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GrandInquisitor
New York City13113 Posts
I've long been a fan of riddling, and so far I've had 8 very favorite ones that I think are extremely difficult. This is a challenge to TL - if they can solve these (WITHOUT GOOGLING OR CHEATING)
Honestly, I think 1 & 2 & 8 are the hardest and I doubt anyone can get them.
(and yes I do know the answers to them. Most of them I solved myself, but not #2 or #6)
EDIT: ALL OF THEM HAVE BEEN SOLVED. =D
1. You have 2 pieces of string of different, unspecified length, and some matches. Each piece of string takes one hour to burn, but the burn rate is not constant. This means that it could take 59 minutes to burn the first 1/4, and 1 minute for the rest. The strings have different burn rates, and of course you don't know the rates anyway.
Using only the matches and the strings, measure 45 minutes.
2. You have a very nice, shiny watch. But this is no ordinary watch. This watch can answer two yes or no questions 100% accurately per day. You ask, and either a blue or yellow light will flash. Unfortunately, you don't know which light means yes and which light means no, and you can never find out because there's a 50-50 chance of the lights switching every night.
You happen to be on a game show. The rules are simple. There are three doors. Behind two are goats. Behind one is a shiny red car. Obviously, you want the car. You can choose any door, after asking the watch two questions.
What are the two yes/no questions you can ask so that you will have a 100% guaranteed chance of getting the car?
3. You are in a room and there are three light switches that are turned off. The light switches are connected to three light bulbs that are in a different room. You do not know which switch is connected to which light bulb. You cannot see into that room. But you are allowed to go into that room exactly once. How can you figure out which switch is connected to which light bulb?
4. A man is in a rowing boat floating on a lake, holding a brick in his hands. He throws the brick over the side of the boat. The brick sinks quickly. Does the water level in the lake go up or down? Explain.
5. In a certain matriarchal town, the women all believe in an old prophecy that says there will come a time when a stranger will visit the town and announce whether any of the men folks are cheating on their wives. The stranger will simply say yes or no, without announcing the number of men implicated or their identities. If the stranger arrives and makes his announcement, the women know that they must follow a particular rule: If on any day following the stranger’s announcement a woman deduces that her husband is not faithful to her, she must kick him out into the street at 10 A.M. the next day. This action is immediately observable by every resident in the town. It is well known that each wife is already observant enough to know whether any man (except her own husband) is cheating on his wife. However, no woman can reveal that information to any other. A cheating husband is also assumed to remain silent about his infidelity.
The time comes, and a stranger arrives. He announces that there are cheating men in the town. On the morning of the 10th day following the stranger’s arrival, some unfaithful men are kicked out into the street for the first time. How many of them are there?
6. You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weigh the same or that the dish that falls lower has heavier contents than the other.
The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more.
Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light.
7. There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person.
Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6, ...). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9, ...). This continues until all 100 people have passed through the room.
How many of the light bulbs are illuminated after the 100th person has passed through the room? More specifically, which light bulbs are still illuminated, and why?
8. You have Some Terminal Condition, which necessitates taking two pills a day: one Pill A and one Pill B. If you neglect to take either pill, you die; if you take more than one A or more than one B, you die. If you don't take them at exactly the same time, you die.
This morning you are going through your usual routine. You pick up your bottle of A Pills and gently tap one into your palm. Then you pick up your bottle of B Pills and tap it, but two pills accidentally fall into your hand. You now hold three pills (one A and two Bs), you don't know which are which, and they are completely indistinguishable from each other. The A Pills are the same color as the B Pills, they are the same shape, same size -- they are identical in every respect. Man, your doctor is a dumbass. But he's a rich dumbass, because he's charging you $10,000,000 a pill! So you dare not throw any away.
Thus, the puzzle: what can you do to ensure that you take only one A Pill and only one B Pill today, without wasting any pills (either today or in the future)?
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3. You are in a room and there are three light switches that are turned off. The light switches are connected to three light bulbs that are in a different room. You do not know which switch is connected to which light bulb. You cannot see into that room. But you are allowed to go into that room exactly once. How can you figure out which switch is connected to which light bulb?
+ Show Spoiler +You turn on one switch, wait for an hour, turn it off again. Then you turn on another switch and go into the room, the lit bulb belongs to the second switch. Then you touch the other two bulbs, the hot one belongs to the first switch 
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#6 is my favorite. LTT posted it here years ago. It took me many months to solve that. I've looked up the answers to #1 and #5 before. #7 I'm sure I've looked up before but it was too lame for me to remember or care. #4 is dumb. #3 is old.
Does #8,
+ Show Spoiler +Have anything to do with cutting the pills in half?
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8
+ Show Spoiler +
Cut each pill into half, and eat each one. So you will have eaten 1/2A, 1/2B,1/2B and left with 1/2A, 1/2B,1/2B. Take another A from bottle and eat 1/2, so total you would have eaten 1A, 1B.
With 1/2A, 1/2B, 1/2B left, next day just eat those and another 1/2A from the bottle.
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So here's the attempt to solve first riddle:
+ Show Spoiler +We burn the 1st string from TWO SIDES and second string from ONE side. After 30 minutes 1st string is down and the second will burn for the next 30 minutes. So basically after 30 minutes we have only one string left and we know that 30 minutes expired, and the second string will burn for 30 minutes. Now we burn it from the second side.of the string. It will be down after 15 minutes. We measured 15 minutes. 30+15 = 45 minutes.
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On September 19 2006 08:24 Ethenielle wrote:Show nested quote +On September 19 2006 08:05 GrandInquisitor wrote:
2. You have a very nice, shiny watch. But this is no ordinary watch. This watch can answer two yes or no questions 100% accurately per day. You ask, and either a blue or yellow light will flash. Unfortunately, you don't know which light means yes and which light means no, and you can never find out because there's a 50-50 chance of the lights switching every night.
You happen to be on a game show. The rules are simple. There are three doors. Behind two are goats. Behind one is a shiny red car. Obviously, you want the car. You can choose any door, after asking the watch two questions.
