has to be false. A didn't eat the cake, that must be true. However, D did eat the cake. since part of his statement was false the statement was a lie.
Why uh...thank you for the rather untoward presentation of lies but what you're disputing has honestly has no relevancy to the riddle at hand. At point blank, you're examples are also a facetious misrepresentation of girl B's statement. She sad neither A nor D ate the cake as one statement. If I were to take examples to your, I suppose flippant, level of abstract, it would be: Neither China nor the US nuked Russia. She made only a single claim that involves both parties, there is no "part" to it. If one of the two nuked Russia, that would have to mean the other nation did too. Or else the girl is lying; ergo there are more than two nukers. Juxtapose that with the cake, and you presto. Hence, your lecture didn't..establish much of anything.
P.S. No offense to Russians, nobody was nuked in the crafting of this explanation.
Taekwon have you taken a sentential logic class before? Just curious. Because I still think Taulon is correct in his logic.
A=I will punch you in the face B=I will take your wallet
Either A or B What makes this sentence true? 1. A (I punch you in the face) (or) 2. B (I take your wallet)
What makes this sentence false? 1. A&B (I both punch you in the face and take your wallet) This makes my bolded statement false because I said I will do either A OR B, not A AND B) (or) 2. Not A&B (I don't punch you in the face, and I also don't take your wallet) This makes the bolded statement false because I said Either A Or B. I will pick one.
-In other words, I can do one or the other to be true. If I do both or neither, it's false. This is because I used the word "or" and not "and"
Now, lets say instead: Neither A nor B What makes this sentence true? 1. Not A and Not B (I don't go to the store, and don't play sc2)
What makes this sentence false? 1. A and Not B (I go to the store and don't play sc2) 2. Not A and B (I go to the store and play sc2) 3. Not A&B (I don't go to the store and play sc2)
Long story short: In sentential logic, the phrase A nor B translates to Not A and Not B, or not (A or B)
For reference: A: "I didn't eat it!" B: "Neither A nor D ate it." C: "I swear on my momma's grave I didn't eat it!" D: "C is telling the truth!"
Lets take B's sentence. Neither A nor D ate it. This means that A did not eat it, AND D did not eat it. They both have to be in conjunction, or in other words, you can't have one without the other.
Lets pretend we KNOW B is lying. We can't 100% deduce that both A and D ate the cake. If A ate the cake, B is lying. If D ate the cake, B is lying. If both ate the cake, B is lying.
Imo the riddle is worded a bit poorly if Not (Not A nor D) is suppose to translate to A&D
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
7 races. Break the group into 5 sets called A-E, and number determine their ranking in that set. Race1-5:A,B,C,D,E. Race6: A1,B1,C1,D1,E1. Say that ABC are the fastest in Race 6. We know A1 is the fastest horse. Race 7: A2,A3,B1,B2,C1
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor D ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!"
lol, since this is the third time B has been selected as the answer (including the same logic), I'll clarify something in regards to the matter. It's a simple logical oversight: if B is the liar, that would also mean that there are two cake eaters, which as indicated in the original riddle, is not the case. Remember to focus on what the riddle is asking.
Not true, A person can be lying without every single thing in their sentence being untrue. If half the sentence is false, it is still a lie. For example: I live in the US and own a computer. That is a true statement. I live in the US, and do not own a computer. That is a lie, even though part of the statement is true.
If you don't like that example, try this one: I am a man who attends college. That is true. I am a woman who attends college. A lie, even though i do attend college. The same applies to this scenario. B's statement is "A didn't eat it" and "D didn't eat it". For him to be lying, only one of those has to be false. A didn't eat the cake, that must be true. However, D did eat the cake. since part of his statement was false the statement was a lie.
Why uh...thank you for the rather untoward presentation of lies but what you're disputing has honestly has no relevancy to the riddle at hand. At point blank, you're examples are also a facetious misrepresentation of girl B's statement. She sad neither A nor D ate the cake as one statement. If I were to take examples to your, I suppose flippant, level of abstract, it would be: Neither China nor the US nuked Russia. She made only a single claim that involves both parties, there is no "part" to it. If one of the two nuked Russia, that would have to mean the other nation did too. Or else the girl is lying; ergo there are more than two nukers. Juxtapose that with the cake, and you presto. Hence, your lecture didn't..establish much of anything.
P.S. No offense to Russians, nobody was nuked in the crafting of this explanation.
Taekwon have you taken a sentential logic class before? Just curious. Because I still think Taulon is correct in his logic.
A=I will punch you in the face B=I will take your wallet
Either A or B What makes this sentence true? 1. A (I punch you in the face) (or) 2. B (I take your wallet)
What makes this sentence false? 1. A&B (I both punch you in the face and take your wallet) This makes my bolded statement false because I said I will do either A OR B, not A AND B) (or) 2. Not A&B (I don't punch you in the face, and I also don't take your wallet) This makes the bolded statement false because I said Either A Or B. I will pick one.
-In other words, I can do one or the other to be true. If I do both or neither, it's false. This is because I used the word "or" and not "and"
Now, lets say instead: Neither A nor B What makes this sentence true? 1. Not A and Not B (I don't go to the store, and don't play sc2)
What makes this sentence false? 1. A and Not B (I go to the store and don't play sc2) 2. Not A and B (I go to the store and play sc2) 3. Not A&B (I don't go to the store and play sc2)
Long story short: In sentential logic, the phrase A nor B translates to Not A and Not B, or not (A or B)
For reference: A: "I didn't eat it!" B: "Neither A nor D ate it." C: "I swear on my momma's grave I didn't eat it!" D: "C is telling the truth!"
Lets take B's sentence. Neither A nor D ate it. This means that A did not eat it, AND D did not eat it. They both have to be in conjunction, or in other words, you can't have one without the other.
Lets pretend we KNOW B is lying. We can't 100% deduce that both A and D ate the cake. If A ate the cake, B is lying. If D ate the cake, B is lying. If both ate the cake, B is lying.
Imo the riddle is worded a bit poorly if Not (Not A nor D) is suppose to translate to A&D
Hmmm the warden problem is difficult. I wrote a really long solution, then before I posted it I realized there was a hole in my theory and scrapped it
The biggest problem is the lack of communication between the prisoners, since they have no way to relay information (unless there's something I'm missing)
EDIT: I missed the part where they were able to collude with each other at the start, before they go into individual cells. So the solution I thought up was correct after all haha.
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
9 Races total 5 races of 5 horses. 4/5 out 15 left Winners of first race race. winner is fastest horse. 4/5 out. 2nd Place of first race race. 3/4/5 out. (if a horse got second place in the first race he CAN'T be fastest horse so only 2 go on) 3rd Place of first race race. 2/3/4/5 out. (if a horse got 3rd place in the first race he can ONLY at best be 3rd fastest horse 1 goes 1) 5 left Race 5. 1/2 are 2nd/3rd fastest.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers. for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.
I address that point explicitly in my original solution in the following sentence: + Show Spoiler +
his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers. for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.
I address that point explicitly in my original solution in the following sentence: + Show Spoiler +
his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
You say that the counter adds 1 if the left lever is up, and flips it down. You also say his count is 1 when he leaves the room for the first time, regardless of the state of the levers. Everyone else flips the left lever up the first chance they get, but not subsequently.
Suppose the left lever starts down. A non-counter enters the room first (unknowingly) and flicks the left lever up. Some time later, the counter enters the room for the first time. The lever is up. You say he flips it down and leaves the room with a count of 1 (himself). But the non-counter who got there first will never flip the left lever again, so will never be counted. The counter can never reach 23.
Nor can the counter assume the above happened and start his count at 2 just because he found the lever up (since the lever might have started in the up position), because in that case he would declare victory one too soon.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers. for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.
I address that point explicitly in my original solution in the following sentence: + Show Spoiler +
his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
You say that the counter adds 1 if the left lever is up, and flips it down. You also say his count is 1 when he leaves the room for the first time, regardless of the state of the levers. Everyone else flips the left lever up the first chance they get, but not subsequently.
Suppose the left lever starts down. A non-counter enters the room first (unknowingly) and flicks the left lever up. Some time later, the counter enters the room for the first time. The lever is up. You say he flips it down and leaves the room with a count of 1 (himself). But the non-counter who got there first will never flip the left lever again, so will never be counted. The counter can never reach 23.
Nor can the counter assume the above happened and start his count at 2 just because he found the lever up (since the lever might have started in the up position), because in that case he would declare victory one too soon.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers. for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.
I address that point explicitly in my original solution in the following sentence: + Show Spoiler +
his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
You say that the counter adds 1 if the left lever is up, and flips it down. You also say his count is 1 when he leaves the room for the first time, regardless of the state of the levers. Everyone else flips the left lever up the first chance they get, but not subsequently.
Suppose the left lever starts down. A non-counter enters the room first (unknowingly) and flicks the left lever up. Some time later, the counter enters the room for the first time. The lever is up. You say he flips it down and leaves the room with a count of 1 (himself). But the non-counter who got there first will never flip the left lever again, so will never be counted. The counter can never reach 23.
Nor can the counter assume the above happened and start his count at 2 just because he found the lever up (since the lever might have started in the up position), because in that case he would declare victory one too soon.
Five coworkers want to know what the average of all their salaries is, but refuse to reveal ANY information about their own salaries to their coworkers. How can they calculate the average?
Five coworkers want to know what the average of all their salaries is, but refuse to reveal ANY information about their own salaries to their coworkers. How can they calculate the average?
OK, let's have a go at this. Gonna try a 3 person version first and see where I get.
Let's call these people a, b and c with salaries A, B and C.
Then a needs to find out (B + C) without ever being able to individually deduce B or C, b needs to find out (A + C), etc.
Then b pretty much has to give some altered (but undecipherable) form of his salary to c, so c can pass the (altered) sum on to a. The complete sum has to come back to b to be deciphered, since otherwise one of the others would be able to use the information to deduce his salary.
So, let's have a, b and c generate some arbitrary numbers and hand them out. - a thinks of a number D and tells b only - b thinks of a number E and tells c only - c thinks of a number F and tells a only
Then each person multiplies their salary by the number and gives it to the third person: - a tells AF to b (who doesn't know F) - b tells BD to c (who doesn't know D) - c tells CE to a (who doesn't know E)
Now we can give salted sums of 2 salaries for the 3rd person to add his. Although the 3rd person knows the multiplier, he's given a sum of 2 salaries so can cannot work out either. - a tells (A + CE) to b - b tells (B + AF) to c - c tells (C + BD) to a
We can now get people to put together all 3 salaries, and tell this sum to the person with the modified salary in the sum to decipher: - a tells (A + BD + C) to b - b tells (A + B + CE) to c - c tells (AF + B + C) to a
Now each of the people can subtract (their own salary * the multiplier they were given at the start), add their own salary, and divide by 3 to get the average.
At no point is anyone able to calculate either of the other two people's salaries.
Let's call these people a, b, c, d and e with salaries A, B, C, D and E.
So, let's have a, b and c generate some arbitrary numbers and hand them out. - a thinks of a number F and tells b only - b thinks of a number G and tells c only - c thinks of a number H and tells d only - d thinks of a number I and tells e only - e thinks of a number J and tells a only
Then each person multiplies their salary by the number and gives it to the next person: - a tells AJ to b (who doesn't know J) - b tells BF to c (who doesn't know F) - c tells CG to d (who doesn't know G) - d tells DH to e (who doesn't know H) - e tells EI to a (who doesn't know I)
Now we can give salted sums of 2 salaries for a 3rd person to add his. - a tells (A + EI) to b - b tells (B + AJ) to c - c tells (C + BF) to d - d tells (D + CG) to e - e tells (E + DH) to a
And again each person adds their salary and passes it around again, for a 4th person to add his. - a tells (A + DH + E) to b - b tells (A + B + EI) to c - c tells (AJ + B + C) to d - d tells (BF + C + D) to e - e tells (CG + D + E) to a
And again. - a tells (A + CG + D + E) to b - b tells (A + B + DH + E) to c - c tells (A + B + C + EI) to d - d tells (AJ + B + C + D) to e - e tells (BF + C + D + E) to a
One final time, to pass the final sum around. The final sum arrives at the person whose own salary is altered, so they can decipher it. - a tells (A + BF + C + D + E) to b - b tells (A + B + CG + D + E) to c - c tells (A + B + C + DH + E) to d - d tells (A + B + C + D + EI) to e - e tells (AJ + B + C + D + E) to a
Now each of the people can subtract (their own salary * the multiplier they were given at the start), add their own salary, and divide by 5 to get the average.
At no point is anyone able to calculate any other the other people's salaries.
EDIT2: That's actually a really convoluted way of doing it wasn't it!! Simple solution below
Thinking about it, was it completely unnecessary to have someone else generate a multiplier in the first step?
All we actually need is for each person to generate a random number, get everyone else to add their salary to it in turn, and then when it gets back to each person only they can subtract their random number and add their salary to get the total :p
Five coworkers want to know what the average of all their salaries is, but refuse to reveal ANY information about their own salaries to their coworkers. How can they calculate the average?
(Working out the average of salaries without telling anyone what your salary is) + Show Spoiler +
The simplest way is to find a trusted 3rd party who promises not to tell anyone what anyone else's salary is, have everyone tell that person their salary, and they work out the average and let everyone know what it is. But if you can't find such a person, there's another way...
Arrange the cow-orkers in a line, and have them all pick a random number and remember it (without telling anyone else). This number is allowed to be negative, and can be as big as they like.
The first person adds his salary and random number together, and tells that sum to the next person (without telling anyone else). They add on their salary and their random number, and tell the new total to the next person, who does the same, through all of them.
The last person, after adding on her salary and random number, tells her result to the first person.
The first person then subtracts his random number from the number he was told, and passes the answer on to the second person. They subtract their random number, pass the result on, down the line again. The last person, after subtracting her random number, announces the number to everybody. That is the total salary of them all, and they just have to divide by the number of people to get the average.
For the first person, the only information they get passed is the sum of everyone's salary, obscured by all their random numbers. He can't figure anything out from that.
For everyone else except the last person, the two numbers they are told are obscured either by the random numbers of the people in front of them (the first time through) or after them (the second time through). So they can't tell anything either.
The last person has nobody after them, so the number they get the second time through tells them something! Well no, not really, since all it tells them is the total salary of everyone else, and everyone else can work out the same thing once they are told the total salary, just by subtracting their own salary. So the system does not reveal any additional information about anyone else's salary to anyone. QED
(however, the system is not perfect. If the people standing on either side of someone get together, they can work out that person's salary without revealing to each other anything more about their own salaries.)
Horse question solution. Haven't looked this up or anything but pretty confident. + Show Spoiler +
9 races. I'll break them into lettered groups.
A: 5 races of 5 horses each. Group stages. B: 1 race of 5 winners of group stages A. Winner is your fastest horse. C: 1 race of runner ups of A. D: 1 race of 3rd place finishers of A. E: 1 race to determine 2nd/3rd fastest horses: -2nd and 3rd place of race B -1st and 2nd place of race C -1st place of race D
Rationale: -Race B winner is fastest horse, so you only have to worry about the other 2 slots. -Race D: to qualify, the horse finished 3rd in its original group, meaning 2 horses were already faster. Therefore only the winner in Race D has a chance to be among the top 3 overall (or it'd be at best 4th, behind the 1st/2nd place in its race A, and 1st in race D). -Similar rationale for Race C: only the 1st/2nd place finishers have a chance to be top 3 overall, or else they'd be at best 4th (behind 1st place in their race A, and 1st/2nd in race C). -You have exactly 5 potential horses remaining for race E to determine your 2nd/3rd fastest horses overall.
EDIT: This assumes you couldn't identify horses across races. If you could, it would be easier, and take 7 races.
Same A and B as above. Then one race of: -2nd and 3rd place of race B -2nd and 3rd place of the race A that had the horse who won race B -2nd place of the race A that had the horse who was 2nd in race B
Reason here being if you finished 2nd/3rd in a race where the winner wasn't the top 3, there's no chance that you're one of the top 3.
I'm sending the detailed answer of Horse Race question via pm to whoever gets close to the solution. It may have been already asked, but here is a new one to keep you busy:
Your enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.
Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?
On April 21 2012 01:36 Aelfric wrote: I'm sending the detailed answer of Horse Race question via pm to whoever gets close to the solution. It may have been already asked, but here is a new one to keep you busy:
Your enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.
Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?
You don't spin the barrel. If you spin it, you have a 1/3 chance of landing on a chamber with a bullet, and 2/3 chance of landing on a chamber without one. If you don't spin it, you have a 1/4 chance of the next chamber having a bullet, and a 3/4 chance of the next chamber being empty. 3/4 > 2/3.
On April 21 2012 01:36 Aelfric wrote: I'm sending the detailed answer of Horse Race question via pm to whoever gets close to the solution. It may have been already asked, but here is a new one to keep you busy:
Your enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.
Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?
Statistically, you should not spin it. The chance is 1/4 that the first shot was in the position near the 2 consecutive bullets. There is a 3/4 or 75% chance that it was in locations 1-3 of the 4 consecutive empty slots. Some people are going to argue for the slots that are filled with bullets should be taken into the equation, but knowing that the first shot was a miss means that you know you are dealing with bullet slots 1-5 as the very worst spot the bullet could be at this moment is in slot number 5. there is only a 25% chance it is there. Because you know as soon as the first shot doesnt go off that it its NOT in slot 1. Therefore it can only be slots 2-5.
If you spin the gun you chance of survival drops from 75% to 66%. Its a 6 round chamber, and 2/6 or 1/3 are bullets. You are giving an opportunity for it to land on any of the 6. Whereas before there is only 1 bullet to 3 empty slots possible.
On April 21 2012 01:36 Aelfric wrote: I'm sending the detailed answer of Horse Race question via pm to whoever gets close to the solution. It may have been already asked, but here is a new one to keep you busy:
Your enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.
Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?
On April 21 2012 01:36 Aelfric wrote: I'm sending the detailed answer of Horse Race question via pm to whoever gets close to the solution. It may have been already asked, but here is a new one to keep you busy:
Your enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.
Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?
On April 21 2012 01:36 Aelfric wrote: I'm sending the detailed answer of Horse Race question via pm to whoever gets close to the solution. It may have been already asked, but here is a new one to keep you busy:
Your enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.
Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?
initial 5, where we eliminate the 2 slowes from each group. 15 horses remain.
then race all the winning horses.. winner is the fastest, the bottom two eliminated, and the 4 remaining horses that lost to the bottom 2 in the first race are eliminated. for the horse that came in 3rd the winners race, we can eliminate the horses that lost to him( 2 horses). we can also elimate the third place horse from the initial race of the horse who came in second in the winners race.. 2+4+2+1 = 9 horses eliminated.
6 horses remain, but the winner of the winners race does not have to race again. so 5 horses, one more race, top 2 are 2nd/third fastest.
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
The winners of the 5 group races will race. (winner's race) (this will be the fastest horse)
In the final race (consolation race) race the 2nd, 3rd from the group of the fastest horse and the 2nd fastest horse from the group of 2nd fastest in the winners race, and the 2nd and 3rd fastest in winner's race. (1st and 2nd will be 2nd and 3rd respectively)
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
Does this really need a [SFW] tag? Imagine this situation:
A man is at work, browsing teamliquid.net. He sees the title "Riddles / Puzzles / Brain Teasers", but dares not click it for fear of the likelyhood of it containing content unsafe for work.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
I'm confused. You listed 3 scenarios. One out of the three scenarios has the second ball being orange. However, one of those scenarios has no orange ball in the first place. Therefore you need only consider Orange, Orange and Orange, White. Therefore you have 1 chance out of 2 choices, 50%.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
I'm confused. You listed 3 scenarios. One out of the three scenarios has the second ball being orange. However, one of those scenarios has no orange ball in the first place. Therefore you need only consider Orange, Orange and Orange, White. Therefore you have 1 chance out of 2 choices, 50%.
This question is similar to the one I posted, and it really depends on how you interpret the question.
Interpretation 1: Say you reach into a random bag and pull out a random ball 30 times. 10 times you will pull an orange ball out of the OO bag, 5 times you will pull an orange ball out of the OW bag, 5 times you will pull a white ball out of the OW bag and 10 times you will pull a white ball out of the WW bag. You can see that in this case that out of the 15 times you pull an orange ball 10 times the other ball is also orange. Answer = 10/15 = 66%. You were more likely to reveal an orange ball form the OO bag than from the OW bag, so you are most likely reaching in the OO bag.
Interpretation 2: You always pull out an orange ball if possible because the question says you do. If you do this 30 times, 10 times you will pull an orange ball lout of the OO bag, 10 times you will pull an orange ball out of the OW bag and 10 times you will pull an White ball out of the WW bag. You will never pull a white ball out of the OW bag first in this interpretation of the question. You can see that out of the 20 times you pull an orange ball, 10 of those have another orange ball inside, so this gives an answer of 10/20 = 50%.
I wish there was an easier way of describing the randomness this type of question. For example, in Monty Hall style questions, it is always stated that the revealed choice is not random (interpretation 2).
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
I'm confused. You listed 3 scenarios. One out of the three scenarios has the second ball being orange. However, one of those scenarios has no orange ball in the first place. Therefore you need only consider Orange, Orange and Orange, White. Therefore you have 1 chance out of 2 choices, 50%.
This question is similar to the one I posted, and it really depends on how you interpret the question.
Interpretation 1: Say you reach into a random bag and pull out a random ball 30 times. 10 times you will pull an orange ball out of the OO bag, 5 times you will pull an orange ball out of the OW bag, 5 times you will pull a white ball out of the OW bag and 10 times you will pull a white ball out of the WW bag. You can see that in this case that out of the 15 times you pull an orange ball the other ball is also orange. Answer = 66%. You were more likely to reveal an orange ball form the OO bag than from the OW bag, so you are most likely reaching in the OO bag.
Interpretation 2: You always pull out an orange ball if possible because the question says you do. If you do this 30 times, 10 times you will pull an orange ball lout of the OO bag, 10 times you will pull an orange ball out of the OW bag and 10 times you will pull an White ball out of the WW bag. You can see that out of the 20 times you pull an orange ball, 10 of those have another orange ball inside, so this gives an answer of 50%.
I wish there was an easier way of describing the randomness this type of question. For example, in Monty Hall style questions, it is always stated that the revealed choice is not random (interpretation 2).
Recently heard a neat twist on an old puzzle. The original puzzle:
Suppose you have 10 jars each with unlimited marbles. 9 of the jars contain marbles weighing 1 gram and one of the jars contains marbles weighing 1.1 grams. You have a digital scale that will tell you the exact weight of what you put on it. How can you tell which jar has the heavier marbles in 1 weighing?
The twist: Now suppose that you don't know how many jars have marbles that weigh 1 gram and how many jars have marbles that weigh 1.1 grams (but you still have 10 really big (infinite) jars total). How can you tell which jars have heavier marbles in just 1 weighing?
EDIT: My mistake, the problem is more interesting if you assume you have fewer than 4000 marbles in each jar. The unlimited case is somewhat trivial.
On April 21 2012 04:59 nq07 wrote: Recently heard a neat twist on an old puzzle. The original puzzle:
Suppose you have 10 jars each with unlimited marbles. 9 of the jars contain marbles weighing 1 gram and one of the jars contains marbles weighing 1.1 grams. You have a digital scale that will tell you the exact weight of what you put on it. How can you tell which jar has the heavier marbles in 1 weighing?
The twist: Now suppose that you don't know how many jars have marbles that weigh 1 gram and how many jars have marbles that weigh 1.1 grams (but you still have 10 really big (infinite) jars total). How can you tell which jars have heavier marbles in just 1 weighing?
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
On April 21 2012 04:59 nq07 wrote: Recently heard a neat twist on an old puzzle. The original puzzle:
Suppose you have 10 jars each with unlimited marbles. 9 of the jars contain marbles weighing 1 gram and one of the jars contains marbles weighing 1.1 grams. You have a digital scale that will tell you the exact weight of what you put on it. How can you tell which jar has the heavier marbles in 1 weighing?
The twist: Now suppose that you don't know how many jars have marbles that weigh 1 gram and how many jars have marbles that weigh 1.1 grams (but you still have 10 really big (infinite) jars total). How can you tell which jars have heavier marbles in just 1 weighing?
On April 21 2012 03:25 BallerIndustries wrote: Does this really need a [SFW] tag? Imagine this situation:
A man is at work, browsing teamliquid.net. He sees the title "Riddles / Puzzles / Brain Teasers", but dares not click it for fear of the likelyhood of it containing content unsafe for work.
The [SFW] tag means it's safe to view at work. If it were not, it would have the [NSFW] tag..
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
But the problem implies that after reaching into a random pocket, you also randomly select one of the two balls in that pocket (otherwise, how would you end up with a ball)? So in fact the problem is implying random ball, in which case the 2/3 interpretation would seem to be correct?
On April 21 2012 03:52 Whitewing wrote: If a chicken and a half can lay an egg and a half in a day and a half, how many eggs could six chickens lay in six days?
6 days is 1.5*4. So 1.5 chickens will lay 1.5*4 eggs in 1.5*4 days.
Now 1.5*4 = 6 eggs
But there are 6 chickens, which is equivalent to 1.5*4. So multiplying 4 "groups" of 1.5 chickens by the output of each chicken group for 6 days, we get:
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
On April 21 2012 03:52 Whitewing wrote: If a chicken and a half can lay an egg and a half in a day and a half, how many eggs could six chickens lay in six days?
I don't think half a chicken is able to lay eggs. So it's just 1 chicken that lays 1.5 eggs in 1.5 days. Which basically means 1 chicken lays 1 egg per day. Based on this 6 chicken would lay 6 eggs in a day and in six days that would be 36 eggs.
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
The group designates a single person the "counter". This person is the only one who will reset switch A to the bottom position. A starts in the down position. If it is your first visit to the room, you move switch A to up. If switch A is in the up position, you don't touch it, you just move B (also, if you have been there before you move B). If the counter finds the switch in the up position, he moves it down and adds +1 to his count. When the counter's count reaches 22, he frees himself and all his co-prisoners!
your version doesn't account for the fact that the switch could be up to begin with^^
all he has to add to ensure that his solution is correct is add a couple conditions: 1) they decide who the counter is beforehand. 2) he doesn't start his count until after the first time he adjusts the switch
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
But the problem implies that after reaching into a random pocket, you also randomly select one of the two balls in that pocket (otherwise, how would you end up with a ball)? So in fact the problem is implying random ball, in which case the 2/3 interpretation would seem to be correct?
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
But the problem implies that after reaching into a random pocket, you also randomly select one of the two balls in that pocket (otherwise, how would you end up with a ball)? So in fact the problem is implying random ball, in which case the 2/3 interpretation would seem to be correct?
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
Agreed. Answer is definitely 1/2 since they stated that you have already pulled out an orange ping pong ball. That means you are either in the pocket (lets call it 1) with 2 oranges or 1 orange and 1 white (lets call this 2). If you already chose 1 ping pong ball, the chances the other 1 is orange depends on which pocket you are in. Either you are in pocket 1 or pocket 2 which will give you a 50% chance of getting another orange.
At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar?
The special is in effect equivalent to buy 2 beers get 3rd free with the caveat that you need to have an empty 3rd glass before you get that free one. The goal is therefore to drink 15 beers based on your initial haul of 10.
So you drink 10 + 3 + 1 and manage 14 beers with two empty glasses. For the 15th beer you grab a friend's full beer, drink it, and turn in the empty 3 glasses to give your friend his beer back.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
But the problem implies that after reaching into a random pocket, you also randomly select one of the two balls in that pocket (otherwise, how would you end up with a ball)? So in fact the problem is implying random ball, in which case the 2/3 interpretation would seem to be correct?
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
The one orange ball that is pull out is the given event. The given event is NOT that you picked a pocket with at least one orange ball.
Based on the given, we count three possibilities, and two chances to be in the pocket with two orange balls and one chance to be in the pocket with the pocket with one of each. The probability is clearly 2/3.
At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar?
We don't have enough information to answer this question.
If you and your friends are male, the group would be able to drink 14 beers split between them, plus any empties you can snatch from other tables while they're not looking.
At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar?
At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar?
You can get 15 out of it. Buy 10 and you can drink 9 of them, getting 3 new. Drink the 3 new ones and get a new-new one. Then borrow a beer, drink these 3 (the new-new, the borrowed and the one from the original 10) and return the last free beer to the person you borrowed from.
At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar?
You can get 15 out of it. Buy 10 and you can drink 9 of them, getting 3 new. Drink the 3 new ones and get a new-new one. Then borrow a beer, drink these 3 (the new-new, the borrowed and the one from the original 10) and return the last free beer to the person you borrowed from.
At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar?
The special is in effect equivalent to buy 2 beers get 3rd free with the caveat that you need to have an empty 3rd glass before you get that free one. The goal is therefore to drink 15 beers based on your initial haul of 10.
So you drink 10 + 3 + 1 and manage 14 beers with two empty glasses. For the 15th beer you grab a friend's full beer, drink it, and turn in the empty 3 glasses to give your friend his beer back.
Your answer assumes that you have 11 beers, not 10. If you can borrow 1 beer from another patron, why not collect empty glasses from them as well or panhandle outside for more money? The whole "riddle" falls apart when you introduce borrowing.
At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar?
The special is in effect equivalent to buy 2 beers get 3rd free with the caveat that you need to have an empty 3rd glass before you get that free one. The goal is therefore to drink 15 beers based on your initial haul of 10.
So you drink 10 + 3 + 1 and manage 14 beers with two empty glasses. For the 15th beer you grab a friend's full beer, drink it, and turn in the empty 3 glasses to give your friend his beer back.
Your answer assumes that you have 11 beers, not 10. If you can borrow 1 beer from another patron, why not collect empty glasses from them as well or panhandle outside for more money? The whole "riddle" falls apart when you introduce borrowing.
The point is that you return to the friend exactly what was borrowed so that he is not affected in any way. Collecting empty glasses or panhandling outside for more money would not be analogous.
On April 21 2012 03:52 Whitewing wrote: If a chicken and a half can lay an egg and a half in a day and a half, how many eggs could six chickens lay in six days?
6 days is 1.5*4. So 1.5 chickens will lay 1.5*4 eggs in 1.5*4 days.
Now 1.5*4 = 6 eggs
But there are 6 chickens, which is equivalent to 1.5*4. So multiplying 4 "groups" of 1.5 chickens by the output of each chicken group for 6 days, we get:
On April 21 2012 03:52 Whitewing wrote: If a chicken and a half can lay an egg and a half in a day and a half, how many eggs could six chickens lay in six days?
I don't think half a chicken is able to lay eggs. So it's just 1 chicken that lays 1.5 eggs in 1.5 days. Which basically means 1 chicken lays 1 egg per day. Based on this 6 chicken would lay 6 eggs in a day and in six days that would be 36 eggs.
No, the riddle explicitly states that a chicken and a half can indeed lay one egg and a half in a day and a half. Your answer is incorrect.
You are trapped in a perfectly cylindrical room with perfectly smooth walls, no friction at all. There is a light at the top of the room so that you can see, but it is a good 10 feet above your head. Your only possessions inside this room are a mirror and a table. The walls are sufficiently strong that you have no hope of damaging them directly.
On April 21 2012 07:53 Whitewing wrote: Pure word riddle for you: good luck!
You are trapped in a perfectly cylindrical room with perfectly smooth walls, no friction at all. There is a light at the top of the room so that you can see, but it is a good 10 feet above your head. Your only possessions inside this room are a mirror and a table. The walls are sufficiently strong that you have no hope of damaging them directly.
A solution from a simpler puzzle should work here - we don't need to use the mirror or the light.
1: Bang your hand on the table until it's sore. 2: Use the saw to cut the table in half. 3: Put the two halves together on the wall to make a whole. 4: Climb out the hole. (optionally) 5: Yell until you're hoarse. 6: Ride off into the sunset.
You have 100 marbles and two empty marble jars. To start you place 1 marble in each. Then you take time putting the marbles one at a time into either of the two jars. For each marble, the probability that the marble goes into a jar is equal to the fraction of number of marbles that the jar holds of the number of marbles in the two jars (example jar A has 2 marbles, jar B has 5 marbles; the probability is 2/7 and 5/7 respectively.)
You're also running a gambling ring. What is your betting line for the number of marbles in the jar with fewer marbles?
On April 21 2012 07:53 Whitewing wrote: Pure word riddle for you: good luck!
You are trapped in a perfectly cylindrical room with perfectly smooth walls, no friction at all. There is a light at the top of the room so that you can see, but it is a good 10 feet above your head. Your only possessions inside this room are a mirror and a table. The walls are sufficiently strong that you have no hope of damaging them directly.
A solution from a simpler puzzle should work here - we don't need to use the mirror or the light.
1: Bang your hand on the table until it's sore. 2: Use the saw to cut the table in half. 3: Put the two halves together on the wall to make a whole. 4: Climb out the hole. (optionally) 5: Yell until you're hoarse. 6: Ride off into the sunset.
There was once a very wise blind man, and the town philosopher was jealous of the man's popularity. One day, he took a full deck of 52 cards, turned all the Aces, Kings, and Queens face up, shuffled the deck well and handed the deck to the blind man. He challenged the wise man to split the cards into piles such that the number of cards facing up in all of the piles was equal. Without pausing for a beat, the wise man counted off a few cards, split the decks into piles, and completed the challenge in a matter of seconds. How did the wise man do it?
On April 21 2012 08:31 TanGeng wrote: Marble Probability
You have 100 marbles and two empty marble jars. To start you place 1 marble in each. Then you take time putting the marbles one at a time into either of the two jars. For each marble, the probability that the marble goes into a jar is equal to the fraction of number of marbles that the jar holds of the number of marbles in the two jars (example jar A has 2 marbles, jar B has 5 marbles; the probability is 2/7 and 5/7 respectively.)
You're also running a gambling ring. What is your betting line for the number of marbles in the jar with fewer marbles?
On April 21 2012 09:40 TanGeng wrote: Blind Man's Deck
There was once a very wise blind man, and the town philosopher was jealous of the man's popularity. One day, he took a full deck of 52 cards, turned all the Aces, Kings, and Queens face up, shuffled the deck well and handed the deck to the blind man. He challenged the wise man to split the cards into piles such that the number of cards facing up in all of the piles was equal. Without pausing for a beat, the wise man counted off a few cards, split the decks into piles, and completed the challenge in a matter of seconds. How did the wise man do it?
Trivial solution: Split the cards into 1 pile. Optionally, flip some of the over. All 1 piles have the same number of face up cards!
2 pile solution: Count off 12 cards into one pile. Leave the other 40 cards in the second pile. Flip the first pile upside-down. If n is the number of aces/kings/queens in the first pile (0 <= n <= 12), there are now 12-n face-up cards in each pile.
n pile solution (n>2): Impossible. Proof left as an exercise for the reader
On April 21 2012 07:53 Whitewing wrote: Pure word riddle for you: good luck!
You are trapped in a perfectly cylindrical room with perfectly smooth walls, no friction at all. There is a light at the top of the room so that you can see, but it is a good 10 feet above your head. Your only possessions inside this room are a mirror and a table. The walls are sufficiently strong that you have no hope of damaging them directly.
On April 21 2012 07:53 Whitewing wrote: Pure word riddle for you: good luck!
You are trapped in a perfectly cylindrical room with perfectly smooth walls, no friction at all. There is a light at the top of the room so that you can see, but it is a good 10 feet above your head. Your only possessions inside this room are a mirror and a table. The walls are sufficiently strong that you have no hope of damaging them directly.
The roof falls down because the walls have no friction. You hide under the table to survive and make sure the mirror is higher than the table. The roof breaks the mirror so gets bad luck and breaks when it hits the table. You stack the bits of roof to make stair and walk out of the hole where the roof used to be.
On April 21 2012 07:53 Whitewing wrote: Pure word riddle for you: good luck!
You are trapped in a perfectly cylindrical room with perfectly smooth walls, no friction at all. There is a light at the top of the room so that you can see, but it is a good 10 feet above your head. Your only possessions inside this room are a mirror and a table. The walls are sufficiently strong that you have no hope of damaging them directly.
Always heard this riddle when I was younger. "You look in the mirror and see what you saw, take the saw and cut the table in half. Two halves make a whole (hole) and you climb out the hole and leave."
On April 21 2012 07:53 Whitewing wrote: Pure word riddle for you: good luck!
You are trapped in a perfectly cylindrical room with perfectly smooth walls, no friction at all. There is a light at the top of the room so that you can see, but it is a good 10 feet above your head. Your only possessions inside this room are a mirror and a table. The walls are sufficiently strong that you have no hope of damaging them directly.
1. Use mirror to focus light on table to laser cut table into ten foot shaped sections. 2. Stack the ten foot shaped sections into a neat pile. 3. Stand on the pile of feet. 4. Unscrew light bulb causing the room to go dark. 5. Yell "LET THERE BE LIGHT" and then quickly screw in the light bulb. 6. Proclaim yourself god. 7. Use your god powers to walk through the perfectly smooth walls.
On April 21 2012 09:40 TanGeng wrote: Blind Man's Deck
There was once a very wise blind man, and the town philosopher was jealous of the man's popularity. One day, he took a full deck of 52 cards, turned all the Aces, Kings, and Queens face up, shuffled the deck well and handed the deck to the blind man. He challenged the wise man to split the cards into piles such that the number of cards facing up in all of the piles was equal. Without pausing for a beat, the wise man counted off a few cards, split the decks into piles, and completed the challenge in a matter of seconds. How did the wise man do it?
On April 21 2012 09:40 TanGeng wrote: Blind Man's Deck
There was once a very wise blind man, and the town philosopher was jealous of the man's popularity. One day, he took a full deck of 52 cards, turned all the Aces, Kings, and Queens face up, shuffled the deck well and handed the deck to the blind man. He challenged the wise man to split the cards into piles such that the number of cards facing up in all of the piles was equal. Without pausing for a beat, the wise man counted off a few cards, split the decks into piles, and completed the challenge in a matter of seconds. How did the wise man do it?
On April 21 2012 09:40 TanGeng wrote: Blind Man's Deck
There was once a very wise blind man, and the town philosopher was jealous of the man's popularity. One day, he took a full deck of 52 cards, turned all the Aces, Kings, and Queens face up, shuffled the deck well and handed the deck to the blind man. He challenged the wise man to split the cards into piles such that the number of cards facing up in all of the piles was equal. Without pausing for a beat, the wise man counted off a few cards, split the decks into piles, and completed the challenge in a matter of seconds. How did the wise man do it?
On April 21 2012 09:40 TanGeng wrote: Blind Man's Deck
There was once a very wise blind man, and the town philosopher was jealous of the man's popularity. One day, he took a full deck of 52 cards, turned all the Aces, Kings, and Queens face up, shuffled the deck well and handed the deck to the blind man. He challenged the wise man to split the cards into piles such that the number of cards facing up in all of the piles was equal. Without pausing for a beat, the wise man counted off a few cards, split the decks into piles, and completed the challenge in a matter of seconds. How did the wise man do it?
On April 21 2012 09:40 TanGeng wrote: Blind Man's Deck
There was once a very wise blind man, and the town philosopher was jealous of the man's popularity. One day, he took a full deck of 52 cards, turned all the Aces, Kings, and Queens face up, shuffled the deck well and handed the deck to the blind man. He challenged the wise man to split the cards into piles such that the number of cards facing up in all of the piles was equal. Without pausing for a beat, the wise man counted off a few cards, split the decks into piles, and completed the challenge in a matter of seconds. How did the wise man do it?
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
But the problem implies that after reaching into a random pocket, you also randomly select one of the two balls in that pocket (otherwise, how would you end up with a ball)? So in fact the problem is implying random ball, in which case the 2/3 interpretation would seem to be correct?
I would still argue that 1/2 is correct. The first ping pong ball pulled out being orange is stated as a given condition, not as a random event.
On April 21 2012 05:23 TanGeng wrote:
On April 21 2012 05:07 LastPrime wrote:
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
The one orange ball that is pull out is the given event. The given event is NOT that you picked a pocket with at least one orange ball.
Based on the given, we count three possibilities, and two chances to be in the pocket with two orange balls and one chance to be in the pocket with the pocket with one of each. The probability is clearly 2/3.
i think the confusion lies in the fact that you dont have equal chances of being the pockets based on the given event. Its more likely that you are in the orange orange pocket when you end up with a orange ball on the first draw. Hence, the higher probability.
On April 21 2012 09:40 TanGeng wrote: Blind Man's Deck
There was once a very wise blind man, and the town philosopher was jealous of the man's popularity. One day, he took a full deck of 52 cards, turned all the Aces, Kings, and Queens face up, shuffled the deck well and handed the deck to the blind man. He challenged the wise man to split the cards into piles such that the number of cards facing up in all of the piles was equal. Without pausing for a beat, the wise man counted off a few cards, split the decks into piles, and completed the challenge in a matter of seconds. How did the wise man do it?
My favourite puzzle is the so-called Einstein Enigma (although apparently it wasn't actually invented by Einstein):
Five men live in five houses of five different color. They smoke five different brands of cigar,drink five different beverages,and keep five different pets.
We know that: * The Norwegian lives in the first house. * The brit lives in the red house. * The Swede keeps dogs as pets. * The Dane drinks tea. * The green house is just on the left of the white house. * The green house owner drinks coffee. * The man who smokes Blend lives next to the one who keeps cats. * The person who smokes Pall Mall rears birds. * The owner of the yellow house smokes Dunhill. * The man living in the house right in the center drinks milk. * The German smokes Prince. * The man who smokes Blend has a neighbor who drinks water. * The Norwegian lives next to the blue house. * The man who keeps horses lives next to the man who smokes Dunhill. * The owner who smokes Blue Master drinks beer.
sore and saw aren't homophones. they are pronounced differently
(that's no homophone!) Well, I guess that explains why your version includes the mirror, as over here + Show Spoiler +
saw and sore
are pronounced the same. The internets inform me that this is because of the difference between rhotic and non-rhotic accents (or more colloquially, because you americans don't speak engligh properly . Of course, I knew that already, but I had never encountered that particular difference before). The things you learn on TL.net...
On April 21 2012 09:40 TanGeng wrote: Blind Man's Deck
There was once a very wise blind man, and the town philosopher was jealous of the man's popularity. One day, he took a full deck of 52 cards, turned all the Aces, Kings, and Queens face up, shuffled the deck well and handed the deck to the blind man. He challenged the wise man to split the cards into piles such that the number of cards facing up in all of the piles was equal. Without pausing for a beat, the wise man counted off a few cards, split the decks into piles, and completed the challenge in a matter of seconds. How did the wise man do it?
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
But the problem implies that after reaching into a random pocket, you also randomly select one of the two balls in that pocket (otherwise, how would you end up with a ball)? So in fact the problem is implying random ball, in which case the 2/3 interpretation would seem to be correct?
I would still argue that 1/2 is correct. The first ping pong ball pulled out being orange is stated as a given condition, not as a random event.
On April 21 2012 05:23 TanGeng wrote:
On April 21 2012 05:07 LastPrime wrote:
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
The one orange ball that is pull out is the given event. The given event is NOT that you picked a pocket with at least one orange ball.
Based on the given, we count three possibilities, and two chances to be in the pocket with two orange balls and one chance to be in the pocket with the pocket with one of each. The probability is clearly 2/3.
i think the confusion lies in the fact that you dont have equal chances of being the pockets based on the given event. Its more likely that you are in the orange orange pocket when you end up with a orange ball on the first draw. Hence, the higher probability.
Because it's a given you have already drawn an orange ball, you can dismiss the pocket with 2 white balls and assume that you are only dealing with 1 of TWO pockets, as opposed to 1 of THREE. It's a big distinction. So you either have a 50% chance of having a 100% chance of drawing another orange ball (2 orange balls), or you have a 50% chance of having a 0% chance of drawing another orange ball (1 orange 1 white). Let's put it together in an equation. (.5)(1.0) + (.5)(0) = .5 = 50%
On April 21 2012 22:23 khaydarin9 wrote: My favourite puzzle is the so-called Einstein Enigma (although apparently it wasn't actually invented by Einstein):
Five men live in five houses of five different color. They smoke five different brands of cigar,drink five different beverages,and keep five different pets.
We know that: * The Norwegian lives in the first house. * The brit lives in the red house. * The Swede keeps dogs as pets. * The Dane drinks tea. * The green house is just on the left of the white house. * The green house owner drinks coffee. * The man who smokes Blend lives next to the one who keeps cats. * The person who smokes Pall Mall rears birds. * The owner of the yellow house smokes Dunhill. * The man living in the house right in the center drinks milk. * The German smokes Prince. * The man who smokes Blend has a neighbor who drinks water. * The Norwegian lives next to the blue house. * The man who keeps horses lives next to the man who smokes Dunhill. * The owner who smokes Blue Master drinks beer.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
But the problem implies that after reaching into a random pocket, you also randomly select one of the two balls in that pocket (otherwise, how would you end up with a ball)? So in fact the problem is implying random ball, in which case the 2/3 interpretation would seem to be correct?