What are the two yes/no questions you can ask so that you will have a 100% guaranteed chance of getting the car? + Show Spoiler +I'm not entirely sure if it's allowed, but you could just ask "are there goats behind these two doors?", if you get a yes, then it's the third door, if you get a no you can ask "is there a goat behind this door(= one of the two doors you chose)", if no then it's the car, if yes then it's the other. If you can't ask about two doors at once, though..
+ Show Spoiler +You don't know which light means yes and which light means no.
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8. You have Some Terminal Condition, which necessitates taking two pills a day: one Pill A and one Pill B. If you neglect to take either pill, you die; if you take more than one A or more than one B, you die. If you don't take them at exactly the same time, you die.
This morning you are going through your usual routine. You pick up your bottle of A Pills and gently tap one into your palm. Then you pick up your bottle of B Pills and tap it, but two pills accidentally fall into your hand. You now hold three pills (one A and two Bs), you don't know which are which, and they are completely indistinguishable from each other. The A Pills are the same color as the B Pills, they are the same shape, same size -- they are identical in every respect. Man, your doctor is a dumbass. But he's a rich dumbass, because he's charging you $10,000,000 a pill! So you dare not throw any away.
Thus, the puzzle: what can you do to ensure that you take only one A Pill and only one B Pill today, without wasting any pills (either today or in the future)?
Not heard of this before so here's my guess:
+ Show Spoiler +
add one more A pill to what's in your hand (so it's a 50-50 mix) then grind it to powder then dissolve it in a known amount of water, shake/mix well before seperating into two containers. drink 1 cup today and the other cup tomorrow
You could also not dissolve in water it and just seperate out the well-mixed powder into 2 piles but that sounds less accurate to me, but may work if they don't dissolve.
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On September 19 2006 08:25 Olmer wrote:So here's the attempt to solve first riddle: + Show Spoiler +We burn the 1st string from TWO SIDES and second string from ONE side. After 30 minutes 1st string is down and the second will burn for the next 30 minutes. So basically after 30 minutes we have only one string left and we know that 30 minutes expired, and the second string will burn for 30 minutes. Now we burn it from the second side.of the string. It will be down after 15 minutes. We measured 15 minutes. 30+15 = 45 minutes.
that's pretty clever! i'd never have thought of it =]
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On September 19 2006 08:33 Resonate wrote:Show nested quote +On September 19 2006 08:25 Olmer wrote:So here's the attempt to solve first riddle: + Show Spoiler +We burn the 1st string from TWO SIDES and second string from ONE side. After 30 minutes 1st string is down and the second will burn for the next 30 minutes. So basically after 30 minutes we have only one string left and we know that 30 minutes expired, and the second string will burn for 30 minutes. Now we burn it from the second side.of the string. It will be down after 15 minutes. We measured 15 minutes. 30+15 = 45 minutes. that's pretty clever! i'd never have thought of it =]
And it only took him all of 10-15 minutes. I think he's a genius
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Really? Thanks - when I started to write this - I thought so, but when I ended, I thought "No way - it's too lame. There must be something that I cannot see". I thought it over, found nothing wrong...This must be it! ;P
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On September 19 2006 08:24 Ethenielle wrote:Show nested quote +On September 19 2006 08:05 GrandInquisitor wrote:
2. You have a very nice, shiny watch. But this is no ordinary watch. This watch can answer two yes or no questions 100% accurately per day. You ask, and either a blue or yellow light will flash. Unfortunately, you don't know which light means yes and which light means no, and you can never find out because there's a 50-50 chance of the lights switching every night.
You happen to be on a game show. The rules are simple. There are three doors. Behind two are goats. Behind one is a shiny red car. Obviously, you want the car. You can choose any door, after asking the watch two questions.
What are the two yes/no questions you can ask so that you will have a 100% guaranteed chance of getting the car? + Show Spoiler +I'm not entirely sure if it's allowed, but you could just ask "are there goats behind these two doors?", if you get a yes, then it's the third door, if you get a no you can ask "is there a goat behind this door(= one of the two doors you chose)", if no then it's the car, if yes then it's the other. If you can't ask about two doors at once, though..
You dont know which colour means yes and which colour means no though. So your not sure if the watch is saying yes or no when you ask your question
If you knew what colour meant yes and no you could just say "is there a car behind door 1?" "Is there a car behind door 2?"
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Korea (South)17174 Posts
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On September 19 2006 08:35 Olmer wrote:
Really? Thanks - when I started to write this - I thought so, but when I ended, I thought "No way - it's too lame. There must be something that I cannot see". I thought it over, found nothing wrong...This must be it! ;P
i dunno if it's the right answer, but it sounds pretty good to me =]
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regarding #1
+ Show Spoiler + plz pm if im right..
for number 1, could you just put a match down for every minute until you have 45.. you could count like
IIII with a cross through for the 5th minute etc etc till u have 9 groups..
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GrandInquisitor
New York City13113 Posts
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8th riddle (twsan inspired me - I wouldn't solve the riddle without his post here which, unfortunately, provides us with not-so-good solution but his IDEA is right)
+ Show Spoiler + Everyday you just eat one pill from each tube as if nothing wrong happened. After several days you have those 3 mixed pills and 1 pill in one of the tubes. Then you take 4 of them, cut etc (as twsan suggested). He just didn't mention that you need to eat all others before. He just wanted you to take "1/2 A pill from the tube" but he doesn't know thich tube contains A pills! When you have 4 pills remaining, it's obvious that you have 2 of each kind. Before - it wasn't. Of course this solution works if before the accident there was equal number of A and B pills.
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how do i add spoiler thingy? I have some for #2 
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Korea (South)17174 Posts
On September 19 2006 08:49 Olmer wrote:8th riddle (twsan inspired me - I wouldn't solve the riddle without his post here which, unfortunately, provides us with not-so-good solution but his IDEA is right) + Show Spoiler + Everyday you just eat one pill from each tube as if nothing wrong happened. After several days you have those 3 mixed pills and 1 pill in one of the tubes. Then you take 4 of them, cut etc (as twsan suggested). He just didn't mention that you need to eat all others before. He just wanted you to take "1/2 A pill from the tube" but he doesn't know thich tube contains A pills! When you have 4 pills remaining, it's obvious that you have 2 of each kind. Before - it wasn't. Of course this solution works if before the accident there was equal number of A and B pills.
Ummm..U do know which pills are in which bottle.
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"2 does the watch change its colours any time any day or just from day to day?
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regarding number 8:
+ Show Spoiler +
pm if i am right plz..
just take two pills
if you die, you obv took the wrong two
if u live u took the right two
either way, you will never waste money because 1. you will have died and money will mean nothing
or
2. you will live and know the other pill is a b
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On September 19 2006 08:05 GrandInquisitor wrote:
7. There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person.
Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6, ...). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9, ...). This continues until all 100 people have passed through the room.
How many of the light bulbs are illuminated after the 100th person has passed through the room? More specifically, which light bulbs are still illuminated, and why?
Number Seven:
+ Show Spoiler +
The number lightbulbs which are left illuminated will all have an odd number of divisors, because the number of divisors relates to the number of times the switch is moved on each lightbulb. These numbers are perfect squares, i.e. 1 4 9 16 25 36 49 64 81 100.
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With #8, couldn't you just take an extra pill from A and then cut every pill in half, which would mean you would have 1 A and 1 B in total and then the next day take the other halves and then you are back to taking one from each pot again?
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That's what twsan did but if you do this before emptying both bottles, you don't know from which bottle you should take additional pill. You have three and you cannot determine, from which bottle you take the fourth (or 1/2 of fourth).
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The riddle says you don't know which pill is which, not which bottle.
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regarding number 2
+ Show Spoiler +
plz pm if i am right (this one is prrrrrrrrrobably wrong)
ask the watch to show you blue if number 2 is correct or yellow if number 3 is correct.
if it does nothing you know its number1, if it does show a colour you know its that particular box
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U cant ask it to show something if thatis correct or not, u can just ask it a Yes or No question man...
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On September 19 2006 09:11 sdpgposd wrote:regarding number 2 + Show Spoiler +
plz pm if i am right (this one is prrrrrrrrrobably wrong)
ask the watch to show you blue if number 2 is correct or yellow if number 3 is correct.
if it does nothing you know its number1, if it does show a colour you know its that particular box
That's not within the capabilities of the watch, because it only answers yes/no.
Regarding #8: You could continue eating the pills until there are 6 left in total, then mix them in water, divide into thirds and drink the solution >_>
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Dunno how to explain  My shitty english... Maybe Inquisitor will explain it ...and if he sees nothing wrong with twsan's solution then nevermind - it's him who solved this riddle even if it's incomplete... If it's only me who complain - then there's no need to discuss it anymore - it won't change anything...
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ook, number 2 was wrong. i'll try think of something else
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Haha i got asked number 4, the brick in a boat thing for my university practice interview.
Erm...hmm someone tell me how to use a spoiler tag dammit!
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what about the dissolving thing, that would work right?
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For #2, What happens if I ask the watch "Does blue mean no?", but blue means yes?
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if u ask that it cant show blue, it will show yellow, but that yellow light will only tell u that it means either yes or no..
And even if it did work and you could find out if blue was yes/no you couldnt make a second question decide wich door held the car..
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haha didn't remember the yes/no thing  anyway Pacifist already picked it out, so I think you can safely delete your post, Fen.
you pick two doors and ask if there are goats behind them, you get either blue or yellow. let's say you pick the two outermost doors. next up you pick either of those doors and the middle one, and ask if there are goats behind them. you can't get yes and yes, because then there would be three goats. you can get yes and no, then the car will be in the middle door. you can also get no and yes, and you can also get no and no, in which case the car must be in the door you didn't pick the second time around. thus: if you get blue/blue it can't be yes/yes, it must be no/no. same goes for yellow, of course. if you get a no/no, problem is pretty much solved. the car must be in the door you picked, because there will be a goat in the middle door. and a "no" gives us that there is a goat and a car behind the doors - car is in the door you picked, goat in the middle and goat in the one you didn't pick. now if you get a blue/yellow or yellow/blue, it's either a yes/no or a no/yes in both cases. in case blue/yellow is a yes/no, the car is in the middle door. in case it's a no/yes, the car is in the door you didn't pick.
but all this doesn't really help either way, because you still can't know if blue/yellow is yes/no or no/yes.
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Calgary25980 Posts
4 is dumb. + Show Spoiler + brick pushing down on boat displaces more water volume than volume of brick; water level falls
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number 2
+ Show Spoiler +Which colour will the light be after i ask this question?
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GrandInquisitor
New York City13113 Posts
Spoiler tags are like bold where you go [ b ] (wihtout spaces) - it's [ spoiler ] without the spaces and the [/ spoiler ] (without spaces) to end it.
Also, twsan's solution is correct.
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You ask is the car in one or two. You get yellow (which is wrong). You ask is the car in two or three. You get blue (which is right). So now you know it's in 3.
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On September 19 2006 09:52 Chill wrote:+ Show Spoiler + brick pushing down on boat displaces more water volume than volume of brick; water level falls
+ Show Spoiler + if it displaces more volume in the boat than out of the boat, why would the water level fall? following your logic I thought it'd rise 
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On September 19 2006 09:59 Ethenielle wrote:+ Show Spoiler + if it displaces more volume in the boat than out of the boat, why would the water level fall? following your logic I thought it'd rise
with my new found spoiler tagging abilities.... :D
+ Show Spoiler + brick is heavier/more dense than water. so whilst in the boat, it displaces the volume of water equal to that of the weight of the brick. when the brick is thrown in, the volume of the brick is displaced. so water level fall :D
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On September 19 2006 09:59 Ethenielle wrote:+ Show Spoiler + if it displaces more volume in the boat than out of the boat, why would the water level fall? following your logic I thought it'd rise
+ Show Spoiler +Spoiler chain.
Ask yourself this; What happens when you get into a bathtub that is nearly full, but not quite going over the edge?
Or, think of a ball that weighs fifty pounds, but is as golf-ball sized. It's not moving much water in it's space, no matter how much it weighs.It's pushing some water up, but not a whole lot. Now, imagine putting that fifty pound ball on a boat, which has a much wider surface area. More water gets pushed up, because of that weight being spread out across more of the water.
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A man is in a rowing boat floating on a lake, holding a brick in his hands. He throws the brick over the side of the boat. The brick sinks quickly. Does the water level in the lake go up or down? Explain.