I would still argue that 1/2 is correct. The first ping pong ball pulled out being orange is stated as a given condition, not as a random event.
On April 21 2012 05:23 TanGeng wrote:
On April 21 2012 05:07 LastPrime wrote:
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
The one orange ball that is pull out is the given event. The given event is NOT that you picked a pocket with at least one orange ball.
Based on the given, we count three possibilities, and two chances to be in the pocket with two orange balls and one chance to be in the pocket with the pocket with one of each. The probability is clearly 2/3.
i think the confusion lies in the fact that you dont have equal chances of being the pockets based on the given event. Its more likely that you are in the orange orange pocket when you end up with a orange ball on the first draw. Hence, the higher probability.
Because it's a given you have already drawn an orange ball, you can dismiss the pocket with 2 white balls and assume that you are only dealing with 1 of TWO pockets, as opposed to 1 of THREE. It's a big distinction. So you either have a 50% chance of having a 100% chance of drawing another orange ball (2 orange balls), or you have a 50% chance of having a 0% chance of drawing another orange ball (1 orange 1 white). Let's put it together in an equation. (.5)(1.0) + (.5)(0) = .5 = 50%
think of how you got the first orange ball. You might pick the white ball from white-white or a white ball from orange-white. You keep going until you pick an orange ball. You can see how you will be more likely to pick the orange ball from orange-orange, right? Since the first orange ball is more likely to be from orange-orange, you have a higher chance of picking a second orange ball
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
But the problem implies that after reaching into a random pocket, you also randomly select one of the two balls in that pocket (otherwise, how would you end up with a ball)? So in fact the problem is implying random ball, in which case the 2/3 interpretation would seem to be correct?
I would still argue that 1/2 is correct. The first ping pong ball pulled out being orange is stated as a given condition, not as a random event.
On April 21 2012 05:23 TanGeng wrote:
On April 21 2012 05:07 LastPrime wrote:
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
The one orange ball that is pull out is the given event. The given event is NOT that you picked a pocket with at least one orange ball.
Based on the given, we count three possibilities, and two chances to be in the pocket with two orange balls and one chance to be in the pocket with the pocket with one of each. The probability is clearly 2/3.
i think the confusion lies in the fact that you dont have equal chances of being the pockets based on the given event. Its more likely that you are in the orange orange pocket when you end up with a orange ball on the first draw. Hence, the higher probability.
Because it's a given you have already drawn an orange ball, you can dismiss the pocket with 2 white balls and assume that you are only dealing with 1 of TWO pockets, as opposed to 1 of THREE. It's a big distinction. So you either have a 50% chance of having a 100% chance of drawing another orange ball (2 orange balls), or you have a 50% chance of having a 0% chance of drawing another orange ball (1 orange 1 white). Let's put it together in an equation. (.5)(1.0) + (.5)(0) = .5 = 50%
Imagine there are 2 jars, jar 1 has 1000 orange marbles and jar 2 has 10 orange marbles and 990 white marbles.
You close your eyes and pick a random marble from a random jar. When you open your eyes, you have an orange marble in your hand. Which jar do you think you reached into? (~99% chance you have a marble from jar 1) If you pick another random marble from the same jar, what is the chance that it is also orange? (~99%)
Now imagine that you pick a random jar, look into it and pull out an orange marble. Which jar did you most likely choose? (50% chance for either jar) If you pick a random marble from the same jar what is the probability that it is orange? (since you are equally likely to be in the orange jar as you are in the white jar, the chance of pulling an orange marble is about 50%).
On April 21 2012 09:40 TanGeng wrote: Blind Man's Deck
There was once a very wise blind man, and the town philosopher was jealous of the man's popularity. One day, he took a full deck of 52 cards, turned all the Aces, Kings, and Queens face up, shuffled the deck well and handed the deck to the blind man. He challenged the wise man to split the cards into piles such that the number of cards facing up in all of the piles was equal. Without pausing for a beat, the wise man counted off a few cards, split the decks into piles, and completed the challenge in a matter of seconds. How did the wise man do it?
there are 12 face up cards in the deck. take 12 cards out and flip it all of em over so the face up ones are now face down and vice versa. for example, if i pick out 4 face up cards in the 12 cards i take out, 8 cards will be face up after i flip em all over. because there are 8 face up cards left in the remaining 40 cards, it should be equal in both piles.
New riddle, i dont know the answer yet (there is an answer and i will know it in few days at most, would help if you geniuses would find it for me) 1cd+2bd+2cd+1bd+3cd+1bd+1ad+2cd+1bd+1ad+2cd=
if you want to know if ur answer is correct, you can type it in here: http://diaknaposfeladat.hostei.com/osszekot/ if tha page changes to a new one, u ve done it (captcha required)
On May 01 2012 06:29 Geo.Rion wrote: New riddle, i dont know the answer yet (there is an answer and i will know it in few days at most, would help if you geniuses would find it for me) 1cd+2bd+2cd+1bd+3cd+1bd+1ad+2cd+1bd+1ad+2cd=
if you want to know if ur answer is correct, you can type it in here: http://diaknaposfeladat.hostei.com/osszekot/ if tha page changes to a new one, u ve done it (captcha required)
On May 01 2012 06:29 Geo.Rion wrote: New riddle, i dont know the answer yet (there is an answer and i will know it in few days at most, would help if you geniuses would find it for me) 1cd+2bd+2cd+1bd+3cd+1bd+1ad+2cd+1bd+1ad+2cd=
if you want to know if ur answer is correct, you can type it in here: http://diaknaposfeladat.hostei.com/osszekot/ if tha page changes to a new one, u ve done it (captcha required)
On April 19 2012 18:04 ]343[ wrote: Haha, I have a vaguely related one, but it's not so much a brain teaser as a math problem (and I think I've posted it here before). I think it's way more counterintuitive ><
Infinitely many prisoners, labeled 1, 2, 3, 4, ... are standing in a line so that person N can see everyone with smaller label (and not himself). Each prisoner k is assigned a hat labeled with a [integer, rational number, or real number; doesn't actually matter (!)], c_k. Now, starting from prisoner 1 and going up, each prisoner will try to guess the number on their own hat; if he fails, he is shot, and if he succeeds, he lives.
Show that, if the prisoners are given (an infinite amount of) time beforehand to formulate a strategy, they can ensure that only finitely many prisoners are shot.
Nice -- but you don't mean an infinite amount of time, you mean an arbitrarily large finite amount of time. If they had an infinite amount of time to prepare, you could ensure that everyone lived by planning forever, and that's a stupid linguistic loophole. I'll have to think on that a bit more to get it down to cofinitely many survivors, but arbitrarily high proportions of survivors is easy (see spoiler as food for thought).
You can code lists of numbers into a single number in any number of ways -- Gödel codes, pairing functions, a bijection from N to N^k (or Q, or C, or whatever), etc. Suppose you want at most one person out of 10,000 to die. Then when you're planning, line everyone up, and have the first person memorize the number that encodes the numbers of the next 9,999 people, and have the 10,001st person do the same thing for the 9,999 people after him, and so on with the (10000k+1)th people for all natural numbers k. Then those people announce that they think their number is the relevant code, and are almost certainly killed, since that would be a ridiculous coincidence, and the next 9,999 people can immediately calculate their numbers with just arithmetic, and so on and so on.
Obviously moving from "arbitrarily close to 100% survival rate" to "all but finitely many survivors" is a huge gap that requires a whole new technique to bridge, but I think it's neat that it's so simple to get that much. I'll get my mental hamsters on their wheels and come back later with a full solution.
Hmm crap, I just realized yesterday (when giving this to a friend) that I typoed and meant "bigger label" when I said "smaller label" (so everyone can see infinitely many other people, i.e. everyone with bigger label than himself.)
Split all sequences of reals into equivalence classes, where the relation is "have the same tail." That is, (a_1 a_2 a_3 ...) ~ (b_1 b_2 b_3 ... ) if there exists N > 0 such that for all n > N, a_n = b_n.
What is the largest natural number you can create with only two 9's you are not allowed any more digits, but you can use any mathematical symbols.
EDIT: Forgot to mention, the number must be finite, and repetitions go under infinite, that is not to say you can't use a symbol more than once of course.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
I didn't get this explanation. If the sports bag had 100 pockets, 98 of which had two white balls, 1 of which had an orange ball and a white ball, and 1 of which had two orange balls, and you reach into a pocket and pull out an orange ball- does the problem change in the least bit? Clearly not- the pockets with no orange balls are irrelevant once we know that the pocket that was picked has at least 1 orange ball.
In order to answer "What's the probability of the other ball in the pocket being orange?", the real question that must be answered is, "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"
Let the event of choosing the pocket with 2 orange balls be known as X, the event of choosing the pocket with an orange ball and a white ball be known as Y, the event of picking out an orange ball be known as A, and the event of picking out a white ball be known as B.
P(X|A) = P(X∩A)/P(A)
If we only consider pockets X and Y,
P(X∩A) = 1/2, P(A) = 3/4
Therefore, P(X|A) = P(X∩A)/P(A) = 2/3
So the probability of the other ball in the pocket being orange is, in fact, 2/3.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
I didn't get this explanation. If the sports bag had 100 pockets, 98 of which had two white balls, 1 of which had an orange ball and a white ball, and 1 of which had two orange balls, and you reach into a pocket and pull out an orange ball- does the problem change in the least bit? Clearly not- the pockets with no orange balls are irrelevant once we know that the pocket that was picked has at least 1 orange ball.
In order to answer "What's the probability of the other ball in the pocket being orange?", the real question that must be answered is, "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"
Let the event of choosing the pocket with 2 orange balls be known as X, the event of choosing the pocket with an orange ball and a white ball be known as Y, the event of picking out an orange ball be known as A, and the event of picking out a white ball be known as B.
P(X|A) = P(X∩A)/P(A)
If we only consider pockets X and Y,
P(X∩A) = 1/2, P(A) = 3/4
Therefore, P(X|A) = P(X∩A)/P(A) = 2/3
So the probability of the other ball in the pocket being orange is, in fact, 2/3.
once you KNOW you drew an orange ball, the math changes.
it does not matter which orange ball you have drawn. the question says only "you drew an orange ball."
so, if you drew an orange ball in the O-O pocket, it's a 100% chance to draw another orange. if you drew an orange ball in the O-W pocket, it's a 0% chance to draw another orange. you cannot ever draw an orange ball in the W-W pocket, so this is discounted for math purposes.
because you have already given the outcome of event #1, you don't multiply two events for probability. it's only one event, and in the given situation, only two outcomes: O-O or O-W pocket. 50/50.
this is wrong because it is NOT 1/2 to get the first orange ball. it is 1/1. 100%. the question has given it already. you do not multiply here for two events because there are not two events. there is only one event.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
I didn't get this explanation. If the sports bag had 100 pockets, 98 of which had two white balls, 1 of which had an orange ball and a white ball, and 1 of which had two orange balls, and you reach into a pocket and pull out an orange ball- does the problem change in the least bit? Clearly not- the pockets with no orange balls are irrelevant once we know that the pocket that was picked has at least 1 orange ball.
In order to answer "What's the probability of the other ball in the pocket being orange?", the real question that must be answered is, "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"
Let the event of choosing the pocket with 2 orange balls be known as X, the event of choosing the pocket with an orange ball and a white ball be known as Y, the event of picking out an orange ball be known as A, and the event of picking out a white ball be known as B.
P(X|A) = P(X∩A)/P(A)
If we only consider pockets X and Y,
P(X∩A) = 1/2, P(A) = 3/4
Therefore, P(X|A) = P(X∩A)/P(A) = 2/3
So the probability of the other ball in the pocket being orange is, in fact, 2/3.
once you KNOW you drew an orange ball, the math changes.
it does not matter which orange ball you have drawn. the question says only "you drew an orange ball."
so, if you drew an orange ball in the O-O pocket, it's a 100% chance to draw another orange. if you drew an orange ball in the O-W pocket, it's a 0% chance to draw another orange. you cannot ever draw an orange ball in the W-W pocket, so this is discounted for math purposes.
because you have already given the outcome of event #1, you don't multiply two events for probability. it's only one event, and in the given situation, only two outcomes: O-O or O-W pocket. 50/50.
You're wrong. As I said, the question that you really want to answer is "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"
Yes, there are only two possible outcomes. But those two outcomes aren't equally likely. Put another way: If you were to play a game of BW with Flash and the only possible outcomes would be for you to win or for you to lose... what is the probability that you will win? There are only two possible outcomes, but they are not equally likely.
In order to determine how likely each possibility is, I used conditional probabilities, and showed that the chance that the pocket with 2 orange balls was chosen is 2/3.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
I didn't get this explanation. If the sports bag had 100 pockets, 98 of which had two white balls, 1 of which had an orange ball and a white ball, and 1 of which had two orange balls, and you reach into a pocket and pull out an orange ball- does the problem change in the least bit? Clearly not- the pockets with no orange balls are irrelevant once we know that the pocket that was picked has at least 1 orange ball.
In order to answer "What's the probability of the other ball in the pocket being orange?", the real question that must be answered is, "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"
Let the event of choosing the pocket with 2 orange balls be known as X, the event of choosing the pocket with an orange ball and a white ball be known as Y, the event of picking out an orange ball be known as A, and the event of picking out a white ball be known as B.
P(X|A) = P(X∩A)/P(A)
If we only consider pockets X and Y,
P(X∩A) = 1/2, P(A) = 3/4
Therefore, P(X|A) = P(X∩A)/P(A) = 2/3
So the probability of the other ball in the pocket being orange is, in fact, 2/3.
once you KNOW you drew an orange ball, the math changes.
it does not matter which orange ball you have drawn. the question says only "you drew an orange ball."
so, if you drew an orange ball in the O-O pocket, it's a 100% chance to draw another orange. if you drew an orange ball in the O-W pocket, it's a 0% chance to draw another orange. you cannot ever draw an orange ball in the W-W pocket, so this is discounted for math purposes.
because you have already given the outcome of event #1, you don't multiply two events for probability. it's only one event, and in the given situation, only two outcomes: O-O or O-W pocket. 50/50.
You're wrong. As I said, the question that you really want to answer is "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"
Yes, there are only two possible outcomes. But those two outcomes aren't equally likely. Put another way: If you were to play a game of BW with Flash and the only possible outcomes would be for you to win or for you to lose... what is the probability that you will win? There are only two possible outcomes, but they are not equally likely.
In order to determine how likely each possibility is, I used conditional probabilities, and showed that the chance that the pocket with 2 orange balls was chosen is 2/3.
do you understand that the fact something was randomly selected is irrelevant to probabilistic calculations IF the question presumes that an outcome has already been determined? the question in the riddle is only concerned with the probability involved with drawing ball 2, given some predetermined guaranteed information.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
I didn't get this explanation. If the sports bag had 100 pockets, 98 of which had two white balls, 1 of which had an orange ball and a white ball, and 1 of which had two orange balls, and you reach into a pocket and pull out an orange ball- does the problem change in the least bit? Clearly not- the pockets with no orange balls are irrelevant once we know that the pocket that was picked has at least 1 orange ball.
In order to answer "What's the probability of the other ball in the pocket being orange?", the real question that must be answered is, "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"
Let the event of choosing the pocket with 2 orange balls be known as X, the event of choosing the pocket with an orange ball and a white ball be known as Y, the event of picking out an orange ball be known as A, and the event of picking out a white ball be known as B.
P(X|A) = P(X∩A)/P(A)
If we only consider pockets X and Y,
P(X∩A) = 1/2, P(A) = 3/4
Therefore, P(X|A) = P(X∩A)/P(A) = 2/3
So the probability of the other ball in the pocket being orange is, in fact, 2/3.
once you KNOW you drew an orange ball, the math changes.
it does not matter which orange ball you have drawn. the question says only "you drew an orange ball."
so, if you drew an orange ball in the O-O pocket, it's a 100% chance to draw another orange. if you drew an orange ball in the O-W pocket, it's a 0% chance to draw another orange. you cannot ever draw an orange ball in the W-W pocket, so this is discounted for math purposes.
because you have already given the outcome of event #1, you don't multiply two events for probability. it's only one event, and in the given situation, only two outcomes: O-O or O-W pocket. 50/50.
You're wrong. As I said, the question that you really want to answer is "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"
Yes, there are only two possible outcomes. But those two outcomes aren't equally likely. Put another way: If you were to play a game of BW with Flash and the only possible outcomes would be for you to win or for you to lose... what is the probability that you will win? There are only two possible outcomes, but they are not equally likely.
In order to determine how likely each possibility is, I used conditional probabilities, and showed that the chance that the pocket with 2 orange balls was chosen is 2/3.
do you understand that the fact something was randomly selected is irrelevant to probabilistic calculations IF the question presumes that an outcome has already been determined? the question in the riddle is only concerned with the probability involved with drawing ball 2, given some predetermined guaranteed information.
Sorry, but your interlocutors are correct. Think of it this way:
Suppose you do 60 trials and the results accord perfectly with the probabilities. Then, you should get 30 orange balls and 30 white balls. 1/3 of the time, you will reach into the double orange pocket, and every time you do so you will get an orange, resulting in 20 orange balls from this pocket. 1/3 of the time you will reach into the one-orange one-white pocket and half of that time you will get an orange, resulting in 10 orange balls from that pocket.
When you pull out an orange ball, you learn that you are in one of the 30 trials where an orange ball was pulled out. 2/3 of these are trials in which the double orange pocket was drawn from, so that's the probability that the other is orange.
edit: here's an analogy. There's a competition between a masters player and DRG. There is 1/3 chance that DRG will automatically get the win, 1/3 that the masters player will automatically get the win, and 1/3 chance that they will play a match of SC2 to determine the winner. If you find out that the masters player won, what's the probability that the middle option was chosen? According to your method, it is 50%, because that it is one of the two ways that the masters player can win.
However, this is obviously false. If you find out that the masters player won, you have a very good reason to believe that the middle option was chosen because the odds of winning against DRG in a match are so low. This is perfectly analogous to the present examples. There are two ways to win (i.e., get the orange ball) with an equal probability of occurring, but one of the ways delivers the winning result a higher percentage of the time than the other. In such scenarios, learning that a win has occurred makes it likely that the higher percentage of winning way happened.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
I didn't get this explanation. If the sports bag had 100 pockets, 98 of which had two white balls, 1 of which had an orange ball and a white ball, and 1 of which had two orange balls, and you reach into a pocket and pull out an orange ball- does the problem change in the least bit? Clearly not- the pockets with no orange balls are irrelevant once we know that the pocket that was picked has at least 1 orange ball.
In order to answer "What's the probability of the other ball in the pocket being orange?", the real question that must be answered is, "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"
Let the event of choosing the pocket with 2 orange balls be known as X, the event of choosing the pocket with an orange ball and a white ball be known as Y, the event of picking out an orange ball be known as A, and the event of picking out a white ball be known as B.
P(X|A) = P(X∩A)/P(A)
If we only consider pockets X and Y,
P(X∩A) = 1/2, P(A) = 3/4
Therefore, P(X|A) = P(X∩A)/P(A) = 2/3
So the probability of the other ball in the pocket being orange is, in fact, 2/3.
once you KNOW you drew an orange ball, the math changes.
it does not matter which orange ball you have drawn. the question says only "you drew an orange ball."
so, if you drew an orange ball in the O-O pocket, it's a 100% chance to draw another orange. if you drew an orange ball in the O-W pocket, it's a 0% chance to draw another orange. you cannot ever draw an orange ball in the W-W pocket, so this is discounted for math purposes.
because you have already given the outcome of event #1, you don't multiply two events for probability. it's only one event, and in the given situation, only two outcomes: O-O or O-W pocket. 50/50.
You're wrong. As I said, the question that you really want to answer is "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"
Yes, there are only two possible outcomes. But those two outcomes aren't equally likely. Put another way: If you were to play a game of BW with Flash and the only possible outcomes would be for you to win or for you to lose... what is the probability that you will win? There are only two possible outcomes, but they are not equally likely.
In order to determine how likely each possibility is, I used conditional probabilities, and showed that the chance that the pocket with 2 orange balls was chosen is 2/3.
do you understand that the fact something was randomly selected is irrelevant to probabilistic calculations IF the question presumes that an outcome has already been determined? the question in the riddle is only concerned with the probability involved with drawing ball 2, given some predetermined guaranteed information.
Ok, new question, just for you: There are 2 pockets. Pocket A has 100 orange marbles, whereas pocket B has 99 white marbles and 1 orange marble. You draw a random marble from a random pocket. Question 1: What is the probability that you drew the marble from pocket A? Question 2: If the marble you drew is orange, then what is the probability that you drew it from pocket A?
The answer to Question 1 is 1/2; there are two, equally likely answers. The answer to Question 2 is greater than 1/2, because the fact that you drew an orange marble means that you either a) You drew a marble from pocket A, or b) You drew a marble from pocket B AND out of the 100 marbles, you happened to pick the only orange one. In this case, the contents of the container give us additional information about how likely it is that we picked one container or the other.
Further proof: Let's go back to the original problem of 3 pockets and the ping pong balls. Let's say that Pocket X has 2 orange balls, Pocket Y has 1 orange ball and 1 white ball, and pocket Z has 2 white balls:
If we pick a random ball from a random pocket, then there's a 50% chance that we'll pick a white ball, and a 50% chance that we'll pick an orange ball. If what you say is true (Probability of second ball being orange given that first ball is orange = 1/2), then the following is also true:
If first ball is orange (50% chance), then the probability of having picked pocket X = 1/2, probability of having picked pocket Y = 1/2. If first ball is white (50% chance), then the probability of having picked pocket Z = 1/2, probability of having picked pocket Y = 1/2.
Therefore, if we pick a random ball from a random pocket, Total probability of having picked a ball from pocket X = (1/2)*(1/2) = 25% Total probability of having picked a ball from pocket Y = (1/2)*(1/2)+(1/2)*(1/2) = 50% Total probability of having picked a ball from pocket Z = (1/2)*(1/2) = 25%
The above makes no sense, there's 3 pockets and there should be a probability of 1/3 or having picked any one of them (if you repeat the exercise assuming that the probability of second ball being orange given that the first ball is orange = 2/3, then the numbers check out correctly)
this is wrong because it is NOT 1/2 to get the first orange ball. it is 1/1. 100%. the question has given it already. you do not multiply here for two events because there are not two events. there is only one event.
Yes, that's the point of conditional probability. The question is "What's the probability of the other ball in the pocket being orange?" The tree takes into account the different ways the first ball can be orange, and then sees is the second ball would also be orange.
But fine, let's look at it your way. The first event already happened. The orange ball was chosen. How many ways could that happen? Three. (circled in orange on the chart). So the first event already happened, and it's one of those three positions. Good! Now, of those three, how many would lead to the second ball being orange? Two. (boxed in blue)
On the marbles question: Many of you have already closed your mind to the fact that the other side might be right. You have stopped reading what your counterpart might be saying, and instead post the same off-topic response countless times. + Show Spoiler +
In the problem does it specifically state that you draw an orange ball from the pocket? Yes! Therefore, drawing the orange ball out of the pocket has a 1/1 chance. It is given in the problem, and as I said before, will happen 100% of the time. Try solving from this point, knowing this information, and see what the answer is.
You have a rectangular chocolate bar marked into m x n squares, and you wish to break up the bar into its constituent squares. At each step, you may pick up one piece and break it along any of its marked vertical or horizontal lines. However, you must not stack two pieces and break them together. Prove that every method finishes in the same number of steps. How many steps?
It takes mn-1 steps to break apart the bar. Each time you break the bar, you increase the number of pieces by one. The method you use is irrelevant, since you always increase the number of pieces by one with each step, and you stop when you finish breaking up the bar.
Riddle 2: Infection
Nine 1x1 squares of a 10x10 board are infected. In one time unit, the cells with at least two infected neighbour cells (having a common side) become infected. Can the infection spread to the entire board?
No, the entire board cannot be infected. The perimeter of all infected areas on the board is invariant. The maximum perimeter achievable with an arrangement of 9 infected cells is 36, whereas the perimeter of the board itself is 40. So, the board cannot become completely infected no matter the arrangement of the starting infected cells.
Riddle 3: Who wins?
A row of minuses are written on a piece of paper. Two players take turns in replacing either a single minus or two adjacent minuses by pluses. The one who cannot make a move loses. Is there a winning strategy?
Yes, there is a winning strategy. The player who has the first move will always win if they play optimally. The strategy is as follows: First turn either the middle minus, if the total number is odd, or the middle two minuses if the total number is even. Now the row of minuses is divided into two sections. To win, exactly copy you opponent's moves onto the opposite section from where he makes them. When he takes the final move in one section, you take the final move in the other, and he will be left with no available moves on his next turn. You win.
You have 10 bottles of unmarked, visually identical pills and a scale. In 1 of the 10 bottles, the pills are lethal. A lethal pill weighs slightly more than a regular pill (1.1g opposed to 1.0g). How do you determine which bottle contains the deadly pills with only one measurement?
Take 1 pill from bottle 1, 2 pills from bottle 2, etc., then weigh the total amount. The amount of additional weight above what the the total should be, assuming all regular pills, will tell you which bottle contains the lethal pills.
On the marbles question: Many of you have already closed your mind to the fact that the other side might be right. You have stopped reading what your counterpart might be saying, and instead post the same off-topic response countless times. + Show Spoiler +
In the problem does it specifically state that you draw an orange ball from the pocket? Yes! Therefore, drawing the orange ball out of the pocket has a 1/1 chance. It is given in the problem, and as I said before, will happen 100% of the time. Try solving from this point, knowing this information, and see what the answer is.
On the marbles question: Many of you have already closed your mind to the fact that the other side might be right. You have stopped reading what your counterpart might be saying, and instead post the same off-topic response countless times. + Show Spoiler +
In the problem does it specifically state that you draw an orange ball from the pocket? Yes! Therefore, drawing the orange ball out of the pocket has a 1/1 chance. It is given in the problem, and as I said before, will happen 100% of the time. Try solving from this point, knowing this information, and see what the answer is.
So we have 2 pockets A and B and each has two balls, A has 2 oragne 1 and 2 and B has white and orange 1 and 2.
That means of the two possible pockets there are these outcomes.
Pocket A Ball 1 = Second ball will be orange Pocket A Ball 2 = Second ball will be orange Pocket B Ball 2 = Second ball will be white
On the marbles question: Many of you have already closed your mind to the fact that the other side might be right. You have stopped reading what your counterpart might be saying, and instead post the same off-topic response countless times. + Show Spoiler +
In the problem does it specifically state that you draw an orange ball from the pocket? Yes! Therefore, drawing the orange ball out of the pocket has a 1/1 chance. It is given in the problem, and as I said before, will happen 100% of the time. Try solving from this point, knowing this information, and see what the answer is.
I can't imagine what compelled you to copy and paste this nonsense a mere 8 posts below its original location.
But what do I know? My mind has probably just been "closed" by actually understanding math and conditional probabilities.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
I didn't get this explanation. If the sports bag had 100 pockets, 98 of which had two white balls, 1 of which had an orange ball and a white ball, and 1 of which had two orange balls, and you reach into a pocket and pull out an orange ball- does the problem change in the least bit? Clearly not- the pockets with no orange balls are irrelevant once we know that the pocket that was picked has at least 1 orange ball.
In order to answer "What's the probability of the other ball in the pocket being orange?", the real question that must be answered is, "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"
Let the event of choosing the pocket with 2 orange balls be known as X, the event of choosing the pocket with an orange ball and a white ball be known as Y, the event of picking out an orange ball be known as A, and the event of picking out a white ball be known as B.
P(X|A) = P(X∩A)/P(A)
If we only consider pockets X and Y,
P(X∩A) = 1/2, P(A) = 3/4
Therefore, P(X|A) = P(X∩A)/P(A) = 2/3
So the probability of the other ball in the pocket being orange is, in fact, 2/3.
once you KNOW you drew an orange ball, the math changes.
it does not matter which orange ball you have drawn. the question says only "you drew an orange ball."
so, if you drew an orange ball in the O-O pocket, it's a 100% chance to draw another orange. if you drew an orange ball in the O-W pocket, it's a 0% chance to draw another orange. you cannot ever draw an orange ball in the W-W pocket, so this is discounted for math purposes.
because you have already given the outcome of event #1, you don't multiply two events for probability. it's only one event, and in the given situation, only two outcomes: O-O or O-W pocket. 50/50.
You are wrong because you are only considering 1 orange ball from the pocket that has two. Once you know you get an orange ball as you said.... There are only three possible outcomes..... 1 ) orange ball from orange-white pocket.... (which will result in the other ball being white)... 2) orange ball from orange orange pocket... which will result in orange being the other ball...... AND 3) the one you forgot............ you get the other ball from the orange orange pocket.... which results on the other ball being orange....
On May 07 2012 14:44 xxpack09 wrote: soccerdude, you should look up the "Monty Hall Problem," and why the solution is to switch. Open your mind; you might learn something.
Very true. There is only one right answer, and how you get to it will be logically demonstrated by this problem. Warning: said problem might blow your mind if you don't understand conditional probabilities.
Rather than lurk in the shadows I have decided to come out and say it: My post was idiotic and wrong.
I was wrong about the problem, and because of the complete sense of hierarchy I compelled through my word choice I definitely understand the backlash. Hopefully in the future I take a little more time reading the problem before posting some radical response.
On a side note, I had already heard and understood the Monty Hall problem before I saw the marbles question, I just made a pretty big error in solving the marbles question that was unrelated to the Monty Hall problem. I also like the responses on opening my mind; they were very creative.
On May 08 2012 12:28 soccerdude wrote: Rather than lurk in the shadows I have decided to come out and say it: My post was idiotic and wrong.
I was wrong about the problem, and because of the complete sense of hierarchy I compelled through my word choice I definitely understand the backlash. Hopefully in the future I take a little more time reading the problem before posting some radical response.
On a side note, I had already heard and understood the Monty Hall problem before I saw the marbles question, I just made a pretty big error in solving the marbles question that was unrelated to the Monty Hall problem. I also like the responses on opening my mind; they were very creative.
Thanks for being so mature about this (certainly more than I was). You should have a bright future here.
Make this sum right by using a single straight line in any way. 5+5+5=550
I don't know the answer Saw this on a children's drawing plate and I was a bit amazed that I actually couldn't figure it out. Maybe the makers made a mistake or I'm just missing something really obvious here.
On May 09 2012 21:03 frietjeman wrote: Make this sum right by using a single straight line in any way. 5+5+5=550
I don't know the answer Saw this on a children's drawing plate and I was a bit amazed that I actually couldn't figure it out. Maybe the makers made a mistake or I'm just missing something really obvious here.
Oh you can turn a + sign into a 4 I suppose. Hadn't thought of that ^^. I don't think you can remove the equals sign though cause then you're not really solving the sum anymore, just removing it. Didn't expect such a fast answer wow.
man that pingpong ball one was really interesting. I had my mind set on + Show Spoiler +
50%
as well for a while, but I think it was because I had just read the riddle about "the family with 2 kids, and a boy answers the door, what's the probability the other child is also a boy." and seeing as that one is actually + Show Spoiler +
50%
I had a hard time changing my mind about the ping pong balls. I'm glad I kept reading all the comments! and to say I actually used to be good at stats... lol
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox The question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Here some riddles for you guys to kill some time...
The Clock: Sam and John was trapped in a room with only a digital clock in front of them. A speaker from nowhere told them that they can get out only if they solved the "Clock Riddle". Here it is: The clock will beep every time it reads 3 consecutive same number(e.g. 01:11). How many beeps will the clock do 2 days from now? At the beginning the clock reads 12:00am.
Numbers: How can you place the numbers 1 through 9 in a 3x3 grid such that every row, column, and the two diagonals all add up to 15?
Assassination Attempt (This one is a bit harder): The King of a small country invites 1000 senators to his annual party. As gifts, each senator brings the King a bottle of wine, for a grand total of 1000 bottles of wine. Each bottle is signed by the senator who gave it.
At the end of the party, the Queen tells the King that one of the senators is trying to assassinate him, and has put deadly poison in the bottle of wine he gave as a gift. Unfortunately, the Queen doesn't know which senator is the traitor (and thus doesn't know which bottle of wine has the poison in it).
The King has 10 servants. He views them as expendable, and does not care if they live or die. He decides to use them to figure out which bottle is poisoned, which will then indicate which senator is trying to assassinate him.
His plan is to make each servant drink from zero or more of the bottles of wine. The King knows that the poison is such that if a servant drinks it, he will feel fine until noon on the next day, at which point he will instantly drop dead.
The King must know for sure who the traitor is by noon on the day after the party, or else the traitor will try to find another way to assassinate him. This essentially means that he has one shot to make his servants drink the wine in order to figure out which is the poison wine.
Note that the King can make any of the servants drink from any of the wine bottles. He does not need to make all of the servants drink wine if he doesn't want to. Any servant who drinks from the poisoned bottle will die the next day at noon.
How can the King figure out for sure who the traitor is by noon on the following day?
Critical Hint (Don't check if you want to solve everything on your own): + Show Spoiler +
Think of a grid, with 1000 rows representing the bottles of wine, and 10 columns representing the servants.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox The question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Well, I guess this is what I'm using my 500th post on.
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random P(B)=0.5 since there are equal number of balls of each color and they are drawn at random. P(B|A)=1 since any draw from the double orange hole yields an orange.
On May 10 2012 05:50 Aelfric wrote: Assassination Attempt (This one is a bit harder): The King of a small country invites 1000 senators to his annual party. As gifts, each senator brings the King a bottle of wine, for a grand total of 1000 bottles of wine. Each bottle is signed by the senator who gave it.
At the end of the party, the Queen tells the King that one of the senators is trying to assassinate him, and has put deadly poison in the bottle of wine he gave as a gift. Unfortunately, the Queen doesn't know which senator is the traitor (and thus doesn't know which bottle of wine has the poison in it).
The King has 10 servants. He views them as expendable, and does not care if they live or die. He decides to use them to figure out which bottle is poisoned, which will then indicate which senator is trying to assassinate him.
His plan is to make each servant drink from zero or more of the bottles of wine. The King knows that the poison is such that if a servant drinks it, he will feel fine until noon on the next day, at which point he will instantly drop dead.
The King must know for sure who the traitor is by noon on the day after the party, or else the traitor will try to find another way to assassinate him. This essentially means that he has one shot to make his servants drink the wine in order to figure out which is the poison wine.
Note that the King can make any of the servants drink from any of the wine bottles. He does not need to make all of the servants drink wine if he doesn't want to. Any servant who drinks from the poisoned bottle will die the next day at noon.
How can the King figure out for sure who the traitor is by noon on the following day?
Critical Hint (Don't check if you want to solve everything on your own): + Show Spoiler +
Think of a grid, with 1000 rows representing the bottles of wine, and 10 columns representing the servants.
I really don't want to do a truth table for 10 servants and 1000 bottles, so i'll do one for 3 and 8. The principle is the same and will always work if the number of bottles is equal or smaller than 2^(number of servants). B is the bottle with the corresponding number, S1-S3 are the servants. 1 means "drink from bottle", 0 "don't drink from bottle". B D1 D2 D3 1 1 1 1 2 1 1 0 3 1 0 1 4 1 0 0 5 0 1 1 6 0 1 0 7 0 0 1 8 0 0 0
If 3 servants die the next day, bottle 1 was poisoned. If D1 and D2 die, bottle 2 was poisoned. And so on... For 1000 bottles and 10 servants, the first servant would have to drink the first 500 bottles, the second would have to drink bottles 0-250 and 500-750, and so on.
edit: writing this down took forever and it still looks kind of bad. Oh well, hope it's understandable anyway.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox The question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Well, I guess this is what I'm using my 500th post on.
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random P(B)=0.5 since there are equal number of balls of each color and they are drawn at random. P(B|A)=1 since any draw from the double orange hole yields an orange.
zzzzzzzz. I'm upset. I spent so much effort convincing myself that the answer was 1/2 that I completely forgot that P(orange-orange) =/= P(orange-white) given orange is chosen, which, of course, is the basis of the problem anyway. herp derp. And I try Bayes'... but completely botched it when I did it earlier cause I got 1/2...
In addition to this answer (as there are many if you count rotational symmetry), here's a quick and easy method to answer that for any odd-numbered square.
start at the top row, in the center with the number 1. Assume rows and columns wrap around (e.g. going "up" from the first row ==> last row, and so on). Go up and right, and place the next number there. If a number exists in that position already, then instead of going up and right, go down once instead. Continue until square is filled.
In this case, this would result in the following square
816 357 492
If you'll notice, that's the same square as frog's except rotated 90* ccw and flipped horizontally!
In addition to this answer (as there are many if you count rotational symmetry), here's a quick and easy method to answer that for any odd-numbered square.
start at the top row, in the center with the number 1. Assume rows and columns wrap around (e.g. going "up" from the first row ==> last row, and so on). Go up and right, and place the next number there. If a number exists in that position already, then instead of going up and right, go down once instead. Continue until square is filled.
In this case, this would result in the following square
816 357 492
If you'll notice, that's the same square as frog's except rotated 90* ccw and flipped horizontally!
There aren't that many solutions, there are only eight 3x3 magic squares. There's only one, really, and then its seven symmetric counterparts (rotate it 90 degrees three times to get the next three of them, then take its mirror image and rotate that four times to get the other ones. For n>3, however, there are lots and lots of ways to arrange the numbers from 1 to n^2 in a grid with row/column/diagonal sums equal. For example, there are 880 (7040 if you count symmetric copies as different) 4x4 such squares, and so on.
On May 10 2012 05:50 Aelfric wrote: Here some riddles for you guys to kill some time...
Assassination Attempt (This one is a bit harder): The King of a small country invites 1000 senators to his annual party. As gifts, each senator brings the King a bottle of wine, for a grand total of 1000 bottles of wine. Each bottle is signed by the senator who gave it.
At the end of the party, the Queen tells the King that one of the senators is trying to assassinate him, and has put deadly poison in the bottle of wine he gave as a gift. Unfortunately, the Queen doesn't know which senator is the traitor (and thus doesn't know which bottle of wine has the poison in it).
The King has 10 servants. He views them as expendable, and does not care if they live or die. He decides to use them to figure out which bottle is poisoned, which will then indicate which senator is trying to assassinate him.
His plan is to make each servant drink from zero or more of the bottles of wine. The King knows that the poison is such that if a servant drinks it, he will feel fine until noon on the next day, at which point he will instantly drop dead.
The King must know for sure who the traitor is by noon on the day after the party, or else the traitor will try to find another way to assassinate him. This essentially means that he has one shot to make his servants drink the wine in order to figure out which is the poison wine.
Note that the King can make any of the servants drink from any of the wine bottles. He does not need to make all of the servants drink wine if he doesn't want to. Any servant who drinks from the poisoned bottle will die the next day at noon.
How can the King figure out for sure who the traitor is by noon on the following day?
Critical Hint (Don't check if you want to solve everything on your own): + Show Spoiler +
Think of a grid, with 1000 rows representing the bottles of wine, and 10 columns representing the servants.
So this problem should be possible for up to 1024 bottles of wine. For each bottle, assign some combination of servants to drink it. (so bottle 1 can have no one drink it, bottle 2 just servant 1, etc. Since each of the 10 servants can either drink or not drink (2 possibilities), the number of unique bottle combinations is 2^10 = 1024. Upon seeing what combination of servants die, the correct bottle can be determined.
A company bus with twenty seats picks up twenty employees at their homes and shuttles them to and from work everyday. The twenty employees each have an assigned seat from the back to the front in the order they are picked up. The first employee on the bus route is an obnoxious manager with unpredictable behavior. Every morning, the manager boards the bus randomly at any one of the twenty stops. The manager cuts in front other employees at the stop, boards the bus, sits in an unoccupied seats at random, and then yaks away on the cell phone.
All other employees roll their eyes at the behavior. When they board the bus, if they sit in their assigned seat if unoccupied. Otherwise, they pick an unoccupied seat at random and sit there. You are Frank, the last employee to be picked up along the bus route. How often do you sit in the frontmost seat?
On May 10 2012 05:50 Aelfric wrote: Here some riddles for you guys to kill some time...
The Clock: Sam and John was trapped in a room with only a digital clock in front of them. A speaker from nowhere told them that they can get out only if they solved the "Clock Riddle". Here it is: The clock will beep every time it reads 3 consecutive same number(e.g. 01:11). How many beeps will the clock do 2 days from now? At the beginning the clock reads 12:00am.
A mountaineer is attempting a climb through a mountain gap from base camp A to base camp B. The route takes six days. At base camp A, the mountaineer finds sherpas for hire. He can get them to carry his supplies and equipment but must feed them and must not allow the sherpas to starve. The mountaineer and the sherpas can carry a maximum of four days supply while climbing. How does the mountaineer get from base camp to base camp in six days? How many days of supply does he buy?
The next year, the mountaineer returns to make the climb but no sherpas are available. How does the mountaineer make the climb on his own?
The mountaineer hires two sherpas, who we will call Tarjee and Harjee, and buys 12 days worth of supplies. The mountaineer with his two companies set off with full packs. After the first day, Tarjee gives up two days' worth of supplies and returns home. Harjee and the mountaineer each receive one day's worth of supplies and continue the climb. After the climb on the second day, Harjee gives up one day's worth of supplies and return home. The mountain again receives one day's worth of supplies. He has four days' supply and four days' climb and makes it to Camp B.
The next year, when the mountaineer has to do it alone, the mountaineer figures out a way to cache and retrieve supplies in the wilderness after a day of climbing. The mountaineer buys 12 days worth of supplies and mimics, in series, the climb and supply exchanges of Tarjee, Harjee, and himself the year before. He climbs as Tarjee first, then as Harjee, then as himself, and makes it to Camp B after 12 days.
On May 10 2012 12:44 TanGeng wrote: The Seat Snatcher Manager
A company bus with twenty seats picks up twenty employees at their homes and shuttles them to and from work everyday. The twenty employees each have an assigned seat from the back to the front in the order they are picked up. The first employee on the bus route is an obnoxious manager with unpredictable behavior. Every morning, the manager boards the bus randomly at any one of the twenty stops. The manager cuts in front other employees at the stop, boards the bus, sits in an unoccupied seats at random, and then yaks away on the cell phone.
All other employees roll their eyes at the behavior. When they board the bus, if they sit in their assigned seat if unoccupied. Otherwise, they pick an unoccupied seat at random and sit there. You are Frank, the last employee to be picked up along the bus route. How often do you sit in the frontmost seat?
Suppose there were only 2 seats. Then clearly the manager either sits in your seat or his own, and you get your own seat 50% of the time. Increase the number of seats by one at a time and watch what happens. 3 seats. We have three equally likely cases. If the manager takes his own seat, everyone (including you) gets their own seat. If the manager takes your seat, then regardless of where the second guy goes, you're not getting your seat. In the third case, the manager takes the second guy's seat, and then the second guy either takes yours (50%) or leaves yours for you (50%). That evens out to a 50-50 overall shot you get your own seat.
Four seats (and after this I trust the pattern will be obvious enough that you can see it holds for any number of seats). We've got four equally likely cases. If the manager takes his own seat, you get your own seat, and so does everyone else. If the manager takes your seat, you obviously get someone else's seat. In the other two cases, the manager takes either the 2nd or 3rd guy's seat. If he had taken the 3rd guy's seat, then the 2nd guy takes his own seat, and the third guy either goes in the manager's seat or in yours (50% each way). If he had taken the 2nd guy's seat, then we're back in the scenario with 3 seats total described above -- we have three empty seats, you're going last, and the next person to get on is taking a random seat, which we already know comes out 50-50 that you get your seat.
This continues. With N seats, there's N equally likely scenarios -- one in which everyone gets their own seat, one in which you definitely get someone else's seat, and N-2 more which all reduce to previous cases that all have a 50% chance of you getting your own seat. Basically, if the manager takes his seat we know what happens, if the manager takes you seat we know what happens, and if he takes anyone else's seat we don't care -- all that does is delay the inevitable arrival of someone who has two seats to choose randomly between; yours and the one corresponding to whoever took his seat.
In addition to this answer (as there are many if you count rotational symmetry), here's a quick and easy method to answer that for any odd-numbered square.
start at the top row, in the center with the number 1. Assume rows and columns wrap around (e.g. going "up" from the first row ==> last row, and so on). Go up and right, and place the next number there. If a number exists in that position already, then instead of going up and right, go down once instead. Continue until square is filled.
In this case, this would result in the following square
816 357 492
If you'll notice, that's the same square as frog's except rotated 90* ccw and flipped horizontally!
Interesting, I did it simply by finding the central number first. If you put, for example, the nine in the centre, you'll find that wherever you put the eight/seven/six in, that line will come to over fifteen. The highest number that you can put in the centre is five, and it makes sense that you want the highest number possible in the centre since you're going to have to make use of one and two etc, and they need 'counter balancing' so to speak.