Ok this is my attempted answer. Also is this really a riddle? It's more of a physics problem...
+ Show Spoiler +
Ok this is pretty simple - The brick displaces more water by displacing weight in the boat than displacing volume in the lake. Therefore the water level actually should drop a little when you throw the brick into the lake. Of course it's a LAKE so I doubt it makes any difference at all. Maybe if you were in a boat in a small tub. Lol.
That's the only one I can do, cause it's pretty simple physics. But the other would probably take me ages to figure out.
ok someone answered it while i was typing my answer... =_=
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I put that a little bit weird
2 again
+ Show Spoiler +. You ask the watch if the next time it flashes the colour will be blue. If the watch flashes the blue light you know blue is true. If you get the yellow light, you know yellow is false.
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day[9] will crack #2 in seconds
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badteeth. even if you know which one is which, you wont be able to find out which of the three is which for certain, it will be a 33% chance for you to get it correct. day is good with this sort of stuff.. hopefully he will come enlighten us all
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On September 19 2006 10:07 badteeth wrote:I put that a little bit weird 2 again + Show Spoiler +. You ask the watch if the next time it flashes the colour will be blue. If the watch flashes the blue light you know blue is true. If you get the yellow light, you know yellow is false.
you could just ask it if 2 + 2 = 4 and see which one flashes. But knowing which light is which doesn't solve the riddle.
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yeah u can find out if blue is right or false, but the next question cant decide then if u waste your first question like that... you can also find out if blue is right or wrong is u ask: Are there any horses behind either door 1 or 2... coz it will show yes. and then u know atleast something more but still not good enough
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On September 19 2006 09:27 123bangbang wrote:
Haha i got asked number 4, the brick in a boat thing for my university practice interview.
Erm...hmm someone tell me how to use a spoiler tag dammit!
edit - nevermind.. my post was rendered totally useless :[
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Ooooh, you're right. Seemed a bit too easy to me .
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#2 riddle
+ Show Spoiler +Hmm this seems very wierd but i cant see what is wrong.
You ask the watch:
If I ask you if there is a horse behind both door 1 and 2, will you then show blue?
Then it can show blue or yellow.
When you ask later if there is a horse behind both those doors it can show blue, and then blue will be yes wich means the car is in door 3. If the watch shows yellow, then it means that blue is NO and that means yellow must be yes, wich means there are 2 horses in those doors aswell and the car is once again in door 3.
It quite clear, but i can't understand it, coz then it means the car is always in door 3. And if u ask about door 2 and 3 then its always in door 1, so you can choose where its gonna be!!
Never mind, I found out what was wrong. If the watch shows yellow the first time, and yellow means no. Then it will show yellow when i ask the second question and the car can be in nr1 OR nr2 so u dont know for sure.. but im getting somewhere i can feel it^^
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Missing one thing for 6... :
+ Show Spoiler +
You make 4 groups of 3 marbles.
you weigh 2 groups.
a) The scale tilts
b) The scale does not tilt
a) take the hevier 3 marbles and weigh them with 3 you haven't weighted.
a1) tilts again -> You know in which pack of 3 the marble is and you know it's hevier.
Take the 3 marbles weigh 2 -> does'n tilt it's the last one
tilts -> take the hevier one
a2) doesn't tilt -> You know in which pack of 3 the marble is and you know it's lighter.
Take the 3 marbles weigh 2 -> does'n tilt it's the last one
tilts -> take the ligher one
b) take 3 of the marbles you havn't weighted and weigh them with 3 or the marbles you have
b1) it tilts -> you know in which pack of 3 the marble is and if it's lighter or hevier
Do like a1 or a2
b2) It doesn't tilt... -> You know in which pac of 3 the marble is but have no clue if it's hevier or lighter... And I'm stuck ;D
Only 82.5% solved :p
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regarding number 5
+ Show Spoiler +
there are 10 cheating men in the town.
if there is only 1 cheating man, then he will be kicked out the first night. the reason is that his wife knows that someone in the town is cheating, but she knows that no one else in the town is cheating, so it must be her husband.
if there is two cheating men, then the they will be both kicked out on day 2. the reason is that their wives know that one other husband is cheating, but he was not kicked out on the first night so they deduce that that wife must know the same thing.
and on and on until the 10th day or 10 men
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Calgary25980 Posts
On September 19 2006 10:04 MTF wrote:Show nested quote +On September 19 2006 09:59 Ethenielle wrote:On September 19 2006 09:52 Chill wrote:+ Show Spoiler + brick pushing down on boat displaces more water volume than volume of brick; water level falls + Show Spoiler + if it displaces more volume in the boat than out of the boat, why would the water level fall? following your logic I thought it'd rise + Show Spoiler +Spoiler chain.
Ask yourself this; What happens when you get into a bathtub that is nearly full, but not quite going over the edge?
Or, think of a ball that weighs fifty pounds, but is as golf-ball sized. It's not moving much water in it's space, no matter how much it weighs.It's pushing some water up, but not a whole lot. Now, imagine putting that fifty pound ball on a boat, which has a much wider surface area. More water gets pushed up, because of that weight being spread out across more of the water.
+ Show Spoiler + Yea. It's just a comparison of densities, right? If you have something denser than water in the boat, the water level will fall when you put it in the water. For example, if the boat was half filled with water, when you bailed all the water out, the water level would be the same. If you had something lighter than water, it takes up more volume than equally-weighing-water, so the water level would rise (except stuff lighter than water floats so you'd have to magically suspend it there somehow)
This seems more like a physics problem than a logic puzzle to me, that's why I think it's the weakest of the 8. I'd still like to hear the answer to the 10am cheating husbands one though, because I have no clue what the answer to that is.
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Here's my try at #2
A bit twisted but I think it's the only way
+ Show Spoiler +Messed up... need to yhink again
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On September 19 2006 10:38 KarlSberg~ wrote:Here's my try at #2 A bit twisted but I think it's the only way + Show Spoiler +does blue light mean "yes" today and is the car beghind door 1 or
does blue light mean "no" today and is the car behind door 2 or 3
All this is just 1 question (with priority given to "and" operator over "or", just classical boolean algebra)
It shows blue if the car is behind 1
If it doesn't, ask the same to know about door2, and you know all you need to know.
i haven't given your answer much thought, but if you are allowed to ask a 2 part question then I don't really care what the answer is :[
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Why should I not be allowed 2 parts questions?
There is only 1 answer to the whole question which is yes/no
I see nothing in the ridle that would forbid it
Anyway I got it wrong
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#4: Water level stays constant. First the water was being pushed up by the draught (?) of the boat containing the brick. Aftter the brick submerging the water is once again pushed by the empty boat and the brick in the water.
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karlsberg u still cant find out the right door that way... even if u can ask that
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On September 19 2006 10:44 Krzych wrote:
#4: Water level stays constant. First the water was being pushed up by the draught (?) of the boat containing the brick. Aftter the brick submerging the water is once again pushed by the empty boat and the brick in the water.
I'll give you a hint. The mass of the brick is greater than that of water since the brick actually sunk when tossed in, right?
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MTF, damn nice explanation, cheers.
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please everyone try solve #2 its so hard
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On September 19 2006 10:52 Luhh wrote:Show nested quote +On September 19 2006 10:44 Krzych wrote:
#4: Water level stays constant. First the water was being pushed up by the draught (?) of the boat containing the brick. Aftter the brick submerging the water is once again pushed by the empty boat and the brick in the water. I'll give you a hint. The mass of the brick is greater than that of water since the brick actually sunk when tossed in, right?
Yeah, I have just realized this and was to edit my post, but I saw yours. Thanks 
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2.
Best i can do is a car 66% of the time hehe.. If i was that guy i would just go with that lol.
This is really hard..
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btw, you are allowed to ask 2 part questions on #2
day brought up this thing in another thread it was like a 3 ghosts question... one is random, one truth, one lie
and uh... he solved it using something like, is my name sean if and only if road 1 is ...
or whatever
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#4 This one is quite clever, but I think this is right:
+ Show Spoiler + It's a trick question, it stays constant. The water level is dependent on the amount of water being displaced. The amount of water displaced is proportional to the weight of the object. (Since ice has a density of 7/8 of seawater or something, only 1/8 of an iceberg is visible). Thus, since the total weight (boat + person + brick) remains unchanged, the water level is also unchanged.
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I'VE SOLVED 2 ANSWER IN SPOILER AND AS MUCH EXPLANATION AS I CAN FIT IN 10 MINS
+ Show Spoiler +
Q1 Are there exactly 2 goats in 1/2
Q2 Is there exactly 1 goat in 2/3
If you get a Blue Blue lights (or yellow yellow irrelevant) they can either mean yes or no. If they mean Yes Yes then the car must be 3. If it's a no no then there must be 1 or 0 goats in 1/2 and 2 or 0 goats in 2/3. If 0 in 1/2 there must be 2 in 2/3 so there must be 1 in 1/2. Impossible. If there's 1 goat in 1/2 it must be either goat 1 or 2. Regardless of the goat there then must be 1 in 2/3.
SO THERE CAN'T BE NO NO IT MUST BE YES YES
Now for BY or YB (doesn't matter)
If it's a yes to 1 and a no to 2 there must be 2 in 1/2. So there must be a 2 in 2/3 SO THERE CAN'T BE A YES NO. It must be no yes.
If it's a no to 1 then there must be 0 or 1 goats in 1/2. If it's 0, there's 2 in 2/3 and thus 1 in 1/2. THERE CAN'T BE 0 thusly. If there's 1 it must be either 1/2. If it's 2, then 3 must be the other goat, and we have 2 and 3 being goats thus 2 goats in 2/3 and impossible. If it's a 1, then the other one can't be 2 so the other goat must be 3 and thus car is 2
A bit might be off because of fast typing in thinking but that's it. I'm so proud!
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Regarding the yes/no watch riddle.
You ask if you're gonna win the car and then ask if there is a car behind gate #1. If you get the same answer (yes/yes or no/no) just choose any other gate and you have a 50% chance of proving this bastard to lie! If you get different answers just pick #1. The watch is fooled once again.
Yes, I know this is not a solution
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On September 19 2006 09:51 Ethenielle wrote:haha didn't remember the yes/no thing anyway Pacifist already picked it out, so I think you can safely delete your post, Fen. you pick two doors and ask if there are goats behind them, you get either blue or yellow. let's say you pick the two outermost doors. next up you pick either of those doors and the middle one, and ask if there are goats behind them. you can't get yes and yes, because then there would be three goats. you can get yes and no, then the car will be in the middle door. you can also get no and yes, and you can also get no and no, in which case the car must be in the door you didn't pick the second time around. thus: if you get blue/blue it can't be yes/yes, it must be no/no. same goes for yellow, of course. if you get a no/no, problem is pretty much solved. the car must be in the door you picked, because there will be a goat in the middle door. and a "no" gives us that there is a goat and a car behind the doors - car is in the door you picked, goat in the middle and goat in the one you didn't pick. now if you get a blue/yellow or yellow/blue, it's either a yes/no or a no/yes in both cases. in case blue/yellow is a yes/no, the car is in the middle door. in case it's a no/yes, the car is in the door you didn't pick. but all this doesn't really help either way, because you still can't know if blue/yellow is yes/no or no/yes.
exactly what i was thinking, didn't realize the blue/yellow though =(
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On September 19 2006 11:35 -_- wrote:I'VE SOLVED 2 ANSWER IN SPOILER AND AS MUCH EXPLANATION AS I CAN FIT IN 10 MINS + Show Spoiler +
Q1 Are there exactly 2 goats in 1/2
Q2 Is there exactly 1 goat in 2/3
If you get a Blue Blue lights (or yellow yellow irrelevant) they can either mean yes or no. If they mean Yes Yes then the car must be 3. If it's a no no then there must be 1 or 0 goats in 1/2 and 2 or 0 goats in 2/3. If 0 in 1/2 there must be 2 in 2/3 so there must be 1 in 1/2. Impossible. If there's 1 goat in 1/2 it must be either goat 1 or 2. Regardless of the goat there then must be 1 in 2/3.