For the rest you can pretty much fill it in quite easily because of the rotational symmetry. Just put a random number elsewhere in the box, complete that line, and continue from there
On May 10 2012 05:50 Aelfric wrote: Here some riddles for you guys to kill some time...
Assassination Attempt (This one is a bit harder): The King of a small country invites 1000 senators to his annual party. As gifts, each senator brings the King a bottle of wine, for a grand total of 1000 bottles of wine. Each bottle is signed by the senator who gave it.
At the end of the party, the Queen tells the King that one of the senators is trying to assassinate him, and has put deadly poison in the bottle of wine he gave as a gift. Unfortunately, the Queen doesn't know which senator is the traitor (and thus doesn't know which bottle of wine has the poison in it).
The King has 10 servants. He views them as expendable, and does not care if they live or die. He decides to use them to figure out which bottle is poisoned, which will then indicate which senator is trying to assassinate him.
His plan is to make each servant drink from zero or more of the bottles of wine. The King knows that the poison is such that if a servant drinks it, he will feel fine until noon on the next day, at which point he will instantly drop dead.
The King must know for sure who the traitor is by noon on the day after the party, or else the traitor will try to find another way to assassinate him. This essentially means that he has one shot to make his servants drink the wine in order to figure out which is the poison wine.
Note that the King can make any of the servants drink from any of the wine bottles. He does not need to make all of the servants drink wine if he doesn't want to. Any servant who drinks from the poisoned bottle will die the next day at noon.
How can the King figure out for sure who the traitor is by noon on the following day?
Critical Hint (Don't check if you want to solve everything on your own): + Show Spoiler +
Think of a grid, with 1000 rows representing the bottles of wine, and 10 columns representing the servants.
So this problem should be possible for up to 1024 bottles of wine. For each bottle, assign some combination of servants to drink it. (so bottle 1 can have no one drink it, bottle 2 just servant 1, etc. Since each of the 10 servants can either drink or not drink (2 possibilities), the number of unique bottle combinations is 2^10 = 1024. Upon seeing what combination of servants die, the correct bottle can be determined.
An alternative explanation that illuminates the coding principle behind the solution:
Number the bottles from 1 to 1000 and each servant from 1 to 10. For each bottle i, let b be i in binary (base 2) with digits b1...b10 (since 2^10 = 1024, you need 10 binary digits (bits) to represent the numbers from 1 to 1000), and let the nth servant drink from the bottle if bn is 1. For example, 75 in binary is 0001001011 so servants 1, 2, 4, and 7 should drink from the 75th bottle.
The ith bottle is poisoned if and only if all the servants that drank from that bottle die. Because there is a unique representation in binary for each digit from 1 to 1000 using 10 bits, there is a unique combination of servants that drink from each bottle.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox The question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Well, I guess this is what I'm using my 500th post on.
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random P(B)=0.5 since there are equal number of balls of each color and they are drawn at random. P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason.
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
On May 10 2012 20:36 TanGeng wrote: Mountaineers and Sherpas
A mountaineer is attempting a climb through a mountain gap from base camp A to base camp B. The route takes six days. At base camp A, the mountaineer finds sherpas for hire. He can get them to carry his supplies and equipment but must feed them and must not allow the sherpas to starve. The mountaineer and the sherpas can carry a maximum of four days supply while climbing. How does the mountaineer get from base camp to base camp in six days? How many days of supply does he buy?
The next year, the mountaineer returns to make the climb but no sherpas are available. How does the mountaineer make the climb on his own?
16 days' worth of supplies, hires 3 sherpas. After the first day, one sherpa gives 1 days' worth of supplies to each of the other two sherpas, then returns home. After the second day, each of the remaining sherpas gives 1 days' worth of supplies to the mountaineer, then return. At this point, there's four days left to go, and the mountaineer has four days' worth of food.
The next year, assuming the supplies don't spoil and he can find them after he stores them along the road, he takes four days' worth of supplies, advances for 1 day, plants 2 days' worth of supplies, then returns. Then he takes four more days' worth of supplies, advances 1 day, retrieves 1 day of supplies, advances another day, plants 2 days' worth of supplies, and returns, retrieving the rest of the cache of supplies left at the day 1 stop. Finally, he takes 4 days' worth of supplies, advances 2 days, retrieves the cache with 2 days' supplies, and a-moves for the win.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox The question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Well, I guess this is what I'm using my 500th post on.
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random P(B)=0.5 since there are equal number of balls of each color and they are drawn at random. P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason.
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
But I've also seen a lot of arguing based on the wording of the riddle, rather than the actual mathematics or process behind solving it. Clearly, if we all agreed on what the question was specifically asking, there would probably be no debate
On May 10 2012 20:36 TanGeng wrote: Mountaineers and Sherpas
A mountaineer is attempting a climb through a mountain gap from base camp A to base camp B. The route takes six days. At base camp A, the mountaineer finds sherpas for hire. He can get them to carry his supplies and equipment but must feed them and must not allow the sherpas to starve. The mountaineer and the sherpas can carry a maximum of four days supply while climbing. How does the mountaineer get from base camp to base camp in six days? How many days of supply does he buy?
The next year, the mountaineer returns to make the climb but no sherpas are available. How does the mountaineer make the climb on his own?
16 days' worth of supplies, hires 3 sherpas. After the first day, one sherpa gives 1 days' worth of supplies to each of the other two sherpas, then returns home. After the second day, each of the remaining sherpas gives 1 days' worth of supplies to the mountaineer, then return. At this point, there's four days left to go, and the mountaineer has four days' worth of food.
The next year, assuming the supplies don't spoil and he can find them after he stores them along the road, he takes four days' worth of supplies, advances for 1 day, plants 2 days' worth of supplies, then returns. Then he takes four more days' worth of supplies, advances 1 day, retrieves 1 day of supplies, advances another day, plants 2 days' worth of supplies, and returns, retrieving the rest of the cache of supplies left at the day 1 stop. Finally, he takes 4 days' worth of supplies, advances 2 days, retrieves the cache with 2 days' supplies, and a-moves for the win.
I think you can do it with 2 sherpas and 12 days worth of supply. The three of them use 3 days worth of supply from sherpa 1 on first day, and send him home with his remaining day worth. The mountaineer and the other sherpa still have full supply, and they use up 2 days worth of supply from sherpa 2 on the second day, and sent him home with his remaining 2 days of supply, while the mountaineer still has his 4 days of supply to get him all the way to camp B.
On May 10 2012 05:50 Aelfric wrote: Assassination Attempt (This one is a bit harder):+ Show Spoiler +
The King of a small country invites 1000 senators to his annual party. As gifts, each senator brings the King a bottle of wine, for a grand total of 1000 bottles of wine. Each bottle is signed by the senator who gave it.
At the end of the party, the Queen tells the King that one of the senators is trying to assassinate him, and has put deadly poison in the bottle of wine he gave as a gift. Unfortunately, the Queen doesn't know which senator is the traitor (and thus doesn't know which bottle of wine has the poison in it).
The King has 10 servants. He views them as expendable, and does not care if they live or die. He decides to use them to figure out which bottle is poisoned, which will then indicate which senator is trying to assassinate him.
His plan is to make each servant drink from zero or more of the bottles of wine. The King knows that the poison is such that if a servant drinks it, he will feel fine until noon on the next day, at which point he will instantly drop dead.
The King must know for sure who the traitor is by noon on the day after the party, or else the traitor will try to find another way to assassinate him. This essentially means that he has one shot to make his servants drink the wine in order to figure out which is the poison wine.
Note that the King can make any of the servants drink from any of the wine bottles. He does not need to make all of the servants drink wine if he doesn't want to. Any servant who drinks from the poisoned bottle will die the next day at noon.
How can the King figure out for sure who the traitor is by noon on the following day?
9: X01-X02, X05-X06, X09-X10, X13-X14, X17-X18, X21-X22, X25-X26, X29-X30, X33-X34, X37-X38, X41-X42, X45-X46, X49-X50, X53-X54, X57-X58, X61-X62, X65-X66, X69-X70, X73-X74, X77-X78, X81-X82, X85-X86, X89-X90, X93-X94, X97-X98, where X is valid for all whole numbers from 0 through 9
10: All odd numbers between 1 and 999
If each servant drinks from the bottles listed above, then they will all drink from between 500 and 512 bottles, between 2 and 10 servants will die, and there's a unique combination of dead servants for every possible number of the poison bottle, which allows the poisoned bottle to be pinpointed precisely.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox The question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Well, I guess this is what I'm using my 500th post on.
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random P(B)=0.5 since there are equal number of balls of each color and they are drawn at random. P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason.
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
Your argument has been presented and refuted repeatedly in this thread. I don't really know what else to say. It matters that the pocket was originally chosen at random, and the fact that we now know with certainty that the drawn ball is orange doesn't change that. If you can express your complaint in the probability calculus, showing my false assumption and demonstrating your result, you should.
Your reasoning leads to absurdity in a number of cases. Suppose there are two pockets, one with a billion orange balls and one with a billion white balls and one orange ball and a third pocket with a billion white balls. You reach into a pocket at random and draw an orange ball. How confident should you be that you drew from the all orange pocket? By your reasoning, 50%. This is because your reasoning has it that the only information you learn is that you drew from one of the two not-all white pockets, and that the previous chances are rendered irrelevant. Don't you understand how absurd this is? If you repeated the trial thousands of times, the answer would always be the all orange pocket, but you claim that it is 50% the mostly white pocket.
Please, just construct the 6 ball version as an experiment and do 30 trials. I guarantee you that the majority of times you have drawn an orange ball first will be times that you drew from the 2 ball pocket.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox The question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Well, I guess this is what I'm using my 500th post on.
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random P(B)=0.5 since there are equal number of balls of each color and they are drawn at random. P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason.
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
But I've also seen a lot of arguing based on the wording of the riddle, rather than the actual mathematics or process behind solving it. Clearly, if we all agreed on what the question was specifically asking, there would probably be no debate
There variants of this type of puzzle that do involve crucial ambiguities, such as the boy or girl paradox, but this simply isn't one of them. Sorry, but your side simply is wrong about how to evaluate the appropriate conditional probabilities and an experiment would demonstrate that after not too many trials.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox The question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Well, I guess this is what I'm using my 500th post on.
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random P(B)=0.5 since there are equal number of balls of each color and they are drawn at random. P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason.
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
But I've also seen a lot of arguing based on the wording of the riddle, rather than the actual mathematics or process behind solving it. Clearly, if we all agreed on what the question was specifically asking, there would probably be no debate
There variants of this type of puzzle that do involve crucial ambiguities, such as the boy or girl paradox, but this simply isn't one of them. Sorry, but your side simply is wrong about how to evaluate the appropriate conditional probabilities and an experiment would demonstrate that after not too many trials.
Because there aren't any conditional probabilities that you need to work out the math for, and so you don't need to invoke Bayes' theorem at all. You're already holding an orange ball (which technically means the probability of selecting an orange ball on the first try is 1, since it already occurred). Regardless, that event is now in the past.
What that event does tell you, however, is that all pockets with zero orange balls are now out of play (as you can't select an orange ball from a pocket that has two white balls). This leaves the pocket with two orange balls, and the pocket with one white and one orange. The probabilities of selecting a particular orange ball is not an issue, as you are already holding an orange ball. The point is now that you have two pockets, one with the second ball being necessarily white, and the other with the second ball being necessarily orange.
The question is not how you got to the first orange ball, but how you get to the second orange ball.
Therefore, the probability of the second ball being orange is 1/2.
There's a difference between saying "What's the probability of pulling 2 orange balls from the same pocket" (given the same pocket set up) and saying "You're already holding an orange ball, what's the probability that the second one from the same pocket is also orange" (given the same pocket set up). The former invokes necessary conditional probability; the latter doesn't require it, and is what the actual question is asking.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox The question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Well, I guess this is what I'm using my 500th post on.
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random P(B)=0.5 since there are equal number of balls of each color and they are drawn at random. P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason.
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
Your argument has been presented and refuted repeatedly in this thread. I don't really know what else to say. It matters that the pocket was originally chosen at random, and the fact that we now know with certainty that the drawn ball is orange doesn't change that. If you can express your complaint in the probability calculus, showing my false assumption and demonstrating your result, you should.
Your reasoning leads to absurdity in a number of cases. Suppose there are two pockets, one with a billion orange balls and one with a billion white balls and one orange ball and a third pocket with a billion white balls. You reach into a pocket at random and draw an orange ball. How confident should you be that you drew from the all orange pocket? By your reasoning, 50%. This is because your reasoning has it that the only information you learn is that you drew from one of the two not-all white pockets, and that the previous chances are rendered irrelevant. Don't you understand how absurd this is? If you repeated the trial thousands of times, the answer would always be the all orange pocket, but you claim that it is 50% the mostly white pocket.
Please, just construct the 6 ball version as an experiment and do 30 trials. I guarantee you that the majority of times you have drawn an orange ball first will be times that you drew from the 2 ball pocket.
It really depends on how you interpret the question. The first way is "What is the probability that the second ball is orange, given that the first ball is orange?" In this scenario, you're right. The second way is "I'm holding an orange ball. What is the probability that the next ball is orange?" In this scenario, it's no longer a case of conditional probability. You're already holding an orange ball. In your exaggerated example, it doesn't matter if there's only a 1/3 probability that you're holding an orange ball on the first try. You're holding it.
The second case that I highlighted in another way. I draw a single ball. If it's white, then fuck this world I'm outta here. If I draw a orange ball, then I ask the question. Now does it feel like it's 1/2?
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox The question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Well, I guess this is what I'm using my 500th post on.
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random P(B)=0.5 since there are equal number of balls of each color and they are drawn at random. P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason.
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
Your argument has been presented and refuted repeatedly in this thread. I don't really know what else to say. It matters that the pocket was originally chosen at random, and the fact that we now know with certainty that the drawn ball is orange doesn't change that. If you can express your complaint in the probability calculus, showing my false assumption and demonstrating your result, you should.
Your reasoning leads to absurdity in a number of cases. Suppose there are two pockets, one with a billion orange balls and one with a billion white balls and one orange ball and a third pocket with a billion white balls. You reach into a pocket at random and draw an orange ball. How confident should you be that you drew from the all orange pocket? By your reasoning, 50%. This is because your reasoning has it that the only information you learn is that you drew from one of the two not-all white pockets, and that the previous chances are rendered irrelevant. Don't you understand how absurd this is? If you repeated the trial thousands of times, the answer would always be the all orange pocket, but you claim that it is 50% the mostly white pocket.
Please, just construct the 6 ball version as an experiment and do 30 trials. I guarantee you that the majority of times you have drawn an orange ball first will be times that you drew from the 2 ball pocket.
It really depends on how you interpret the question. The first way is "What is the probability that the second ball is orange, given that the first ball is orange?" In this scenario, you're right. The second way is "I'm holding an orange ball. What is the probability that the next ball is orange?" In this scenario, it's no longer a case of conditional probability. You're already holding an orange ball. In your exaggerated example, it doesn't matter if there's only a 1/3 probability that you're holding an orange ball on the first try. You're holding it.
The second case that I highlighted in another way. I draw a single ball. If it's white, then fuck this world I'm outta here. If I draw a orange ball, then I ask the question. Now does it feel like it's 1/2?
Just to be clear, you are accepting the ludicrous result that if you draw an orange ball in the billion-ball-pockets case you think that there's 50% chance it came from the pocket with only one orange ball, right? Same question to you darkplasma.
What on earth does it mean to say "I'm holding an orange ball?" Well, where did you get that orange ball? Did you go out to 7-11 and buy it? Well in that case it tells us nothing. Oh, it turns out that you drew it at random from one of the three pockets. Thanks, that gives me added information about which pocket it likely came from because one of them delivers orange balls more frequently than the other and is hence more likely to be the source. Is that so hard to understand?
The only question that 1/2 is the right answer to is the question of what the probability of the other ball being white is given that you are in a pocket with at least one orange ball (with 0 added information!). This is explicitly not the question. The question tells us that the drawing procedure was random, and that tells us additional information about likelihoods that cannot be ignored.
I'll try one more time to turn this into a more common sense example. You're on the other side of an outfield wall and know that your three friends are taking turns batting but don't know which is up. One of them is incapable of hitting the ball over the fence, another hits the ball over the fence every time, and another hits it over the fence once every 100 swings. You see the ball come over the fence (i.e., you are given that it was in fact a home run). Isn't it obvious that this is fantastic evidence that the middle friend was at the plate. You two think it is only evidence that one of the latter two is at the plate. Don't you see how silly that is?
On May 11 2012 03:08 Heh_ wrote: The second case that I highlighted in another way. I draw a single ball. If it's white, then fuck this world I'm outta here. If I draw a orange ball, then I ask the question. Now does it feel like it's 1/2?
Yikes, this is still going on. The possibility of drawing the white ball from the mixed pocket is precisely why the answer is 2/3. Because you are drawing a ball and could have drawn the white ball but given the fact that you drew an orange ball invokes conditional probability.
If the question did not involve chances of drawing a white ball. For example: Your friend picks a pocket with at least one orange ball. He then proves it to you by showing a orange ball. What are the chances that the other ball in the pocket is orange?
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox The question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Wow looks like a lot of people are still arguing over this problem Given the wording of this problem though,
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
My method for testing was this: 1) Make the bag and pocket and balls, and set two counters (trials && success) at 0. 2) Pick a pocket at random. 3) Pick a ball from that pocket at random. 4) If the ball is white, then we can't include this as part of our trials, so don't change the value of trials or successes. 5) If the ball is orange, then this is a valid trial so increase the trial counter by 1. 6) If the second ball in the pocket is also orange, increase the success counter by 1. 7) Do steps 2-6 1,000,000 times 8) Print out the probability as success/trials
public class Ball { public String color; public Ball (String color) { this.color = color; } public String getColor() { return color; } } public class Pocket { private Ball[] balls; private java.util.Random generator = new java.util.Random(); private int first; public Pocket(Ball a, Ball b) { balls = new Ball[2]; balls[0] = a; balls[1] = b; } public Ball getFirst() { first = generator.nextInt(2); return balls[first]; } public Ball getLast() { if (first == 0) return balls[1]; return balls[0]; } } public class Bag { private Pocket[] pockets; private java.util.Random generator = new java.util.Random(); public Bag() { pockets = new Pocket[3]; pockets[0] = new Pocket(new Ball("white"), new Ball("white")); pockets[1] = new Pocket(new Ball("white"), new Ball("orange")); pockets[2] = new Pocket(new Ball("orange"), new Ball("orange")); } public int runTrial() { // returns 0 if no orange ball was drawn first (can't be counted) // returns 1 if an orange ball was drawn first but a white ball was drawn second // returns 2 if an orange ball was drawn first and an orange ball was drawn second int pocketSelect = generator.nextInt(3); Ball first = pockets[pocketSelect].getFirst(); if (!first.getColor().equals("orange")) return 0; Ball last = pockets[pocketSelect].getLast(); if (!last.getColor().equals("orange")) return 1; return 2; } } public class Tester { public static void main(String[] args) { Bag bag = new Bag(); int trials = 0; int orangeOrange = 0; for(int i=0; i < 1000000; i++) { int result = bag.runTrial(); if (result == 0) continue; else if (results == 1) trials++; else { trials++; orangeOrange++; } } System.out.println("There were " + trials + " trials and " + orangeOrange + " orange-orange combinations"); System.out.println("The probability is: " + ((double)orangeOrange/(double)trials)); } }
There were 500170 trials and 332801 orange-orange combinations The probability is: 0.6653757722374393 There were 500559 trials and 333707 orange-orange combinations The probability is: 0.6666686644331637 There were 500068 trials and 333559 orange-orange combinations The probability is: 0.6670272842893367 There were 500562 trials and 333987 orange-orange combinations The probability is: 0.667224040178839 There were 500334 trials and 333134 orange-orange combinations The probability is: 0.665823230082305
And expanding to 100,000,000 times instead of 1,000,000 gave + Show Spoiler +
There were 49988348 trials and 33328647 orange-orange combinations The probability is: 0.6667283143663799
I think this speaks for itself.
The flaw in these steps is that our picker is GUARANTEED to pick an orange ball. In your program, you set it such that there are three distinct orange balls to pick from. In reality, if you are guaranteed an orange ball, it's the same thing as having a second person look inside the pockets, and give you an orange ball, in which case it doesn't matter which ball you took from the "orange-orange" pocket
On May 11 2012 03:43 Akari Takai wrote: Time for science. Computer science that is.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
My method for testing was this: 1) Make the bag and pocket and balls, and set two counters (trials && success) at 0. 2) Pick a pocket at random. 3) Pick a ball from that pocket at random. 4) If the ball is white, then we can't include this as part of our trials, so don't change the value of trials or successes. 5) If the ball is orange, then this is a valid trial so increase the trial counter by 1. 6) If the second ball in the pocket is also orange, increase the success counter by 1. 7) Do steps 2-6 1,000,000 times 8) Print out the probability as success/trials
public class Ball { public String color; public Ball (String color) { this.color = color; } public String getColor() { return color; } } public class Pocket { private Ball[] balls; private java.util.Random generator = new java.util.Random(); private int first; public Pocket(Ball a, Ball b) { balls = new Ball[2]; balls[0] = a; balls[1] = b; } public Ball getFirst() { first = generator.nextInt(2); return balls[first]; } public Ball getLast() { if (first == 0) return balls[1]; return balls[0]; } } public class Bag { private Pocket[] pockets; private java.util.Random generator = new java.util.Random(); public Bag() { pockets = new Pocket[3]; pockets[0] = new Pocket(new Ball("white"), new Ball("white")); pockets[1] = new Pocket(new Ball("white"), new Ball("orange")); pockets[2] = new Pocket(new Ball("orange"), new Ball("orange")); } public int runTrial() { // returns 0 if no orange ball was drawn first (can't be counted) // returns 1 if an orange ball was drawn first but a white ball was drawn second // returns 2 if an orange ball was drawn first and an orange ball was drawn second int pocketSelect = generator.nextInt(3); Ball first = pockets[pocketSelect].getFirst(); if (!first.getColor().equals("orange")) return 0; Ball last = pockets[pocketSelect].getLast(); if (!last.getColor().equals("orange")) return 1; return 2; } } public class Tester { public static void main(String[] args) { Bag bag = new Bag(); int trials = 0; int orangeOrange = 0; for(int i=0; i < 1000000; i++) { int result = bag.runTrial(); if (result == 0) continue; else if (results == 1) trials++; else { trials++; orangeOrange++; } } System.out.println("There were " + trials + " trials and " + orangeOrange + " orange-orange combinations"); System.out.println("The probability is: " + ((double)orangeOrange/(double)trials)); } }
There were 500170 trials and 332801 orange-orange combinations The probability is: 0.6653757722374393 There were 500559 trials and 333707 orange-orange combinations The probability is: 0.6666686644331637 There were 500068 trials and 333559 orange-orange combinations The probability is: 0.6670272842893367 There were 500562 trials and 333987 orange-orange combinations The probability is: 0.667224040178839 There were 500334 trials and 333134 orange-orange combinations The probability is: 0.665823230082305
And expanding to 100,000,000 times instead of 1,000,000 gave + Show Spoiler +
There were 49988348 trials and 33328647 orange-orange combinations The probability is: 0.6667283143663799
I think this speaks for itself.
The flaw in these steps is that our picker is GUARANTEED to pick an orange ball. In your program, you set it such that there are three distinct orange balls to pick from. In reality, if you are guaranteed an orange ball, it's the same thing as having a second person look inside the pockets, and give you an orange ball, in which case it doesn't matter which ball you took from the "orange-orange" pocket
The problem says that we pick a RANDOM pocket (1/3). You can't just throw out a pocket because it includes no orange ball. It's that we picked a random pocket and the first ball we get turns out to be orange. If THAT has happened, what is the probability that our second ball is also orange.
Also, the problem is written in precisely the same way here: Bertran's box paradox. There are a lot of scholarly papers about this problem exactly as it is written. A quick googling should help you find them.
...How is it random if we are guaranteeing a result -.-
On May 11 2012 03:43 Akari Takai wrote: Time for science. Computer science that is.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
My method for testing was this: 1) Make the bag and pocket and balls, and set two counters (trials && success) at 0. 2) Pick a pocket at random. 3) Pick a ball from that pocket at random. 4) If the ball is white, then we can't include this as part of our trials, so don't change the value of trials or successes. 5) If the ball is orange, then this is a valid trial so increase the trial counter by 1. 6) If the second ball in the pocket is also orange, increase the success counter by 1. 7) Do steps 2-6 1,000,000 times 8) Print out the probability as success/trials
public class Ball { public String color; public Ball (String color) { this.color = color; } public String getColor() { return color; } } public class Pocket { private Ball[] balls; private java.util.Random generator = new java.util.Random(); private int first; public Pocket(Ball a, Ball b) { balls = new Ball[2]; balls[0] = a; balls[1] = b; } public Ball getFirst() { first = generator.nextInt(2); return balls[first]; } public Ball getLast() { if (first == 0) return balls[1]; return balls[0]; } } public class Bag { private Pocket[] pockets; private java.util.Random generator = new java.util.Random(); public Bag() { pockets = new Pocket[3]; pockets[0] = new Pocket(new Ball("white"), new Ball("white")); pockets[1] = new Pocket(new Ball("white"), new Ball("orange")); pockets[2] = new Pocket(new Ball("orange"), new Ball("orange")); } public int runTrial() { // returns 0 if no orange ball was drawn first (can't be counted) // returns 1 if an orange ball was drawn first but a white ball was drawn second // returns 2 if an orange ball was drawn first and an orange ball was drawn second int pocketSelect = generator.nextInt(3); Ball first = pockets[pocketSelect].getFirst(); if (!first.getColor().equals("orange")) return 0; Ball last = pockets[pocketSelect].getLast(); if (!last.getColor().equals("orange")) return 1; return 2; } } public class Tester { public static void main(String[] args) { Bag bag = new Bag(); int trials = 0; int orangeOrange = 0; for(int i=0; i < 1000000; i++) { int result = bag.runTrial(); if (result == 0) continue; else if (results == 1) trials++; else { trials++; orangeOrange++; } } System.out.println("There were " + trials + " trials and " + orangeOrange + " orange-orange combinations"); System.out.println("The probability is: " + ((double)orangeOrange/(double)trials)); } }
There were 500170 trials and 332801 orange-orange combinations The probability is: 0.6653757722374393 There were 500559 trials and 333707 orange-orange combinations The probability is: 0.6666686644331637 There were 500068 trials and 333559 orange-orange combinations The probability is: 0.6670272842893367 There were 500562 trials and 333987 orange-orange combinations The probability is: 0.667224040178839 There were 500334 trials and 333134 orange-orange combinations The probability is: 0.665823230082305
And expanding to 100,000,000 times instead of 1,000,000 gave + Show Spoiler +
There were 49988348 trials and 33328647 orange-orange combinations The probability is: 0.6667283143663799
I think this speaks for itself.
The flaw in these steps is that our picker is GUARANTEED to pick an orange ball. In your program, you set it such that there are three distinct orange balls to pick from. In reality, if you are guaranteed an orange ball, it's the same thing as having a second person look inside the pockets, and give you an orange ball, in which case it doesn't matter which ball you took from the "orange-orange" pocket
The problem says that we pick a RANDOM pocket (1/3). You can't just throw out a pocket because it includes no orange ball. It's that we picked a random pocket and the first ball we get turns out to be orange. If THAT has happened, what is the probability that our second ball is also orange.
Also, the problem is written in precisely the same way here: Bertran's box paradox. There are a lot of scholarly papers about this problem exactly as it is written. A quick googling should help you find them.
...How is it random if we are guaranteeing a result -.-
he picked a pocket at random to begin with, looking at the wording of the original problem. You can think of the problem as you have 3 pockets OO, OW, and WW, you randomly pick out an O, what is the probability that the other one is O. This is equivalent to asking you what is the probability of randomly picking the OO pocket (that is the only choice that results in you picking O and the one left over is also O. It can be seen that this probability is obviously 1/3 since there is only one OO pocket out of 3.
On May 11 2012 03:43 Akari Takai wrote: Time for science. Computer science that is.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
My method for testing was this: 1) Make the bag and pocket and balls, and set two counters (trials && success) at 0. 2) Pick a pocket at random. 3) Pick a ball from that pocket at random. 4) If the ball is white, then we can't include this as part of our trials, so don't change the value of trials or successes. 5) If the ball is orange, then this is a valid trial so increase the trial counter by 1. 6) If the second ball in the pocket is also orange, increase the success counter by 1. 7) Do steps 2-6 1,000,000 times 8) Print out the probability as success/trials
public class Ball { public String color; public Ball (String color) { this.color = color; } public String getColor() { return color; } } public class Pocket { private Ball[] balls; private java.util.Random generator = new java.util.Random(); private int first; public Pocket(Ball a, Ball b) { balls = new Ball[2]; balls[0] = a; balls[1] = b; } public Ball getFirst() { first = generator.nextInt(2); return balls[first]; } public Ball getLast() { if (first == 0) return balls[1]; return balls[0]; } } public class Bag { private Pocket[] pockets; private java.util.Random generator = new java.util.Random(); public Bag() { pockets = new Pocket[3]; pockets[0] = new Pocket(new Ball("white"), new Ball("white")); pockets[1] = new Pocket(new Ball("white"), new Ball("orange")); pockets[2] = new Pocket(new Ball("orange"), new Ball("orange")); } public int runTrial() { // returns 0 if no orange ball was drawn first (can't be counted) // returns 1 if an orange ball was drawn first but a white ball was drawn second // returns 2 if an orange ball was drawn first and an orange ball was drawn second int pocketSelect = generator.nextInt(3); Ball first = pockets[pocketSelect].getFirst(); if (!first.getColor().equals("orange")) return 0; Ball last = pockets[pocketSelect].getLast(); if (!last.getColor().equals("orange")) return 1; return 2; } } public class Tester { public static void main(String[] args) { Bag bag = new Bag(); int trials = 0; int orangeOrange = 0; for(int i=0; i < 1000000; i++) { int result = bag.runTrial(); if (result == 0) continue; else if (results == 1) trials++; else { trials++; orangeOrange++; } } System.out.println("There were " + trials + " trials and " + orangeOrange + " orange-orange combinations"); System.out.println("The probability is: " + ((double)orangeOrange/(double)trials)); } }
There were 500170 trials and 332801 orange-orange combinations The probability is: 0.6653757722374393 There were 500559 trials and 333707 orange-orange combinations The probability is: 0.6666686644331637 There were 500068 trials and 333559 orange-orange combinations The probability is: 0.6670272842893367 There were 500562 trials and 333987 orange-orange combinations The probability is: 0.667224040178839 There were 500334 trials and 333134 orange-orange combinations The probability is: 0.665823230082305
And expanding to 100,000,000 times instead of 1,000,000 gave + Show Spoiler +
There were 49988348 trials and 33328647 orange-orange combinations The probability is: 0.6667283143663799
I think this speaks for itself.
The flaw in these steps is that our picker is GUARANTEED to pick an orange ball. In your program, you set it such that there are three distinct orange balls to pick from. In reality, if you are guaranteed an orange ball, it's the same thing as having a second person look inside the pockets, and give you an orange ball, in which case it doesn't matter which ball you took from the "orange-orange" pocket
The problem says that we pick a RANDOM pocket (1/3). You can't just throw out a pocket because it includes no orange ball. It's that we picked a random pocket and the first ball we get turns out to be orange. If THAT has happened, what is the probability that our second ball is also orange.
Also, the problem is written in precisely the same way here: Bertran's box paradox. There are a lot of scholarly papers about this problem exactly as it is written. A quick googling should help you find them.
...How is it random if we are guaranteeing a result -.-
It doesn't say guaranteed anywhere.
You reach into one random pocket and pull out an orange ping pong ball.
You reach into one RANDOM pocket and [happen to] pull out an orange ping pong ball. [Now that you have done that,] what's the probability of the other ball in the pocket being orange?
It's all in the grammar. As I've already said, there's two ways to interpret it. The first way is the conditional probability thing, which accounts for the probability that the first ball is orange, which is 2/3.
There's another way to visualize the second argument. The three bags remain the same. One person LOOKS at the contents of the bags, and pulls out one orange ball. Now what is the probability that the other ball is orange? 1/2.
It's really not all in the grammar. you pull and orange first so that is given. there are three orange balls. 2 of the 3 are in the same pocket so the odds that you picked the OO pocket are 2/3. if you picked that pocket the other ball will be orange. The answer to this question is 2/3 regardless of how crazily you read the problem
Agreed - Two ways to read this puzzle, one gives 1/2 the other 2/3. I would say 1/2 because the possibility of you picking a white ball first is not even mentioned. BUT this is a riddle, and it is too simple math to say 1/2, while the 2/3 case actually requires thinking and considering more than: "well there are only two options".
I think this is a poorly worded riddle with an ambigous solotion that should have been 2/3, but the wording supports 1/2 as a more "correct" answer.
On May 11 2012 03:43 Akari Takai wrote: Time for science. Computer science that is.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
My method for testing was this: 1) Make the bag and pocket and balls, and set two counters (trials && success) at 0. 2) Pick a pocket at random. 3) Pick a ball from that pocket at random. 4) If the ball is white, then we can't include this as part of our trials, so don't change the value of trials or successes. 5) If the ball is orange, then this is a valid trial so increase the trial counter by 1. 6) If the second ball in the pocket is also orange, increase the success counter by 1. 7) Do steps 2-6 1,000,000 times 8) Print out the probability as success/trials
public class Ball { public String color; public Ball (String color) { this.color = color; } public String getColor() { return color; } } public class Pocket { private Ball[] balls; private java.util.Random generator = new java.util.Random(); private int first; public Pocket(Ball a, Ball b) { balls = new Ball[2]; balls[0] = a; balls[1] = b; } public Ball getFirst() { first = generator.nextInt(2); return balls[first]; } public Ball getLast() { if (first == 0) return balls[1]; return balls[0]; } } public class Bag { private Pocket[] pockets; private java.util.Random generator = new java.util.Random(); public Bag() { pockets = new Pocket[3]; pockets[0] = new Pocket(new Ball("white"), new Ball("white")); pockets[1] = new Pocket(new Ball("white"), new Ball("orange")); pockets[2] = new Pocket(new Ball("orange"), new Ball("orange")); } public int runTrial() { // returns 0 if no orange ball was drawn first (can't be counted) // returns 1 if an orange ball was drawn first but a white ball was drawn second // returns 2 if an orange ball was drawn first and an orange ball was drawn second int pocketSelect = generator.nextInt(3); Ball first = pockets[pocketSelect].getFirst(); if (!first.getColor().equals("orange")) return 0; Ball last = pockets[pocketSelect].getLast(); if (!last.getColor().equals("orange")) return 1; return 2; } } public class Tester { public static void main(String[] args) { Bag bag = new Bag(); int trials = 0; int orangeOrange = 0; for(int i=0; i < 1000000; i++) { int result = bag.runTrial(); if (result == 0) continue; else if (results == 1) trials++; else { trials++; orangeOrange++; } } System.out.println("There were " + trials + " trials and " + orangeOrange + " orange-orange combinations"); System.out.println("The probability is: " + ((double)orangeOrange/(double)trials)); } }
There were 500170 trials and 332801 orange-orange combinations The probability is: 0.6653757722374393 There were 500559 trials and 333707 orange-orange combinations The probability is: 0.6666686644331637 There were 500068 trials and 333559 orange-orange combinations The probability is: 0.6670272842893367 There were 500562 trials and 333987 orange-orange combinations The probability is: 0.667224040178839 There were 500334 trials and 333134 orange-orange combinations The probability is: 0.665823230082305
And expanding to 100,000,000 times instead of 1,000,000 gave + Show Spoiler +
There were 49988348 trials and 33328647 orange-orange combinations The probability is: 0.6667283143663799
I think this speaks for itself.
The flaw in these steps is that our picker is GUARANTEED to pick an orange ball. In your program, you set it such that there are three distinct orange balls to pick from. In reality, if you are guaranteed an orange ball, it's the same thing as having a second person look inside the pockets, and give you an orange ball, in which case it doesn't matter which ball you took from the "orange-orange" pocket
The problem says that we pick a RANDOM pocket (1/3). You can't just throw out a pocket because it includes no orange ball. It's that we picked a random pocket and the first ball we get turns out to be orange. If THAT has happened, what is the probability that our second ball is also orange.
Also, the problem is written in precisely the same way here: Bertran's box paradox. There are a lot of scholarly papers about this problem exactly as it is written. A quick googling should help you find them.
...How is it random if we are guaranteeing a result -.-
It doesn't say guaranteed anywhere.
You reach into one random pocket and pull out an orange ping pong ball.
You reach into one RANDOM pocket and [happen to] pull out an orange ping pong ball. [Now that you have done that,] what's the probability of the other ball in the pocket being orange?
It's all in the grammar. As I've already said, there's two ways to interpret it. The first way is the conditional probability thing, which accounts for the probability that the first ball is orange, which is 2/3.
There's another way to visualize the second argument. The three bags remain the same. One person LOOKS at the contents of the bags, and pulls out one orange ball. Now what is the probability that the other ball is orange? 1/2.
On May 11 2012 05:09 Zaphod Beeblebrox wrote: Agreed - Two ways to read this puzzle, one gives 1/2 the other 2/3. I would say 1/2 because the possibility of you picking a white ball first is not even mentioned. BUT this is a riddle, and it is too simple math to say 1/2, while the 2/3 case actually requires thinking and considering more than: "well there are only two options".
I think this is a poorly worded riddle with an ambigous solotion that should have been 2/3, but the wording supports 1/2 as a more "correct" answer.
No offense, but did either of you read the original problem? Yes, there are two ways the problem could have been asked. It could have been asked such that it was possible that your friend did not get the orange ball using a random procedure but instead guaranteed it, and in that case (if you suitably fill it out) the answer will be 1/2.
The problem is that this is unambiguously not what was asked. The question unambiguously stated that the first orange ball was drawn at random. This explicitly rules out the scenario that you need to make 1/2 correct. There may have been an ambiguity in your head, but there was none in the question.
OK here's another way for you people who are convinced it is 50% to look at it:
Instead of orange and white balls let's replace them with money. Gray and brown coins. (For you non-Americans all coins are gray colored except for pennies, which are brown.)
One pocket has two gray coins. A quarter and a dime. One pocket has 1 gray coin and a brown coin. A nickel and a penny One pocket has 2 brown coins. 2 pennies.
You pick a coin out of one of the pockets and it's gray. The reason I'm not stating the value of the coin is because you have to treat each orange ball as a separate entity, likewise each gray coin as a separate entity.
Here are your possibilities:
You drew a quarter from the quarter/dime pair. You drew a dime from the dime/quarter You drew a nickel from the nickel/penny pair.
What are the odds that the other coin is gray?
2/3rds of these possibilities will be a gray coin.
These three possibilities are eliminated: You drew a penny from the nickel/penny pair. You drew a penny from the penny/penny pair. You drew the OTHER penny from the penny/penny pair.
On May 11 2012 07:07 zaikantos wrote: Perhaps a short drawing can help explain why I think it is a 50% chance.
Your assumption that "because a and b are in the same pocket we can say you picked a or c" is your mistake. You say there are 2 possibilities, but there are actually 3 possibilities.
You picked a, so the other ball is b. You picked b, so the other ball is a. You picked c, so the other ball is d.
You are a russian bomber, and you want to send a bomb on a nuclear submarine. The submarine is located on a line and has an integer position (can be negative). It moves at a constant integer speed (can be negative too) each second. The good part is that you have an unlimited amount of bombs, and you can send one each second at a any position. The problem is that you don't know where is the submarine at time 0, and what its speed is.
Is there a way to ensure that you can send hit the submarine in a finite amount of time ? If yes, what is your strategy ?
Yes! Consider the problem as finding two parameters initial position (p) and speed (s). After t seconds, the submarine is at position p + t*s. Try and find a way to explore all possible initial positions and speeds !
Your assumption that "because a and b are in the same pocket we can say you picked a or c" is your mistake. You say there are 2 possibilities, but there are actually 3 possibilities.
You picked a, so the other ball is b. You picked b, so the other ball is a. You picked c, so the other ball is d.
2/3
That's what I thought untill i looked closer at the wording of the riddle.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
You do not pick a random orange ball, which would lead to your 3 possibilites and a 2/3rd chance, but you reach into a random pocket.
If we label the two pockets with at least one orange ball X and Y,
You then "reach into one random pocket and pull out an orange ping pong ball". You pick one of the two pockets, X or Y.
Let's say X has two orange balls, and Y has one orange ball and one white ball.
Two options: You picked pocket X and the second ball is also orange. You picked pocket Y and the second ball is white.
50% or ½.
If the wording came down to "you pick a random orange ball" instead of "you pick a random pocket and take a orange ball out", it would indeed be 2/3.
I guess I'll just throw one of those lateral thinking problems into this mix of relatively intelligent arguing about probability and wording
A bridge has a weight limit of 10,000 KG. A truck weighing exactly 10,000 KG drives onto the bridge, and stops at the center, where a bird weighing 30g lands on it. Why doesn't the bridge collapse?
wow, Last time i commented it seemed like we were done with the ping pong balls.. originally I thought 1/2, but I realized shortly after I was wrong.
I think it's quite simple, tree pockets O1+O2, O3+W1, W2+W3. You drew an O for sure, so either you drew:
- O1 which means your next ball is O2 (Success!) - O2 which means your next ball is O1 (Success!) - O3 which means your next ball is W1 (failure)
it's 2/3, there's really no other way... if the both pockets had same number of Oranges in them, then it would 50% cuz they have the same probabilities of being picked, but in this case there's more chances of u getting a ball from the OO pocket than the OW pocket.
On May 11 2012 23:30 AndyGB4 wrote: wow, Last time i commented it seemed like we were done with the ping pong balls.. originally I thought 1/2, but I realized shortly after I was wrong.
I think it's quite simple, tree pockets O1+O2, O3+W1, W2+W3. You drew an O for sure, so either you drew:
- O1 which means your next ball is O2 (Success!) - O2 which means your next ball is O1 (Success!) - O3 which means your next ball is W1 (failure)
it's 2/3, there's really no other way... if the both pockets had same number of Oranges in them, then it would 50% cuz they have the same probabilities of being picked, but in this case there's more chances of u getting a ball from the OO pocket than the OW pocket.
On May 11 2012 07:13 Slithe wrote: Your assumption that "because a and b are in the same pocket we can say you picked a or c" is your mistake. You say there are 2 possibilities, but there are actually 3 possibilities.
You picked a, so the other ball is b. You picked b, so the other ball is a. You picked c, so the other ball is d.
2/3
That's what I thought untill i looked closer at the wording of the riddle.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
You do not pick a random orange ball, which would lead to your 3 possibilites and a 2/3rd chance, but you reach into a random pocket.
If we label the two pockets with at least one orange ball X and Y,
You then "reach into one random pocket and pull out an orange ping pong ball". You pick one of the two pockets, X or Y.
Let's say X has two orange balls, and Y has one orange ball and one white ball.
Two options: You picked pocket X and the second ball is also orange. You picked pocket Y and the second ball is white.
50% or ½.
If the wording came down to "you pick a random orange ball" instead of "you pick a random pocket and take a orange ball out", it would indeed be 2/3.
On May 11 2012 23:30 AndyGB4 wrote: wow, Last time i commented it seemed like we were done with the ping pong balls.. originally I thought 1/2, but I realized shortly after I was wrong.
I think it's quite simple, tree pockets O1+O2, O3+W1, W2+W3. You drew an O for sure, so either you drew:
- O1 which means your next ball is O2 (Success!) - O2 which means your next ball is O1 (Success!) - O3 which means your next ball is W1 (failure)
it's 2/3, there's really no other way... if the both pockets had same number of Oranges in them, then it would 50% cuz they have the same probabilities of being picked, but in this case there's more chances of u getting a ball from the OO pocket than the OW pocket.
On May 11 2012 07:13 Slithe wrote: Your assumption that "because a and b are in the same pocket we can say you picked a or c" is your mistake. You say there are 2 possibilities, but there are actually 3 possibilities.
You picked a, so the other ball is b. You picked b, so the other ball is a. You picked c, so the other ball is d.
2/3
That's what I thought untill i looked closer at the wording of the riddle.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
You do not pick a random orange ball, which would lead to your 3 possibilites and a 2/3rd chance, but you reach into a random pocket.
If we label the two pockets with at least one orange ball X and Y,
You then "reach into one random pocket and pull out an orange ping pong ball". You pick one of the two pockets, X or Y.
Let's say X has two orange balls, and Y has one orange ball and one white ball.
Two options: You picked pocket X and the second ball is also orange. You picked pocket Y and the second ball is white.
50% or ½.
If the wording came down to "you pick a random orange ball" instead of "you pick a random pocket and take a orange ball out", it would indeed be 2/3.
I read your last post before posting mine, and I disagree with your reasoning.