SO THERE CAN'T BE NO NO IT MUST BE YES YES
Now for BY or YB (doesn't matter)
If it's a yes to 1 and a no to 2 there must be 2 in 1/2. So there must be a 2 in 2/3 SO THERE CAN'T BE A YES NO. It must be no yes.
If it's a no to 1 then there must be 0 or 1 goats in 1/2. If it's 0, there's 2 in 2/3 and thus 1 in 1/2. THERE CAN'T BE 0 thusly. If there's 1 it must be either 1/2. If it's 2, then 3 must be the other goat, and we have 2 and 3 being goats thus 2 goats in 2/3 and impossible. If it's a 1, then the other one can't be 2 so the other goat must be 3 and thus car is 2
A bit might be off because of fast typing in thinking but that's it. I'm so proud!
+ Show Spoiler +sorry.. there can be a no no (car in 1, goat in 2, goat in 3 - 1 goat in 1/2, 2 goats in 2/3)
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2.
I think there will always be the yes/no no/yes dillema. So you gota phrase a question so that you get one relevent information along with the color code.
Then again, i may be going about this all wrong
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#6 is a classic one, which is a right bitch to explain.
+ Show Spoiler +Break up the marbles into 3 groups of 4: group 1, 2, and 3.
1st weighing: Find weight of group 1 compared to group 2. Two cases can occur: either group 1's weight is not equal to group 2's, or group 1's weight and group 2's weight are the same, and therefore the marble is in group 3.
If group 1 = group 2, then their weight is known. You can then take one marble from group 1 (which you know is normal, which I'll call a), and 3 marbles from group 3(b, c, d). 2nd weighing: a+b vs c+d. If they are the same, the unknown marble is the remaining marble from group 3, and you can weigh this against a to find heavier/lighter.
If they are different: the remaining marble is a normal marble, which means that either b, c, or d is the different one. 3rd weighing: c vs d. If c = d, then b was different, and you know if it's heavier depending on the 2nd weighing ab> cd or ab < cd. If c != d, (which means b is normal) then you again know which one was different; if ab > cd, then the lighter one in the 3rd weighing (c or d), is the different one. If ab < cd, then the heavier one in the 3rd weighing is the different one.
NOW FOR THE OTHER HALF (and more) OF THE SOLUTION
Back to the beginning! 1st weighing: group 1 vs group 2, and they were DIFFFERENT. So, group 3 had normal marbles. I'm going to say there are marbles a, b, c, and m in each group. The 2nd weighing will take one marble from each group (1m, 2m, 3m) vs TWO marbles from the first group (1a, 1b) and one marble from the second group (2a). 3m is normal. So, 1m2m3m vs 1a1b2a. If 1m2m3m = 1a1b2a, that means the different marble is in the remaining marbles of groups 1 and 2; 1c, 2b, or 2c.
3rd weighing: 2b vs 2c. if they are same, 1c was the different marble, and the weight heavier/lighter is known from the first weighing (1a1b1c1m vs 2a2b2c2m). If 2b != 2c, then again, heavier/lighter is determined by first weighing (normal group 1a1b1c1m vs diff group 2b2c2m heavier/lighter) and which one of 2b/2c was heavier/lighter.
Okay. 1st weighing, group 1 vs group 2, different weights. 2nd weighing, 1m2m3m vs 1a1b2a, now DIFFERENT weights. Since 3m is normal, that means different marble is among 1m, 2m, 1a, 1b, 2a. 4 different scenarios here.
#1: group 1 > group 2 (1a1b1c1m > 2a2b2c2m), and 1m2m3m > 1a1b2a. 2a, or 1m are the different ones. This is found by crossing out each one that does not fit. 2m can't be the one, because it'll mean 2m is heavier, and in the first weighing, group 1 was heavier. 3m can't, because it's normal. 1a can't, because in 1st weighing, 1a's group was heavier. Same with 1b. So, 2a or 1m. Pick one (2a) and weigh it against normal 3m. If same, different was 1m. If different, well, 2a's the different one then. 2a would be lighter; 1m heavier.
#2: group 1 > group 2 (1a1b1c1m > 2a2b2c2m), 1m2m3m < 1a1b2a. Crossing out choices 3m and 1m (1m is larger in 1st weighing but smaller the next), leaves 1a, 1b, and 2m. 2m, if different, is lighter. But also, 1a or 1b could be the different one, in which case it's heavier. 3rd weighing just compares 1a and 1b; the heavier one is different, if both the same, then 2m must have been lighter.
#3: 1a1b1c1m < 2a2b2c2m, 1m2m3m > 1a1b2a. 2m is heavy, or 1a or 1b is light. Again, compare 1a and 1b in the 3rd weighing to find out the lighter, and so different one. If same, 2m is the heavy, different one.
#4: 1a1b1c1m < 2a2b2c2m, 1m2m3m < 1a1b2a. Similar to #1, either 2a is heavy or 1m is lighter. Compare 2a to known 3m in the 3rd weighing; if 2a is heavier, then it's the different one. If 2a = 3m, then 1m is lighter and different.
This should be it I think :V
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#3:
+ Show Spoiler + Turn one of the switches on for awhile (5 minutes or so should be ok), then turn that switch off and turn another one on. Now go to the other room. Obviously you know which switch controls the one that is lighted up. Now feel the other two bulbs. The one that is warmer than the other bulb is the one that you turned on originally. So now you know all three.
damn, already solved, somehow I missed the 2nd post >< all well.
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Hahah, I love No 5.
+ Show Spoiler +It's really cool, although I knew it in another version. The answer is 10
I guess you may wanna know how to solve it. Well..
Let us assume there is only one man cheating on his wife. That wife saw none of the other men to cheat on their wives, yet she hears the stranger say yes. It's obvious that her husband is the only cheater, so she will kick him out.
Case no2: 2 cheaters. Wife 1 will see wife 2 being cheated and will wait for her to kick her husband out. Wife 2 will do the same. However, when Wife 1 sees wife 2 not doing anything she will know that wife 2 is waiting for wife 1 to kick his husband out. Same with wife 2. So on the second day 2 men will be kicked out.
Case no3: 3 cheaters. Following the same line of thought as in case 2, Wife 3 will wait for wife 1 and 2 to kick their husbands out on the second day. Since that won't happen, since w1 and w2 will be doing the same, she will know that she is being cheated on. On the third day, 3 men are kicked out.
Case no10: 10 men are kicked out on the 10th day.
As for the boat and brick one:
+ Show Spoiler +it stays the same, because of Archimede's Law, like someone else said before. ^^
LOL@ the 100 bulbs one. I solved it during a Math's Contest when I was like 13? Can't remember the exact solution, but I'll try to resolve it.
+ Show Spoiler +Obviously no1 stays on. This is the exception.
Now, it's all about the number of dividers (except 1 and the no itself) that the no of the bulb has.
If it has no dividers, as in being prime, it's off, because it's state changes 2 times, once for the first person and again for the nth person, n=bulb no.
If the number has an odd number of dividers, it stays on./ The only question now is how to find all the numbers up to 100 that have an odd number of dividers. Taking each one and checking the divider's number is too time consuming, so, we'll have to go around this somehow.
Now, we can take all the possible forms a number can take when broken down in it's prime factors:
x*y means it can divide by x and y, not good.
x^2 only 1 divider, x, meaning: 4, 9, 25, 49.
x*y*z it can divide by x, y, z, x*y, x*z, y*x, not good.
x^2*y it has 4 dividers: x, y, x*y, x^2, not good.
x^3 has as dividers x, x^2, not good.
x*y*z*w has x,y,z,w,x*y,x*z,x*w,y*z,y*w,z*w, x*y*z, x*y*w, x*z*w, y*z*w, not good.
x^2*y*z has x, y, z, x^2, x*y, x*z, x^2*y, x^2*z, 7 dividers, which is good. This means: 60, 84, 90.
x^3*y has x, y, x^2, x*y, x^3, x^2*y, not good.
x^2*y^2 has x, y, x*y, x^2, y^2, x^2*y, x*y^2 7 dividers, meaning: 36, 100.
x^4 has x, x^2, x^3, meaning: 16, 81.
x^5 has x, x^2, x^3, x^4, not good.
x^4*y has x, y, x*y, x^2, x^2*y, x^3, x^3*y, x^4, 8 dividers ftw.
x^3*y^2 has x, y, x*y, x^2, y^2, x^3, x^2*y, y^2*x, x^3*y, x^2*y^2 10 dividers. Can't go on, because the min value that x^3*y*z can take is 2^3*3*5=120>100
x^6 has x, x^2, x^3, x^4, x^5: 64
x^5*y has 10 dividers, gay...
Can't go on, because the lowest value that x^4*y^3 can take is 16*9>100
From x^7 can't go on, because it's lowest value is 2^7>100.
So, the bulbs that are on: 4, 9, 25, 49, 60, 84, 90, 16, 81, 64.
Well, taht was time consuming aswell, pretty sure you can work around that aswell. I might have messed up aswell at some point.
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I have a question for #6. Do you get to keep track of individual marbles when you group them? Or do they get mixed up when you put em in a group.
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Watch riddle solution(aka #2):
+ Show Spoiler +Q1: If I ask is there a car behind 1 will you light in blue? Q2: If I ask is there a car behind 3 will you light in blue? This way the answers will always be the same no matter which color means yes and which no. Car at 1 - Blue, Yellow Car at 2 - Yellow, Yellow Car at 3 - Yellow, Blue Edit: Blue, Blue - There are 2 cars behind 1 and 3 )) Ofc, it could be for cars at 1/2 or 2/3 or goats behind 2 door combinations or yellow instead of blue, it doesn't matter.
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#2
+ Show Spoiler +
Ask this:
Would you flash yellow if i asked you if there was a car behind door 1?
Would you flash yellow if i asked you if there was a car behind door 2?
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Asking it if it will flash a certain way sounds kind of retarded and will probably just end up creating a paradox that it cannot answer.
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Braavos36374 Posts
Two quarters are randomly flipped in the air.
If you had to choose whether they land "same" (heads/heads or tails/tails) or "different" (heads/tails) which would you choose and why?
PS This is a pretty easy riddle.
answer below:
+ Show Spoiler +"same," because at worst, "same" is 50%, while variantion in anything, wind, coin, whatever, can cause "different" to be less than 50%
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On September 19 2006 12:45 BlackJack wrote:
Asking it if it will flash a certain way sounds kind of retarded and will probably just end up creating a paradox that it cannot answer.
It's no paradox, you're not asking it what will flash for your current question, but a specific simple question, so there is a sure answer.
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On September 19 2006 12:45 Hot_Bid wrote:
Two quarters are randomly flipped in the air.
If you had to choose whether they land "same" (heads/heads or tails/tails) or "different" (heads/tails) which would you choose and why?
PS This is a pretty easy riddle.
+ Show Spoiler + doesn't matter. There are 4 possible outcomes:
1- heads, heads
2- heads, tails
3- tails, heads
4- tails, tails
so you have a 1/2 chance of getting the same, and a 1/2 chance of getting different.
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lololol u edited your question to my answer >:O why why why^^ coz im right
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Braavos36374 Posts
to AcrossFiveJulys:
+ Show Spoiler +it does matter, as you are not accounting for the possibility that the coins are not perfect
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On September 19 2006 12:52 AnOth3rDAy wrote:
lololol u edited your question to my answer >:O why why why^^ coz im right
I'm not sure what you mean, because we wrote the same answers, I was just first, you could've just copied and changed it a bit to look convincing
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I remembered your answer differently, and when i updated it was Edited.. :O anyway i dont care, i hated that riddle took me 3 hours to figure out, my day is ruined. thnx alot mr riddlerer .!._Ò.ó_.!.
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#2
+ Show Spoiler +We want a set of questions where there are three unique answers that are independant of the watches settings.
Consider the questions: Is the Car Behind door 1 or 2? and Is the Car Behind door 2 or 3?
C = car, G = goat. Doors are presented in order from 1 to 3
Case:
CGG Yes, No
GCG Yes, Yes
GGC No, Yes
Now, for our questions to ask the watch, use the following:
Is the answer to the question "Is the car behind door 1 or 2?" blue?
Is the answer to the question "Is the car behind door 2 or 3?" blue?