You can't just say its 50% because they say "you reach into a random pocket"... The fact that you grabbed an Orange ball AFTER picking a random pocket affects the probability of which pocket you picked at random.
Here are some perfectly good examples to prove my point:
If the two pockets had 99 balls in each: P1 = 99 Orange balls P2 = 1 Orange ball, 98 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 99% chance you drew your ball from P1, There's a 1% chance you drew your ball from P2, therefore a 99%(99/100) chance the next ball will also be Orange. and a 1%(1/100) chance the next ball will be White.
If the two pockets had 9 balls in each: P1 = 9 Orange balls P2 = 1 Orange ball, 8 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 90% chance you drew your ball from P1, There's a 10% chance you drew your ball from P2, therefore a 90%(9/10) chance the next ball will also be Orange. and a 10%(1/10) chance the next ball will be White.
If the two pockets had 4 balls in each: P1 = 4 Orange balls P2 = 1 Orange ball, 3 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 80% chance you drew your ball from P1, There's a 20% chance you drew your ball from P2, therefore a 80%(4/5) chance the next ball will also be Orange. and a 20%(1/5) chance the next ball will be White.
If the two pockets had 3 balls in each: P1 = 3 Orange balls P2 = 1 Orange ball, 2 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 75% chance you drew your ball from P1, There's a 25% chance you drew your ball from P2, therefore a 75%(3/4) chance the next ball will also be Orange. and a 25%(1/4) chance the next ball will be White.
*** And this is the ACTUAL riddle: *** If the two pockets had 2 balls in each: P1 = 2 Orange balls P2 = 1 Orange ball, 1 White ball And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 66% chance you drew your ball from P1, There's a 33% chance you drew your ball from P2, therefore a 66%(2/3) chance the next ball will also be Orange. and a 33%(1/3) chance the next ball will be White.
Under your logic from earlier, all these examples would still be 1/2 to you, correct? Don't you see how silly that logic looks on a larger scale though?
Honestly, for anyone who still believes its 1/2, please try it out yourself, recreate this riddle with real objects. Take bottle caps, or pencils, or whatever, and split them up like in the problem, and record each draw. You will see that about 66% of the time, you will draw another Orange ball. The more times you do it, the clearer the probability will be. PLEASEEEEEE try it out
On May 11 2012 23:30 AndyGB4 wrote: wow, Last time i commented it seemed like we were done with the ping pong balls.. originally I thought 1/2, but I realized shortly after I was wrong.
I think it's quite simple, tree pockets O1+O2, O3+W1, W2+W3. You drew an O for sure, so either you drew:
- O1 which means your next ball is O2 (Success!) - O2 which means your next ball is O1 (Success!) - O3 which means your next ball is W1 (failure)
it's 2/3, there's really no other way... if the both pockets had same number of Oranges in them, then it would 50% cuz they have the same probabilities of being picked, but in this case there's more chances of u getting a ball from the OO pocket than the OW pocket.
On May 11 2012 07:13 Slithe wrote: Your assumption that "because a and b are in the same pocket we can say you picked a or c" is your mistake. You say there are 2 possibilities, but there are actually 3 possibilities.
You picked a, so the other ball is b. You picked b, so the other ball is a. You picked c, so the other ball is d.
2/3
That's what I thought untill i looked closer at the wording of the riddle.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
You do not pick a random orange ball, which would lead to your 3 possibilites and a 2/3rd chance, but you reach into a random pocket.
If we label the two pockets with at least one orange ball X and Y,
You then "reach into one random pocket and pull out an orange ping pong ball". You pick one of the two pockets, X or Y.
Let's say X has two orange balls, and Y has one orange ball and one white ball.
Two options: You picked pocket X and the second ball is also orange. You picked pocket Y and the second ball is white.
50% or ½.
If the wording came down to "you pick a random orange ball" instead of "you pick a random pocket and take a orange ball out", it would indeed be 2/3.
I read your last post before posting mine, and I disagree with your reasoning.
You can't just say its 50% because they say "you reach into a random pocket"... The fact that you grabbed an Orange ball AFTER picking a random pocket affects the probability of which pocket you picked at random.
Here are some perfectly good examples to prove my point:
If the two pockets had 99 balls in each: P1 = 99 Orange balls P2 = 1 Orange ball, 98 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 99% chance you drew your ball from P1, There's a 1% chance you drew your ball from P2, therefore a 99%(99/100) chance the next ball will also be Orange. and a 1%(1/100) chance the next ball will be White.
If the two pockets had 9 balls in each: P1 = 9 Orange balls P2 = 1 Orange ball, 8 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 90% chance you drew your ball from P1, There's a 10% chance you drew your ball from P2, therefore a 90%(9/10) chance the next ball will also be Orange. and a 10%(1/10) chance the next ball will be White.
If the two pockets had 4 balls in each: P1 = 4 Orange balls P2 = 1 Orange ball, 3 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 80% chance you drew your ball from P1, There's a 20% chance you drew your ball from P2, therefore a 80%(4/5) chance the next ball will also be Orange. and a 20%(1/5) chance the next ball will be White.
If the two pockets had 3 balls in each: P1 = 3 Orange balls P2 = 1 Orange ball, 2 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 75% chance you drew your ball from P1, There's a 25% chance you drew your ball from P2, therefore a 75%(3/4) chance the next ball will also be Orange. and a 25%(1/4) chance the next ball will be White.
*** And this is the ACTUAL riddle: *** If the two pockets had 2 balls in each: P1 = 2 Orange balls P2 = 1 Orange ball, 1 White ball And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 66% chance you drew your ball from P1, There's a 33% chance you drew your ball from P2, therefore a 66%(2/3) chance the next ball will also be Orange. and a 33%(1/3) chance the next ball will be White.
Under your logic from earlier, all these examples would still be 1/2 to you, correct? Don't you see how silly that logic looks on a larger scale though?
Honestly, for anyone who still believes its 1/2, please try it out yourself, recreate this riddle with real objects. Take bottle caps, or pencils, or whatever, and split them up like in the problem, and record each draw. You will see that about 66% of the time, you will draw another Orange ball. The more times you do it, the clearer the probability will be. PLEASEEEEEE try it out
You didn't understand what I meant.
You do not draw a random orange ball, you open a random pocket and take a orange ball out.
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
If the two pockets had 2 balls in each: P1 = 2 Orange balls P2 = 1 Orange ball, 1 White ball If you randomly open one of the two pockets and take an orange ball out, what are the chances your next ball is Orange? There's a 50% chance you drew your ball from P1, There's a 50% chance you drew your ball from P2, therefore a 50%(1/2) chance the next ball will also be Orange. and a 50%(1/2) chance the next ball will be White.
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
If the two pockets had 2 balls in each: P1 = 2 Orange balls P2 = 1 Orange ball, 1 White ball If you randomly open one of the two pockets and take an orange ball out, what are the chances your next ball is Orange? There's a 50% chance you drew your ball from P1, There's a 50% chance you drew your ball from P2, therefore a 50%(1/2) chance the next ball will also be Orange. and a 50%(1/2) chance the next ball will be White.
I understand what you mean, but the fact that you pulled out an orange ball gives you some info on what Pocket you may have picked.
here's another way to look at it: You know how we both agree that the WW pocket is out of the question? WHY? because we drew an orange ball. So you NEED to take into account the orange ball that was picked. So basically this is what it looks like..
Since we KNOW that we drew an orange ball from our Random Pocket, and there's 3 orange balls in all: P1 = 2 orange balls = (2/3 orange balls) = 66% P2 = 1 orange ball = (1/3 orange balls) = 33% P3 = 0 orange balls = (0/3 orange balls) = 0%
P3 is 0% because it has no orange balls. thats why we rule it out. The fact that P1 has more orange balls in it, gives it better odds of being that pocket chosen than being P2.
Ok let's try something else. If we start by ONLY opening a pocket at random (dont choose a ball yet). (since you said we're just picking a pocket at random) This means there's a 1/3 chance it can be P1, P2 and P3 respectively, correct?
Ok I think we all agree on that. Now once inside that random pocket chosen, you pulled out an Orange Ball. There are 3 possible orange balls that you may now be holding. You may have picked P1 and got O1, You may have picked P1 and got O2, You may have picked P2 and got O3.
There are 2 events here, (1) You picked a pocket. (2) You drew an orange ball. Both events must be taken into consideration. Do you see what I mean? you can't just ignore the fact that you pulled an orange ball after you picked a pocket.
If you want, here are ALL the possibilities BEFORE picking a ball. (With possible outcomes from that pocket after picking a ball)
You picked P1 - and picked ball O1. - and picked ball O2.
You picked P2 - and picked ball O3. - and picked ball W1.
You picked P3 - and picked ball W2. - and picked ball W3.
Now obviously any chances of picking a white ball are out of the question because we KNOW we got an orange ball, right?
So all that remains is this:
You picked P1 - and picked ball O1. - and picked ball O2.
You picked P2 - and picked ball O3.
Picking an Orange ball is the info we have to figure out which pocket we're in. You can't ignore the possibilites listed. The only way the pockets are all equal is if we DIDNT pick a ball yet, but we did, and it was Orange, so that tells us that theres 2 chances that ball in our hands came from P1, and 1 chance it came from P2.
Oh I just thought of another example I really like (Sorry I'm writing so much! my mind just gets going lol):
List the balls as A,B,C,D,E,F and they're grouped, as usual, like so: A+B, C+D, E+F. If you picked a random pocket and i tell you the ball in your hands is either A, B or C, (which is like our Orange/White problem above)
Do you still think that there's a 50/50 chance of being in P1 and P2? I'm actually really curious to hear your answer to this last one.
Nothing will happen regardless of how this poorly punctuated sentence is understood.
The difference comes in the form of the distinction between a falsehood and a lie. A falsehood may be a belief one sincerely believes is true, such as an individual of the middle ages stating the falsehood: "the sun revolves around the earth". Pinochio is ignorant in much the same way, as he has no knowledge of whether or not my own or his nose will grow "right now". The statement is one of a falsehood proclaimed in ignorance, not a lie spoken in the face of what is known to be the truth.
You do not draw a random orange ball, you open a random pocket and take a orange ball out.
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
If the two pockets had 2 balls in each: P1 = 2 Orange balls P2 = 1 Orange ball, 1 White ball If you randomly open one of the two pockets and take an orange ball out, what are the chances your next ball is Orange? There's a 50% chance you drew your ball from P1, There's a 50% chance you drew your ball from P2, therefore a 50%(1/2) chance the next ball will also be Orange. and a 50%(1/2) chance the next ball will be White.
Ok let's try something else. If we start by ONLY opening a pocket at random (dont choose a ball yet). (since you said we're just picking a pocket at random) This means there's a 1/3 chance it can be P1, P2 and P3 respectively, correct?
Ok I think we all agree on that. Now once inside that random pocket chosen, you pulled out an Orange Ball. There are 3 possible orange balls that you may now be holding. You may have picked P1 and got O1, You may have picked P1 and got O2, You may have picked P2 and got O3.
There are 2 events here, (1) You picked a pocket. (2) You drew an orange ball. Both events must be taken into consideration. Do you see what I mean? you can't just ignore the fact that you pulled an orange ball after you picked a pocket.
If you want, here are ALL the possibilities BEFORE picking a ball. (With possible outcomes from that pocket after picking a ball)
You picked P1 - and picked ball O1. - and picked ball O2.
You picked P2 - and picked ball O3. - and picked ball W1.
You picked P3 - and picked ball W2. - and picked ball W3.
Now obviously any chances of picking a white ball are out of the question because we KNOW we got an orange ball, right?
So all that remains is this:
You picked P1 - and picked ball O1. - and picked ball O2.
You picked P2 - and picked ball O3.
Picking an Orange ball is the info we have to figure out which pocket we're in. You can't ignore the possibilites listed. The only way the pockets are all equal is if we DIDNT pick a ball yet, but we did, and it was Orange, so that tells us that theres 2 chances that ball in our hands came from P1, and 1 chance it came from P2.
Oh I just thought of another example I really like (Sorry I'm writing so much! my mind just gets going lol):
List the balls as A,B,C,D,E,F and they're grouped, as usual, like so: A+B, C+D, E+F. If you picked a random pocket and i tell you the ball in your hands is either A, B or C, (which is like our Orange/White problem above)
Do you still think that there's a 50/50 chance of being in P1 and P2? I'm actually really curious to hear your answer to this last one.
I left the first part of your post out, as that is just the same as before.
My view: We can simplify the question to "are there two orange balls in the pocket you picked". You first pick one of two pockets without needing knowledge of there even being balls in there. This will give you a 50% chance of picking either pocket. P1 gives you the correct answer, both orange P2 gives you the wrong answer, one orange one white. 50/50.
Your explanation of why there is a 2/3 chance of picking P1 comes down to once again, picking a random orange ball and not picking a random pocket and THEN taking orange out of that pocket.
List the balls as A,B,C,D,E,F and they're grouped, as usual, like so: A+B, C+D, E+F. If you picked a random pocket and i tell you the ball in your hands is either A, B or C, (which is like our Orange/White problem above)
Do you still think that there's a 50/50 chance of being in P1 and P2? I'm actually really curious to hear your answer to this last one.
I don't think that this is the same as our situation. In your example, we start with P1:AB P2:CD P3:EF and then rule out P3:EF leaving us with P1:AB and P2:CD. Then we pick a A, B or C at random, this means that instead of picking a random pocket like in our riddle, you pick a random orange ball.
In our riddle, we start with P1:OO P2:OW P3: WW and then also rule out P3 leaving us with P1:OO P2:OW. Then we randomly pick one of these two pockets, and see if both balls are orange.
Like I said before, picking a random orange ball means a 66% chance of the other being orange as well, but picking a random pocket that has at least one orange ball means a 50% chance of the other being orange, and I read the riddle as:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
My version:
A friend packed four ping pong balls for you, 3 orange, 1 white. He's placed them in two pockets of your sports bag. One pocket has two orange balls, the other has one white one orange. You pick a random pocket and take out an orange ping pong ball. What's the probability of the pocket having two orange balls?
You do not draw a random orange ball, you open a random pocket and take a orange ball out.
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
If the two pockets had 2 balls in each: P1 = 2 Orange balls P2 = 1 Orange ball, 1 White ball If you randomly open one of the two pockets and take an orange ball out, what are the chances your next ball is Orange? There's a 50% chance you drew your ball from P1, There's a 50% chance you drew your ball from P2, therefore a 50%(1/2) chance the next ball will also be Orange. and a 50%(1/2) chance the next ball will be White.
Ok let's try something else. If we start by ONLY opening a pocket at random (dont choose a ball yet). (since you said we're just picking a pocket at random) This means there's a 1/3 chance it can be P1, P2 and P3 respectively, correct?
Ok I think we all agree on that. Now once inside that random pocket chosen, you pulled out an Orange Ball. There are 3 possible orange balls that you may now be holding. You may have picked P1 and got O1, You may have picked P1 and got O2, You may have picked P2 and got O3.
There are 2 events here, (1) You picked a pocket. (2) You drew an orange ball. Both events must be taken into consideration. Do you see what I mean? you can't just ignore the fact that you pulled an orange ball after you picked a pocket.
If you want, here are ALL the possibilities BEFORE picking a ball. (With possible outcomes from that pocket after picking a ball)
You picked P1 - and picked ball O1. - and picked ball O2.
You picked P2 - and picked ball O3. - and picked ball W1.
You picked P3 - and picked ball W2. - and picked ball W3.
Now obviously any chances of picking a white ball are out of the question because we KNOW we got an orange ball, right?
So all that remains is this:
You picked P1 - and picked ball O1. - and picked ball O2.
You picked P2 - and picked ball O3.
Picking an Orange ball is the info we have to figure out which pocket we're in. You can't ignore the possibilites listed. The only way the pockets are all equal is if we DIDNT pick a ball yet, but we did, and it was Orange, so that tells us that theres 2 chances that ball in our hands came from P1, and 1 chance it came from P2.
Oh I just thought of another example I really like (Sorry I'm writing so much! my mind just gets going lol):
List the balls as A,B,C,D,E,F and they're grouped, as usual, like so: A+B, C+D, E+F. If you picked a random pocket and i tell you the ball in your hands is either A, B or C, (which is like our Orange/White problem above)
Do you still think that there's a 50/50 chance of being in P1 and P2? I'm actually really curious to hear your answer to this last one.
I left the first part of your post out, as that is just the same as before.
My view: We can simplify the question to "are there two orange balls in the pocket you picked". You first pick one of two pockets without needing knowledge of there even being balls in there. This will give you a 50% chance of picking either pocket. P1 gives you the correct answer, both orange P2 gives you the wrong answer, one orange one white. 50/50.
Your explanation of why there is a 2/3 chance of picking P1 comes down to once again, picking a random orange ball and not picking a random pocket and THEN taking orange out of that pocket.
List the balls as A,B,C,D,E,F and they're grouped, as usual, like so: A+B, C+D, E+F. If you picked a random pocket and i tell you the ball in your hands is either A, B or C, (which is like our Orange/White problem above)
Do you still think that there's a 50/50 chance of being in P1 and P2? I'm actually really curious to hear your answer to this last one.
I don't think that this is the same as our situation. In your example, we start with P1:AB P2:CD P3:EF and then rule out P3:EF leaving us with P1:AB and P2:CD. Then we pick a A, B or C at random, this means that instead of picking a random pocket like in our riddle, you pick a random orange ball.
In our riddle, we start with P1:OO P2:OW P3: WW and then also rule out P3 leaving us with P1:OO P2:OW. Then we randomly pick one of these two pockets, and see if both balls are orange.
Like I said before, picking a random orange ball means a 66% chance of the other being orange as well, but picking a random pocket that has at least one orange ball means a 50% chance of the other being orange, and I read the riddle as:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
My version:
A friend packed four ping pong balls for you, 3 orange, 1 white. He's placed them in two pockets of your sports bag. One pocket has two orange balls, the other has one white one orange. You pick a random pocket and take out an orange ping pong ball. What's the probability of the pocket having two orange balls?
But I dont understand how you can use the info of having an orange ball to rule out the WW pocket, but you will not use it to calculate the probability of which pocket was picked...
Before you post another post, can you PLEASE try it in person. take ping pong balls and try it. If you dont have ping pong balls, take pieces of paper, it doesnt really matter. And post your results here when you try. The results won't lie. you will pick from the OO pocket more often than the OW pocket.
This isn't a word play riddle. This is a logical riddle, a mathematical riddle if you will.
And for the record, the ABCDEF example can still be exactly like our riddle now, you just changed the wording of it. let me merge the 2 examples and you'll see they're still the same:
In our riddle, we start with P1:AB P2:CD P3: EF and then also rule out P3 leaving us with P1:AB P2:CD. Then we randomly pick one of these two pockets, and see if both balls AB or CD.
Because think about it, each ball is it's own object, just like each letter in this case. And when you draw an orange ball, we don't know which ball it is, correct? all we know is that it's orange. So like in the case above, all we know is that its etiher A, B or C. (just like its either O1, O2, O3 but we don't know which)
You just twisted the words around. Because in the end, were just trying to find out if its pocket AB or pocket CD. Both problems are the same (like you said before, -.
But that's not that important i guess. I disagree with this following part of your answer.
In our riddle, we start with P1:OO P2:OW P3: WW and then also rule out P3 leaving us with P1:OO P2:OW. Then we randomly pick one of these two pockets, and see if both balls are orange.
How could you rule out P3 if you haven't picked a ball yet?
The only way your 50% would be correct was if the question was if we did not pick a ball at all. It would have to be "There are 2 pockets (OO,OW) Pick one of the pockets. Whats the probability of getting OO pocket." That would be 50%.
But the fact that you draw an Orange ball after you choose a Pocket in our real example, means there are 2 chances it was from OO and 1 chance it was from OW. You just can't ignore the orange ball. We are NOT just picking a pocket.
here's a diagram I drew up showing our problem, let me know if you agree with it:
And here's my solution, going through it step by step, starting by picking a random pocket, like you advise.
You simply can't ignore the fact that AFTER picking a random pocket, you PULL a ball from it, and that ball happens to be orange. Theres a 2/3 chance you're in Pocket 1 and a 1/3 chance you're in Pocket 2, because Pocket 1 has more orange balls that Pocket 2.
Answer me this as well, If one pocket has 100 orange balls, and the other pocket has 1 orange ball and 99 white balls. And then you pick a random pocket, and then draw an Orange ball, is it still a 1/2 chance that your next ball is Orange?
EDIT: And I found the flaw in your logic from an image you posted earlier on. (I also made my post a bit cleaner with quotes)
On May 11 2012 07:14 CptZouglou wrote: Enjoy this new (quite difficult) riddle:
You are a russian bomber, and you want to send a bomb on a nuclear submarine. The submarine is located on a line and has an integer position (can be negative). It moves at a constant integer speed (can be negative too) each second. The good part is that you have an unlimited amount of bombs, and you can send one each second at a any position. The problem is that you don't know where is the submarine at time 0, and what its speed is.
Is there a way to ensure that you can send hit the submarine in a finite amount of time ? If yes, what is your strategy ?
Yes! Consider the problem as finding two parameters initial position (p) and speed (s). After t seconds, the submarine is at position p + t*s. Try and find a way to explore all possible initial positions and speeds !
I'm actually curious about the answer for this... Is there another hint?
Ok, I think I realize how I got myself confused on why the answer is 1/2 and why some people are saying that the answer is 1/2. Suppose that instead of you, your friend is the one in the problem, and you are being posed the problem. Now, your friend chooses a random pocket. Your friend looks into the pocket (so he knows the combination of the two balls). If, it's WW, then discard the trial. However, if it's a pocket with at least one orange ball- count the trial. Your friend looks into the pocket and pulls out an orange ball. Now you are asked what the probability of the other ball is orange. In this case the answer is 1/2.
However, I think the way the answer is posed, is that you don't know that you are pulling an orange ball until you pull it out, which is crucial to the reason why the answer is 2/3.
On May 12 2012 03:11 Misder wrote: On the ball problem- + Show Spoiler +
Ok, I think I realize how I got myself confused on why the answer is 1/2 and why some people are saying that the answer is 1/2. Suppose that instead of you, your friend is the one in the problem, and you are being posed the problem. Now, your friend chooses a random pocket. Your friend looks into the pocket (so he knows the combination of the two balls). If, it's WW, then discard the trial. However, if it's a pocket with at least one orange ball- count the trial. Your friend looks into the pocket and pulls out an orange ball. Now you are asked what the probability of the other ball is orange. In this case the answer is 1/2.
However, I think the way the answer is posed, is that you don't know that you are pulling an orange ball until you pull it out, which is crucial to the reason why the answer is 2/3.
The first is indeed what I think.
You pick a pocket, then pull out an/the orange ball.
I think the question is posed so that you always pull out an/the orange ball, and never the white ball in the OW pocket.
You reach into one random pocket and pull out an orange ping pong ball.
I think that this means that you randomly pick a pocket, but then must pick an/the orange ball if possible.
If it said "one of the two balls", "randomly pick a ball out of the pocket" or something with the same meaning i would agree with it being 2/3rd, but it doesn't.
I guess that this is dependant on how you read it though, and I'm not too confident we'll agree soon. I say that we just leave this riddle for what it is now, and don't clutter the thread up about it anymore.
I won't comment on this any further, but the problem is very explicit in wording. It never mentions prioritizing orange, or looking into the bag after picking one.
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
A different way to phrase the same question is, someone else "reached into one random pocket and pulled out an orange ping pong ball." If you were now to place money on the color of the other ball in the bag, the bet with the higher statistical probability, at 2/3, is orange.
The question does not need to say that he picked a random ball in the bag, because he does not know the contents of the bag and therefore the ball selection must be random. And the result was orange. Change the wording if you like, but then you're not answering the same problem. I don't see how you can possibly think the question is saying, "you pick a bag, you look into it, you see at least one orange ball, you take an orange ball out, and now what's the probability of the other ball being orange [even though you plainly can see it]"?
On May 11 2012 07:14 CptZouglou wrote: Enjoy this new (quite difficult) riddle:
You are a russian bomber, and you want to send a bomb on a nuclear submarine. The submarine is located on a line and has an integer position (can be negative). It moves at a constant integer speed (can be negative too) each second. The good part is that you have an unlimited amount of bombs, and you can send one each second at a any position. The problem is that you don't know where is the submarine at time 0, and what its speed is.
Is there a way to ensure that you can send hit the submarine in a finite amount of time ? If yes, what is your strategy ?
Yes! Consider the problem as finding two parameters initial position (p) and speed (s). After t seconds, the submarine is at position p + t*s. Try and find a way to explore all possible initial positions and speeds !
I'm actually curious about the answer for this... Is there another hint?
It would be hard for him to give another hint without giving away the answer.
x-axis is position, y-axis is speed. You start in the middle of the spiral, and try bombing a location based on each combination of position and speed.
come on guys, did we all forget what the OP stated already?
WARNING: Do not argue - about anything - in this thread. Use PMs. Only Post Riddles or answeres, all answeres must be in spoilers. Try to include the answere for every riddle you write.
On May 11 2012 07:14 CptZouglou wrote: Enjoy this new (quite difficult) riddle:
You are a russian bomber, and you want to send a bomb on a nuclear submarine. The submarine is located on a line and has an integer position (can be negative). It moves at a constant integer speed (can be negative too) each second. The good part is that you have an unlimited amount of bombs, and you can send one each second at a any position. The problem is that you don't know where is the submarine at time 0, and what its speed is.
Is there a way to ensure that you can send hit the submarine in a finite amount of time ? If yes, what is your strategy ?
Yes! Consider the problem as finding two parameters initial position (p) and speed (s). After t seconds, the submarine is at position p + t*s. Try and find a way to explore all possible initial positions and speeds !
I'm actually curious about the answer for this... Is there another hint?
It would be hard for him to give another hint without giving away the answer.
x-axis is position, y-axis is speed. You start in the middle of the spiral, and try bombing a location based on each combination of position and speed.
Very nice! Good job! You described one of the solutions. Would you also solve it if the the submarine was on a N-dimensional plane with a N-dimensional speed?
If you find it, I have another generalized riddle ready for you...
Maybe I'm just blind but I don't think this famous riddle has been posted yet Einstein's riddle 1. In a street there are five houses, painted five different colours. 2. In each house lives a person of different nationality 3. These five homeowners each drink a different kind of beverage, smoke different brand of cigar and keep a different pet.
Who owns the FISH?
HINTS
1. The Brit lives in a red house. 2. The Swede keeps dogs as pets. 3. The Dane drinks tea. 4. The Green house is next to, and on the left of the White house. 5. The owner of the Green house drinks coffee. 6. The person who smokes Pall Mall rears birds. 7. The owner of the Yellow house smokes Dunhill. 8. The man living in the centre house drinks milk. 9. The Norwegian lives in the first house. 10. The man who smokes Blends lives next to the one who keeps cats. 11. The man who keeps horses lives next to the man who smokes Dunhill. 12. The man who smokes Blue Master drinks beer. 13. The German smokes Prince. 14. The Norwegian lives next to the blue house. 15. The man who smokes Blends has a neighbour who drinks water.
On May 13 2012 03:23 StoRm_res wrote: Maybe I'm just blind but I don't think this famous riddle has been posted yet Einstein's riddle 1. In a street there are five houses, painted five different colours. 2. In each house lives a person of different nationality 3. These five homeowners each drink a different kind of beverage, smoke different brand of cigar and keep a different pet.
Who owns the FISH?
HINTS
1. The Brit lives in a red house. 2. The Swede keeps dogs as pets. 3. The Dane drinks tea. 4. The Green house is next to, and on the left of the White house. 5. The owner of the Green house drinks coffee. 6. The person who smokes Pall Mall rears birds. 7. The owner of the Yellow house smokes Dunhill. 8. The man living in the centre house drinks milk. 9. The Norwegian lives in the first house. 10. The man who smokes Blends lives next to the one who keeps cats. 11. The man who keeps horses lives next to the man who smokes Dunhill. 12. The man who smokes Blue Master drinks beer. 13. The German smokes Prince. 14. The Norwegian lives next to the blue house. 15. The man who smokes Blends has a neighbour who drinks water.
On May 11 2012 21:35 CyDe wrote: I guess I'll just throw one of those lateral thinking problems into this mix of relatively intelligent arguing about probability and wording
A bridge has a weight limit of 10,000 KG. A truck weighing exactly 10,000 KG drives onto the bridge, and stops at the center, where a bird weighing 30g lands on it. Why doesn't the bridge collapse?
Very nice! Good job! You described one of the solutions. Would you also solve it if the the submarine was on a N-dimensional plane with a N-dimensional speed?
If you find it, I have another generalized riddle ready for you...
If you let its initial position be arbitrarily large, and it's speed be arbitrarily large, then you're never going to find it by starting in the middle and moving out at a rate of 1/1, because the ship'll outrun the bombs forever. On the other hand, if the boat has a speed of 0, and you're skipping integers, you have a chance of never finding it ever. Even if bombs are persistent, you have to hope you manage to choose a integer arbitrarily large enough on your first couple guesses that it makes the position/speed of the sub look arbitrarily small. Which is impossible to guess, even if we start talking about grahm's numbers - tiny compared to infinity..
Very nice! Good job! You described one of the solutions. Would you also solve it if the the submarine was on a N-dimensional plane with a N-dimensional speed?
If you find it, I have another generalized riddle ready for you...
If you let its initial position be arbitrarily large, and it's speed be arbitrarily large, then you're never going to find it by starting in the middle and moving out at a rate of 1/1, because the ship'll outrun the bombs forever. On the other hand, if the boat has a speed of 0, and you're skipping integers, you have a chance of never finding it ever. Even if bombs are persistent, you have to hope you manage to choose a integer arbitrarily large enough on your first couple guesses that it makes the position/speed of the sub look arbitrarily small. Which is impossible to guess, even if we start talking about grahm's numbers - tiny compared to infinity..
No matter how large its initial position and velocity are, they are still finite once decided, and constant for all time >= 0. So what you are essentially doing is not searching for the submarine, but searching for the submarine's initial condition in the x_0 v_0 phasespace, which will terminate in finite time. The solution is again NOT searching for the submarine, but searching for the submarines initial position and velocity. The N dimensional problem is the exact same, except you need to find an onto mapping from Z (the integers) to Z^(2N), which exists since they have the same cardinality. If you let the submarine change its speed or direction... that's obviously intractable for an infinite line.
Very nice! Good job! You described one of the solutions. Would you also solve it if the the submarine was on a N-dimensional plane with a N-dimensional speed?
If you find it, I have another generalized riddle ready for you...
If you let its initial position be arbitrarily large, and it's speed be arbitrarily large, then you're never going to find it by starting in the middle and moving out at a rate of 1/1, because the ship'll outrun the bombs forever. On the other hand, if the boat has a speed of 0, and you're skipping integers, you have a chance of never finding it ever. Even if bombs are persistent, you have to hope you manage to choose a integer arbitrarily large enough on your first couple guesses that it makes the position/speed of the sub look arbitrarily small. Which is impossible to guess, even if we start talking about grahm's numbers - tiny compared to infinity..
No matter how large its initial position and velocity are, they are still finite once decided, and constant for all time >= 0. So what you are essentially doing is not searching for the submarine, but searching for the submarine's initial condition in the x_0 v_0 phasespace, which will terminate in finite time. The solution is again NOT searching for the submarine, but searching for the submarines initial position and velocity. The N dimensional problem is the exact same, except you need to find an onto mapping from Z (the integers) to Z^(2N), which exists since they have the same cardinality. If you let the submarine change its speed or direction... that's obviously intractable for an infinite line.
How are you going to detect the initial position though? It's just empty space, and a bomb isn't some fancy 'ooh, a submarine was here an hour ago' detector. The submarine isn't everywhere on the line at once, just one point on the line. It seems like if you drew a graph of the rate of the submarine's possible movement vs the rate of testing every possible location, it would be at least equal for all non-zero values of the submarine's velocity. And that's even ignoring a negative value for movement, which means the submarine could at any time cross over into your already searched space. Yes, at infinity, they will converge, but not before then.
Very nice! Good job! You described one of the solutions. Would you also solve it if the the submarine was on a N-dimensional plane with a N-dimensional speed?
If you find it, I have another generalized riddle ready for you...
If you let its initial position be arbitrarily large, and it's speed be arbitrarily large, then you're never going to find it by starting in the middle and moving out at a rate of 1/1, because the ship'll outrun the bombs forever. On the other hand, if the boat has a speed of 0, and you're skipping integers, you have a chance of never finding it ever. Even if bombs are persistent, you have to hope you manage to choose a integer arbitrarily large enough on your first couple guesses that it makes the position/speed of the sub look arbitrarily small. Which is impossible to guess, even if we start talking about grahm's numbers - tiny compared to infinity..
No matter how large its initial position and velocity are, they are still finite once decided, and constant for all time >= 0. So what you are essentially doing is not searching for the submarine, but searching for the submarine's initial condition in the x_0 v_0 phasespace, which will terminate in finite time. The solution is again NOT searching for the submarine, but searching for the submarines initial position and velocity. The N dimensional problem is the exact same, except you need to find an onto mapping from Z (the integers) to Z^(2N), which exists since they have the same cardinality. If you let the submarine change its speed or direction... that's obviously intractable for an infinite line.
That isnt what the problem said at all. The way the problem is written is impossible...plus how would you use the bombs to find out the speed of the submarine?
On May 11 2012 21:35 CyDe wrote: I guess I'll just throw one of those lateral thinking problems into this mix of relatively intelligent arguing about probability and wording
A bridge has a weight limit of 10,000 KG. A truck weighing exactly 10,000 KG drives onto the bridge, and stops at the center, where a bird weighing 30g lands on it. Why doesn't the bridge collapse?
The truck would have burned more than 30g of gasoline by the time it reached the center of the bridge.
Wow this is a really clever riddle, which unfortunately I had to look at the answer for
There's another civil engineering way of thinking about this: weight limits aren't the actual load that the bridge can bear. It's actually higher but not shown in case of poor construction/idiots who try to cross the line. The actual load that the bridge can bear is about 2-3x higher. So a flock of birds can land on the truck no problem.
Very nice! Good job! You described one of the solutions. Would you also solve it if the the submarine was on a N-dimensional plane with a N-dimensional speed?
If you find it, I have another generalized riddle ready for you...
If you let its initial position be arbitrarily large, and it's speed be arbitrarily large, then you're never going to find it by starting in the middle and moving out at a rate of 1/1, because the ship'll outrun the bombs forever. On the other hand, if the boat has a speed of 0, and you're skipping integers, you have a chance of never finding it ever. Even if bombs are persistent, you have to hope you manage to choose a integer arbitrarily large enough on your first couple guesses that it makes the position/speed of the sub look arbitrarily small. Which is impossible to guess, even if we start talking about grahm's numbers - tiny compared to infinity..
No matter how large its initial position and velocity are, they are still finite once decided, and constant for all time >= 0. So what you are essentially doing is not searching for the submarine, but searching for the submarine's initial condition in the x_0 v_0 phasespace, which will terminate in finite time. The solution is again NOT searching for the submarine, but searching for the submarines initial position and velocity. The N dimensional problem is the exact same, except you need to find an onto mapping from Z (the integers) to Z^(2N), which exists since they have the same cardinality. If you let the submarine change its speed or direction... that's obviously intractable for an infinite line.
That isnt what the problem said at all. The way the problem is written is impossible...plus how would you use the bombs to find out the speed of the submarine?
It's not impossible. If the submarine has, say, initial position 3 and constant speed 2, it will be at point 3, then 5, then 7, and so on for time 0, 1, 2. So once you have a function from n=1...inf to the plane of points (the initial conditions phasespace), call it x(n) and v(n) if you want, then all you need to do is chose the functions so that you cover every possible position in the Z^2 plane. If you have say x(10) = 3 and v(10) = 2, corresponding to the correct initial conditions, then you will send your bomb to 3 + 10*2 = 23, which is exactly where the submarine will be at time 10 for starting location 3 and velocity 2. Doing it this way you try every possible set of initial conditions. This is guaranteed to terminate in a finite way. If the initial submarine position was 8276 with speed -99810, you will still eventually get to that set of points on the phasespace in a finite time.
On May 11 2012 07:14 CptZouglou wrote: Enjoy this new (quite difficult) riddle:
You are a russian bomber, and you want to send a bomb on a nuclear submarine. The submarine is located on a line and has an integer position (can be negative). It moves at a constant integer speed (can be negative too) each second. The good part is that you have an unlimited amount of bombs, and you can send one each second at a any position. The problem is that you don't know where is the submarine at time 0, and what its speed is.
Is there a way to ensure that you can send hit the submarine in a finite amount of time ? If yes, what is your strategy ?
Yes! Consider the problem as finding two parameters initial position (p) and speed (s). After t seconds, the submarine is at position p + t*s. Try and find a way to explore all possible initial positions and speeds !
I'm actually curious about the answer for this... Is there another hint?
It would be hard for him to give another hint without giving away the answer.
x-axis is position, y-axis is speed. You start in the middle of the spiral, and try bombing a location based on each combination of position and speed.
Very nice! Good job! You described one of the solutions. Would you also solve it if the the submarine was on a N-dimensional plane with a N-dimensional speed?
If you find it, I have another generalized riddle ready for you...
Here's a solution to the original problem that suggests the generalization. Let f(n) = pn + r be the position of the submarine at time n. Let g(n) be the bombed position at time n. The problem is to find a function g(n) that for any integers p, r there exists an integer N>0, such that f(N)=g(N).
First let's map any pair (p,r) from Z^2 to an integer N>0 bijectively*. Now let g(N) = pN+r. So for any pair p,r there exists a number N such that f(N)=g(N).
I didn't prove that such a bijection exists. I'll refer to the theorem that the Cartesian product of finitely many countable sets is countable. As long as the position of the submarine can be described with a function of a finite number of parameters taken from countable sets we can hit the submarine. Just map every possible function to a certain time N where the bomb will hit at exactly at the position that the function takes.
In the extreme case we can assure a hit even if the submarine has arbitrary rational speed and starting position in N dimensions.
A princess will be as old as a prince will be when the princess is twice as old as the prince was when the princess was half the sum of their current age. Which one of these statements is true? A) the prince is 20 and the princess is 30 B)the prince is 30 and the princess is 20 C) the prince is 30 and the princess is 40 D) the prince is 40 and the princess is 30 E) they are both the same age.
On May 13 2012 03:23 StoRm_res wrote: Maybe I'm just blind but I don't think this famous riddle has been posted yet Einstein's riddle 1. In a street there are five houses, painted five different colours. 2. In each house lives a person of different nationality 3. These five homeowners each drink a different kind of beverage, smoke different brand of cigar and keep a different pet.
Who owns the FISH?
HINTS
1. The Brit lives in a red house. 2. The Swede keeps dogs as pets. 3. The Dane drinks tea. 4. The Green house is next to, and on the left of the White house. 5. The owner of the Green house drinks coffee. 6. The person who smokes Pall Mall rears birds. 7. The owner of the Yellow house smokes Dunhill. 8. The man living in the centre house drinks milk. 9. The Norwegian lives in the first house. 10. The man who smokes Blends lives next to the one who keeps cats. 11. The man who keeps horses lives next to the man who smokes Dunhill. 12. The man who smokes Blue Master drinks beer. 13. The German smokes Prince. 14. The Norwegian lives next to the blue house. 15. The man who smokes Blends has a neighbour who drinks water.
these are my favorite type of puzzles/riddles. you can logically come to a conclusion and it doesn't take wordplay, thinking outside the box, etc.
I'll try to explain the submarine one for non math-majors with limited math jargon: + Show Spoiler +
First we are going to use the initial hint, and say that we can express the submarine's position at any time t by saying it's: current position = initial position + time * speed. So there are two numbers, initial position and speed that we don't know but if we did, we could locate the submarine at any time we wanted. So the problem is, how can we account for (or count up) every possible initial position and speed without skipping any. Then we can just test them in that order until we eventually hit on the right one (which we will since speed and initial position are both finite once they are set). So we have to come up with a strategy to count every possible combination of two integers.
That's what this image that Slithe posted is doing. Highlighting his strategy for hitting every combination of two integers (the x coordinate and the y coordinate of each of those points on the graph) that doesn't skip any, and you can continue following until you hit on the right two numbers for initial position and velocity. That's one system that works. The system I thought of before I saw his is to first count all the combinations that add up to 0--that is (0,0), then all the ones that add up to 1: (1,0) and (0,1) and then the corresponding combinations of those values with negative coordinates (-1,0) and (0,-1), then the ones that sum to 2: (2,0), (1,1), (0,2), and (-2,0), (-1,1), (-1,-1), (1,-1), (0,-2), then 3, etc...
Given infinite time, this will hit on every combination of two integers and not skip any. So now we have an order of initial positions and velocities to test, we just have to say how we go about picking our bombing spot for each one. So we use the current position equation we started with: current position = initial position + time * speed. So our first test with my system is (0,0)--0 initial position and 0 speed, and we'll start at time 0. So it's current position if it started with those characteristics is 0 + 0*0 = 0, so we bomb position 0. Then we move on to (1, 0)--initial position 1 and speed 0, and now time has progressed one second so t = 1. Current position = 1 + 0*1 = 1. So we bomb position one to test for those starting characteristics. Next, (0,1) at t = 2 now, so current position = 0 + 1*2 = 2, bomb at 2. Next (-1,0) at t=3, so current position = -1 + 0*3 = -1, bomb at -1. Next (0,-1) at t = 4, so its current position = 0 + -1*4 = -4, so bomb at -4. And so on. Eventually we will land on the right combination and we will know exactly where it is because we know how much time has passed and we will hit it.
Now we ask, does this work if the submarine is in a 2-dimensional space, or a 3-dimensional space, etc.. The answer is yes. We just have to notice that we need more numbers to describe the initial position and velocity. For 2 dimensions we can say: Current horizontal (x) position = initial x + time * horizontal velocity and current vertical (y) position = initial y + time * vertical velocity. So now there are 4 integers which describe each possible starting state instead of two (initial x, initial y, x velocity, y velocity). And we can use the same counting strategy to count all of them without skipping. All combinations of adding up to 0--(0,0,0,0) which we test at time 0, (1,0,0,0) which we test at time 1, (0,1,0,0) at time 2, etc., systematically hitting the negatives for each coordinate as well as we go. And we can do the same thing for any number of dimensions.
On May 25 2012 01:34 Darkcaster wrote: A princess will be as old as a prince will be when the princess is twice as old as the prince was when the princess was half the sum of their current age. Which one of these statements is true? A) the prince is 20 and the princess is 30 B)the prince is 30 and the princess is 20 C) the prince is 30 and the princess is 40 D) the prince is 40 and the princess is 30 E) they are both the same age.
On May 25 2012 01:34 Darkcaster wrote: A princess will be as old as a prince will be when the princess is twice as old as the prince was when the princess was half the sum of their current age. Which one of these statements is true? A) the prince is 20 and the princess is 30 B)the prince is 30 and the princess is 20 C) the prince is 30 and the princess is 40 D) the prince is 40 and the princess is 30 E) they are both the same age.
It's not a 1:1 (bijective) mapping between X and f(X,t) (some positions can come up twice or not at all), it's between X and t (aka the natural numbers), but otherwise yes.
On May 25 2012 01:34 Darkcaster wrote: A princess will be as old as a prince will be when the princess is twice as old as the prince was when the princess was half the sum of their current age. Which one of these statements is true? A) the prince is 20 and the princess is 30 B)the prince is 30 and the princess is 20 C) the prince is 30 and the princess is 40 D) the prince is 40 and the princess is 30 E) they are both the same age.
You and your competitors are walking down the road with rocks and you can only carry one rock at a time. You cannot go backwards to pick a rock at the back. You exit the road whenever you want and the timer stops. You want to be the one with the biggest rock in the shortest amount of time. What is the statistical solution for the problem?
I cannot find it on google for the life of me. Thanks,
On May 11 2012 23:30 AndyGB4 wrote: wow, Last time i commented it seemed like we were done with the ping pong balls.. originally I thought 1/2, but I realized shortly after I was wrong.
I think it's quite simple, tree pockets O1+O2, O3+W1, W2+W3. You drew an O for sure, so either you drew:
- O1 which means your next ball is O2 (Success!) - O2 which means your next ball is O1 (Success!) - O3 which means your next ball is W1 (failure)
it's 2/3, there's really no other way... if the both pockets had same number of Oranges in them, then it would 50% cuz they have the same probabilities of being picked, but in this case there's more chances of u getting a ball from the OO pocket than the OW pocket.
On May 11 2012 07:13 Slithe wrote: Your assumption that "because a and b are in the same pocket we can say you picked a or c" is your mistake. You say there are 2 possibilities, but there are actually 3 possibilities.
You picked a, so the other ball is b. You picked b, so the other ball is a. You picked c, so the other ball is d.
2/3
That's what I thought untill i looked closer at the wording of the riddle.
On April 21 2012 03:07 TanGeng wrote: A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
You do not pick a random orange ball, which would lead to your 3 possibilites and a 2/3rd chance, but you reach into a random pocket.
If we label the two pockets with at least one orange ball X and Y,
You then "reach into one random pocket and pull out an orange ping pong ball". You pick one of the two pockets, X or Y.
Let's say X has two orange balls, and Y has one orange ball and one white ball.
Two options: You picked pocket X and the second ball is also orange. You picked pocket Y and the second ball is white.
50% or ½.
If the wording came down to "you pick a random orange ball" instead of "you pick a random pocket and take a orange ball out", it would indeed be 2/3.
I read your last post before posting mine, and I disagree with your reasoning.
You can't just say its 50% because they say "you reach into a random pocket"... The fact that you grabbed an Orange ball AFTER picking a random pocket affects the probability of which pocket you picked at random.
Here are some perfectly good examples to prove my point:
If the two pockets had 99 balls in each: P1 = 99 Orange balls P2 = 1 Orange ball, 98 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 99% chance you drew your ball from P1, There's a 1% chance you drew your ball from P2, therefore a 99%(99/100) chance the next ball will also be Orange. and a 1%(1/100) chance the next ball will be White.
If the two pockets had 9 balls in each: P1 = 9 Orange balls P2 = 1 Orange ball, 8 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 90% chance you drew your ball from P1, There's a 10% chance you drew your ball from P2, therefore a 90%(9/10) chance the next ball will also be Orange. and a 10%(1/10) chance the next ball will be White.
If the two pockets had 4 balls in each: P1 = 4 Orange balls P2 = 1 Orange ball, 3 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 80% chance you drew your ball from P1, There's a 20% chance you drew your ball from P2, therefore a 80%(4/5) chance the next ball will also be Orange. and a 20%(1/5) chance the next ball will be White.
If the two pockets had 3 balls in each: P1 = 3 Orange balls P2 = 1 Orange ball, 2 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 75% chance you drew your ball from P1, There's a 25% chance you drew your ball from P2, therefore a 75%(3/4) chance the next ball will also be Orange. and a 25%(1/4) chance the next ball will be White.
*** And this is the ACTUAL riddle: *** If the two pockets had 2 balls in each: P1 = 2 Orange balls P2 = 1 Orange ball, 1 White ball And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 66% chance you drew your ball from P1, There's a 33% chance you drew your ball from P2, therefore a 66%(2/3) chance the next ball will also be Orange. and a 33%(1/3) chance the next ball will be White.
Under your logic from earlier, all these examples would still be 1/2 to you, correct? Don't you see how silly that logic looks on a larger scale though?
Honestly, for anyone who still believes its 1/2, please try it out yourself, recreate this riddle with real objects. Take bottle caps, or pencils, or whatever, and split them up like in the problem, and record each draw. You will see that about 66% of the time, you will draw another Orange ball. The more times you do it, the clearer the probability will be. PLEASEEEEEE try it out
To take your first one with 99 balls in each pocket. You open P1, and now your chance for an orange ball is 100%, we seem to agree there, where you fall of is when we instead pick P2, and now the chance for an orange ball is also 100%
After picking a pocket, if it contains an orange ball, we will pick it. Thus whetever it is 1 or a billion oranges in it is irrelevant as we are guaranteed to pick an orange one, since that is given in the riddle.
Thus
If the two pockets had 99 balls in each: P1 = 99 Orange balls P2 = 1 Orange ball, 98 White balls And you picket a RANDOM pocket and from it drew an orange ball, what is the chance for the remaining balls to be orange? There's a 50% chance you drew your ball from P1, There's a 50% chance you drew your ball from P2, therefore a 50% chance the remaining balls are Orange. and a 50% chance the remaining balls are White.
Either you picked P1 or you picked P2, both will give you an orange ball. The chance of picking either P1 or P2 is 50%.
In the original example we can exclude the double white one because a ball was drawn, if he had picked the double white pocket no balls would be drawn as there would be no orange ball to draw from it.
if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?
On May 13 2012 03:23 StoRm_res wrote: Maybe I'm just blind but I don't think this famous riddle has been posted yet Einstein's riddle 1. In a street there are five houses, painted five different colours. 2. In each house lives a person of different nationality 3. These five homeowners each drink a different kind of beverage, smoke different brand of cigar and keep a different pet.
Who owns the FISH?
HINTS
1. The Brit lives in a red house. 2. The Swede keeps dogs as pets. 3. The Dane drinks tea. 4. The Green house is next to, and on the left of the White house. 5. The owner of the Green house drinks coffee. 6. The person who smokes Pall Mall rears birds. 7. The owner of the Yellow house smokes Dunhill. 8. The man living in the centre house drinks milk. 9. The Norwegian lives in the first house. 10. The man who smokes Blends lives next to the one who keeps cats. 11. The man who keeps horses lives next to the man who smokes Dunhill. 12. The man who smokes Blue Master drinks beer. 13. The German smokes Prince. 14. The Norwegian lives next to the blue house. 15. The man who smokes Blends has a neighbour who drinks water.
On June 29 2012 12:34 calderon wrote: if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?
yes the plane will take off. Since the speed of the wheels have nothing to do with the rest of the plane, the plane will continue to travel forward. Mythbusters tested this in a episode.
On June 29 2012 12:34 calderon wrote: if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?
At first I thought, no the plane will not take off. I assumed that it was the wheels propelling the plane forward but in reality the wheels just roll freely. This would mean that the plan would take off and the wheels woud just spin a lot faster than normal when the wheels are still touching the ground. Maybe it would take more power from the airplane to get it to lift than if the ground wasn't spinning underneath?
On June 29 2012 12:34 calderon wrote: if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?
No, but the conveyor belt will need to accelerate much faster than you think to satisfy the "match the speed PERFECTLY" condition, because you'll be spinning the wheels up with a couple of jet turbines' thrust worth of force
Edit: Plexa, if you measure the speed of the "conveyor belts" the mythbusters use, you'll find they're always slower than the plane's wheels
On June 29 2012 12:34 calderon wrote: if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?
No, but the conveyor belt will need to accelerate much faster than you think to satisfy the "match the speed PERFECTLY" condition, because you'll be spinning the wheels up with a couple of jet turbines' thrust worth of force
Edit: Plexa, if you measure the speed of the "conveyor belts" the mythbusters use, you'll find they're always slower than the plane's wheels
Yeah and its because the wheels just sit there. Theres no power going to the wheels to accelerate the plane, its all in the engines/propellers thrusting the plane forward. No matter how quick the conveyer belt is going, the plane will still go forward
On June 29 2012 12:34 calderon wrote: if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?
No, but the conveyor belt will need to accelerate much faster than you think to satisfy the "match the speed PERFECTLY" condition, because you'll be spinning the wheels up with a couple of jet turbines' thrust worth of force
Edit: Plexa, if you measure the speed of the "conveyor belts" the mythbusters use, you'll find they're always slower than the plane's wheels
Yeah and its because the wheels just sit there. Theres no power going to the wheels to accelerate the plane, its all in the engines/propellers thrusting the plane forward. No matter how quick the conveyer belt is going, the plane will still go forward
You can, in theory, apply enough force to the wheels with the conveyor belt to keep the aeroplane in place, it just corresponds to an astronomically fast acceleration of the conveyor belt. There's no upper limit on how hard you can pull on the bottom of the wheels, but it's a bit misleading to talk about matching "speeds" instead of "forces"
The wheels have mass, so there is a force associated with accelerating them - this is normally much smaller than the force associated with accelerating the aeroplane, but if you just set your imaginary conveyor belt to accelerate, say, 1000 times faster than the aeroplane would accelerate when it was unhindered, you could keep the aeroplane in place.
don't people realise that the conveyor belt works against it self? the plane accelerates through it's turbines wheels start spinning conveyor belt matches the speed of the spinning in the oppisite direction now the wheels are spinning twice as fast without slowing the plane down
On June 29 2012 12:34 calderon wrote: if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?
No, but the conveyor belt will need to accelerate much faster than you think to satisfy the "match the speed PERFECTLY" condition, because you'll be spinning the wheels up with a couple of jet turbines' thrust worth of force
Edit: Plexa, if you measure the speed of the "conveyor belts" the mythbusters use, you'll find they're always slower than the plane's wheels
Yeah and its because the wheels just sit there. Theres no power going to the wheels to accelerate the plane, its all in the engines/propellers thrusting the plane forward. No matter how quick the conveyer belt is going, the plane will still go forward
You can, in theory, apply enough force to the wheels with the conveyor belt to keep the aeroplane in place, it just corresponds to an astronomically fast acceleration of the conveyor belt. There's no upper limit on how hard you can pull on the bottom of the wheels, but it's a bit misleading to talk about matching "speeds" instead of "forces"
The wheels have mass, so there is a force associated with accelerating them - this is normally much smaller than the force associated with accelerating the aeroplane, but if you just set your imaginary conveyor belt to accelerate, say, 1000 times faster than the aeroplane would accelerate when it was unhindered, you could keep the aeroplane in place.
Actually, there will be a point where the conveyor belt will create enough wind for the plane to lift off, and this will probably occur faster than any upper limits on the wheel speed are approached - the only one I can think of is wind drag, which I imagine will always be insignificant compared to conveyor belt wind drag.
It's a tricky question, the supposed "answer" disagrees with the conditions set by the question on the grounds that it's physically impossible (which it isn't, it's just harder to achieve than with a powered wheel type vehicle), but if you start disagreeing on account of physics then you have to ask just how much physics do you want to consider?
This is because the plane accelerates with its engines through the AIR, not with its wheels through the ground.
You could compare it with a very aerodynamical car driving in heavy headwind, since the car accelerates through the ground, and due to its aerodynamical shape is isn't affected much by the wind and thus would drive just fine.
On June 30 2012 00:13 klicken wrote: Oh my god, not the airplane again.
Yes it will take off.
This is because the plane accelerates with its engines through the AIR, not with its wheels through the ground.
You could compare it with a very aerodynamical car driving in heavy headwind, since the car accelerates through the ground, and due to its aerodynamical shape is isn't affected much by the wind and thus would drive just fine.
???
the question states that the conveyor belt is moving at the exact same speed as the aeroplanes tyres. the aeroplanes tyres are the only thing propelling the plane forward. the THRUST from the ENGINES are TRANSLATED to the TYRES. if the TYRES are doing ZERO work then the PLANE is not TAKING OFF.
LIFT is not generated unless the plane is moving at an appropriate speed. lift is not generated by the engines. the engines serve as a means to propel the plane forward. lift is created from the pressure difference between the top and bottom of the wings. this lift is not generated if the air is not moving over the wings of the plane (which would not be the case if the plane was on a conveyor belt and NOT moving relative to the air).
i am not sure why this is so puzzling to some.
and that mythbusters episode as above didn't have the conveyor belt moving at the same speed as the plane, if that was the case, the plane would not be moving relative to the observer at any point.
On June 30 2012 04:26 Phael wrote: A plane generates lift when air flows over its wings.
Here, the plane does not move relative to the wind (except the small breeze generated by the belt itself).
Therefor, no lift is produced and the plane does not take off.
The plane does move relative to the wind. The plane moves forward at the same speed no matter whether it is on a runway or a conveyor belt. The wheels are not powered, so the conveyor belt simply makes the wheels spin twice as fast while the plane moves at the same speed through the air.
The only way that a conveyor belt is going to stop a plane taking off is if it causes the whiles to spin so fast that the plane catches fire (or an equally stupid scenario).
Ok maybe we are talking about two different things.
In that picture, it's implied that the plane stays still on the conveyor belt - where its forward speed is matched by the belt. As in, does not move relative to a person sitting nearby. Yes, it only has the friction of the wheels to work on, but if the plane moves at all, even at a tiny speed, it will run off the edge of the treadmill rendering the scenario moot. So if it's moving forward at 100mph, the belt is also running at a high enough speed to push the airplane back, and their forces cancel each other out so the plane does not move.
If there is no wind blowing, then the plane is also stationary relative to the wind. It's stationary to you, wind is stationary to you, commutative property, etc.
In the mythbuster's scenario, the plane was NOT kept still. It was still moving relative to the viewer, and thus, the wind. So it was able to take off.
A plane is standing on a runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane's speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction). Can the plane take off?
A plane is standing on a runway that can move (some sort of band conveyer). The conveyer belt exactly matches the speed of the wheels at any given time, moving in the opposite direction of rotation. Can the plane take off?
It all depends on the wording of the question. As long as the wording of the question does not specifically restrict the translational motion (i.e. conveyer belt exactly matches the speed of the wheels at any given time), then the plane will take off. This is true of the real world - there is no conveyer belt out there that can be made to prevent the plane from taking off.
A conveyer belt or treadmill that can keep a plane with frictionless wheels from taking off will require to run to infinite speeds (constant force over time).
And what would the effect be if you reversed the direction the conveyor belt is going? I mean a plane going forward has it's wheels spinning clockwise right? So if you had the conveyor belt turning in the direction the plane was going wouldn't that cancel out the movement of the wheels?
I mean in the mythbusters video the truck with Jaimy and the plane move in opposite directions and I can see how that would make the wheels move twice as fast. So what would the effect be if both the truck pulling the conveyor belt and the plane moved in the same direction?
On June 30 2012 04:51 Phael wrote: Ok maybe we are talking about two different things.
In that picture, it's implied that the plane stays still on the conveyor belt - where its forward speed is matched by the belt. As in, does not move relative to a person sitting nearby. So if it's moving forward at 100mph, the belt is also running at 100mph, and their speeds cancel out.
If there is no wind blowing, then the plane is also stationary relative to the wind. It's stationary to you, wind is stationary to you, commutative property, etc.
In the mythbuster's scenario, the plane was NOT kept still. It was still moving relative to the viewer, and thus, the wind. So it was able to take off.
You are completely missing the point.
Because the plane's wheels are free (not powered), no matter how fast the conveyor belt moves, the plan will never be stationary relative to a person standing beside the conveyor belt. The planes engines cause it to accelerate through the air, not over the ground. It will always move forward.
A plane is standing on a runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane's speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction). Can the plane take off?
A plane is standing on a runway that can move (some sort of band conveyer). The conveyer belt exactly matches the speed of the wheels at any given time, moving in the opposite direction of rotation. Can the plane take off?
It all depends on the wording of the question. As long as the wording of the question does not specifically restrict the translational motion (i.e. conveyer belt exactly matches the speed of the wheels at any given time), then the plane will take off. This is true of the real world - there is no conveyer belt out there that can be made to prevent the plane from taking off.
A conveyer belt or treadmill that can keep a plane with frictionless wheels from taking off will require to run to infinite speeds (constant force over time).
It doesn't matter if the belt is going 1/2 the speed of the plane, or exactly, or twice.
The air over the wings is completely independent of how fast the wheels are moving. Completely. The plane could be moving backward on a conveyor belt and it would still take off.
On June 30 2012 04:52 Golden Ghost wrote: And what would the effect be if you reversed the direction the conveyor belt is going? I mean a plane going forward has it's wheels spinning clockwise right? So if you had the conveyor belt turning in the direction the plane was going wouldn't that cancel out the movement of the wheels?
I mean in the mythbusters video the truck with Jaimy and the plane move in opposite directions and I can see how that would make the wheels move twice as fast. So what would the effect be if both the truck pulling the conveyor belt and the plane moved in the same direction?
Correct. If you reversed the treadmill then the plane's wheels would not turn at all...but it would still take off normally.
On June 30 2012 04:51 Phael wrote: Ok maybe we are talking about two different things.
In that picture, it's implied that the plane stays still on the conveyor belt - where its forward speed is matched by the belt. As in, does not move relative to a person sitting nearby. So if it's moving forward at 100mph, the belt is also running at 100mph, and their speeds cancel out.
If there is no wind blowing, then the plane is also stationary relative to the wind. It's stationary to you, wind is stationary to you, commutative property, etc.
In the mythbuster's scenario, the plane was NOT kept still. It was still moving relative to the viewer, and thus, the wind. So it was able to take off.
You are completely missing the point.
Because the plane's wheels are free (not powered), no matter how fast the conveyor belt moves, the plan will never be stationary relative to a person standing beside the conveyor belt. The planes engines cause it to accelerate through the air, not over the ground. It will always move forward.
This is only true in a frictionless scenario. In the real world (eg. that picture), a conveyor belt moving ANYTHING will exert a force on the object.
Example: put a toy car on a tablecloth and slowly pull the tablecloth. The toy car moves relative to the observer.
On June 30 2012 04:51 Phael wrote: Ok maybe we are talking about two different things.
In that picture, it's implied that the plane stays still on the conveyor belt - where its forward speed is matched by the belt. As in, does not move relative to a person sitting nearby. So if it's moving forward at 100mph, the belt is also running at 100mph, and their speeds cancel out.
If there is no wind blowing, then the plane is also stationary relative to the wind. It's stationary to you, wind is stationary to you, commutative property, etc.
In the mythbuster's scenario, the plane was NOT kept still. It was still moving relative to the viewer, and thus, the wind. So it was able to take off.
You are completely missing the point.
Because the plane's wheels are free (not powered), no matter how fast the conveyor belt moves, the plan will never be stationary relative to a person standing beside the conveyor belt. The planes engines cause it to accelerate through the air, not over the ground. It will always move forward.
This is only true in a frictionless scenario. In the real world (eg. that picture), a conveyor belt moving ANYTHING will exert a force on the object.
Hence my original post saying that the plane would catch fire before it was actually stopped.
A plane is standing on a runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane's speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction). Can the plane take off?
A plane is standing on a runway that can move (some sort of band conveyer). The conveyer belt exactly matches the speed of the wheels at any given time, moving in the opposite direction of rotation. Can the plane take off?
It all depends on the wording of the question. As long as the wording of the question does not specifically restrict the translational motion (i.e. conveyer belt exactly matches the speed of the wheels at any given time), then the plane will take off. This is true of the real world - there is no conveyer belt out there that can be made to prevent the plane from taking off.
A conveyer belt or treadmill that can keep a plane with frictionless wheels from taking off will require to run to infinite speeds (constant force over time).
It doesn't matter if the belt is going 1/2 the speed of the plane, or exactly, or twice.
The air over the wings is completely independent of how fast the wheels are moving. Completely. The plane could be moving backward on a conveyor belt and it would still take off.
Again - this is up to the wording of the question. In the real world, there is no conveyer that can keep a plan from taking off. On a completely theoretical treadmill that exactly matches the speed of the wheels at any given time, a plane with wheels of no friction or durability to consider will not be able to take off - there is no translational motion. And given that the plane is producing force (acceleration), the treadmill must also be producing the opposite but equal force translated at the wheel - thus it will eventually reach infinite speeds (as there is no friction).
However, in the real world, there is NO treadmill that can prevent a plane from taking off. The reason why I am bringing up that it is a wording issue is because if the question does not explicitly state that no translational motion can take place, then we can assume that the treadmill has limited powers, the wheels have friction, and things break.
On June 30 2012 00:13 klicken wrote: Oh my god, not the airplane again.
Yes it will take off.
This is because the plane accelerates with its engines through the AIR, not with its wheels through the ground.
You could compare it with a very aerodynamical car driving in heavy headwind, since the car accelerates through the ground, and due to its aerodynamical shape is isn't affected much by the wind and thus would drive just fine.
???
the question states that the conveyor belt is moving at the exact same speed as the aeroplanes tyres. the aeroplanes tyres are the only thing propelling the plane forward. the THRUST from the ENGINES are TRANSLATED to the TYRES. if the TYRES are doing ZERO work then the PLANE is not TAKING OFF.
LIFT is not generated unless the plane is moving at an appropriate speed. lift is not generated by the engines. the engines serve as a means to propel the plane forward. lift is created from the pressure difference between the top and bottom of the wings. this lift is not generated if the air is not moving over the wings of the plane (which would not be the case if the plane was on a conveyor belt and NOT moving relative to the air).
i am not sure why this is so puzzling to some.
and that mythbusters episode as above didn't have the conveyor belt moving at the same speed as the plane, if that was the case, the plane would not be moving relative to the observer at any point.
This is the bit that is incorrect. The tyres are not propelling the plane forward, it's the engines. If you removed the tyres and placed the plane on a frictionless surface it would be able to take off. Indeed, if tyres propelling the plane were correct then how does a plane fly through air without the tyres propelling it forward? The tyres are just a way for the plane to move while on the ground, and in a way which reduces as much extra drag as possible. Think of it this way, say you were extremely high up in the air (sationary) and dropped a plane from this height. Assuming the plane stays upright (i.e. doesn't begin rotating as it falls), if the plane turns on its engine will it be able to begin flying mid air?
The difficulty in understanding this problem (or maybe why the problem is ill posed) is that plane do not use their tyres to generate forward motion. If they did, they would take off and be unable to sustain their speed with their engines and immediately crash. Thus its impossible for a tyre propelled plane to even fly, let alone take off from a treadmill.
On June 30 2012 04:51 Phael wrote: Ok maybe we are talking about two different things.
In that picture, it's implied that the plane stays still on the conveyor belt - where its forward speed is matched by the belt. As in, does not move relative to a person sitting nearby. So if it's moving forward at 100mph, the belt is also running at 100mph, and their speeds cancel out.
If there is no wind blowing, then the plane is also stationary relative to the wind. It's stationary to you, wind is stationary to you, commutative property, etc.
In the mythbuster's scenario, the plane was NOT kept still. It was still moving relative to the viewer, and thus, the wind. So it was able to take off.
You are completely missing the point.
Because the plane's wheels are free (not powered), no matter how fast the conveyor belt moves, the plan will never be stationary relative to a person standing beside the conveyor belt. The planes engines cause it to accelerate through the air, not over the ground. It will always move forward.
This is only true in a frictionless scenario. In the real world (eg. that picture), a conveyor belt moving ANYTHING will exert a force on the object.
Hence my original post saying that the plane would catch fire before it was actually stopped.
State 1: both belt and plane at rest. State 2: plane starts up its engine, belt starts spinning. Forces cancel each other out and plane does not move. State 3: plane keeps building up thrust, belt keeps spinning faster to apply enough force on the wheels to keep the plane stationary. State 4: plane applies enough thrust to force the belt to spin so fast something catches on fire. Experiment is nullified.
Until something actually breaks, the plane does not achieve lift-off, how does that not satisfy the question?
Plane question is pretty stupid, since it's a giant wad of misunderstanding on both sides.
Technically, since we don't have frictionless wheels, it's theoretically possible for a treadmill to move so fast that the miniscule friction of the wheels exactly matches the force of the engines. Of course we're talking something like a treadmill moving at c or something. I don't even know if that'll be enough..
Someone should calculate how fast a treadmill has to move given typical friction of the wheels and typical thrust of the engines. Of course we'd have to assume indestructible wheels.
Because a standard plane would take off long before the friction caused it to catch fire.
State 1: both belt and plane at rest
State 2: plane starts up its engine, belt starts spinning. The plane begins to move forward because the force causing it to move forward have nothing to do with it's wheels, and the small amount of friction generated is not enough to stop it.
State 3: Plane is in the air long before the friction generated enough heat to cause a problem.
A plane is standing on a runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane's speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction). Can the plane take off?
A plane is standing on a runway that can move (some sort of band conveyer). The conveyer belt exactly matches the speed of the wheels at any given time, moving in the opposite direction of rotation. Can the plane take off?
It all depends on the wording of the question. As long as the wording of the question does not specifically restrict the translational motion (i.e. conveyer belt exactly matches the speed of the wheels at any given time), then the plane will take off. This is true of the real world - there is no conveyer belt out there that can be made to prevent the plane from taking off.
A conveyer belt or treadmill that can keep a plane with frictionless wheels from taking off will require to run to infinite speeds (constant force over time).
It doesn't matter if the belt is going 1/2 the speed of the plane, or exactly, or twice.
The air over the wings is completely independent of how fast the wheels are moving. Completely. The plane could be moving backward on a conveyor belt and it would still take off.
Again - this is up to the wording of the question. In the real world, there is no conveyer that can keep a plan from taking off. On a completely theoretical treadmill that exactly matches the speed of the wheels at any given time, a plane with wheels of no friction or durability to consider will not be able to take off - there is no translational motion. And given that the plane is producing force (acceleration), the treadmill must also be producing the opposite but equal force translated at the wheel - thus it will eventually reach infinite speeds (as there is no friction).
However, in the real world, there is NO treadmill that can prevent a plane from taking off. The reason why I am bringing up that it is a wording issue is because if the question does not explicitly state that no translational motion can take place, then we can assume that the treadmill has limited powers, the wheels have friction, and things break.
I get what you're saying now but it's a ridiculous point.
On June 30 2012 05:09 JeeJee wrote: Plane question is pretty stupid, since it's a giant wad of misunderstanding on both sides.
Technically, since we don't have frictionless wheels, it's theoretically possible for a treadmill to move so fast that the miniscule friction of the wheels exactly matches the force of the engines. Of course we're talking something like a treadmill moving at c or something. I don't even know if that'll be enough..
Someone should calculate how fast a treadmill has to move given typical friction of the wheels and typical thrust of the engines. Of course we'd have to assume indestructible wheels.
See the link in my post - the calculations are there. It's not "speed" - it's "acceleration".
A plane is standing on a runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane's speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction). Can the plane take off?
A plane is standing on a runway that can move (some sort of band conveyer). The conveyer belt exactly matches the speed of the wheels at any given time, moving in the opposite direction of rotation. Can the plane take off?
It all depends on the wording of the question. As long as the wording of the question does not specifically restrict the translational motion (i.e. conveyer belt exactly matches the speed of the wheels at any given time), then the plane will take off. This is true of the real world - there is no conveyer belt out there that can be made to prevent the plane from taking off.
A conveyer belt or treadmill that can keep a plane with frictionless wheels from taking off will require to run to infinite speeds (constant force over time).
It doesn't matter if the belt is going 1/2 the speed of the plane, or exactly, or twice.
The air over the wings is completely independent of how fast the wheels are moving. Completely. The plane could be moving backward on a conveyor belt and it would still take off.
Again - this is up to the wording of the question. In the real world, there is no conveyer that can keep a plan from taking off. On a completely theoretical treadmill that exactly matches the speed of the wheels at any given time, a plane with wheels of no friction or durability to consider will not be able to take off - there is no translational motion. And given that the plane is producing force (acceleration), the treadmill must also be producing the opposite but equal force translated at the wheel - thus it will eventually reach infinite speeds (as there is no friction).
However, in the real world, there is NO treadmill that can prevent a plane from taking off. The reason why I am bringing up that it is a wording issue is because if the question does not explicitly state that no translational motion can take place, then we can assume that the treadmill has limited powers, the wheels have friction, and things break.
I get what you're saying now but it's a ridiculous point.
I don't see why it's a ridiculous point - the answer to the original post
"If a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?"
Is a solid No because it restricts the translational motion of the plane. However, the other correct answer is (and some posters mention this) "The plane cannot lift off unless the treadmill/wheel/bearing/etc breaks".
The statement "The airplane will lift off regardless" does not keep to the wording of the question. If you play with riddles and not keep to the wording of the question, then you are not playing correctly at all.
On June 30 2012 05:10 hzflank wrote: State 2: plane starts up its engine, belt starts spinning. The plane begins to move forward because the force causing it to move forward have nothing to do with it's wheels, and the small amount of friction generated is not enough to stop it.
Yes, the wheels rolling over the treadmill will cause a backwards force. If the treadmill spins fast enough, this force cancels out the forward thrust of the plane's engines.
And it's not a "small" amount of friction. Since the wheels on belt situation is rubber on rubber, I'd guesstimate that the Crr is roughly 0.03, so the belt only has to spin at about 30 times the speed of what the plane would be moving at (so, in this case, if the plane needs 100 mph to take off and applies the thrust necessary to reach 100mph on the ground, the treadmill only has to spin at around 3000 mph - totally possible).
Edit: You must be aware of this if you've ever driven anywhere. When you release the gas pedal, your car decelerates. It decelerates precisely because due to rolling friction. It might not be that strong, but then my car isn't traveling all that fast. If the belt can spin fast enough without breaking, and the wheels don't get burned out, then you can keep the plane stationary.
On June 30 2012 05:09 JeeJee wrote: Plane question is pretty stupid, since it's a giant wad of misunderstanding on both sides.
Technically, since we don't have frictionless wheels, it's theoretically possible for a treadmill to move so fast that the miniscule friction of the wheels exactly matches the force of the engines. Of course we're talking something like a treadmill moving at c or something. I don't even know if that'll be enough..
Someone should calculate how fast a treadmill has to move given typical friction of the wheels and typical thrust of the engines. Of course we'd have to assume indestructible wheels.
See the link in my post - the calculations are there. It's not "speed" - it's "acceleration".
A plane is standing on a runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane's speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction). Can the plane take off?
A plane is standing on a runway that can move (some sort of band conveyer). The conveyer belt exactly matches the speed of the wheels at any given time, moving in the opposite direction of rotation. Can the plane take off?
It all depends on the wording of the question. As long as the wording of the question does not specifically restrict the translational motion (i.e. conveyer belt exactly matches the speed of the wheels at any given time), then the plane will take off. This is true of the real world - there is no conveyer belt out there that can be made to prevent the plane from taking off.
A conveyer belt or treadmill that can keep a plane with frictionless wheels from taking off will require to run to infinite speeds (constant force over time).
It doesn't matter if the belt is going 1/2 the speed of the plane, or exactly, or twice.
The air over the wings is completely independent of how fast the wheels are moving. Completely. The plane could be moving backward on a conveyor belt and it would still take off.
Again - this is up to the wording of the question. In the real world, there is no conveyer that can keep a plan from taking off. On a completely theoretical treadmill that exactly matches the speed of the wheels at any given time, a plane with wheels of no friction or durability to consider will not be able to take off - there is no translational motion. And given that the plane is producing force (acceleration), the treadmill must also be producing the opposite but equal force translated at the wheel - thus it will eventually reach infinite speeds (as there is no friction).
However, in the real world, there is NO treadmill that can prevent a plane from taking off. The reason why I am bringing up that it is a wording issue is because if the question does not explicitly state that no translational motion can take place, then we can assume that the treadmill has limited powers, the wheels have friction, and things break.
I get what you're saying now but it's a ridiculous point.
I don't see why it's a ridiculous point - the answer to the original post
"If a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?"
Is a solid No because it restricts the translational motion of the plane. However, the other correct answer is (and some posters mention this) "The plane cannot lift off unless the treadmill/wheel/bearing/etc breaks".
The statement "The airplane will lift off regardless" does not keep to the wording of the question. If you play with riddles and not keep to the wording of the question, then you are not playing correctly at all.
Wut, you're back to being wrong again.
The belt matching the speed perfectly doesn't remove the friction from the wheels to the belt. You saying frictionless wheels removes the friction.
On June 30 2012 05:10 hzflank wrote: State 2: plane starts up its engine, belt starts spinning. The plane begins to move forward because the force causing it to move forward have nothing to do with it's wheels, and the small amount of friction generated is not enough to stop it.
Yes, the wheels rolling over the treadmill will cause a backwards force. If the treadmill spins fast enough, this force cancels out the forward thrust of the plane's engines.
And it's not a "small" amount of friction. Since the wheels on belt situation is rubber on rubber, I'd guesstimate that the Crr is roughly 0.03, so the belt only has to spin at about 30 times the speed of what the plane would be moving at (so, in this case, if the plane needs 100 mph to take off and applies the thrust necessary to reach 100mph on the ground, the treadmill only has to spin at around 3000 mph - totally possible).
It happens because the engine creates a force that causes the plane to accelerate forwards through the air. The engine does not make the wheels turn. The wheels only turn because of friction against the ground. In theory, a plane does not need wheels at all to take off, the wheels just make it easier by massively reducing the friction against the ground.
Wait, wait, I'm back to thinking my original thought that you're talking math teacher uselessness.
The plane gets its thrust not from the friction between the wheels and the ground but from the engine and the air. It's completely independent of the ground.
On June 30 2012 05:09 JeeJee wrote: Plane question is pretty stupid, since it's a giant wad of misunderstanding on both sides.
Technically, since we don't have frictionless wheels, it's theoretically possible for a treadmill to move so fast that the miniscule friction of the wheels exactly matches the force of the engines. Of course we're talking something like a treadmill moving at c or something. I don't even know if that'll be enough..
Someone should calculate how fast a treadmill has to move given typical friction of the wheels and typical thrust of the engines. Of course we'd have to assume indestructible wheels.
See the link in my post - the calculations are there. It's not "speed" - it's "acceleration".
A plane is standing on a runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane's speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction). Can the plane take off?
A plane is standing on a runway that can move (some sort of band conveyer). The conveyer belt exactly matches the speed of the wheels at any given time, moving in the opposite direction of rotation. Can the plane take off?
It all depends on the wording of the question. As long as the wording of the question does not specifically restrict the translational motion (i.e. conveyer belt exactly matches the speed of the wheels at any given time), then the plane will take off. This is true of the real world - there is no conveyer belt out there that can be made to prevent the plane from taking off.
A conveyer belt or treadmill that can keep a plane with frictionless wheels from taking off will require to run to infinite speeds (constant force over time).
It doesn't matter if the belt is going 1/2 the speed of the plane, or exactly, or twice.
The air over the wings is completely independent of how fast the wheels are moving. Completely. The plane could be moving backward on a conveyor belt and it would still take off.
Again - this is up to the wording of the question. In the real world, there is no conveyer that can keep a plan from taking off. On a completely theoretical treadmill that exactly matches the speed of the wheels at any given time, a plane with wheels of no friction or durability to consider will not be able to take off - there is no translational motion. And given that the plane is producing force (acceleration), the treadmill must also be producing the opposite but equal force translated at the wheel - thus it will eventually reach infinite speeds (as there is no friction).
However, in the real world, there is NO treadmill that can prevent a plane from taking off. The reason why I am bringing up that it is a wording issue is because if the question does not explicitly state that no translational motion can take place, then we can assume that the treadmill has limited powers, the wheels have friction, and things break.
I get what you're saying now but it's a ridiculous point.
I don't see why it's a ridiculous point - the answer to the original post
"If a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?"
Is a solid No because it restricts the translational motion of the plane. However, the other correct answer is (and some posters mention this) "The plane cannot lift off unless the treadmill/wheel/bearing/etc breaks".
The statement "The airplane will lift off regardless" does not keep to the wording of the question. If you play with riddles and not keep to the wording of the question, then you are not playing correctly at all.
Wut, you're back to being wrong again.
The belt matching the speed perfectly doesn't remove the friction from the wheels to the belt. You saying frictionless wheels removes the friction.
Wheel bearing friction does not matter in the scenario:
The plane's propeller provides forward thrust. This is balanced by a rearward force from the wheels onto the axles. This rearward force must be provided by the rearward force of the conveyer belt on the wheels at the point of contact. These forces are applied at an offset equal to the wheels' radius. That produces a torque on each wheel, causing the wheel to spin. That is, the torque on the wheels is balanced by the angular acceleration of the wheels. Later, the friction of a non-ideal wheel may increase enough to help balance the torque, but it is negligible at the outset. Note (and this is important) that the forces will balance, and the plane will not take off even if the wheels are frictionless, because the inertia and angular acceleration of the wheels will balance the thrust from the propellers and the force of the conveyer on the wheels.
On June 30 2012 05:10 hzflank wrote: State 2: plane starts up its engine, belt starts spinning. The plane begins to move forward because the force causing it to move forward have nothing to do with it's wheels, and the small amount of friction generated is not enough to stop it.
Yes, the wheels rolling over the treadmill will cause a backwards force. If the treadmill spins fast enough, this force cancels out the forward thrust of the plane's engines.
And it's not a "small" amount of friction. Since the wheels on belt situation is rubber on rubber, I'd guesstimate that the Crr is roughly 0.03, so the belt only has to spin at about 30 times the speed of what the plane would be moving at (so, in this case, if the plane needs 100 mph to take off and applies the thrust necessary to reach 100mph on the ground, the treadmill only has to spin at around 3000 mph - totally possible).
It happens because the engine creates a force that causes the plane to accelerate forwards through the air. The engine does not make the wheels turn. The wheels only turn because of friction against the ground. In theory, a plane does not need wheels at all to take off, the wheels just make it easier by massively reducing the friction against the ground.
Are you saying there is NO friction between the wheel and the conveyor belt then? So if the plane was coasting along, and turned off its engine, it would not roll to a stop?
There's always some friction. Magnify that friction with a conveyor belt, and you could get enough friction force to keep the plane stationary.
I don't disagree that the engine provides force. Engine provides F -> thataway. Conveyor belt spins <- thataway, applying a frictional rolling force on the wheels <- thataway. Spin the belt and wheels fast enough ... and you'll get a force equal to F <- thataway.
So the plane has a F -> that direction from the engine, and a F <- that direction from the wheels, they cancel each other out and it stays still.
On June 30 2012 04:51 Phael wrote: Ok maybe we are talking about two different things.
In that picture, it's implied that the plane stays still on the conveyor belt - where its forward speed is matched by the belt. As in, does not move relative to a person sitting nearby. So if it's moving forward at 100mph, the belt is also running at 100mph, and their speeds cancel out.
If there is no wind blowing, then the plane is also stationary relative to the wind. It's stationary to you, wind is stationary to you, commutative property, etc.
In the mythbuster's scenario, the plane was NOT kept still. It was still moving relative to the viewer, and thus, the wind. So it was able to take off.
You are completely missing the point.
Because the plane's wheels are free (not powered), no matter how fast the conveyor belt moves, the plan will never be stationary relative to a person standing beside the conveyor belt. The planes engines cause it to accelerate through the air, not over the ground. It will always move forward.
This is only true in a frictionless scenario. In the real world (eg. that picture), a conveyor belt moving ANYTHING will exert a force on the object.
Hence my original post saying that the plane would catch fire before it was actually stopped.
State 1: both belt and plane at rest. State 2: plane starts up its engine, belt starts spinning. Forces cancel each other out and plane does not move. State 3: plane keeps building up thrust, belt keeps spinning faster to apply enough force on the wheels to keep the plane stationary. State 4: plane applies enough thrust to force the belt to spin so fast something catches on fire. Experiment is nullified.
Until something actually breaks, the plane does not achieve lift-off, how does that not satisfy the question?
you've laid out the scenario but missed the point. Nothing says the plane remains stationary. The riddle only says the belt matches the speed of the wheels. You are making that leap in logic, that the belt matching the speed of the wheels forces the plane to stay stationary. Yes, the belt will accelerate faster and faster as the plane begins to move, but since planes do not transfer the force of the engines through the landing gear no force from the belt slows its forward motion (assuming no friction).
As it has been said already, the engines pulling the plane through the air are what allow it to fly. Only the relative airspeed of the wing is important when generating lift, not ground speed.
Put a spin on it: If the plane sits on an accelerating conveyor (assuming no friction) will it it take off? No. The conveyor will accelerate and spin the wheels, but the plane's inertia will keep it in place relative to the air and it will not generate lift.
On June 30 2012 05:10 hzflank wrote: State 2: plane starts up its engine, belt starts spinning. The plane begins to move forward because the force causing it to move forward have nothing to do with it's wheels, and the small amount of friction generated is not enough to stop it.
Yes, the wheels rolling over the treadmill will cause a backwards force. If the treadmill spins fast enough, this force cancels out the forward thrust of the plane's engines.
And it's not a "small" amount of friction. Since the wheels on belt situation is rubber on rubber, I'd guesstimate that the Crr is roughly 0.03, so the belt only has to spin at about 30 times the speed of what the plane would be moving at (so, in this case, if the plane needs 100 mph to take off and applies the thrust necessary to reach 100mph on the ground, the treadmill only has to spin at around 3000 mph - totally possible).
It happens because the engine creates a force that causes the plane to accelerate forwards through the air. The engine does not make the wheels turn. The wheels only turn because of friction against the ground. In theory, a plane does not need wheels at all to take off, the wheels just make it easier by massively reducing the friction against the ground.
Are you saying there is NO friction between the wheel and the conveyor belt then? So if the plane was coasting along, and turned off its engine, it would not roll to a stop?
There's always some friction. Magnify that friction with a conveyor belt, and you could get enough friction force to keep the plane stationary.
You are stretching it dude. The idea is that the wheels spin freely.
On June 30 2012 04:51 Phael wrote: Ok maybe we are talking about two different things.
In that picture, it's implied that the plane stays still on the conveyor belt - where its forward speed is matched by the belt. As in, does not move relative to a person sitting nearby. So if it's moving forward at 100mph, the belt is also running at 100mph, and their speeds cancel out.
If there is no wind blowing, then the plane is also stationary relative to the wind. It's stationary to you, wind is stationary to you, commutative property, etc.
In the mythbuster's scenario, the plane was NOT kept still. It was still moving relative to the viewer, and thus, the wind. So it was able to take off.
You are completely missing the point.
Because the plane's wheels are free (not powered), no matter how fast the conveyor belt moves, the plan will never be stationary relative to a person standing beside the conveyor belt. The planes engines cause it to accelerate through the air, not over the ground. It will always move forward.
This is only true in a frictionless scenario. In the real world (eg. that picture), a conveyor belt moving ANYTHING will exert a force on the object.
Hence my original post saying that the plane would catch fire before it was actually stopped.
State 1: both belt and plane at rest. State 2: plane starts up its engine, belt starts spinning. Forces cancel each other out and plane does not move. State 3: plane keeps building up thrust, belt keeps spinning faster to apply enough force on the wheels to keep the plane stationary. State 4: plane applies enough thrust to force the belt to spin so fast something catches on fire. Experiment is nullified.
Until something actually breaks, the plane does not achieve lift-off, how does that not satisfy the question?
you've laid out the scenario but missed the point. Nothing says the plane remains stationary. The riddle only says the belt matches the speed of the wheels. You are making that leap in logic, that the belt matching the speed of the wheels forces the plane to stay stationary. Yes, the belt will accelerate faster and faster as the plane begins to move, but since planes do not transfer the force of the engines through the landing gear no force from the belt slows its forward motion (assuming no friction).
As it has been said already, the engines pulling the plane through the air are what allow it to fly. Only the relative airspeed of the wing is important when generating lift, not ground speed.
Put a spin on it: If the plane sits on an accelerating conveyor (assuming no friction) will it it take off? No. The conveyor will accelerate and spin the wheels, but the plane's inertia will keep it in place relative to the air and it will not generate lift.
If the plane does not move relative to the conveyor belt, then the wheels and belt are spinning at the exact same speed whether it's .5mph or 500000 mph.
Furthermore, the only scenario in which the belt and wheel move at the same speed is IF the plane is stationary. If there is any discrepancy in their speed, the plane is moving, which fails the premise of the riddle and obviously allows it to lift-off.
You are stretching it dude. The idea is that the wheels spin freely.
How is applying something that is ever-present stretching it? Spinning freely only ever occurs in frictionless -ie made up - scenarios.
On June 30 2012 05:32 SimDawg wrote: Wait, wait, I'm back to thinking my original thought that you're talking math teacher uselessness.
The plane gets its thrust not from the friction between the wheels and the ground but from the engine and the air. It's completely independent of the ground.
This is exactly where you are misreading the question.
If the wheel is touching the treadmill, and the wheel spinning forwards at a certain speed due to the propulsion from the aircraft - and the treadmill is moving backwards at the exact same speed as the wheel, then there cannot be any forward motion. That's how the question is phrased.
You are much better off arguing that the question is phrased in such a way that cannot exist in the real world.
The treadmill will not be moving at the same speed as the wheels. That is physically impossible because the plane's engine does not provide any power to the wheels. The faster the treadmill moves, the faster the wheels spin and the wheels will always spin faster than the treadmill, unless friction gets sufficiently high to cause the wheels (or treadmill) to break due to the heat being generated.
The treadmill spins at the speed that the plane is moving, not at the speed that the wheels are spinning. Plane speed + treadmill speed = wheel speed, and the question states that the treadmill and the plane move at the same speed. The wheels will always be spinning twice as fast as the plane or treadmill are moving.
Then that invalidates the original premise of the question, which I quote here:
if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?
The only way that this condition is satisfied is if the plane is stationary on the treadmill, no other possible scenarios.
if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?
The only way that this condition is satisfied is if the plane is stationary on the treadmill, no other possible scenarios.
Sorry, I did not even notice but in this case the question is written incorrectly, which is why people have been saying it is an impossible scenario.
As soon as the engine generates force, the wheel speed will never match the treadmill speed. This can only happen if thee engine is not generating any force at all, and a plane cant take off without an engine. So, I guess you were correct in that the plane will not take off.
if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the plane PERFECTLY in the opposite direction, would the plane ever take off?
You have a thread which has posting guidelines to post riddles and answers to riddles only. There is a separate guideline that forbids arguing. How many posts does it take before posters begin breaking every rule simultaneously?
You have a toy radio controlled plane, and it's sitting in a patch of ice. You have a leash attached to it, and you're standing outside the ice patch.
You turn the plane on, going forward, but pull the leash tight, so the plane is struggling against the leash. You are obviously significantly stronger than the plane so it doesn't move.
You pulling _________ plane's engine <---------------------- Plane ------------------------> Both forces cancel each other out, so plane doesn't move.
Obviously this is a possible scenario.
---
Now lets extend it to the treadmill. If the wheels are rolling, there's going to be some frictional rolling resistance. The faster the wheels roll, the more resistance there is.
Imagine, if you will, you're holding a toy car (no engine) over a treadmill, holding it steady as the treadmill runs. At a low enough speed, it's easy. As the treadmill speeds up, there's more and more force trying to push the car back.
This is the same with the plane. Speed the conveyor belt up enough and you will get this scenario:
Rolling resistance ___ plane's engine <---------------------- Plane ------------------------>
Where the forces cancel each other out perfectly, and bam, a plane with engine on but cannot fly.
The question can be worded in many ways. The one in which the riddle was posed gives the answer: no, it would not fly. The one that you posed would give the answer: yes, it would fly. Both are possible, neither question is "wrong" by itself ... it's definitely harder to meet the requirements of the original question in the real world, however.
On June 30 2012 07:09 semioldguy wrote: I've got a riddle
You have a thread which has posting guidelines to post riddles and answers to riddles only. There is a separate guideline that forbids arguing. How many posts does it take before posters begin breaking every rule simultaneously?
On June 30 2012 00:13 klicken wrote: Oh my god, not the airplane again.
Yes it will take off.
This is because the plane accelerates with its engines through the AIR, not with its wheels through the ground.
You could compare it with a very aerodynamical car driving in heavy headwind, since the car accelerates through the ground, and due to its aerodynamical shape is isn't affected much by the wind and thus would drive just fine.
???
the question states that the conveyor belt is moving at the exact same speed as the aeroplanes tyres. the aeroplanes tyres are the only thing propelling the plane forward. the THRUST from the ENGINES are TRANSLATED to the TYRES. if the TYRES are doing ZERO work then the PLANE is not TAKING OFF.
LIFT is not generated unless the plane is moving at an appropriate speed. lift is not generated by the engines. the engines serve as a means to propel the plane forward. lift is created from the pressure difference between the top and bottom of the wings. this lift is not generated if the air is not moving over the wings of the plane (which would not be the case if the plane was on a conveyor belt and NOT moving relative to the air).
i am not sure why this is so puzzling to some.
and that mythbusters episode as above didn't have the conveyor belt moving at the same speed as the plane, if that was the case, the plane would not be moving relative to the observer at any point.
This is the bit that is incorrect. The tyres are not propelling the plane forward, it's the engines. If you removed the tyres and placed the plane on a frictionless surface it would be able to take off. Indeed, if tyres propelling the plane were correct then how does a plane fly through air without the tyres propelling it forward? The tyres are just a way for the plane to move while on the ground, and in a way which reduces as much extra drag as possible. Think of it this way, say you were extremely high up in the air (stationary) and dropped a plane from this height. Assuming the plane stays upright (i.e. doesn't begin rotating as it falls), if the plane turns on its engine will it be able to begin flying mid air?
The difficulty in understanding this problem (or maybe why the problem is ill posed) is that plane do not use their tyres to generate forward motion. If they did, they would take off and be unable to sustain their speed with their engines and immediately crash. Thus its impossible for a tyre propelled plane to even fly, let alone take off from a treadmill.
The plane structure functions as a whole. The velocity of the plane is increased via force produced by the engines. The lift generated is a function of velocity - not force exerted by the engines. The plane will not take off unless the plane is moving relative to the air at a speed high enough to achieve lift. The tyres are a way for the plane to achieve this speed on ground efficiently (without dragging its chassis along the ground). The only losses between the thrust force induced by the engine is from heat and sound (in general) and friction losses due to slip in bearings/gears and in main point the wheels.
We are assuming, in this question, that the plane is moving on a treadmill which matches its ground speed exactly. The question was stated this way.
Planes require a headwind to lift. As above. And as stated by others.
The only condition in which this plane-conveyor belt leaves with the plane taking off are if: - The conveyor belt moves so fast it begins dragging air backward into the plane (like in an air tunnel) - The conveyor belt is only moving at the speed at which a plane would normally take off
In the case of your hypothetical question - the plane will begin to move again as the thrust generated by the plane is not encountering any resistance at all during its acceleration to an appropriate speed.
Think of it this way: A plane is taking off - and half way along it's journey, the tarmac is dragged backward halting the motion of the plane completely. Would it take off? No, as the plane is no longer moving relative to the air and therefore no lift is generated. The conveyor belt just does this constantly (assuming the conveyor belt is spinning freely with the wheels).
I'm not sure why this is so confusing to some. A planes engines don't generate lift, they generate thrust. The wings generate lift, and this is only achieved when there is enough air passing across the wings to remove the ground from the surface of the earth. After that, the only friction loss is due to air resistance.
A frictionless surface would be great, as then the planes wouldn't need any wheels at all.
On June 30 2012 07:09 semioldguy wrote: I've got a riddle
You have a thread which has posting guidelines to post riddles and answers to riddles only. There is a separate guideline that forbids arguing. How many posts does it take before posters begin breaking every rule simultaneously?
Nobody knows whether Switch A starts out as UP or DOWN.
We also do not know who of the twenty three inmates will go in first.
What if the first person was not the counter and saw that Switch A was down? He would then turn it back on, thinking the counter had turned it down already. Then, the counter would come in and see that the switch is on. He would never know whether or not he was the first one to enter the room or not, nor would he know that the switch had started down and someone had turned it up before him. This would screw up his count. The person who had gone first would never touch switch A again, leaving the count at 21 forever....
Nobody knows whether Switch A starts out as UP or DOWN.
We also do not know who of the twenty three inmates will go in first.
What if the first person was not the counter and saw that Switch A was down? He would then turn it back on, thinking the counter had turned it down already. Then, the counter would come in and see that the switch is on. He would never know whether or not he was the first one to enter the room or not, nor would he know that the switch had started down and someone had turned it up before him. This would screw up his count. The person who had gone first would never touch switch A again, leaving the count at 21 forever....
The solution is that you "count" through everybody twice, or if the switch started in the "count" position then you count everybody twice except for one person who you count once. You will always reach this number of counts, and you will never reach it before everyone enters at least once.
I am too lazy to get my words right to go into more detail.
Nobody knows whether Switch A starts out as UP or DOWN.
We also do not know who of the twenty three inmates will go in first.
What if the first person was not the counter and saw that Switch A was down? He would then turn it back on, thinking the counter had turned it down already. Then, the counter would come in and see that the switch is on. He would never know whether or not he was the first one to enter the room or not, nor would he know that the switch had started down and someone had turned it up before him. This would screw up his count. The person who had gone first would never touch switch A again, leaving the count at 21 forever....
Every non-counter flips the switch twice. There can be at most 1 flip before counter enters. The counter counts 43 resets after his first visit, regardless of what he does on his first visit. After 43 counts and no one has flipped before he enters 21 people would have flipped twice and 1 flipped once. After 43 counts and someone flipped before he enters 22 would have flipped twice.
Every non-counter flips the switch twice. There can be at most 1 flip before counter enters. The counter counts 43 resets after his first visit, regardless of what he does on his first visit. After 43 counts and no one has flipped before he enters 21 people would have flipped twice and 1 flipped once. After 43 counts and someone flipped before he enters 22 would have flipped twice.
Ah, I see. So rather than pulling once, each person pulls the lever twice to negate the problem caused by the specific scenario........very smart. Thanks
On June 30 2012 06:28 Phael wrote:The question was, if the wheels were spinning at the exact same perfect speed of the treadmill.
This can ONLY happen if the plane remains stationary relative to the air/observer.
Only true for linear speed if you consider the speed of the wheel matches the speed of the plane, which is not a given. Let's say you block the wheels, so that they cannot move at all (park brakes). Your treadmill does not move either, so you fulfill your requirements. Is the power of the engines enough to reach takeoff speed on blocked wheels ? (not taking into account vertical-takeoff planes :p)
To get a "real" answer, you have to look at the max propulsion force vs max braking force of the plane. If brakes can resist maximum thrust, your plane will stay stationary. If not, it will move and eventually take off.
Answer will vary from one plane to the next. The regulation on brakes for civil planes (JAR 25.735d) states:
The aeroplane must have a parking control that, when set by the pilot, will without further attention, prevent the aeroplane from rolling on a paved, level runway with take-off power on the critical engine
As far as I know, the main issue is with the "critical engine" part: This has been interpreted for most multi-reactor planes as a braking force sufficient for 1 full thrust engine. Meaning there are multi-reactor planes that could move on parking control with all reactors at full thrust (although there are systems to prevent that).
I would say: - all 1 reactor civil planes with park brakes will not move on a static treadmill even with full thrust. - some multi-reactors will start to move even with stopped wheels. Will they reach take-off speed ? Probably : as soon as speed creates a lift, maximum braking power diminishes - most military jets will take off
A prisoner in prison is forced to carry a heavy sack of sand across the courtyard and back every day. After a while he decides he cant manage it anymore, so he puts something in the sack to make it lighter. What does he put in?
On July 11 2012 23:01 ThatGuy89 wrote: probably been posted before but here goes anyway
A prisoner in prison is forced to carry a heavy sack of sand across the courtyard and back every day. After a while he decides he cant manage it anymore, so he puts something in the sack to make it lighter. What does he put in?
On July 11 2012 23:01 ThatGuy89 wrote: probably been posted before but here goes anyway
A prisoner in prison is forced to carry a heavy sack of sand across the courtyard and back every day. After a while he decides he cant manage it anymore, so he puts something in the sack to make it lighter. What does he put in?
Ok heres another stupid one. What one word fills both gaps in this sentence
My _________ is _________than my car
I'm assuming it has to do with a brand of car? I'm not sure since there is almost nothing to go off of.
On June 30 2012 07:09 semioldguy wrote: I've got a riddle
You have a thread which has posting guidelines to post riddles and answers to riddles only. There is a separate guideline that forbids arguing. How many posts does it take before posters begin breaking every rule simultaneously?
I'm looking for riddles with a theme of "confusing" wording, where just hearing the riddle can give you a headache. I'll give an example below. If anyone knows similar riddles, please post them as well.
What day would yesterday be if Thusday was four days before the day after tomorrow?
On July 11 2012 23:01 ThatGuy89 wrote: probably been posted before but here goes anyway
A prisoner in prison is forced to carry a heavy sack of sand across the courtyard and back every day. After a while he decides he cant manage it anymore, so he puts something in the sack to make it lighter. What does he put in?
If I move the hour hand forward by 5 hours it is at 5 O'clock, if I do this again it is at 10 O'clock, then 3 O'clock, then 8, 1, 6, 11, 4, 9, 2, 7 and then back again to 12 O'clock. Thus landing on every number on the clock once and only once.
Q1: Which other numbers will this work for and what is the pattern?
Q2: what about for a clock with M different positions where I move the hand by N positions each turn? prove that the pattern for Q1 is necessary and sufficient for the general case
Answers: I forget... But I know that I have worked it out once. So if need be I could probably figure it out again.
On September 10 2012 05:58 turtles wrote: I have a clock with the hour hand pointing to 12.
If I move the hour hand forward by 5 hours it is at 5 O'clock, if I do this again it is at 10 O'clock, then 3 O'clock, then 8, 1, 6, 11, 4, 9, 2, 7 and then back again to 12 O'clock. Thus landing on every number on the clock once and only once.
Q1: Which other numbers will this work for and what is the pattern?
Q2: what about for a clock with M different positions where I move the hand by N positions each turn? prove that the pattern for Q1 is necessary and sufficient for the general case
Answers: I forget... But I know that I have worked it out once. So if need be I could probably figure it out again.
Ahh, this brings me back to my abstract algebra days/cyclic groups. All numbers that are relatively prime to 12 will do this, so 1,5,7, and 11. (to be relatively prime is to not share any divisors other than 1). The answer is the same for any M.
pf/ It follows easily that if they are relatively prime then it will go to every number. Here's the proof that it won't go to every number if they are not relatively prime.
You return to the original position in the clock whenever the number of moves you've made times the number you move by, n, is divisible by m, the total number of clock positions. Suppose m and n (everything here is a natural numbers s.t. m>n) have a common divisor, d. There exists d2 and d3 s.t. m=d(d2) and n=d(d3). Therefore, d2(n)=d(d2)(d3), which is divisible by m. Since d2<m, n returns to the original position in fewer than m moves meaning it couldn't have possibly gone to all m positions.
On September 10 2012 05:58 turtles wrote: I have a clock with the hour hand pointing to 12.
If I move the hour hand forward by 5 hours it is at 5 O'clock, if I do this again it is at 10 O'clock, then 3 O'clock, then 8, 1, 6, 11, 4, 9, 2, 7 and then back again to 12 O'clock. Thus landing on every number on the clock once and only once.
Q1: Which other numbers will this work for and what is the pattern?
Any prime number that is also not a non-trivial divisor (non-trivial divisor being a number 'n' that divides 'm' without leaving a remainder and isn't equal to 1 or m) of 12. That is: 1, 5, 7, 11 work. 2, 3 are prime but are non-trivial divisors of 12, so they do not work. 4, 6, 8, 9, 10, 12 are not prime, so they do not work.
On second thought, if you want to consider numbers outside of the range 1-M, then take the number modulo M then apply the above rule. 13 and 17 will work, as they mod 12 are 1 and 5, respectively, which both work.
On September 10 2012 05:58 turtles wrote:Q2: what about for a clock with M different positions where I move the hand by N positions each turn? prove that the pattern for Q1 is necessary and sufficient for the general case
Same as above, but I don't have a proof. First take N mod M, then check to see if the remainder is a prime number that is not a non-trivial divisor of M
Edit: Fixed typo. Also got ninja'd by someone with an actual proof :D
On September 10 2012 05:58 turtles wrote: I have a clock with the hour hand pointing to 12.
If I move the hour hand forward by 5 hours it is at 5 O'clock, if I do this again it is at 10 O'clock, then 3 O'clock, then 8, 1, 6, 11, 4, 9, 2, 7 and then back again to 12 O'clock. Thus landing on every number on the clock once and only once.
Q1: Which other numbers will this work for and what is the pattern?
Q2: what about for a clock with M different positions where I move the hand by N positions each turn? prove that the pattern for Q1 is necessary and sufficient for the general case
Answers: I forget... But I know that I have worked it out once. So if need be I could probably figure it out again.
On June 29 2012 12:34 calderon wrote: if a plane was on a conveyor belt, trying to take off, but the conveyor belt would match the speed of the planes wheels PERFECTLY in the opposite direction, would the plane ever take off?
Yeah, this has been done to death, but I had to answer it anyway. Feel free to ignore if you're sick of the topic.
It's a trick question. My answer is it depends, but if the plane did take off, it would not be in a good way.
First, let's note that the mythbusters clip is not directly relevent. In their formulation, they move the treadmill at the plane's takeoff speed, not at the wheel speed. The plane's engine provides thrust pushing against the air, the thrust easily overcomes the friction of the wheels on the treadmill (which, while higher than the usual friction of wheels on stationary ground, is still orders of magnitude smaller than the thrust the engines provide), the plane begins to move forward (at which point, the wheel speed exceeds the treadmill's speed), and when the plane reaches the necessary takeoff speed through the air, it takes off (at that point, the wheel speed is roughly double the treadmill's speed).
So we're going to need a treadmill that can go faster - several orders of magnitude faster - if it's to match the plane's wheel speed. The constraint that the treadmill's speed match the plane's wheel speed means that, as long as the wheels do not skid, the plane must remain stationary relative to the ground. The plane's thrust remains the same, and the thrust is provided by pushing against the air. So the treadmill must move so fast that the rolling resistance is equal to the thrust.
But there are physical limits to what the wheels can take (ignoring the physical limits of the conveyer belt setup, since that's already magical and arbitary). Those wheels aren't designed to turn at thousands of times their usual takeoff speeds. And with a treadmill moving that fast, the airspeed (relative to the stationary ground) isn't going to remain zero either. Let's consider both these effects, starting with the wheels.
Now, if the wheels fall off, then while they are still spinning (and the treadmill still moving backwards at super speed), the plane will collapse onto it's now wheelless undercarridge onto the treadmill. You will effectively be dropping your plane onto a spinning grinder. Bad news for the plane, and bad news for anyone standing anywhere nearby, as bits go flying. The plane will take off, but in pieces (and going backwards).
How might we make our wheels survive? Ordinarily, we would try to minimise rolling friction to allow for higher speeds, but here, that makes things worse. The lower the friction, the faster the wheels have to spin to generate enough backwards pull on the plane to match the engine's thrust (which we have to do in order to keep the plane stationary relative to the ground, which was shown earlier to be necessary to satisfy the problem's requirements). So lowering friction just leads to an even faster treadmill.
No, the trick is to go the other way. If we apply the plane's wheel brakes (and they are sufficiently strong), then the plane's wheel speed will remain 0 (and thus, the treadmill will also remain stationary). The problem then comes down to this: does the plane have sufficient thrust to skid the wheels? If so, can it reach takeoff speed with the wheels locked? If so, it will take off (after burning through a lot of wheel rubber - don't try this at home folks!). Most likely, it will not.
What about air movement? Could that make the plane take off? Well, that depends on something not specified in the problem - how big si the treadmill? There could be a short piece of treadmill under each wheel, or a massive treadmill stretching for miles ahead of the plane. Under short treadmill conditions (as are depicted), the treadmill speed required for this would not be reached until well after the wheels fell off or caught fire. But what about a large treadmill? You'd probably still suffer wheel failure before the airspeed got high enough, but let's suppose the wheels somehow survived; what would happen?
Well, the plane would liftoff. But the plane is still stationary relative to the ground at the moment of takeoff, and all of its airspeed comes from the treadmill-induced wind. This windspeed falls off rapidly with height. A few meters further up, the air is barely moving - and so is the plane. No airspeed means no lift, and also no control over the plane. It would gently nose-down and dive back into the wind, and into the ground. The best you could hope for would be a bumpy landing (slightly further forward from your starting point). The worst would be.. worse.
Figured I'd resurrect this rather than make a new thread. This is an inductive type problem I had pop up recently. I admit defeat. I don't know the answer so if you post be sure to provide a valid explanation!
edit: For clarification, the images are read left to right, and from the choices you are supposed to say what comes next.
At first I thought it might be that all the top ones can be drawn without removing pen from paper or tracing over an already drawn line, like this sort of thing:
and that only one of the answers can also be done like that, but clearly that's wrong. Maybe I'm just missing something and someone else can carry this idea.
On November 29 2012 09:56 Jonoman92 wrote: Figured I'd resurrect this rather than make a new thread. This is an inductive type problem I had pop up recently. I admit defeat. I don't know the answer so if you post be sure to provide a valid explanation!
On November 29 2012 09:56 Jonoman92 wrote: Figured I'd resurrect this rather than make a new thread. This is an inductive type problem I had pop up recently. I admit defeat. I don't know the answer so if you post be sure to provide a valid explanation!
Not really sure... Is it a "what comes next" kind of problem, or "which is of the same type" kind? With "inductive", do you mean "what comes next"?
Only thing that comes to my mind at first is to count the number of cells in each figure: 4, 5, 1, 2, 3 after which I guess you would like a figure with 4 cells, ie B. But not convinced that is what they are after... You can essentially pick any of the alternatives and come up with some kind of motivation. are you supposed to be able to predict the shape of the next figure without seeing the alternatives? Or is it only "which of these alternatives makes sense"?
On November 29 2012 09:56 Jonoman92 wrote: Figured I'd resurrect this rather than make a new thread. This is an inductive type problem I had pop up recently. I admit defeat. I don't know the answer so if you post be sure to provide a valid explanation!
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
My guess is that it's a series of N shapes with N%5+1 (1, 2, 3, 4, 5, 1, 2, 3, 4, 5...) interior cells, and the segment we're seeing is starting at 4. So then we see 5, 1, 2, 3, and we should pick B, which has 4 interior cells.
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
I think you're missing some required information. As described, it's possible that every single Indian is marked with the same symbol, and no amount of information about the symbols marking other Indians is useful for deducing any given Indian's mark.
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
If I understand the riddle correctly, they will do the following: First one will say symbol of person before him, Second one will say his own symbol to save himself. This goes on in pairs to the end, so 50 can be saved for sure. The indians sacrificing themselves however have a 1/12 chance that their symbol is the same as the one before them in the queue, so after them in the queue.
So the expected amount that is saved would be 50+ 50/12, this last part can obviously range from 1-50
There probably is a more optimal strategy, let me think about this some more .
They are all standing in line, and start talking from the back forwards. They say the symbol of the one in front. If they die, the one in front knows what their symbol is and says it, thus lives. The next one starts again, and says the one in front, etc etc. Thus 50% are instantly saved and the rest have a 1/12 chance of survival.
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
I think you're missing some required information. As described, it's possible that every single Indian is marked with the same symbol, and no amount of information about the symbols marking other Indians is useful for deducing any given Indian's mark.
Hmm, I believe it doesn't matter if every indian has the same symbol. Still the indians know from the start that there are 12 different possible symbols and know all the symbols (or maybe I'm missing something?)
There is a much better solution than 50%. All indians but one can be saved!
On November 29 2012 09:56 Jonoman92 wrote: Figured I'd resurrect this rather than make a new thread. This is an inductive type problem I had pop up recently. I admit defeat. I don't know the answer so if you post be sure to provide a valid explanation!
edit: For clarification, the images are read left to right, and from the choices you are supposed to say what comes next.
The way I see it is that you need to look at how many triangles you can identify in each figure: the first has 4, the second 2, and the third and fourth have zero triangles. The fifth one also has 4 triangles implying a restart of the pattern (4,2,0,0). If the above logic makes any sense, than the correct answer should be D, as it has 2 triangles in it.
On November 29 2012 09:56 Jonoman92 wrote: Figured I'd resurrect this rather than make a new thread. This is an inductive type problem I had pop up recently. I admit defeat. I don't know the answer so if you post be sure to provide a valid explanation!
edit: For clarification, the images are read left to right, and from the choices you are supposed to say what comes next.
I think the answer is B. The reason being that I think that you have to count the number of shapes in the picture in the first picture its 4 then 5 then 1 then 2 then 3 it would follow if you have a recouring pattern of 1 to 5 then the next number of shapes would be 4. B has four shapes. At least that's what i think :3
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
They can try something like the following, although i'm not sure it's the easiest way: They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .
By the way, great thread, although we should really be able to filter stuff. Posts commenting with the white/orange ball and threadmill/airplane problem are so annoying.
On November 29 2012 09:56 Jonoman92 wrote: Figured I'd resurrect this rather than make a new thread. This is an inductive type problem I had pop up recently. I admit defeat. I don't know the answer so if you post be sure to provide a valid explanation!
edit: For clarification, the images are read left to right, and from the choices you are supposed to say what comes next.
Problem with these is that there is no way of knowing what parameter you're supposed to look for. The 4,5,1,2,3 answer seems ok, but it's not very convincing. The mod(5) + 1 comes out of nowhere, only there happen to be as many examples and answers.
It could be something completely different, like: Does the shape have straight lines in non 45x degree angles? The current situation would be 1,1,0,0,1 so the correct answer would be E. Which is probably not the answer .
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
They can try something like the following, although i'm not sure it's the easiest way: They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .
By the way, great thread, although we should really be able to filter stuff. Posts commenting with the white/orange ball and threadmill/airplane problem are so annoying.
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
They can try something like the following, although i'm not sure it's the easiest way: They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .
By the way, great thread, although we should really be able to filter stuff. Posts commenting with the white/orange ball and threadmill/airplane problem are so annoying.
Correct, congrats!
I quote:
- The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way).
The solution requires that the Indians are in fact able to communicate prior to the massacre.
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
They can try something like the following, although i'm not sure it's the easiest way: They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .
By the way, great thread, although we should really be able to filter stuff. Posts commenting with the white/orange ball and threadmill/airplane problem are so annoying.
- The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way).
The solution requires that the Indians are in fact able to communicate prior to the massacre.
True, but it also stated in the problem that they've known this solution for generations. You never know when your tribe may end up with a situation where they paint zodiacs on your back and then ask you to name them or they kill you!
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
They can try something like the following, although i'm not sure it's the easiest way: They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .
By the way, great thread, although we should really be able to filter stuff. Posts commenting with the white/orange ball and threadmill/airplane problem are so annoying.
Correct, congrats!
I quote:
- The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way).
The solution requires that the Indians are in fact able to communicate prior to the massacre.
True, but it also stated in the problem that they've known this solution for generations. You never know when your tribe may end up with a situation where they paint zodiacs on your back and then ask you to name them or they kill you!
"From their long riddle experience" they have a pre-arranged communications protocoll for handling situations like this?
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
They can try something like the following, although i'm not sure it's the easiest way: They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .
By the way, great thread, although we should really be able to filter stuff. Posts commenting with the white/orange ball and threadmill/airplane problem are so annoying.
Correct, congrats!
I quote:
- The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way).
The solution requires that the Indians are in fact able to communicate prior to the massacre.
True, but it also stated in the problem that they've known this solution for generations. You never know when your tribe may end up with a situation where they paint zodiacs on your back and then ask you to name them or they kill you!
They just happen to have a number assigned to each zodiac symbol?
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
They can try something like the following, although i'm not sure it's the easiest way: They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .
By the way, great thread, although we should really be able to filter stuff. Posts commenting with the white/orange ball and threadmill/airplane problem are so annoying.
Correct, congrats!
I quote:
- The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way).
The solution requires that the Indians are in fact able to communicate prior to the massacre.
True, but it also stated in the problem that they've known this solution for generations. You never know when your tribe may end up with a situation where they paint zodiacs on your back and then ask you to name them or they kill you!
So suppose everyone has month 1 on his back, lets say its The Ram. The total is thus 99. Which zodiac name would they say, or would they say the mod12 of 99, whatever that would be?
What I meant was: they cannot communicate once the symbols are carved onto their backs. But they can agree on the strategy to adopt beforehand, and they are told the rules.
Each of the 12 signs is represented by an integer from 0 to 11. The indian at the back adds up all the signs modulo 12: this gives a number C between 0 and 11. He can then communicate this number by saying the associated sign. The next indian counts all the signs in front of him mod 12: this gives a number C2. In order to know his number, he only has to find the difference between C and C2 mod 12. This gives his sign: he is saved All the next indian also count the signs in front of them, and add the signs told by all the previous indians (except the first). This sum is called Cn (for the n-th indian). Their sign is the difference between C and Cn mod 12. They are all saved.
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
They can try something like the following, although i'm not sure it's the easiest way: They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .
By the way, great thread, although we should really be able to filter stuff. Posts commenting with the white/orange ball and threadmill/airplane problem are so annoying.
Correct, congrats!
I quote:
- The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way).
The solution requires that the Indians are in fact able to communicate prior to the massacre.
True, but it also stated in the problem that they've known this solution for generations. You never know when your tribe may end up with a situation where they paint zodiacs on your back and then ask you to name them or they kill you!
They just happen to have a number assigned to each zodiac symbol?
And they all somehow randomly assigned the same number to each symbol? And knowledge persists for generations without communication? Wooow.
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
They can try something like the following, although i'm not sure it's the easiest way: They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .
By the way, great thread, although we should really be able to filter stuff. Posts commenting with the white/orange ball and threadmill/airplane problem are so annoying.
Correct, congrats!
I quote:
- The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way).
The solution requires that the Indians are in fact able to communicate prior to the massacre.
True, but it also stated in the problem that they've known this solution for generations. You never know when your tribe may end up with a situation where they paint zodiacs on your back and then ask you to name them or they kill you!
So suppose everyone has month 1 on his back, lets say its The Ram. The total is thus 99. Which zodiac name would they say, or would they say the mod12 of 99, whatever that would be?
Actually, I liked a riddle that is somewhat similar to this, maybe a bit too easy but still fun:
4 gnomes are captured by a wizard who casts a spell, making them unable to move (but they can speak, ofc). The wizard arranges them like this: He puts 3 of them on a staircase, looking in the downwards direction, so the back one can see 2 others and the middle one 1. The fourth one is left in the cave, unable to see the others, and the others can't see him either. All of the gnome hats, two white ones and two red ones, are now redistributed, and the red hats are given to the top and bottom gnomes on the stairs, the white hats to the other 2. The gnomes have no idea what color hat they receive.
The wizard promises to release the gnomes as soon as one correctly states the color of his own hat. If any other word is spoken, or one guesses wrong, all are killed. Which one calls his color to free himself and his friends?
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
They can try something like the following, although i'm not sure it's the easiest way: They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .
By the way, great thread, although we should really be able to filter stuff. Posts commenting with the white/orange ball and threadmill/airplane problem are so annoying.
Correct, congrats!
I quote:
- The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way).
The solution requires that the Indians are in fact able to communicate prior to the massacre.
True, but it also stated in the problem that they've known this solution for generations. You never know when your tribe may end up with a situation where they paint zodiacs on your back and then ask you to name them or they kill you!
So suppose everyone has month 1 on his back, lets say its The Ram. The total is thus 99. Which zodiac name would they say, or would they say the mod12 of 99, whatever that would be?
When would you ever have to? If the person behind you says 8, then it's easy to distinguish between scenarios. If you now see 7, then you know you are a 1. If you now see 4, you know that you are 4. (numbers in mod 12 of course).
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
They can try something like the following, although i'm not sure it's the easiest way: They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .
By the way, great thread, although we should really be able to filter stuff. Posts commenting with the white/orange ball and threadmill/airplane problem are so annoying.
Correct, congrats!
I quote:
- The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way).
The solution requires that the Indians are in fact able to communicate prior to the massacre.
True, but it also stated in the problem that they've known this solution for generations. You never know when your tribe may end up with a situation where they paint zodiacs on your back and then ask you to name them or they kill you!
So suppose everyone has month 1 on his back, lets say its The Ram. The total is thus 99. Which zodiac name would they say, or would they say the mod12 of 99, whatever that would be?
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
They can try something like the following, although i'm not sure it's the easiest way: They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .
By the way, great thread, although we should really be able to filter stuff. Posts commenting with the white/orange ball and threadmill/airplane problem are so annoying.
Correct, congrats!
I quote:
- The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way).
The solution requires that the Indians are in fact able to communicate prior to the massacre.
True, but it also stated in the problem that they've known this solution for generations. You never know when your tribe may end up with a situation where they paint zodiacs on your back and then ask you to name them or they kill you!
So suppose everyone has month 1 on his back, lets say its The Ram. The total is thus 99. Which zodiac name would they say, or would they say the mod12 of 99, whatever that would be?
How do you distguish between 7 + 1 and 4 + 4 in such a modded sum?
You can't distinguish between the two, but you don't have to, since you only have to subtract one number at a time (the difference between the last total and yours). How often you passed the 12 (or how much the total is) doesn't matter to you. Edit: answered already.
On November 29 2012 23:19 aseq wrote: Actually, I liked a riddle that is somewhat similar to this, maybe a bit too easy but still fun:
4 gnomes are captured by a wizard who casts a spell, making them unable to move (but they can speak, ofc). The wizard arranges them like this: He puts 3 of them on a staircase, looking in the downwards direction, so the back one can see 2 others and the middle one 1. The fourth one is left in the cave, unable to see the others, and the others can't see him either. All of the gnome hats, two white ones and two red ones, are now redistributed, and the red hats are given to the top and bottom gnomes on the stairs, the white hats to the other 2. The gnomes have no idea what color hat they receive.
The wizard promises to release the gnomes as soon as one correctly states the color of his own hat. If any other word is spoken, or one guesses wrong, all are killed. Which one calls his color to free himself and his friends?
I'd say the one in the middle. The gnome that sees the two others doesn't see two reds or two white, therefore he doesn't speak. As he doesn't the second gnome on the stairs knows that his hat and the next hat have a different color. He can then see the color of the next gnome (red) and decide he has the other (white)
On November 29 2012 23:19 aseq wrote: Actually, I liked a riddle that is somewhat similar to this, maybe a bit too easy but still fun:
4 gnomes are captured by a wizard who casts a spell, making them unable to move (but they can speak, ofc). The wizard arranges them like this: He puts 3 of them on a staircase, looking in the downwards direction, so the back one can see 2 others and the middle one 1. The fourth one is left in the cave, unable to see the others, and the others can't see him either. All of the gnome hats, two white ones and two red ones, are now redistributed, and the red hats are given to the top and bottom gnomes on the stairs, the white hats to the other 2. The gnomes have no idea what color hat they receive.
The wizard promises to release the gnomes as soon as one correctly states the color of his own hat. If any other word is spoken, or one guesses wrong, all are killed. Which one calls his color to free himself and his friends?
I'd say the one in the middle. The gnome that sees the two others doesn't see two reds or two white, therefore he doesn't speak. As he doesn't the second gnome on the stairs knows that his hat and the next hat have a different color. He can then see the color of the next gnome (red) and decide he has the other (white)
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
I think you're missing some required information. As described, it's possible that every single Indian is marked with the same symbol, and no amount of information about the symbols marking other Indians is useful for deducing any given Indian's mark.
Represent the symbols as numbers n(i) for indian i (first one is 0, last one 99, all values are in [0..11]). Let the first indian answer with N=sum(a(1)...a(99)) mod 12.
Second indian sees a(2) ... a(99), so he can sum them up to compute A(1) = N - a(2) (mod 12) and therefore he will answer a(2)=N-A(1) (mod 12)
Recurse on indians, indian j knows a(1)...a(j-1) (correct values given by the preceding ones) and sees a(j+1)...a(99). Therefore indian j can compute A(j)=(N - a(j)) (mod 12) and his symbol is: a(j)=(N - A(j)) mod 12
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
I think you're missing some required information. As described, it's possible that every single Indian is marked with the same symbol, and no amount of information about the symbols marking other Indians is useful for deducing any given Indian's mark.
Represent the symbols as numbers n(i) for indian i (first one is 0, last one 99, all values are in [0..11]). Let the first indian answer with N=sum(a(1)...a(99)) mod 12.
Second indian sees a(2) ... a(99), so he can sum them up to compute A(1) = N - a(2) (mod 12) and therefore he will answer a(2)=N-A(1) (mod 12)
Recurse on indians, indian j knows a(1)...a(j-1) (correct values given by the preceding ones) and sees a(j+1)...a(99). Therefore indian j can compute A(j)=(N - a(j)) (mod 12) and his symbol is: a(j)=(N - A(j)) mod 12
Yeah, again: The description of the riddle implied that the Indians couldn't communicate prior to having to solve the problem. Thus my problem. Once you allow them to establish a communications protocoll the solution is pretty trivial.
On November 29 2012 21:15 CptZouglou wrote: Generalized version of a classic one:
A hundred indians are captured by cowboys. The sadistic cowboys then play a game to decide which indians to kill: They carve one of the 12 zodiac symbols on each indian's back. Then, during the massacre day, the indians are put in line, and starting from the end of the queue, each indian is asked his symbol: if he can answer correctly, he will be saved.
From their long riddle experience, the indians knew a strategy to try and save as many as they could. How many can they save, and how ?
Notes: - The indians cannot see their own symbol, and can only see the symbols of all the indians in front of them in the queue. - The indians can only say one word when asked: the name of their symbol (they cannot communicate in any other way). - When an indian say something, everyone else hears it - Indians are smart
I think you're missing some required information. As described, it's possible that every single Indian is marked with the same symbol, and no amount of information about the symbols marking other Indians is useful for deducing any given Indian's mark.
Represent the symbols as numbers n(i) for indian i (first one is 0, last one 99, all values are in [0..11]). Let the first indian answer with N=sum(a(1)...a(99)) mod 12.
Second indian sees a(2) ... a(99), so he can sum them up to compute A(1) = N - a(2) (mod 12) and therefore he will answer a(2)=N-A(1) (mod 12)
Recurse on indians, indian j knows a(1)...a(j-1) (correct values given by the preceding ones) and sees a(j+1)...a(99). Therefore indian j can compute A(j)=(N - a(j)) (mod 12) and his symbol is: a(j)=(N - A(j)) mod 12
Yeah, again: The description of the riddle implied that the Indians couldn't communicate prior to having to solve the problem. Thus my problem. Once you allow them to establish a communications protocoll the solution is pretty trivial.
I think it's not trivial. You can see it that way: The cowboys give the rules to the indians. The indians discuss which strategy to use. Then they carve the symbols and from then the indians can't communicate. It's like any other strategy based riddle. If they couldn't communicate at all before knowing the problem, then the best they can do is random or hope the previous one just gave your carving, but why would he as you haven't established any strategy ?
A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Don't you have to specify how many white/black hats there are (or some other relation betwee the hats)? Cause this way, there is no correlation between what hat a gnome is wearing and the hats other gnomes wear right?
Bahamuth November 30 2012 00:14. Posts 70 PM Profile Report Quote # On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Don't you have to specify how many white/black hats there are (or some other relation betwee the hats)? Cause this way, there is no correlation between what hat a gnome is wearing and the hats other gnomes wear right?
No correlation is needed between the hats' colours.
gnomes are smarter than indians. I could see how an infinite number of gnomes could find their hats, but not how only a finite number of gnome answer wrong...
They all answer at the same time: this means their only available information is the hats before them ?
Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
If there is a finite number of whites ahead, either all answer white to get an infinite false or all answer black to get infinite true. If there is a finite number of blacks ahead, either all answer black to get an infinite false or all answer white to get infinite true. It both are infinite, all answer white by convention to get both an infinite true and an infinite false.
Err... There must be something I'm not getting here.
Edit: ok - was finite false. No solution then appart from all gnomes staying silent.
On November 30 2012 00:46 Oshuy wrote: If there is a finite number of whites ahead, either all answer white to get an infinite false or all answer black to get infinite true. If there is a finite number of blacks ahead, either all answer black to get an infinite false or all answer white to get infinite true. It both are infinite, all answer white by convention to get both an infinite true and an infinite false.
Err... There must be something I'm not getting here.
Edit: ok - was finite false. No solution then appart from all gnomes staying silent.
That's right for the infinite amount of right, but the finite wrong is much trickier one ^^
A holy man on his pilgrimage stumbles upon a small town where 100 couples live. This village has a unusual cerimony: every night all the men gather around the town square and each publically either praises his wife imensely if he thinks she's been faithful in their marriage or curses her for her infidelity.
While spending his day there, the holy man notices that all 100 wives are cheating on their husbands, and that every man knows about every single infidelity, except his own! Though every man notices that his companions are being cheated by their respective wives, he assumes with absolute certainty that his wife is and always has been faithful and also knows that all other men think the same way about their own respective wives.
In the holy man's 1st night there, before the men begin to praise their wives, he declares: "one woman in this village is being unfaithful" and leaves the village. None of the men look surprised and every single one begins to praise his wife. The same thing happens for another 98 nights.
In the 100th night, however, they all know with certainty that they've been cheated and forevermore curse their wives.
How did this happen, if the holy man said something that everyone already knew (and better yet, knew that everybody knew)?
Assume that each man only thinks over what happened in the night after the cerimony is over (in other words, if he is to realise something from the cerimony, he will only do so after it's over). Also assume everyone is smart in game theory, etc.
There's a mathematical explanation for this, but honestly I don't remember it all to well, so here's a simplified explanation: Since each man knew about every single infidelity but his own, he assumed that on the 99th night everyone except himself would curse their wives. Since this didn't happen, he knows he was being cheated on.
It's easier to explain with fewer people:
Imagine there are only 2 men in the same situation, and that only one of them are being cheated on (John is being cheated and Mike isn't). Once the holy man says there is one wife cheating on her husband, John will be sure it is his because he doesn't know about any other wife cheating on her husband. In turn, if both men are being cheated, then John expects Mike to curse his wife on the first night and Mike expects John to do the same. Therefore, once the night is over and neither wife has been cursed, the both realise that there must be another cheating wife, and that it must be their own. Thus, on night 2 they would curse their wifes.
Imagine now with 3 men, all being cheated (John, Mike and Paul). From Paul's point of view, there are only two women cheating and their husbands don't know about it, thus he expects it to play out like it did above: with Mike and John cursing their wives on night 2, since he thinks that both John and Mike only see one cheating wife: each other's. His surprise comes on night 2 when they don't curse their wives, indicating that there is a third cheating wife, which must be his.
This keeps going until you have 100 men, cursing on the 100th night.
The funny thing is that the holy man said something everyone already knew: that there was a cheating wife in their mist (everyone knew about 99 of them in fact)! What changed was that at the end of the chain of expectations is always someone who is expected to think there are no cheating wives. In the 2 man example, John thinks Mike thinks there are no cheating women. In the 3 man example, John thinks Mike thinks Paul thinks there are no cheating women. This is impossible once someone declares there is a single cheating wife.
On November 30 2012 00:46 Oshuy wrote: If there is a finite number of whites ahead, either all answer white to get an infinite false or all answer black to get infinite true. If there is a finite number of blacks ahead, either all answer black to get an infinite false or all answer white to get infinite true. It both are infinite, all answer white by convention to get both an infinite true and an infinite false.
Err... There must be something I'm not getting here.
Edit: ok - was finite false. No solution then appart from all gnomes staying silent.
That's right for the infinite amount of right, but the finite wrong is much trickier one ^^
There are lots of tricks that can be pulled : buying a lot of black paint would also work
If we put a hat on each gnome by tossing a coin to decide the color, each draw is independant and the probability of white or black for any given gnome are both 1/2, whatever they see. So they need a form of communication.
One strange part of the riddle is that you state the gnomes cannot see the hats behind them. Even if they see those all other hats, it doesn't provide them with any meaningful information if they all answer at the same time.
On November 30 2012 01:18 Sbrubbles wrote: A variant of the hat puzzle, from game theory:
A holy man on his pilgrimage stumbles upon a small town where 100 couples live. This village has a unusual cerimony: every night all the men gather around the town square and each publically either praises his wife imensely if he thinks she's been faithful in their marriage or curses her for her infidelity.
While spending his day there, the holy man notices that all 100 wives are cheating on their husbands, and that every man knows about every single infidelity, except his own! Though every man notices that his companions are being cheated by their respective wives, he assumes with absolute certainty that his wife is and always has been faithful and also knows that all other men think the same way about their own respective wives.
In the holy man's 1st night there, before the men begin to praise their wives, he declares: "one woman in this village is being unfaithful" and leaves the village. None of the men look surprised and every single one begins to praise his wife. The same thing happens for another 98 nights.
In the 100th night, however, they all know with certainty that they've been cheated and forevermore curse their wives.
How did this happen, if the holy man said something that everyone already knew (and better yet, knew that everybody knew)?
Assume that each man only thinks over what happened in the night after the cerimony is over (in other words, if he is to realise something from the cerimony, he will only do so after it's over). Also assume everyone is smart in game theory, etc.
There's a mathematical explanation for this, but honestly I don't remember it all to well, so here's a simplified explanation: Since each man knew about every single infidelity but his own, he assumed that on the 99th night everyone except himself would curse their wives. Since this didn't happen, he knows he was being cheated on.
It's easier to explain with fewer people:
Imagine there are only 2 men in the same situation, and that only one of them are being cheated on (John is being cheated and Mike isn't). Once the holy man says there is one wife cheating on her husband, John will be sure it is his because he doesn't know about any other wife cheating on her husband. In turn, if both men are being cheated, then John expects Mike to curse his wife on the first night and Mike expects John to do the same. Therefore, once the night is over and neither wife has been cursed, the both realise that there must be another cheating wife, and that it must be their own. Thus, on night 2 they would curse their wifes.
Imagine now with 3 men, all being cheated (John, Mike and Paul). From Paul's point of view, there are only two women cheating and their husbands don't know about it, thus he expects it to play out like it did above: with Mike and John cursing their wives on night 2, since he thinks that both John and Mike only see one cheating wife: each other's. His surprise comes on night 2 when they don't curse their wives, indicating that there is a third cheating wife, which must be his.
This keeps going until you have 100 men, cursing on the 100th night.
The funny thing is that the holy man said something everyone already knew: that there was a cheating wife in their mist (everyone knew about 99 of them in fact)! What changed was that at the end of the chain of expectations is always someone who is expected to think there are no cheating wives. In the 2 man example, John thinks Mike thinks there are no cheating women. In the 3 man example, John thinks Mike thinks Paul thinks there are no cheating women. This is impossible once someone declares there is a single cheating wife.
No offense, but I'm familiar with the actual problem and you seriously misdescribed both it and the mechanism of its solution to the point of it making no sense. Shame given how cool the original is.
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
On November 30 2012 00:46 Oshuy wrote: If there is a finite number of whites ahead, either all answer white to get an infinite false or all answer black to get infinite true. If there is a finite number of blacks ahead, either all answer black to get an infinite false or all answer white to get infinite true. It both are infinite, all answer white by convention to get both an infinite true and an infinite false.
Err... There must be something I'm not getting here.
Edit: ok - was finite false. No solution then appart from all gnomes staying silent.
Also I would add since the problem doesn't specify, if there is a finite number of both in front of you (ie the line extends infinitely behind you but not in front) it's impossible to guarantee infinite correct. Stated simply though, guess a color you see an infinite amount of in front of you and there will be infinite correct.
Although I also have a question which has to do with the theory of random numbers, which I dont' really understand. If each gnome is able to produce an actual unpredictable random guess wouldn't they get infinite correct even in the line only stretches infinitely behind you and not in front of you case? There would be an infinitely small chance that the hats were assigned in a direct contrary pattern to what they end up guessing?
Finite wrong: This seems impossible to guarantee in almost all cases. Unless the line only extends infinitely to the front and there's a finite number of one of the colors. I want to hear the solution to this.
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
So you mean that they prepared an infinite number of classes of infinite sequences of possible patterns, and for each one they have one reference sequence ? So one gnome could actually answer for all the others no ? I don't really get this answer maybe...
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
So you mean that they prepared an infinite number of classes of infinite sequences of possible patterns, and for each one they have one reference sequence ? So one gnome could actually answer for all the others no ? I don't really get this answer maybe...
Each gnome knows what all the others will say, so in that sense could answer for them. Of course, the puzzle states that they must each answer with their own hat color, which they will do by selecting the color of their position in the sequence (from that equivalence class) that they agreed to all associate with whatever equivalence class they happen to find themselves in. This will result in a finite number of deaths, though you can make that finite number arbitrarily large.
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
So you mean that they prepared an infinite number of classes of infinite sequences of possible patterns, and for each one they have one reference sequence ? So one gnome could actually answer for all the others no ? I don't really get this answer maybe...
Each gnome knows what all the others will say, so in that sense could answer for them. Of course, the puzzle states that they must each answer with their own hat color, which they will do by selecting the color of their position in the sequence (from that equivalence class) that they agreed to all associate with that equivalence class. This will result in a finite number of deaths, though you can make that finite number arbitrarily large.
Ok. That would require an infinite amount of preparation, capacity to memorize infinite number of infinite sequences and infinite sight (to see all the gnomes in front) =) Gnomes are indeed very smart!
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
So you mean that they prepared an infinite number of classes of infinite sequences of possible patterns, and for each one they have one reference sequence ? So one gnome could actually answer for all the others no ? I don't really get this answer maybe...
Each gnome knows what all the others will say, so in that sense could answer for them. Of course, the puzzle states that they must each answer with their own hat color, which they will do by selecting the color of their position in the sequence (from that equivalence class) that they agreed to all associate with that equivalence class. This will result in a finite number of deaths, though you can make that finite number arbitrarily large.
Ok. That would require an infinite amount of preparation, capacity to memorize infinite number of infinite sequences and infinite sight (to see all the gnomes in front) =) Gnomes are indeed very smart!
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
So you mean that they prepared an infinite number of classes of infinite sequences of possible patterns, and for each one they have one reference sequence ? So one gnome could actually answer for all the others no ? I don't really get this answer maybe...
Each gnome knows what all the others will say, so in that sense could answer for them. Of course, the puzzle states that they must each answer with their own hat color, which they will do by selecting the color of their position in the sequence (from that equivalence class) that they agreed to all associate with that equivalence class. This will result in a finite number of deaths, though you can make that finite number arbitrarily large.
Ok. That would require an infinite amount of preparation, capacity to memorize infinite number of infinite sequences and infinite sight (to see all the gnomes in front) =) Gnomes are indeed very smart!
They must have generations and generations of riddle experience.
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
So you mean that they prepared an infinite number of classes of infinite sequences of possible patterns, and for each one they have one reference sequence ? So one gnome could actually answer for all the others no ? I don't really get this answer maybe...
Each gnome knows what all the others will say, so in that sense could answer for them. Of course, the puzzle states that they must each answer with their own hat color, which they will do by selecting the color of their position in the sequence (from that equivalence class) that they agreed to all associate with that equivalence class. This will result in a finite number of deaths, though you can make that finite number arbitrarily large.
Ok. That would require an infinite amount of preparation, capacity to memorize infinite number of infinite sequences and infinite sight (to see all the gnomes in front) =) Gnomes are indeed very smart!
They must have generations and generations of riddle experience.
If gnomes reproduction follows any known pattern, they have been reproducing for quite some time to reach infinity indeed.
Misleading data for spoilered answer is the definition of "smart". Usually would mean any average gnome can do before answering any task humanly feasible in a finite amount of time. Here they need "a little" more.
There are 3 people who were taken hostage. Their captors bury them each into the ground with only their head sticking out, in a line (column), so that all of them were facing in the same direction (They can't turn their head to look back). The captors then took out 4 bandanas, 2 black, 2 white and said to the hostages that they would put on one of the bandanas onto each hostage's head. If the hostage guesses what colour of bandana is on their head, they will be freed and the rest will be killed. Which hostage got out?
The 2nd person. He can see the first person and since the 3rd person hasn't called out yet, he knows that he must have a different bandana than the first person.
On November 30 2012 03:46 betaflame wrote: There are 3 people who were taken hostage. Their captors bury them each into the ground with only their head sticking out, in a line (column), so that all of them were facing in the same direction (They can't turn their head to look back). The captors then took out 4 bandanas, 2 black, 2 white and said to the hostages that they would put on one of the bandanas onto each hostage's head. If the hostage guesses what colour of bandana is on their head, they will be freed and the rest will be killed. Which hostage got out?
The 2nd person. He can see the first person and since the 3rd person hasn't called out yet, he knows that he must have a different bandana than the first person.
We already had this one with gnomes !
Here's another then, almost like the indians one: 7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ?
Some info as always - The gnomes can discuss strategy before getting their hats - The gnomes cannot communicate after - Gnomes are smart
On November 30 2012 03:46 betaflame wrote: There are 3 people who were taken hostage. Their captors bury them each into the ground with only their head sticking out, in a line (column), so that all of them were facing in the same direction (They can't turn their head to look back). The captors then took out 4 bandanas, 2 black, 2 white and said to the hostages that they would put on one of the bandanas onto each hostage's head. If the hostage guesses what colour of bandana is on their head, they will be freed and the rest will be killed. Which hostage got out?
The 2nd person. He can see the first person and since the 3rd person hasn't called out yet, he knows that he must have a different bandana than the first person.
What's the downside to guessing wrong? The way I see it, the first person guesses immediately on receiving a bandana. As the person with no information and a flat 50% chance, the first person must take that chance before the other two hostages use their information advantage to reach a conclusion.
On November 30 2012 03:46 betaflame wrote: There are 3 people who were taken hostage. Their captors bury them each into the ground with only their head sticking out, in a line (column), so that all of them were facing in the same direction (They can't turn their head to look back). The captors then took out 4 bandanas, 2 black, 2 white and said to the hostages that they would put on one of the bandanas onto each hostage's head. If the hostage guesses what colour of bandana is on their head, they will be freed and the rest will be killed. Which hostage got out?
The 2nd person. He can see the first person and since the 3rd person hasn't called out yet, he knows that he must have a different bandana than the first person.
We already had this one with gnomes !
Here's another then, almost like the indians one: 7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ?
Some info as always - The gnomes can discuss strategy before getting their hats - The gnomes cannot communicate after - Gnomes are smart
Gnomes assign numbers from 0 to 6 to colors. Each gnome is assigned a different number from 0 to 6. Gnomes guess their hat color by assuming that the sum of all hats is equal to their number (mod 7) so at least one of them gets it right.
On November 30 2012 03:46 betaflame wrote: There are 3 people who were taken hostage. Their captors bury them each into the ground with only their head sticking out, in a line (column), so that all of them were facing in the same direction (They can't turn their head to look back). The captors then took out 4 bandanas, 2 black, 2 white and said to the hostages that they would put on one of the bandanas onto each hostage's head. If the hostage guesses what colour of bandana is on their head, they will be freed and the rest will be killed. Which hostage got out?
The 2nd person. He can see the first person and since the 3rd person hasn't called out yet, he knows that he must have a different bandana than the first person.
We already had this one with gnomes !
Here's another then, almost like the indians one: 7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ?
Some info as always - The gnomes can discuss strategy before getting their hats - The gnomes cannot communicate after - Gnomes are smart
Gnomes assign numbers from 0 to 6 to colors. Each gnome is assigned a different number from 0 to 6. Gnomes guess their hat color by assuming that the sum of all hats is equal to their number (mod 7) so at least one of them gets it right.
I don't know if this one has been posted already, but I like it. Instead of writing it myself, I found it on the internet so that it's better written.
You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).
You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter).
You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?
On November 30 2012 03:46 betaflame wrote: There are 3 people who were taken hostage. Their captors bury them each into the ground with only their head sticking out, in a line (column), so that all of them were facing in the same direction (They can't turn their head to look back). The captors then took out 4 bandanas, 2 black, 2 white and said to the hostages that they would put on one of the bandanas onto each hostage's head. If the hostage guesses what colour of bandana is on their head, they will be freed and the rest will be killed. Which hostage got out?
The 2nd person. He can see the first person and since the 3rd person hasn't called out yet, he knows that he must have a different bandana than the first person.
We already had this one with gnomes !
Here's another then, almost like the indians one: 7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ?
Some info as always - The gnomes can discuss strategy before getting their hats - The gnomes cannot communicate after - Gnomes are smart
Gnomes assign numbers from 0 to 6 to colors. Each gnome is assigned a different number from 0 to 6. Gnomes guess their hat color by assuming that the sum of all hats is equal to their number (mod 7) so at least one of them gets it right.
Nice and clean answer!
I understand the answer, but I don't understand how it saves the gnomes.
If the gnomes already know which seven colors are present, then each gnome can deduce the answer trivially; they can see the other six colors, so they have the color they don't see. If they just know that some arbitrary seven colors are used, then they have no way of assigning the available colors to numbers in order for Geiko's solution to be applied.
On November 30 2012 03:46 betaflame wrote: There are 3 people who were taken hostage. Their captors bury them each into the ground with only their head sticking out, in a line (column), so that all of them were facing in the same direction (They can't turn their head to look back). The captors then took out 4 bandanas, 2 black, 2 white and said to the hostages that they would put on one of the bandanas onto each hostage's head. If the hostage guesses what colour of bandana is on their head, they will be freed and the rest will be killed. Which hostage got out?
The 2nd person. He can see the first person and since the 3rd person hasn't called out yet, he knows that he must have a different bandana than the first person.
We already had this one with gnomes !
Here's another then, almost like the indians one: 7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ?
Some info as always - The gnomes can discuss strategy before getting their hats - The gnomes cannot communicate after - Gnomes are smart
Gnomes assign numbers from 0 to 6 to colors. Each gnome is assigned a different number from 0 to 6. Gnomes guess their hat color by assuming that the sum of all hats is equal to their number (mod 7) so at least one of them gets it right.
Nice and clean answer!
I understand the answer, but I don't understand how it saves the gnomes.
If the gnomes already know which seven colors are present, then each gnome can deduce the answer trivially; they can see the other six colors, so they have the color they don't see. If they just know that some arbitrary seven colors are used, then they have no way of assigning the available colors to numbers in order for Geiko's solution to be applied.
What am I missing?
The colors could be anything (there could be 3 red, 4 blue for instance), they have no information on the color distribution. But to save all the gnomes, only one good answer is required.
On November 30 2012 03:46 betaflame wrote: There are 3 people who were taken hostage. Their captors bury them each into the ground with only their head sticking out, in a line (column), so that all of them were facing in the same direction (They can't turn their head to look back). The captors then took out 4 bandanas, 2 black, 2 white and said to the hostages that they would put on one of the bandanas onto each hostage's head. If the hostage guesses what colour of bandana is on their head, they will be freed and the rest will be killed. Which hostage got out?
The 2nd person. He can see the first person and since the 3rd person hasn't called out yet, he knows that he must have a different bandana than the first person.
We already had this one with gnomes !
Here's another then, almost like the indians one: 7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ?
Some info as always - The gnomes can discuss strategy before getting their hats - The gnomes cannot communicate after - Gnomes are smart
Gnomes assign numbers from 0 to 6 to colors. Each gnome is assigned a different number from 0 to 6. Gnomes guess their hat color by assuming that the sum of all hats is equal to their number (mod 7) so at least one of them gets it right.
Nice and clean answer!
I understand the answer, but I don't understand how it saves the gnomes.
If the gnomes already know which seven colors are present, then each gnome can deduce the answer trivially; they can see the other six colors, so they have the color they don't see. If they just know that some arbitrary seven colors are used, then they have no way of assigning the available colors to numbers in order for Geiko's solution to be applied.
What am I missing?
The colors could be anything (there could be 3 red, 4 blue for instance), they have no information on the color distribution. But to save all the gnomes, only one good answer is required.
So the gnomes meet, and they decide on the following codes:
0 - red 1 - blue 2 - green 3 - yellow 4 - black 5 - white
Now I'm a gnome, and I look around at my six buddies. I see two orange hats, one pink, one grey, and two purple. What do I answer?
Edit: Wait, I get it. The "out of seven different colors" in the question is meant to imply that the gnomes know the set of available colors. I withdraw my objection.
On November 30 2012 04:05 CptZouglou wrote: Here's another then, almost like the indians one: 7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ?
Some info as always - The gnomes can discuss strategy before getting their hats - The gnomes cannot communicate after - Gnomes are smart
The Gnomes could just decide to write the color of one specific gnomes hat, before they get their hats. So they would have 6 times the color of his hat writen down and he just says that color. -> one gnome knows his hatcolor
And if every gnome needs to know the color of his hat to be freed they could number themselfs 1-7 and decide that gnome 1 writes the color of gnome 2's hat (2->3, 3->4, ...). Therefore they would have every hatcolor writen down and every gnome just needs to sort out the colors that he sees and the one that is left is the color of his hat. -> every gnome knows his hatcolor
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
Is being able to see the hats in front enough to determine which equivalence class the set of all gnomes belongs to? Couldn't you have two sets which don't differ in front of one gnome but differ at an infinite number of points behind and are therefore in different classes?
A relatively easy puzzle (I don't remember the exact wording so I'll paraphrase) Two pirates are captured and kept in separate cells with no communication between them. Every day, two coins are flipped - one in front of each pirate, where he can see it land but not influence it. Each pirate then states their guess for the other pirate's coin. If they're both wrong, they're put to death. How long will they last on average?
edit: they had a chance to discuss strategy before they were captured just in case something like this happened, and both pirates are smart
On November 30 2012 04:39 CptZouglou wrote: I don't know if this one has been posted already, but I like it. Instead of writing it myself, I found it on the internet so that it's better written.
You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).
You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter).
You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?
This can only be solved if you know the odd ball is lighter or heavier.
Lets say its heavier.
1st weigh: weigh 4 balls against 4 balls. If one set of 4 balls is heavier, the heavy ball is in that group. If scales are equal, heavy ball is in the leftover group of 4.
2nd weigh: From the 4 ball group which includes the heavy ball, weigh 1 ball against 1 ball. If 1 ball is heavier, its that ball. Game over. If scales are equal, the heavy ball is one of the two remaining balls.
3rd weigh: weigh the two remaining balls to find the heavier one. Game over.
On November 30 2012 04:39 CptZouglou wrote: I don't know if this one has been posted already, but I like it. Instead of writing it myself, I found it on the internet so that it's better written.
You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).
You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter).
You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?
This can only be solved if you know the odd ball is lighter or heavier.
Lets say its heavier.
1st weigh: weigh 4 balls against 4 balls. If one set of 4 balls is heavier, the heavy ball is in that group. If scales are equal, heavy ball is in the leftover group of 4.
2nd weigh: From the 4 ball group which includes the heavy ball, weigh 1 ball against 1 ball. If 1 ball is heavier, its that ball. Game over. If scales are equal, the heavy ball is one of the two remaining balls.
3rd weigh: weigh the two remaining balls to find the heavier one. Game over.
No ! There's a way to find the odd ball and determine if it's lighter or heavier, but it's not trivial !
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
Is being able to see the hats in front enough to determine which equivalence class the set of all gnomes belongs to? Couldn't you have two sets which don't differ in front of one gnome but differ at an infinite number of points behind and are therefore in different classes?
Solution supposes you work in N and not Z (gnome line has a start), therefore there is always a finite number of gnomes behind the one answering.
A relatively easy puzzle (I don't remember the exact wording so I'll paraphrase) Two pirates are captured and kept in separate cells with no communication between them. Every day, two coins are flipped - one in front of each pirate, where he can see it land but not influence it. Each pirate then states their guess for the other pirate's coin. If they're both wrong, they're put to death. How long will they last on average?
edit: they had a chance to discuss strategy before they were captured just in case something like this happened, and both pirates are smart
If both answer at the same time, basically 1/4 of both wrong at each try and an average survival would be of 3 days ? Once again if there is no communication it's answering on random independant events, which doesn't leave much room for strategy ...
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
Is being able to see the hats in front enough to determine which equivalence class the set of all gnomes belongs to? Couldn't you have two sets which don't differ in front of one gnome but differ at an infinite number of points behind and are therefore in different classes?
Solution supposes you work in N and not Z (gnome line has a start), therefore there is always a finite number of gnomes behind the one answering.
Oh, that explains it. I wonder if it's unsolvable in Z then?
If both answer at the same time, basically 1/4 of both wrong at each try and an average survival would be of 3 days ? Once again if there is no communication it's answering on random independant events, which doesn't leave much room for strategy ...
There is a way to do better than chance ^.^
Actually, while I was looking for where I first saw this problem I found another nice one (credit goes to mindyourdecisions.com). Hopefully it hasn't been posted, there are a lot of hat riddles here...
Three gnomes are in a room and a red or blue hat (with equal probability) is placed on each one's head. They must simultaneously either guess the colour of their hat or pass. There is only one round. The gnomes win if at least one gnome guesses correctly and no one guesses incorrectly. What's the best strategy?
standard stuff about gnomes being smart and having planning time applies
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
Is being able to see the hats in front enough to determine which equivalence class the set of all gnomes belongs to? Couldn't you have two sets which don't differ in front of one gnome but differ at an infinite number of points behind and are therefore in different classes?
A relatively easy puzzle (I don't remember the exact wording so I'll paraphrase) Two pirates are captured and kept in separate cells with no communication between them. Every day, two coins are flipped - one in front of each pirate, where he can see it land but not influence it. Each pirate then states their guess for the other pirate's coin. If they're both wrong, they're put to death. How long will they last on average?
edit: they had a chance to discuss strategy before they were captured just in case something like this happened, and both pirates are smart
haha took me ten minutes I think. I counted the number of non-prime factors of each number in each series, and added that up., In the case where there were no non-prime factors and a zero, increased the count by 1 for each zero (There were two numbers). Still worked
If both answer at the same time, basically 1/4 of both wrong at each try and an average survival would be of 3 days ? Once again if there is no communication it's answering on random independant events, which doesn't leave much room for strategy ...
There is a way to do better than chance ^.^
With 3 pirates and seeing the 2 others, its trivial (identical to gnomes bellow), still have trouble with only 2.
Three gnomes are in a room and a red or blue hat (with equal probability) is placed on each one's head. They must simultaneously either guess the colour of their hat or pass. There is only one round. The gnomes win if at least one gnome guesses correctly and no one guesses incorrectly. What's the best strategy?
standard stuff about gnomes being smart and having planning time applies
haha took me ten minutes I think. I counted the number of non-prime factors of each number in each series, and added that up., In the case where there were no non-prime factors and a zero, increased the count by 1 for each zero (There were two numbers). Still worked
On November 30 2012 04:39 CptZouglou wrote: I don't know if this one has been posted already, but I like it. Instead of writing it myself, I found it on the internet so that it's better written.
You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).
You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter).
You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?
This can only be solved if you know the odd ball is lighter or heavier.
Lets say its heavier.
1st weigh: weigh 4 balls against 4 balls. If one set of 4 balls is heavier, the heavy ball is in that group. If scales are equal, heavy ball is in the leftover group of 4.
2nd weigh: From the 4 ball group which includes the heavy ball, weigh 1 ball against 1 ball. If 1 ball is heavier, its that ball. Game over. If scales are equal, the heavy ball is one of the two remaining balls.
3rd weigh: weigh the two remaining balls to find the heavier one. Game over.
No ! There's a way to find the odd ball and determine if it's lighter or heavier, but it's not trivial !
Split the twelve balls into three piles of four. You weigh two of the three piles with each other. Now their are two different progressions depending on the result.
If they're equal you weigh three of the now known balls with three of the unknown ones. If the scales are equal then the ball is the last remaining and you need to weigh it with a known ball in order to determine if it's heavier or lighter. If the scales are unequal then you swap a ball with one from the other side and then remove a different ball from each side, you then weigh the scales one last time, if a different side is now heavier than the odd is the unknown ball you swapped, if the sides are now equal than it's the removed ball, otherwise it's the unknown ball you didn't do anything to. In order to determine whether it was lighter or heavier than the rest of the balls you simply need to check which side of the uneven scales it was on.
If they're unequal you swap three of the balls with the other side (the third by adding one of the balls you know isn't odd) and remove three balls (the third is once again substituted with a known ball). Then you weigh the two sides again, if a different side is now heavier than the odd ball is one of the balls you swapped, if the sides are now equal than it's among the removed balls, otherwise it's one of the two remaining unknown balls. After that you replace one of the balls and swap it with one of the others and weigh it a final time. If the scales are equal the odd ball was the replaced one, if a different side is now heavier the odd ball was the swapped one. If nothing happened with the weight distribution the odd ball was the one remaining unknown ball you didn't do anything to. In order to determine whether it was lighter or heavier than the rest of the balls you simply need to check which side of the uneven scales it was on.
haha took me ten minutes I think. I counted the number of non-prime factors of each number in each series, and added that up., In the case where there were no non-prime factors and a zero, increased the count by 1 for each zero (There were two numbers). Still worked
the number of non-prime factors for every digit is equal to the number of enclosed spaces they have, other than zero which is a special case anyway cause it doesn't have factors. Well in writing 2's and 4's sometimes are written differently though.
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
Is being able to see the hats in front enough to determine which equivalence class the set of all gnomes belongs to? Couldn't you have two sets which don't differ in front of one gnome but differ at an infinite number of points behind and are therefore in different classes?
A relatively easy puzzle (I don't remember the exact wording so I'll paraphrase) Two pirates are captured and kept in separate cells with no communication between them. Every day, two coins are flipped - one in front of each pirate, where he can see it land but not influence it. Each pirate then states their guess for the other pirate's coin. If they're both wrong, they're put to death. How long will they last on average?
edit: they had a chance to discuss strategy before they were captured just in case something like this happened, and both pirates are smart
Is the guess simultaneous ?
Yes, but they can't hear the other's guess so it doesn't really matter. Let's say the cells are really far apart, in different buildings.
On December 01 2012 00:59 Oshuy wrote: With 3 pirates and seeing the 2 others, its trivial (identical to gnomes bellow), still have trouble with only 2.
Yeah, the way it's worded is intentionally a bit misleading but it's actually quite different from the hat problems.
Haven't seen the gnome version. A few similar ones generated a bit of traffic a while back ^^ + Show Spoiler +
Name opposite color if I see 2 hats of the same color, pass if 2 different
Yup! I used gnomes just because it's a hat problem :p
On November 30 2012 00:05 Sienionelain wrote: A similar to the indian one: An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right? What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle: -no communicating in the line (they can discuss the strategy beforehand) -every gnome answers at the same time ! -gnomes are smart
Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
Is being able to see the hats in front enough to determine which equivalence class the set of all gnomes belongs to? Couldn't you have two sets which don't differ in front of one gnome but differ at an infinite number of points behind and are therefore in different classes?
A relatively easy puzzle (I don't remember the exact wording so I'll paraphrase) Two pirates are captured and kept in separate cells with no communication between them. Every day, two coins are flipped - one in front of each pirate, where he can see it land but not influence it. Each pirate then states their guess for the other pirate's coin. If they're both wrong, they're put to death. How long will they last on average?
edit: they had a chance to discuss strategy before they were captured just in case something like this happened, and both pirates are smart
Is the guess simultaneous ?
Yes, but they can't hear the other's guess so it doesn't really matter. Let's say the cells are really far apart, in different buildings.
On December 01 2012 00:59 Oshuy wrote: With 3 pirates and seeing the 2 others, its trivial (identical to gnomes bellow), still have trouble with only 2.
Yeah, the way it's worded is intentionally a bit misleading but it's actually quite different from the hat problems.
Haven't seen the gnome version. A few similar ones generated a bit of traffic a while back ^^ + Show Spoiler +
Name opposite color if I see 2 hats of the same color, pass if 2 different
Yup! I used gnomes just because it's a hat problem :p
The pirates in the cell is the exact same as the 7 different color hats with 7 gnomes. This is just the case of 2 instead of 7. + Show Spoiler +
And with 2 seeing your coin is the exact same amount of information as seeing all the other individuals' results except yours. Also you can word the solution differently with 2 without needing to invoke mod. One pirate always guesses what's on his coin, the other always guesses the opposite of his coin. They last indefinitely.
On April 18 2012 22:47 yeaR wrote: If Pinochio said my nose will grow right now, what would happen ?
Short solution: Pinocchio is God. Long solution: + Show Spoiler +
Which leads us to another riddle: if pinochio by virtue of said argument is omnipotent, what happens if he says "either my nose will grow, or my nose growing mechanism never existed!"?
I saw this many years ago, hopefully you find it interesting....
"Last night I went to a dinner party with my wife and four other couples (total of 10 people present). As people arrived, handshakes were exchanged. However, not everyone shook hands. Some did and some didn't. No one shook hands with his or her spouse, and no one shook hands with him or her self. After the party I asked all the others how many hands they shook, and to my surprise each person told me a different number.
The puzzle to solve is: how many hands did my wife shake?"
On December 11 2012 12:49 ziggurat wrote: I saw this many years ago, hopefully you find it interesting....
"Last night I went to a dinner party with my wife and four other couples (total of 10 people present). As people arrived, handshakes were exchanged. However, not everyone shook hands. Some did and some didn't. No one shook hands with his or her spouse, and no one shook hands with him or her self. After the party I asked all the others how many hands they shook, and to my surprise each person told me a different number.
The puzzle to solve is: how many hands did my wife shake?"
There are 9 possible numbers of hands shaken (0-8) for 9 possible shakers (10 minus you). The person that shakes hands with 8 must be married to the person that shakes hands with 0 because everyone outside of this couple in the group is shaken at least once (by the 8-shaker). Similarly, the 7-shaker must be married to the 1-shaker, since, from the prior sentence, we know that the 1-shaker was shaken only by the 8-shaker, and the 7-shaker must shake hands with everyone else except the 0-shaker and 1-shaker to get to 7. Similar reasoning applies to 6-shaker/2-shaker and 5-shaker/3-shaker. Only possibility left is 4.
edit: Not sure how clear the above reasoning was. Maybe this will be better.
Every shaker position in {0,...,8} is occupied. The 8-shaker must shake hands with everyone not in spouse-group to reach 8. So everyone outside it has at least one, meaning the spouse is the only candidate for 0-shaker. The 7-shaker must shake hands with everyone not in spouse-group or 0-shaker, meaning everyone not in either of these spouse groups is at least a 2-shaker. So the 7-shaker spouse is the only candidate for 1-shaker. The 6-shaker must shake hands with everyone not in spouse-group or 0-shaker or 1-shaker (we already know the 1 person that shakes hands with this person), meaning everyone not in the first 3 spouse-groups shakes hands with at least 3 people. So the 6-shaker spouse is the only candidate for 2-shaker...
On November 30 2012 04:39 CptZouglou wrote: I don't know if this one has been posted already, but I like it. Instead of writing it myself, I found it on the internet so that it's better written.
You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).
You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter).
You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?
Weigh 3 balls vs. 3 balls. If they're uneven, then the odd one is in the set of 6 you weighed. If they're even, then the odd one is in the set of 6 you didn't weigh.
Then weigh 3 of the balls in the set of normal ones vs. 3 in the set you know the odd one is in. If they're uneven, you know the odd one is in one of the 3 you weighed. If they're even, then they're in the 3 you didn't weigh.
So now you have 3 balls you know the odd one is in. 3 is the lowest amount I've managed to reduce the chances to in 2 weighs.
Just to be clear, the solution doesn't involve chopping balls in half, does it?
On November 30 2012 04:39 CptZouglou wrote: I don't know if this one has been posted already, but I like it. Instead of writing it myself, I found it on the internet so that it's better written.
You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).
You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter).
You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?
Weigh 3 balls vs. 3 balls. If they're uneven, then the odd one is in the set of 6 you weighed. If they're even, then the odd one is in the set of 6 you didn't weigh.
Then weigh 3 of the balls in the set of normal ones vs. 3 in the set you know the odd one is in. If they're uneven, you know the odd one is in one of the 3 you weighed. If they're even, then they're in the 3 you didn't weigh.
So now you have 3 balls you know the odd one is in. 3 is the lowest amount I've managed to reduce the chances to in 2 weighs.
Just to be clear, the solution doesn't involve chopping balls in half, does it?
On December 11 2012 12:49 ziggurat wrote: I saw this many years ago, hopefully you find it interesting....
"Last night I went to a dinner party with my wife and four other couples (total of 10 people present). As people arrived, handshakes were exchanged. However, not everyone shook hands. Some did and some didn't. No one shook hands with his or her spouse, and no one shook hands with him or her self. After the party I asked all the others how many hands they shook, and to my surprise each person told me a different number.
The puzzle to solve is: how many hands did my wife shake?"
There are 9 possible numbers of hands shaken (0-8) for 9 possible shakers (10 minus you). The person that shakes hands with 8 must be married to the person that shakes hands with 0 because everyone outside of this couple in the group is shaken at least once (by the 8-shaker). Similarly, the 7-shaker must be married to the 1-shaker, since, from the prior sentence, we know that the 1-shaker was shaken only by the 8-shaker, and the 7-shaker must shake hands with everyone else except the 0-shaker and 1-shaker to get to 7. Similar reasoning applies to 6-shaker/2-shaker and 5-shaker/3-shaker. Only possibility left is 4.
edit: Not sure how clear the above reasoning was. Maybe this will be better.
Every shaker position in {0,...,8} is occupied. The 8-shaker must shake hands with everyone not in spouse-group to reach 8. So everyone outside it has at least one, meaning the spouse is the only candidate for 0-shaker. The 7-shaker must shake hands with everyone not in spouse-group or 0-shaker, meaning everyone not in either of these spouse groups is at least a 2-shaker. So the 7-shaker spouse is the only candidate for 1-shaker. The 6-shaker must shake hands with everyone not in spouse-group or 0-shaker or 1-shaker (we already know the 1 person that shakes hands with this person), meaning everyone not in the first 3 spouse-groups shakes hands with at least 3 people. So the 6-shaker spouse is the only candidate for 2-shaker...
that seems less clear if anything.
I feel like I'm missing something here, but I just woke up so I got an excuse. How do you know the person telling the riddle isn't the person who shook hands with 8 people, and his wife shook hands with 0? I just missed how you determined which couple the story teller is
On November 30 2012 00:41 aseq wrote: Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
On November 30 2012 00:41 aseq wrote: Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
if you swim in a zigzag line away from the wolf's starting position, wouldnt the wolf get stuck oscillating back and fourth across his initial position?
On November 30 2012 00:41 aseq wrote: Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
Assume the lake is 100m across. Its circumference is thus (via C/100m = PI) about 314m, so the wolf can reach any point from any other point in at most 157m. Your goal is thus to get within 39m of the shore while the wolf is exactly opposite your current position.
Swim toward the wolf until you get 39m from the shore. Now swim in a circle such that you stay 39m from the shore at all times. Your path is describing a circle that is 22m in diameter and thus about 69m in circumference, which means you're swimming about 14% faster than the wolf in terms of degrees. Swim until you're exactly opposite the wolf, then swim to shore.
On December 11 2012 12:49 ziggurat wrote: I saw this many years ago, hopefully you find it interesting....
"Last night I went to a dinner party with my wife and four other couples (total of 10 people present). As people arrived, handshakes were exchanged. However, not everyone shook hands. Some did and some didn't. No one shook hands with his or her spouse, and no one shook hands with him or her self. After the party I asked all the others how many hands they shook, and to my surprise each person told me a different number.
The puzzle to solve is: how many hands did my wife shake?"
There are 9 possible numbers of hands shaken (0-8) for 9 possible shakers (10 minus you). The person that shakes hands with 8 must be married to the person that shakes hands with 0 because everyone outside of this couple in the group is shaken at least once (by the 8-shaker). Similarly, the 7-shaker must be married to the 1-shaker, since, from the prior sentence, we know that the 1-shaker was shaken only by the 8-shaker, and the 7-shaker must shake hands with everyone else except the 0-shaker and 1-shaker to get to 7. Similar reasoning applies to 6-shaker/2-shaker and 5-shaker/3-shaker. Only possibility left is 4.
edit: Not sure how clear the above reasoning was. Maybe this will be better.
Every shaker position in {0,...,8} is occupied. The 8-shaker must shake hands with everyone not in spouse-group to reach 8. So everyone outside it has at least one, meaning the spouse is the only candidate for 0-shaker. The 7-shaker must shake hands with everyone not in spouse-group or 0-shaker, meaning everyone not in either of these spouse groups is at least a 2-shaker. So the 7-shaker spouse is the only candidate for 1-shaker. The 6-shaker must shake hands with everyone not in spouse-group or 0-shaker or 1-shaker (we already know the 1 person that shakes hands with this person), meaning everyone not in the first 3 spouse-groups shakes hands with at least 3 people. So the 6-shaker spouse is the only candidate for 2-shaker...
that seems less clear if anything.
I feel like I'm missing something here, but I just woke up so I got an excuse. How do you know the person telling the riddle isn't the person who shook hands with 8 people, and his wife shook hands with 0? I just missed how you determined which couple the story teller is
Due to the line that says he asked and everyone gave a different response.
The groups of couples go 8 0 7 1 6 2 5 3 4 4
As you can see, there are two people who shook 4 peoples hands. As the question asker said that everyone gave different answers, the duplicate must be between him and his wife as he didnt ask himself.
On November 30 2012 00:41 aseq wrote: Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
Swim close to the edge of the lake. The wolf will be waiting for you. Swim very slowly towards the opposite side of the lake. When the wolf arrives at the point you're slowly swimming towards, do a 180 and swim full speed to the shore. The wolf won't get there in time to eat you.
On November 30 2012 00:41 aseq wrote: Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
Swim close to the edge of the lake. The wolf will be waiting for you. Swim very slowly towards the opposite side of the lake. When the wolf arrives at the point you're slowly swimming towards, do a 180 and swim full speed to the shore. The wolf won't get there in time to eat you.
This doesn't work. The wolf has to run PI*(the lake's radius), to get from one side of the lake to the other. PI is less than 4, so he can always get from any point on the edge of the lake to any other point faster than you can swim there from the center.
On November 30 2012 00:41 aseq wrote: Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
if you swim in a zigzag line away from the wolf's starting position, wouldnt the wolf get stuck oscillating back and fourth across his initial position?
Once you start moving, the wolf will start moving toward the edge closest to you. If you zig toward any point in the circle ahead of the wolf's position, he'll just keep running that direction. If you zig toward any point in the circle behind his position, he'll turn around when you get closer to that point than to the point you initially swam toward. Every zig and zag will end with the wolf closer to your current position, unless you swim back toward the center, in which case you've gained nothing.
On November 30 2012 00:41 aseq wrote: Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
start swimming in the opposite direction of the wolfs starting position. swim slightly circular in the same direction like the wolf. You can have a positional advantage until you are at 0.25*r at which point your relative speed will be the same as the wolves since your circumference is 1/4th of his. Make sure to keep the wolf at the opposite side until you reach 0.25 r. Now make a straight line to the shore. If r = 1 and human speed = 1m/s it will take the human 0.75s to reach the shore. The wolf needs (pi*r)/(4m/s) which is 0.78s.
You have a circle with radius 4R. You have unlimited semicircles with radius R. How many semicircles maximum can you place inside the big circle with no overlap? PM for an answer.
On November 30 2012 04:39 CptZouglou wrote: I don't know if this one has been posted already, but I like it. Instead of writing it myself, I found it on the internet so that it's better written.
You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).
You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter).
You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?
Weigh 3 balls vs. 3 balls. If they're uneven, then the odd one is in the set of 6 you weighed. If they're even, then the odd one is in the set of 6 you didn't weigh.
Then weigh 3 of the balls in the set of normal ones vs. 3 in the set you know the odd one is in. If they're uneven, you know the odd one is in one of the 3 you weighed. If they're even, then they're in the 3 you didn't weigh.
So now you have 3 balls you know the odd one is in. 3 is the lowest amount I've managed to reduce the chances to in 2 weighs.
Just to be clear, the solution doesn't involve chopping balls in half, does it?
I remember working this out YEARS ago, in about 2003 or so during high school (ha ha...) Never confirmed if this is true or not. This was a pain to type up in text form; a flowchart or binomial/trinomial tree is much easier!
Balls are numbered 1-12. First, weigh them 1,2,3,4 // 5,6,7,8
1-a) If original weighing is balanced, weigh 1,9 // 10,11 (I.e.3 new balls, 1 good one.) => We know all 8 in first weighing are good.
1-a-i) If [1-a] balanced, weigh 1 // 12. We know 12 is bad, and weighing it against a good one to see if it's lighter or heavier.
1-a-ii) If [1-a] tips heavier left, weigh 10 // 11. => We know either 9 is heavier, OR 10 or 11 are lighter. Weighing 10 vs 11 lets us know which is lighter, or if they’re the same, then 9 is heavier. (Same logic if weighing tipped other way; flip heavier/lighter in the sentence.)
2-a) If original weighing is not balanced (let’s say heavier left), weigh 1,5,9 // 6,7,2. (I.e. remove 2 from one side, remove 1 from the other side and replace with a good ball, and swap the sides of 2 balls.)
2-a-i) If [2-a] balanced, weigh 3 // 4. => One of the 3 removed balls (3,4,8) is bad.
(2-a-i-1) If [2-a-i] balanced, we know 8, the other ball removed, is the heavier/lighter ball (based on which way original weighing tipped).
(2-a-i-2) if [2-a-i] didn’t balance, we know 3 or 4 are bad. You know original weighing tipped left, meaning it was heavier, so whichever side remains heavier is the heavier ball out. (Same logic if original weighing tipped other way, except you know 3 or 4 are lighter.)
2-a-ii) if [2-a] tips the other way from 1st weighing (in this case, heavier right) weigh 5 // 12. We know one of the swapped balls (2 or 5) are bad, so you weigh one of them against a good ball. If it doesn’t balance, we know it’s bad and if it’s heavier or lighter; if it balances, then you know the other ball is bad, and either heavier or lighter.
2-a-iii) if [2-a] tips same as 1st weighing (in this case, left), weigh 6 // 7. We know either 1, 6, or 7 are bad. If 6 and 7 balance, then 1 is bad and we know which way as well. If 6 vs 7 don't balance, we know if it's heavier or lighter from previous weighing since they were on the same side, so we know the ball that tips the same way is is bad.
TL;DR: Split 4 vs 4. Following this, the harder scenario is when they don’t balance. For the 2nd weighing remove a pair, swap a pair, and replace one ball with a good ball. From there you can always determine the bad ball, and if it’s heavier or lighter, with a 3rd weighing.
On November 30 2012 00:41 aseq wrote: Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
Swim close to the edge of the lake. The wolf will be waiting for you. Swim very slowly towards the opposite side of the lake. When the wolf arrives at the point you're slowly swimming towards, do a 180 and swim full speed to the shore. The wolf won't get there in time to eat you.
This doesn't work. The wolf has to run PI*(the lake's radius), to get from one side of the lake to the other. PI is less than 4, so he can always get from any point on the edge of the lake to any other point faster than you can swim there from the center.
I don't think you got my solution. It has nothing to do with the center. + Show Spoiler +
Swim to anywhere on the edge of the lake. The wolf will get there before you and will be waiting. Now SLOWLY swim towards the other side. As the wolf runs at full speed and you swim slowly he'll get there while you've only covered a tiny distance. You now reverse and swim at full speed. The wolf won't be able to cover the distance in the time it takes you to once again reach the shore since (wolf speed/slow swim speed) > (wolf speed /max swim speed) and the distances are the same.
On November 30 2012 00:41 aseq wrote: Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
Swim close to the edge of the lake. The wolf will be waiting for you. Swim very slowly towards the opposite side of the lake. When the wolf arrives at the point you're slowly swimming towards, do a 180 and swim full speed to the shore. The wolf won't get there in time to eat you.
This doesn't work. The wolf has to run PI*(the lake's radius), to get from one side of the lake to the other. PI is less than 4, so he can always get from any point on the edge of the lake to any other point faster than you can swim there from the center.
I don't think you got my solution. It has nothing to do with the center. + Show Spoiler +
Swim to anywhere on the edge of the lake. The wolf will get there before you and will be waiting. Now SLOWLY swim towards the other side. As the wolf runs at full speed and you swim slowly he'll get there while you've only covered a tiny distance. You now reverse and swim at full speed. The wolf won't be able to cover the distance in the time it takes you to once again reach the shore.
"He will always run towards the point on the shore closest to your current position"
The wolf doesn't run to your destination, as your method requires. He runs to the point on the shore closest to your current position. If you're swimming directly away from the wolf, however slowly you do so, that point will remain in the same place until you pass the center of the circle.
On November 30 2012 00:41 aseq wrote: Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
Swim close to the edge of the lake. The wolf will be waiting for you. Swim very slowly towards the opposite side of the lake. When the wolf arrives at the point you're slowly swimming towards, do a 180 and swim full speed to the shore. The wolf won't get there in time to eat you.
This doesn't work. The wolf has to run PI*(the lake's radius), to get from one side of the lake to the other. PI is less than 4, so he can always get from any point on the edge of the lake to any other point faster than you can swim there from the center.
I don't think you got my solution. It has nothing to do with the center. + Show Spoiler +
Swim to anywhere on the edge of the lake. The wolf will get there before you and will be waiting. Now SLOWLY swim towards the other side. As the wolf runs at full speed and you swim slowly he'll get there while you've only covered a tiny distance. You now reverse and swim at full speed. The wolf won't be able to cover the distance in the time it takes you to once again reach the shore.
"He will always run towards the point on the shore closest to your current position"
The wolf doesn't run to your destination, as your method requires. He runs to the point on the shore closest to your current position. If you're swimming directly away from the wolf, however slowly you do so, that point will remain in the same place until you pass the center of the circle.
Question - what does the wolf do if you're equidistant from the shores at multiple points? Let's say you're in a center, and the wolf is at the east (0 degrees). You swim due north (90 degrees), and the wolf is there before you are, so you swim directly back towards the center. When you hit the center, is he still at the same spot until you pass the center?
On December 11 2012 12:49 ziggurat wrote: I saw this many years ago, hopefully you find it interesting....
"Last night I went to a dinner party with my wife and four other couples (total of 10 people present). As people arrived, handshakes were exchanged. However, not everyone shook hands. Some did and some didn't. No one shook hands with his or her spouse, and no one shook hands with him or her self. After the party I asked all the others how many hands they shook, and to my surprise each person told me a different number.
The puzzle to solve is: how many hands did my wife shake?"
There are 9 possible numbers of hands shaken (0-8) for 9 possible shakers (10 minus you). The person that shakes hands with 8 must be married to the person that shakes hands with 0 because everyone outside of this couple in the group is shaken at least once (by the 8-shaker). Similarly, the 7-shaker must be married to the 1-shaker, since, from the prior sentence, we know that the 1-shaker was shaken only by the 8-shaker, and the 7-shaker must shake hands with everyone else except the 0-shaker and 1-shaker to get to 7. Similar reasoning applies to 6-shaker/2-shaker and 5-shaker/3-shaker. Only possibility left is 4.
edit: Not sure how clear the above reasoning was. Maybe this will be better.
Every shaker position in {0,...,8} is occupied. The 8-shaker must shake hands with everyone not in spouse-group to reach 8. So everyone outside it has at least one, meaning the spouse is the only candidate for 0-shaker. The 7-shaker must shake hands with everyone not in spouse-group or 0-shaker, meaning everyone not in either of these spouse groups is at least a 2-shaker. So the 7-shaker spouse is the only candidate for 1-shaker. The 6-shaker must shake hands with everyone not in spouse-group or 0-shaker or 1-shaker (we already know the 1 person that shakes hands with this person), meaning everyone not in the first 3 spouse-groups shakes hands with at least 3 people. So the 6-shaker spouse is the only candidate for 2-shaker...
that seems less clear if anything.
I feel like I'm missing something here, but I just woke up so I got an excuse. How do you know the person telling the riddle isn't the person who shook hands with 8 people, and his wife shook hands with 0? I just missed how you determined which couple the story teller is
There are 10 people but only 9 possible variants for the number of handshakes. That means that everyone has a different number of handshakes except YOU. So you just analyze the other 9 people. One more note is that your wife cant be a 8-shaker, because than everybody will have 1 shake and none will have 0 shakes. By that pattern she also cant be a 7,6,5-shaker. In the end you get all places for number of shakes filled in and there cant be anymore handshakes or the rules will be broken. Just count the number of shakes you and your wife got and you'll have a 4.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
Ooh, I love evasion puzzles. Previous posters were on the right track, but not quite precise enough. The following works (apologies for the wordiness): + Show Spoiler +
To make this easier to talk about, let's make the lake the unit circle -- you're at (0,0) and the wolf is at some arbitrary point on the circle, say (0,1). Previously posters correctly said that if we want to get out at (0,1) safely, we need to somehow arrange for the wolf to be at the opposite point, (0,-1) when we're close enough to the circle to make a beeline for safety without the wolf catching up. Other posters just didn't correctly see how close "close enough" is.
First, notice that if we swim around the circle of radius r, for r near 1 (we're near the edge of the lake), the wolf will always easily keep up with us, staying right at the point on the circle closest to us, since for every distance d we swim in a circle, the wolf runs pi*d (arc length), which is less than the 4d he could cover going at top speed. However, if we stay close to the middle of the lake and swim in tiny circles around the origin, the situation is reversed... If we swim in circles of radius r, for r near 0, our angular velocity increases without limit.
So if we're faster near the center but the wolf is faster near the edge, where's the breakpoint? Other than the fact that 4>pi, this is the only significance of the wolf running at 4 times our speed. If we swim around the circle r=1/4 (a quarter of the way out to the edge), suppose we swim an arc of length theta. Then the distance we cover is s=r*theta=theta/4, so in the same time the wolf can cover 4 times that distance, or exactly theta distance. But the wolf is out at r=1, so for him, arc length s=r*theta=theta, and we see that you and the wolf cover exactly the same arc length.
Now we're nearly done. Anywhere inside the circle r=1/4, we can outpace the wolf's angular velocity, but anywhere outside, he can outpace ours. To get the wolf around to the opposite point, swim out to just slightly under radius 1/4 (within an arbitrarily small but nonzero distance). Start swimming in circles. Every time we go around the circle, the wolf lags behind us a little bit more, so keep doing this until he's half a circle behind. Depending on how close to 1/4 you are, this will take more or fewer trips around the circle, but our angular velocity is always higher so it's clear this happens eventually.
Now we're in the desired situation. The wolf is at (0,-1) opposite our exit point, and we're at (0,1/4), a quarter of the way there. Now stop circling and head to safety. We have to cover distance 3/4, so in that time the wolf can cover distance 3. But he's half a circle away -- he has to cover distance pi, which is more than the 3 he can manage at top speed. Huzzah!
Our angular velocity is higher than the wolf's if and only if we're less than a quarter of the way out from the center of the lake. Swim out slightly less than a quarter of the way and start circling. The wolf will gradually lag behind our position more and more -- when he's on the far point of the lake opposite our position, head straight for the closest point on the bank; you'll make it because 3<pi<4.
Edit: Sandster, that's correct. At any given time, draw a line from the center of the lake through your current position. The wolf heads towards where that line hits the edge of the lake, which doesn't change if you're moving directly towards or away from him without passing the center of the lake.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
Ooh, I love evasion puzzles. Previous posters were on the right track, but not quite precise enough. The following works (apologies for the wordiness): + Show Spoiler +
To make this easier to talk about, let's make the lake the unit circle -- you're at (0,0) and the wolf is at some arbitrary point on the circle, say (0,1). Previously posters correctly said that if we want to get out at (0,1) safely, we need to somehow arrange for the wolf to be at the opposite point, (0,-1) when we're close enough to the circle to make a beeline for safety without the wolf catching up. Other posters just didn't correctly see how close "close enough" is.
First, notice that if we swim around the circle of radius r, for r near 1 (we're near the edge of the lake), the wolf will always easily keep up with us, staying right at the point on the circle closest to us, since for every distance d we swim in a circle, the wolf runs pi*d (arc length), which is less than the 4d he could cover going at top speed. However, if we stay close to the middle of the lake and swim in tiny circles around the origin, the situation is reversed... If we swim in circles of radius r, for r near 0, our angular velocity increases without limit.
So if we're faster near the center but the wolf is faster near the edge, where's the breakpoint? Other than the fact that 4>pi, this is the only significance of the wolf running at 4 times our speed. If we swim around the circle r=1/4 (a quarter of the way out to the edge), suppose we swim an arc of length theta. Then the distance we cover is s=r*theta=theta/4, so in the same time the wolf can cover 4 times that distance, or exactly theta distance. But the wolf is out at r=1, so for him, arc length s=r*theta=theta, and we see that you and the wolf cover exactly the same arc length.
Now we're nearly done. Anywhere inside the circle r=1/4, we can outpace the wolf's angular velocity, but anywhere outside, he can outpace ours. To get the wolf around to the opposite point, swim out to just slightly under radius 1/4 (within an arbitrarily small but nonzero distance). Start swimming in circles. Every time we go around the circle, the wolf lags behind us a little bit more, so keep doing this until he's half a circle behind. Depending on how close to 1/4 you are, this will take more or fewer trips around the circle, but our angular velocity is always higher so it's clear this happens eventually.
Now we're in the desired situation. The wolf is at (0,-1) opposite our exit point, and we're at (0,1/4), a quarter of the way there. Now stop circling and head to safety. We have to cover distance 3/4, so in that time the wolf can cover distance 3. But he's half a circle away -- he has to cover distance pi, which is more than the 3 he can manage at top speed. Huzzah!
Our angular velocity is higher than the wolf's if and only if we're less than a quarter of the way out from the center of the lake. Swim out slightly less than a quarter of the way and start circling. The wolf will gradually lag behind our position more and more -- when he's on the far point of the lake opposite our position, head straight for the closest point on the bank; you'll make it because 3<pi<4.
Edit: Sandster, that's correct. At any given time, draw a line from the center of the lake through your current position. The wolf heads towards where that line hits the edge of the lake, which doesn't change if you're moving directly towards or away from him without passing the center of the lake.
I don't know what wasn't precise enough for you. My answer is as correct as yours but uses actual units for readability. The other correct answer was the same unitless form as yours but arrived at the same answer without being nearly as wordy. I give you extra credit for using "angular velocity", but I think a correct answer is a correct answer :-P.
Edit: My point is, the riddle doesn't ask for a mathematically precise model of the problem, it asks how you get out of the lake. There are literally an infinite number of correct answers, because you have greater angular velocity than the wolf at an infinite number of points that are less than PI from the edge of the lake, the riddle doesn't give bonus points for precision, just for not getting eaten by a hungry canid.
^ Fair enough, I think I mistook your earlier post to be the wrong one you quoted from someone else. Looking more carefully at the last page, your solution is the same idea and that works just as well. I hadn't double-checked the numbers the first time I read it, and it's not obvious from the way you wrote it why 39 is a "safe" value (just over 3/4 of the radius away from the shore). Redbluegreen posted more or less the same thing as I did but in confusing words ("slightly circular", "relative speed").
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
Ooh, I love evasion puzzles. Previous posters were on the right track, but not quite precise enough. The following works (apologies for the wordiness): + Show Spoiler +
To make this easier to talk about, let's make the lake the unit circle -- you're at (0,0) and the wolf is at some arbitrary point on the circle, say (0,1). Previously posters correctly said that if we want to get out at (0,1) safely, we need to somehow arrange for the wolf to be at the opposite point, (0,-1) when we're close enough to the circle to make a beeline for safety without the wolf catching up. Other posters just didn't correctly see how close "close enough" is.
First, notice that if we swim around the circle of radius r, for r near 1 (we're near the edge of the lake), the wolf will always easily keep up with us, staying right at the point on the circle closest to us, since for every distance d we swim in a circle, the wolf runs pi*d (arc length), which is less than the 4d he could cover going at top speed. However, if we stay close to the middle of the lake and swim in tiny circles around the origin, the situation is reversed... If we swim in circles of radius r, for r near 0, our angular velocity increases without limit.
So if we're faster near the center but the wolf is faster near the edge, where's the breakpoint? Other than the fact that 4>pi, this is the only significance of the wolf running at 4 times our speed. If we swim around the circle r=1/4 (a quarter of the way out to the edge), suppose we swim an arc of length theta. Then the distance we cover is s=r*theta=theta/4, so in the same time the wolf can cover 4 times that distance, or exactly theta distance. But the wolf is out at r=1, so for him, arc length s=r*theta=theta, and we see that you and the wolf cover exactly the same arc length.
Now we're nearly done. Anywhere inside the circle r=1/4, we can outpace the wolf's angular velocity, but anywhere outside, he can outpace ours. To get the wolf around to the opposite point, swim out to just slightly under radius 1/4 (within an arbitrarily small but nonzero distance). Start swimming in circles. Every time we go around the circle, the wolf lags behind us a little bit more, so keep doing this until he's half a circle behind. Depending on how close to 1/4 you are, this will take more or fewer trips around the circle, but our angular velocity is always higher so it's clear this happens eventually.
Now we're in the desired situation. The wolf is at (0,-1) opposite our exit point, and we're at (0,1/4), a quarter of the way there. Now stop circling and head to safety. We have to cover distance 3/4, so in that time the wolf can cover distance 3. But he's half a circle away -- he has to cover distance pi, which is more than the 3 he can manage at top speed. Huzzah!
Our angular velocity is higher than the wolf's if and only if we're less than a quarter of the way out from the center of the lake. Swim out slightly less than a quarter of the way and start circling. The wolf will gradually lag behind our position more and more -- when he's on the far point of the lake opposite our position, head straight for the closest point on the bank; you'll make it because 3<pi<4.
Edit: Sandster, that's correct. At any given time, draw a line from the center of the lake through your current position. The wolf heads towards where that line hits the edge of the lake, which doesn't change if you're moving directly towards or away from him without passing the center of the lake.
I don't know what wasn't precise enough for you. My answer is as correct as yours but uses actual units for readability. The other correct answer was the same unitless form as yours but arrived at the same answer without being nearly as wordy. I give you extra credit for using "angular velocity", but I think a correct answer is a correct answer :-P.
Edit: My point is, the riddle doesn't ask for a mathematically precise model of the problem, it asks how you get out of the lake. There are literally an infinite number of correct answers, because you have greater angular velocity than the wolf at an infinite number of points that are less than PI from the edge of the lake, the riddle doesn't give bonus points for precision, just for not getting eaten by a hungry canid.
You could also swim under water as well. The wolf won't see you and then you could pop up and sneak off.
You could also call the eagles to rescue you.
You could also use a square/diamond instead of a circle and get there quicker.
EDIT: I hope this doesn't come across as sarcastic because it isn't!
On April 18 2012 23:02 kochanfe wrote: You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?
Second Problem: You need exactly four gallons. How do you do it?
On April 18 2012 23:02 kochanfe wrote: You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?
Second Problem: You need exactly four gallons. How do you do it?
Fill up the 5 gallon one Pour 3 gallons into the 3 gallon tank Empty 3 gallon tank Pour remaining 2 gallons from 5 gallon tank into 3 gallon one Fill up 5 gallon tank
You will now have a full 5 gallon tank + a 3 gallon tank with 2 gallons...
Fill 5 Gallon Pour into three gallons in 3-gallon, so two gallons are remaining in the 5-gallon. Empty the 3-gallon, and then pour the two gallons from the 5-gallon into the 3-gallon. Fill the 5-gallon. Pour one gallon into the 3-gallon, making the 3-gallon full. There are four gallons left in the 5-gallon.
Fill 5 Gallon Pour into three gallons in 3-gallon, so two gallons are remaining in the 5-gallon. Empty the 3-gallon, and then pour the two gallons from the 5-gallon into the 3-gallon. Fill the 5-gallon. Pour one gallon into the 3-gallon, making the 3-gallon full. There are four gallons left in the 5-gallon.
Lol I did this a longer way but got the same thing. Fill the 3 gallon tank, pour into the 5 gallon tank. Fill the 3 gallon tank again, pour 2 gallons into the 5 gallon tank (1 gallon left in 3 gallon tank). Empty 5 gallon tank. Pour 1 gallon from 3 gallon tank into 5 gallon tank. Fill 3 gallon tank. 4 gallons.
My way is much more round about >.> same answer though!
On December 11 2012 21:39 eluv wrote: Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore?
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend?
On December 11 2012 21:39 eluv wrote: Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore?
Yikes, the harder version is too much for me, but I've got an idea for the 4x challenge
If you imagine the pond has radius 'x', then the wolf has to run pi*x units to get to the opposite side. If you are swimming in a circle of radius x/4 or less from the center of the pond, then for you the distance to travel from one side to the other is less than 25% of the amount the wolf has to travel, so you can change angles faster than it.
So therefore if you swim in a circle at a radius a little less than x/4 for long enough, you'll end up across from the wolf. At this point, it's a little over 3/4*x for you to get to the center, whereas the wolf will need to travel pi*x. Even if he is going 4 times faster than you, he'll only make it 3*x of the way to where you are landing, so he won't make it in time.
Now if he's going 4.5 times your speed, this won't work. What a conundrum...
Edit: My instinct is that you would do something similar to what I posted to start, then spiral towards the pond edge to escape, but I have no idea how to show whether that would be better or how much better it would be.
Here's one I just made up. I was hunting around for interesting number-theoretic coincidences, so I'm not sure this will actually make a good puzzle, but I think it's neat... Also I'll be pretty impressed if people come up with a solution, since unlike most of the rest of this thread, I doubt you can find a solution via 30 seconds of Google-fu.
Take a piece of graph paper, color in a bunch of non-overlapping squares of varying sizes, and keep track of the total area (number of grid sections) you just covered. For example, you might color in a 2x2 square, three 5x5 squares, a 6x6 square, and a 10x10 square, for a total area of 4+3*25+36+100=215.
If you're not allowed to use squares of size 1, then clearly some total areas will be impossible to achieve. For example... + Show Spoiler +
Total area 1,2,3: obviously impossible Total area 4: one 2x2 Total area 5,6,7: also obviously impossible Total area 8: two 2x2s Total area 9: one 3x3 Total area 10,11: impossible, slightly less obvious Total area 12: three 2x2s Total area 13: one 2x2 and one 3x3 Total area 14: impossible, slightly still less obvious ...and so on...
If I give you a specified total volume to try and cover, like 797, it's pretty difficult to decide whether or not this is possible without writing a program to do it for you. It turns out to be possible (two 2x2s, three 3x3s, three 4x4s, four 6x6s, three 7x7s, a 9x9, a 10x10, and two 11x11s do the trick).
Question: Are there infinitely many "impossible" areas, or will we eventually reach a point where every total area from there on can be covered exactly by drawing a bunch of non-overlapping squares? If the latter, find this point.
Reworded in terms of pure numbers: find the largest number that cannot be represented as a sum of squares other than 1, or explain why no such number exists. Once you understand things the right way, you can solve this by hand pretty quickly. (Note: there's a ton of information out there about representing numbers as a sum of squares, but excluding the square 1 makes this somewhat tricky. Obviously everything can be represented as a sum of squares if you allow using 1, so the sum-of-squares problem typically includes 1 but restricts the number of squares you're allowed to use. I'm excluding 1 but letting you use as many squares as you want.)
Bonus 1: (Easier) What if you're allowed to let the squares overlap? So for example, 7 is now possible if you take two 2x2s and have them intersect at a corner...
Bonus 2: (Harder) How does the answer change if instead of using two-dimensional squares and looking for the last "impossible" area, use three-dimensional cubes and look for the last "impossible" volume?
On December 11 2012 21:39 eluv wrote: Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore?
Yikes, the harder version is too much for me, but I've got an idea for the 4x challenge
If you imagine the pond has radius 'x', then the wolf has to run pi*x units to get to the opposite side. If you are swimming in a circle of radius x/4 or less from the center of the pond, then for you the distance to travel from one side to the other is less than 25% of the amount the wolf has to travel, so you can change angles faster than it.
So therefore if you swim in a circle at a radius a little less than x/4 for long enough, you'll end up across from the wolf. At this point, it's a little over 3/4*x for you to get to the center, whereas the wolf will need to travel pi*x. Even if he is going 4 times faster than you, he'll only make it 3*x of the way to where you are landing, so he won't make it in time.
Now if he's going 4.5 times your speed, this won't work. What a conundrum...
Edit: My instinct is that you would do something similar to what I posted to start, then spiral towards the pond edge to escape, but I have no idea how to show whether that would be better or how much better it would be.
On December 12 2012 09:36 TanGeng wrote: Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend?
Not sure if this is optimal, but eleven is sufficient. + Show Spoiler +
Using payments of 1, 2, 4, 8, 16, up to 2^n, you can make any number up to 2^(n+1)-1. So to get up to 79, the merchant can cut the chain up into pieces of length (1, 2, 4, 8, 16, 32), (1, 2, 4, 8), and 1 link left over. That's 79 total, cutting the circular chain into 11 pieces, so the optimal solution is at most 11.
Any price up to 63 can be covered with some combination of the first set of pieces listed above. Any price between 64 and 78 can be covered by the entire first set plus some combination of the second set. And 79 is obviously all 11 pieces.
On December 12 2012 10:18 Iranon wrote: Question: Are there infinitely many "impossible" areas, or will we eventually reach a point where every total area from there on can be covered exactly by drawing a bunch of non-overlapping squares? If the latter, find this point.
23 use modulo 4(smallest 2x2 square). 3x3 is equivalent to 1. The biggest "impossible" is therefore the one that requires 3 3x3 squares to complete, and that happens at 27.
Bonus 1: (Easier) What if you're allowed to let the squares overlap? So for example, 7 is now possible if you take two 2x2s and have them intersect at a corner... + Show Spoiler +
6
use modulo 3 instead of modulo 4 - and the 2x2 can be used as 3 (equivalent to 0) or 4 (equivalent to 1).
modulo 1 is covered at 4 (one 2x2) module 2 is covered at 8 (two 2x2) modulo 3 is covered at 9 (one 3x3)
On December 12 2012 09:36 TanGeng wrote: Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend?
Not sure if this is optimal, but eleven is sufficient. + Show Spoiler +
Using payments of 1, 2, 4, 8, 16, up to 2^n, you can make any number up to 2^(n+1)-1. So to get up to 79, the merchant can cut the chain up into pieces of length (1, 2, 4, 8, 16, 32), (1, 2, 4, 8), and 1 link left over. That's 79 total, cutting the circular chain into 11 pieces, so the optimal solution is at most 11.
Any price up to 63 can be covered with some combination of the first set of pieces listed above. Any price between 64 and 78 can be covered by the entire first set plus some combination of the second set. And 79 is obviously all 11 pieces.
Here's what happens when the friend makes the first cut.
The link is cut, and the merchant pull the link out of the chain. Basically, the link can be disconnected from the chain on both sides. It leaves behind an orphan link of one length and a straight chain of 78 in length.
On December 12 2012 10:18 Iranon wrote: Question: Are there infinitely many "impossible" areas, or will we eventually reach a point where every total area from there on can be covered exactly by drawing a bunch of non-overlapping squares? If the latter, find this point.
23 use modulo 4(smallest 2x2 square). 3x3 is equivalent to 1. The biggest "impossible" is therefore the one that requires 3 3x3 squares to complete, and that happens at 27.
Bonus 1: (Easier) What if you're allowed to let the squares overlap? So for example, 7 is now possible if you take two 2x2s and have them intersect at a corner... + Show Spoiler +
6
use modulo 3 instead of modulo 4 - and the 2x2 can be used as 3 (equivalent to 0) or 4 (equivalent to 1).
modulo 1 is covered at 4 (one 2x2) module 2 is covered at 8 (two 2x2) modulo 3 is covered at 9 (one 3x3)
On December 12 2012 10:18 Iranon wrote: Here's one I just made up. I was hunting around for interesting number-theoretic coincidences, so I'm not sure this will actually make a good puzzle, but I think it's neat... Also I'll be pretty impressed if people come up with a solution, since unlike most of the rest of this thread, I doubt you can find a solution via 30 seconds of Google-fu.
Take a piece of graph paper, color in a bunch of non-overlapping squares of varying sizes, and keep track of the total area (number of grid sections) you just covered. For example, you might color in a 2x2 square, three 5x5 squares, a 6x6 square, and a 10x10 square, for a total area of 4+3*25+36+100=215.
If you're not allowed to use squares of size 1, then clearly some total areas will be impossible to achieve. For example... + Show Spoiler +
Total area 1,2,3: obviously impossible Total area 4: one 2x2 Total area 5,6,7: also obviously impossible Total area 8: two 2x2s Total area 9: one 3x3 Total area 10,11: impossible, slightly less obvious Total area 12: three 2x2s Total area 13: one 2x2 and one 3x3 Total area 14: impossible, slightly still less obvious ...and so on...
If I give you a specified total volume to try and cover, like 797, it's pretty difficult to decide whether or not this is possible without writing a program to do it for you. It turns out to be possible (two 2x2s, three 3x3s, three 4x4s, four 6x6s, three 7x7s, a 9x9, a 10x10, and two 11x11s do the trick).
Question: Are there infinitely many "impossible" areas, or will we eventually reach a point where every total area from there on can be covered exactly by drawing a bunch of non-overlapping squares? If the latter, find this point.
Reworded in terms of pure numbers: find the largest number that cannot be represented as a sum of squares other than 1, or explain why no such number exists. Once you understand things the right way, you can solve this by hand pretty quickly. (Note: there's a ton of information out there about representing numbers as a sum of squares, but excluding the square 1 makes this somewhat tricky. Obviously everything can be represented as a sum of squares if you allow using 1, so the sum-of-squares problem typically includes 1 but restricts the number of squares you're allowed to use. I'm excluding 1 but letting you use as many squares as you want.)
Bonus 1: (Easier) What if you're allowed to let the squares overlap? So for example, 7 is now possible if you take two 2x2s and have them intersect at a corner...
Bonus 2: (Harder) How does the answer change if instead of using two-dimensional squares and looking for the last "impossible" area, use three-dimensional cubes and look for the last "impossible" volume?
Nope, there's a very small finite number of impossible areas.
The key to the problem is figuring out that you have an object with an area of 4, so if you ever encounter 4 consecutive numbers you can reach, you will have every possible area. To get 1 above the last of the numbers, add 4 to the first of the series. To get 2 above it, add 4 to the second of the series, and so forth for 3 and 4. And then you can do the same process to get the next 4 afterward. (Hope that was clear, since I don't feel like writing out the PMI proof right now).
The first place with four consecutive reachable areas I encountered was: 24,25,26,27.
edit: Well, that was a while after the answer was posted. I thought I'd refreshed more recently...
On December 12 2012 09:36 TanGeng wrote: Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend?
Not sure if this is optimal, but eleven is sufficient. + Show Spoiler +
Using payments of 1, 2, 4, 8, 16, up to 2^n, you can make any number up to 2^(n+1)-1. So to get up to 79, the merchant can cut the chain up into pieces of length (1, 2, 4, 8, 16, 32), (1, 2, 4, 8), and 1 link left over. That's 79 total, cutting the circular chain into 11 pieces, so the optimal solution is at most 11.
Any price up to 63 can be covered with some combination of the first set of pieces listed above. Any price between 64 and 78 can be covered by the entire first set plus some combination of the second set. And 79 is obviously all 11 pieces.
Here's what happens when the friend makes the first cut.
The link is cut, and the merchant pull the link out of the chain. Basically, the link can be disconnected from the chain on both sides. It leaves behind an orphan link of one length and a straight chain of 78 in length.
On December 12 2012 09:36 TanGeng wrote: Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend?
Not sure if this is optimal, but eleven is sufficient. + Show Spoiler +
Using payments of 1, 2, 4, 8, 16, up to 2^n, you can make any number up to 2^(n+1)-1. So to get up to 79, the merchant can cut the chain up into pieces of length (1, 2, 4, 8, 16, 32), (1, 2, 4, 8), and 1 link left over. That's 79 total, cutting the circular chain into 11 pieces, so the optimal solution is at most 11.
Any price up to 63 can be covered with some combination of the first set of pieces listed above. Any price between 64 and 78 can be covered by the entire first set plus some combination of the second set. And 79 is obviously all 11 pieces.
Here's what happens when the friend makes the first cut.
The link is cut, and the merchant pull the link out of the chain. Basically, the link can be disconnected from the chain on both sides. It leaves behind an orphan link of one length and a straight chain of 78 in length.
On December 11 2012 12:49 ziggurat wrote: I saw this many years ago, hopefully you find it interesting....
"Last night I went to a dinner party with my wife and four other couples (total of 10 people present). As people arrived, handshakes were exchanged. However, not everyone shook hands. Some did and some didn't. No one shook hands with his or her spouse, and no one shook hands with him or her self. After the party I asked all the others how many hands they shook, and to my surprise each person told me a different number.
The puzzle to solve is: how many hands did my wife shake?"
There are 9 possible numbers of hands shaken (0-8) for 9 possible shakers (10 minus you). The person that shakes hands with 8 must be married to the person that shakes hands with 0 because everyone outside of this couple in the group is shaken at least once (by the 8-shaker). Similarly, the 7-shaker must be married to the 1-shaker, since, from the prior sentence, we know that the 1-shaker was shaken only by the 8-shaker, and the 7-shaker must shake hands with everyone else except the 0-shaker and 1-shaker to get to 7. Similar reasoning applies to 6-shaker/2-shaker and 5-shaker/3-shaker. Only possibility left is 4.
edit: Not sure how clear the above reasoning was. Maybe this will be better.
Every shaker position in {0,...,8} is occupied. The 8-shaker must shake hands with everyone not in spouse-group to reach 8. So everyone outside it has at least one, meaning the spouse is the only candidate for 0-shaker. The 7-shaker must shake hands with everyone not in spouse-group or 0-shaker, meaning everyone not in either of these spouse groups is at least a 2-shaker. So the 7-shaker spouse is the only candidate for 1-shaker. The 6-shaker must shake hands with everyone not in spouse-group or 0-shaker or 1-shaker (we already know the 1 person that shakes hands with this person), meaning everyone not in the first 3 spouse-groups shakes hands with at least 3 people. So the 6-shaker spouse is the only candidate for 2-shaker...
that seems less clear if anything.
I feel like I'm missing something here, but I just woke up so I got an excuse. How do you know the person telling the riddle isn't the person who shook hands with 8 people, and his wife shook hands with 0? I just missed how you determined which couple the story teller is
Due to the line that says he asked and everyone gave a different response.
The groups of couples go 8 0 7 1 6 2 5 3 4 4
As you can see, there are two people who shook 4 peoples hands. As the question asker said that everyone gave different answers, the duplicate must be between him and his wife as he didnt ask himself.
On December 12 2012 09:36 TanGeng wrote: Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend?
On December 12 2012 09:36 TanGeng wrote: Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend?
On December 12 2012 09:36 TanGeng wrote: Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend?
i take it it's unacceptable to just line up everything such that 1 cut will cut through it all?
Ha! Will cost 79 feet of silk. The cost by number of cut links and not number of cuts.
Another one: possibly very simple.
A dark priest has a chalice that holds slightly more than 1/2 liter of fluid. The priest is looking to make a sacrificial offering and must pour exactly 1/3 liter of blood into the chalice.
The priest has 10cm x 10cm x 10cm cubic container with an open top. If you know your metric system, the volume of the cube is 1 liter. How does the priest pour out the 1/3 liter of blood to make the offering?
On November 30 2012 00:41 aseq wrote: Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
if you swim in a zigzag line away from the wolf's starting position, wouldnt the wolf get stuck oscillating back and fourth across his initial position?
Once you start moving, the wolf will start moving toward the edge closest to you. If you zig toward any point in the circle ahead of the wolf's position, he'll just keep running that direction. If you zig toward any point in the circle behind his position, he'll turn around when you get closer to that point than to the point you initially swam toward. Every zig and zag will end with the wolf closer to your current position, unless you swim back toward the center, in which case you've gained nothing.
i dunno... im not doing any mathematical calculations but just thinking of a hypothetical situation. Lets say you are at the center and the wolf is at the north shore relative to you (0 degree). You make a small zig southeasterly. This will make the wolf run clockwise since the closest shore to you is somewhere in the 90~180 degree quadrant. Now you make a zag southwesterly. If you think about it, the wolf will only have moved a tiny bit clockwise if your zigzags are small. When you cross over the midline of the circle, suddenly the closest shore ends up at the 180~270 degree quadrant. Assuming the wolf only ran a small distance from its starting point, he would have to double back and run the other way since that is the shortest path towards the closest shore. You keep doing this until you pass the threshold.
My deductions might be way off though. Does this make sense to anyone else? haha
You start off like this, moving some distance (1) from the center of the lake. It doesn't matter how far. The wolf moves four times as far along the edge toward the bank you're moving toward.
Now you're here. The green point is the spot the wolf wants to move toward.
Now you zig away.
But the dot won't move past the spot 180 degrees across from the wolf and get him to turn around.
On December 12 2012 09:36 TanGeng wrote: Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend?
Cut it into segments of 1, 2, 4, 8, 16, 32 and 47. You can then pay any number between 1 and 79.
I am sure most people here thought of binary, but the little catch is that you only need to be able to make any number between 1 and 40 with your base 2 segments, as you can use the remaining segment to work backwards, if you need to pay more than 63.
On December 11 2012 21:39 eluv wrote: Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore?
Yikes, the harder version is too much for me, but I've got an idea for the 4x challenge
If you imagine the pond has radius 'x', then the wolf has to run pi*x units to get to the opposite side. If you are swimming in a circle of radius x/4 or less from the center of the pond, then for you the distance to travel from one side to the other is less than 25% of the amount the wolf has to travel, so you can change angles faster than it.
So therefore if you swim in a circle at a radius a little less than x/4 for long enough, you'll end up across from the wolf. At this point, it's a little over 3/4*x for you to get to the center, whereas the wolf will need to travel pi*x. Even if he is going 4 times faster than you, he'll only make it 3*x of the way to where you are landing, so he won't make it in time.
Now if he's going 4.5 times your speed, this won't work. What a conundrum...
Edit: My instinct is that you would do something similar to what I posted to start, then spiral towards the pond edge to escape, but I have no idea how to show whether that would be better or how much better it would be.
On December 11 2012 21:39 eluv wrote: Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore?
Yikes, the harder version is too much for me, but I've got an idea for the 4x challenge
If you imagine the pond has radius 'x', then the wolf has to run pi*x units to get to the opposite side. If you are swimming in a circle of radius x/4 or less from the center of the pond, then for you the distance to travel from one side to the other is less than 25% of the amount the wolf has to travel, so you can change angles faster than it.
So therefore if you swim in a circle at a radius a little less than x/4 for long enough, you'll end up across from the wolf. At this point, it's a little over 3/4*x for you to get to the center, whereas the wolf will need to travel pi*x. Even if he is going 4 times faster than you, he'll only make it 3*x of the way to where you are landing, so he won't make it in time.
Now if he's going 4.5 times your speed, this won't work. What a conundrum...
Edit: My instinct is that you would do something similar to what I posted to start, then spiral towards the pond edge to escape, but I have no idea how to show whether that would be better or how much better it would be.
That obviously doesn't work. The wolf then needs to cover pi distance, the girl 1. And the wolf is 4x faster. I get the solution up to 4.14, I can't get the solution for 4.59 though.
On December 11 2012 21:39 eluv wrote: Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore?
Yikes, the harder version is too much for me, but I've got an idea for the 4x challenge
If you imagine the pond has radius 'x', then the wolf has to run pi*x units to get to the opposite side. If you are swimming in a circle of radius x/4 or less from the center of the pond, then for you the distance to travel from one side to the other is less than 25% of the amount the wolf has to travel, so you can change angles faster than it.
So therefore if you swim in a circle at a radius a little less than x/4 for long enough, you'll end up across from the wolf. At this point, it's a little over 3/4*x for you to get to the center, whereas the wolf will need to travel pi*x. Even if he is going 4 times faster than you, he'll only make it 3*x of the way to where you are landing, so he won't make it in time.
Now if he's going 4.5 times your speed, this won't work. What a conundrum...
Edit: My instinct is that you would do something similar to what I posted to start, then spiral towards the pond edge to escape, but I have no idea how to show whether that would be better or how much better it would be.
That obviously doesn't work. The wolf then needs to cover pi distance, the girl 1. And the wolf is 4x faster. I get the solution up to 4.14, I can't get the solution for 4.59 though.
Sorry, was thinking 2 pi, have not slept in too long :p I can see the theory, but I don't think I have the mental capacity to calculate it.
A dark priest has a chalice that holds slightly more than 1/2 liter of fluid. The priest is looking to make a sacrificial offering and must pour exactly 1/3 liter of blood into the chalice.
The priest has 10cm x 10cm x 10cm cubic container with an open top. If you know your metric system, the volume of the cube is 1 liter. How does the priest pour out the 1/3 liter of blood to make the offering?
gave this a try, thought i found something interesting and then realized my math sucked. Still, might be someone sees a use for my stuff so will post it here in a spoiler. + Show Spoiler +
chalice=x; container=c Fill the container and then, from the container fill the chalise. The container now contains c-x. empty the chalise, refill it from the container. Since the chalise can hold more than ½ of the container, it wont completely fill, so the difference beteen competely filled (x) and what it is now will be called d. (x-d=current state of chalice) Refill the container completely and from the container, start filling the chalice again. Once the chalice is full, the container holds c-(x-d), lets call this z. Now empty the chalise again, and fill it from the container. This means the container holds z-x now, or c-2x+d. So if we empty the chalice again, we have an empty chalice (obviously) and a container that's filled as c-2x+d. we know c=1L, x is "slightly" more than ½L and d is the difference between them. Should we fill the chalice from the container, we're not gonna manage to competely fill the chalice (just insert .51 as x and calculate, it wont work, if it would fit the outcome would be x again). So we have an almost full chalice, call it a, and an empty container. Refill the container, fill the chalise with it, the container now holds c-a. a was x-(what was left in c; c-2x+d) => the container holds c-(x-(c-2x+d))=c-(x-c+2x-d)=c-x+c-2x+d=2c-3x+d. at this point i gave up cuz it got pretty complicated to do it on my screen without writing it down, and i had no idea if im heading the right direction. Ofc i tried 2c-3x+d=1/3c (working it out is 6c-9x+3d=c, 5c-9x+3d=0, 5L=9x-3d and i didnt see what that would be in an instant so left it there.
if someone thinks he can use (part of) it, feel free, hope i was going the right way so i didnt get a headache for nothing :D
On December 12 2012 13:31 TanGeng wrote: Another one: possibly very simple.
A dark priest has a chalice that holds slightly more than 1/2 liter of fluid. The priest is looking to make a sacrificial offering and must pour exactly 1/3 liter of blood into the chalice.
The priest has 10cm x 10cm x 10cm cubic container with an open top. If you know your metric system, the volume of the cube is 1 liter. How does the priest pour out the 1/3 liter of blood to make the offering?
Using the cubic container, if you look at one of the base corners and the 3 corners adjacent to it, this forms a pyramid shape, which has a volume of 1/6 liter. You can tilt the container onto the base corner and fill it up to the height of the 3 other corners. Repeat twice to get 1/3 liter.
On December 12 2012 13:31 TanGeng wrote: Another one: possibly very simple.
A dark priest has a chalice that holds slightly more than 1/2 liter of fluid. The priest is looking to make a sacrificial offering and must pour exactly 1/3 liter of blood into the chalice.
The priest has 10cm x 10cm x 10cm cubic container with an open top. If you know your metric system, the volume of the cube is 1 liter. How does the priest pour out the 1/3 liter of blood to make the offering?
Using the cubic container, if you look at one of the base corners and the 3 corners adjacent to it, this forms a pyramid shape, which has a volume of 1/6 liter. You can tilt the container onto the base corner and fill it up to the height of the 3 other corners. Repeat twice to get 1/3 liter.
On December 11 2012 21:39 eluv wrote: Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore?
It takes a bit of thought to arrive at the starting point. Then after that it is provable that the tangent swim-out of the inner circle is the optimal strategy.
Plugging in the all the numbers it's possible to show that velocity ratios larger than 4 and as high as 4.6 are possible.
On December 11 2012 21:39 eluv wrote: Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore?
It takes a bit of thought to arrive at the starting point. Then after that it is provable that the tangent swim-out of the inner circle is the optimal strategy.
Plugging in the all the numbers it's possible to show that velocity ratios larger than 4 and as high as 4.6 are possible.
The way you draw that the wolf is going to run clockwise at the wolfstart/swimstart positions. However if you adjust the swim start position clockwise a little or the wolf start counter clockwise a little to get him going on your red path I think it could work.
I'm trying to think about the calculations but they seem really complicated. You need his run distance to be 1.44*pi*r at least.
On December 11 2012 21:39 eluv wrote: Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore?
It takes a bit of thought to arrive at the starting point. Then after that it is provable that the tangent swim-out of the inner circle is the optimal strategy.
Plugging in the all the numbers it's possible to show that velocity ratios larger than 4 and as high as 4.6 are possible.
The way you draw that the wolf is going to run clockwise at the wolfstart/swimstart positions.
At the instant you change paths from circular to tangent, the wolf is running counterclockwise (since you had been running counterclockwise) targeted at the easternmost point of the lake. If we assume you spiral out correctly, at the point where you change paths, your angular velocity is equal to that of the wolf's. After a short time-step, your radius is increasing, but your linear velocity remains constant, so your angular velocity is decreasing. The wolf, however, runs at constant angular velocity--so he will be less than pi radians away from the point on the shoreline you're closest to.
That explanation was probably shitty and hard to understand so if someone can make a better one by all means do so.
On December 11 2012 21:39 eluv wrote: Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore?
It takes a bit of thought to arrive at the starting point. Then after that it is provable that the tangent swim-out of the inner circle is the optimal strategy.
Plugging in the all the numbers it's possible to show that velocity ratios larger than 4 and as high as 4.6 are possible.
The way you draw that the wolf is going to run clockwise at the wolfstart/swimstart positions.
At the instant you change paths from circular to tangent, the wolf is running counterclockwise (since you had been running counterclockwise) targeted at the easternmost point of the lake. If we assume you spiral out correctly, at the point where you change paths, your angular velocity is equal to that of the wolf's. After a short time-step, your radius is increasing, but your linear velocity remains constant, so your angular velocity is decreasing. The wolf, however, runs at constant angular velocity--so he will be less than pi radians away from the point on the shoreline you're closest to.
That explanation was probably shitty and hard to understand so if someone can make a better one by all means do so.
that point is the point where it switches over and the wolf is indifferent, so in reality, the wolf start has to be an infinitesimally small distance further counterclockwise to guarantee he goes counterclockwise.
On December 11 2012 21:39 eluv wrote: Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore?
It takes a bit of thought to arrive at the starting point. Then after that it is provable that the tangent swim-out of the inner circle is the optimal strategy.
Plugging in the all the numbers it's possible to show that velocity ratios larger than 4 and as high as 4.6 are possible.
The way you draw that the wolf is going to run clockwise at the wolfstart/swimstart positions.
At the instant you change paths from circular to tangent, the wolf is running counterclockwise (since you had been running counterclockwise) targeted at the easternmost point of the lake. If we assume you spiral out correctly, at the point where you change paths, your angular velocity is equal to that of the wolf's. After a short time-step, your radius is increasing, but your linear velocity remains constant, so your angular velocity is decreasing. The wolf, however, runs at constant angular velocity--so he will be less than pi radians away from the point on the shoreline you're closest to.
That explanation was probably shitty and hard to understand so if someone can make a better one by all means do so.
that point is the point where it switches over and the wolf is indifferent, so in reality, the wolf start has to be an infinitesimally small distance further counterclockwise to guarantee he goes counterclockwise.
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain.
Day 1: There is at least one person that has green eyes.
Start from base case-- 1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes.
Though not sure if correct.
Edit: hmm I feel slightly better after reading other posts in this thread about this.
sorry for bringing up this old one, i just found it, thought about it and i really dont see why there wouldnt be a much more simpler solution, i m not sure if i have to spoiler it since, it's not the actual solution: + Show Spoiler +
Since they have perfect logic, and they see how everyone has either blue or green eyes, they will simply pair people who have identical eye colors. In other words, you dont know ur own eye-colour but u know all the rest, so you can help them and in return they will help you. It's not communication, u just grab two guys with same eye-color and pair them. Even if they wouldnt guess what u are doing, when they see u paired others (or grouped them up for that matter) they could undertand that their partner(s) have the same eye color as they do, since they can see pairs or groups with identical eyes. You will be paired with someone by another person, who sees your eyecolor.
Ok, u dont actually know if u have one of the two eye-colors, u could have purple eyes, but that's the worst case scenario, and still 199 will be saved. I know it's not as fancy, but i think it's valid, since it's not mentioned that they canot interact, only that they canot communicate.
On April 19 2012 03:24 Bahamuth wrote: Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes: - They are all perfect logicians - Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour. - This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain.
Day 1: There is at least one person that has green eyes.
Start from base case-- 1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes.
Though not sure if correct.
Edit: hmm I feel slightly better after reading other posts in this thread about this.
sorry for bringing up this old one, i just found it, thought about it and i really dont see why there wouldnt be a much more simpler solution, i m not sure if i have to spoiler it since, it's not the actual solution: + Show Spoiler +
Since they have perfect logic, and they see how everyone has either blue or green eyes, they will simply pair people who have identical eye colors. In other words, you dont know ur own eye-colour but u know all the rest, so you can help them and in return they will help you. It's not communication, u just grab two guys with same eye-color and pair them. Even if they wouldnt guess what u are doing, when they see u paired others (or grouped them up for that matter) they could undertand that their partner(s) have the same eye color as they do, since they can see pairs or groups with identical eyes. You will be paired with someone by another person, who sees your eyecolor.
Ok, u dont actually know if u have one of the two eye-colors, u could have purple eyes, but that's the worst case scenario, and still 199 will be saved. I know it's not as fancy, but i think it's valid, since it's not mentioned that they canot interact, only that they canot communicate.
That is a form of communication. Communication does not entail the verbal use of conventional signs. Providing others information in a goal directed manner suffices.
when I work it out, which is the reverse of what it needs to be unless I am messing up. Can anyone help me with this? I use the - to denote between rule applications
do you guys have any clue? Feel free to PM me, ill know the answer eventually and post it if nobody else figures it out
Solution, well not that cool, but it was pretty hard to figure out. Actually really really hard> + Show Spoiler +
The numbers in the link were the first 14 fibonacci numbers, mixed up, and the latin quote has 14 letters. To solve this puzzle you had to take the proper order of the Fibonacci numbers and identify each number with one letter. So D-0, I-1, V-1, I-2 , D-3, E-5 etc, and re-arrange the letters so the order is the mixed up order of the numbers in the link. The solution was in Hungarian, "Idei-medve-parti" which translates raughly to "This year's bear party". Pretty sick one, and this is only the 2nd from a 10-20 long bonus-task list.
do you guys have any clue? Feel free to PM me, ill know the answer eventually and post it if nobody else figures it out
sry but, i don't even know what im looking at, can you maybe explain it a bit?
there is no more explanation. It has something to do with the Fibonacci numbers, and the quote is in latin, it translates to "divide and concqure" i think
EPT![[image loading]](http://ompldr.org/vN2JkOA/Pinocchio-paradox-solved.jpg)
![[image loading]](http://i.imgur.com/LYebe.png)
![[image loading]](http://i.imgur.com/Vv01y.jpg)
![[image loading]](http://i.imgur.com/V9XrH.jpg)
![[image loading]](http://imgur.com/V9XrH][IMG]http://i.imgur.com/V9XrH.jpg)
.
"language". 
![[image loading]](http://images.wikia.com/wowwiki/images/a/a4/Warden_artwork.jpg)
![[image loading]](http://img848.imageshack.us/img848/2559/dbc090b232454ab3a38f236.png)
![[image loading]](http://imgur.com/v3XEC)
![[image loading]](http://i.imgur.com/UwVYh.png)
Said logic and way of thinking can help with a lot of riddles
![[image loading]](http://i.imgur.com/DhiBS.png)
![[image loading]](http://img600.imageshack.us/img600/1813/riddle.png)
![[image loading]](http://img823.imageshack.us/img823/6989/pocketxory.png)
![[image loading]](http://img819.imageshack.us/img819/212/pocketxaaandyab.png)
![[image loading]](http://i48.tinypic.com/jqmb7r.jpg)
![[image loading]](http://i48.tinypic.com/24brib7.jpg)
![[image loading]](http://i48.tinypic.com/2hzn85h.png)
![[image loading]](http://img167.imageshack.us/img167/24/infinityyd1.jpg)
![[image loading]](http://msbrooks.com/blog/wp-content/uploads/2008/01/will_it_take_off.jpg)
![[image loading]](http://i.imgur.com/OGPR8.png)
![[image loading]](http://i.imgur.com/QMKlR.jpg)
![[image loading]](http://i.imgur.com/dqfqH.png)
![[image loading]](http://i.imgur.com/gEvOC.png)
![[image loading]](http://i.imgur.com/Eng9q.png)
![[image loading]](http://i.imgur.com/iXGLi.png)