Blue = Yes, Yellow = No
cgg Blue, Yellow
gcg Blue, Blue
ggc Yellow, Blue
Blue = No, Yellow = Yes
cgg Blue, Yellow
gcg Yellow, Yellow
ggc Yellow, Blue
We get the same results irregardless of the truth values of the watch. If Blue, Yellow pick door 1, If the answers are the same pick door 2, If Yellow, Blue pick door 3.
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On September 19 2006 12:45 Hot_Bid wrote:
Two quarters are randomly flipped in the air.
If you had to choose whether they land "same" (heads/heads or tails/tails) or "different" (heads/tails) which would you choose and why?
PS This is a pretty easy riddle.
+ Show Spoiler +if you had to choose... well, i'd take the heads/heads, mainly because the sky is green and the sun is blue.
+ Show Spoiler +fun with a thread where you can actually abuse the spoiler tags without going offtopic
I remembered your answer differently, and when i updated it was Edited.. :O anyway i dont care, i hated that riddle took me 3 hours to figure out, my day is ruined. thnx alot mr riddlerer .!._Ò.ó_.!.
+ Show Spoiler + you don't care but your day is ruined? :/ what a paradox!
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Why everyone started posting answers to #2 right after me posting it?
All riddles are now solved
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On September 19 2006 12:55 AnOth3rDAy wrote:
I remembered your answer differently, and when i updated it was Edited.. :O anyway i dont care, i hated that riddle took me 3 hours to figure out, my day is ruined. thnx alot mr riddlerer .!._Ò.ó_.!.
I just changed the wording a bit and cleaned it up, because my english needs some work The question was for a car at one and a goat at 2, but it didn't need the second part anyway, it was part of my previous attempt at finding solution, so I removed it, because it doesn't matter.
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Quarter answer
+ Show Spoiler +It will most likely be Heads/tails or Tails/head H = heads T = tails also | means arrow down and in the second throw the first two arrow downs are from the "H" in the first throw and the 3rd and 4th arrow down is from the first throws "T".. its so damn hard to sketch something First throw | .... | H....T Second throw |...|..|..| H T H T you see that there are 4 possible outcomes now. it can be HH HT TH or TT. So HH has only 25% chanse and TT has also only 25% chanse.. this means TH and HT has 25% each and together they ahve 50%...
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On September 19 2006 12:45 BlackJack wrote:
Asking it if it will flash a certain way sounds kind of retarded and will probably just end up creating a paradox that it cannot answer.
So what, you won't think about whether it works but instead choose to ridicule people who are correct when you're too dumb to figure it out yourself?
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On September 19 2006 12:45 Hot_Bid wrote:Two quarters are randomly flipped in the air. If you had to choose whether they land "same" (heads/heads or tails/tails) or "different" (heads/tails) which would you choose and why? PS This is a pretty easy riddle. answer below: + Show Spoiler +"same," because at worst, "same" is 50%, while variantion in anything, wind, coin, whatever, can cause "different" to be less than 50%
+ Show Spoiler +Don't you mean the other way around? The same will be less than the different, because one coin may land in the extreme on it's edge, albeit they're both still close to 50/50.
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Calgary25980 Posts
Hot_Bid, your logic is quite flawed.
First off, you didn't say the quarters are identical, so one could have a bias to heads while the other to tails. Because the bias could be either way on either coin, there's no reason to even consider this.
Secondly, you say they are randomly flipped, but then you bring up factors like wind. So, say for the average height of a flipper, this wind is causing a large distribution of 29 rotations. So what? They were randomly flipped, so if they started on heads, they're more likely to land on tails. Because they are randomly flipped, there's a 50% chance they will start in either orientation, and therefore your wind bias makes no difference.
So we have coins with random bias starting in a random orientation. No matter what outside bias you put on that system, it's going to be random. Your answer of "same" doesn't make sense to me at all.
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Braavos36374 Posts
i meant not perfect as in heads or tails can be favored due to the grooves in the coin
they are both quarters and are identical coins, its just that not heads or tails is necessarily 50% chance to land
sorry if i was unclear...
if all quarters were uniform, but you were unsure which side (heads or tails) lands more often, "same" would be correct
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4. A man is in a rowing boat floating on a lake, holding a brick in his hands. He throws the brick over the side of the boat. The brick sinks quickly. Does the water level in the lake go up or down? Explain.
+ Show Spoiler +The water level of the lake does not go up or down. The brick going into the water does not displace any more water than it did when being supported by the canoe.
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Hot Bid.. Im right aint I? its 25% chanse to get head head or Tail tail and its 50% chanse to get tail head or head tail..
Wich means its easier to get 2 different sides faced up than for example head head.
But if u say, the chanse of getting head+head OR tails+tails, then u sum up those 25% and u get 50% aswell...
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Korea (South)17174 Posts
On September 19 2006 12:57 LTT wrote:#2 + Show Spoiler +We want a set of questions where there are three unique answers that are independant of the watches settings.
Consider the questions: Is the Car Behind door 1 or 2? and Is the Car Behind door 2 or 3?
C = car, G = goat. Doors are presented in order from 1 to 3
Case:
CGG Yes, No
GCG Yes, Yes
GGC No, Yes
Now, for our questions to ask the watch, use the following:
Is the answer to the question "Is the car behind door 1 or 2?" blue?
Is the answer to the question "Is the car behind door 2 or 3?" blue?
Blue = Yes, Yellow = No
cgg Blue, Yellow
gcg Blue, Blue
ggc Yellow, Blue
Blue = No, Yellow = Yes
cgg Blue, Yellow
gcg Yellow, Yellow
ggc Yellow, Blue
We get the same results irregardless of the truth values of the watch. If Blue, Yellow pick door 1, If the answers are the same pick door 2, If Yellow, Blue pick door 3.
No. This still leaves you with a 50/50 cause you won't know if it's door one or three considering the fact that you don't know which light means no or yes.
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GrandInquisitor
New York City13113 Posts
Wow, all the riddles have been solved! =D
Mad props to those who got #2 - the hardest part about it is that the solution:
+ Show Spoiler +Would you flash yellow if i asked you if there was a car behind door 1?
Would you flash yellow if i asked you if there was a car behind door 2?
I picked this cause it was best phrasing
isn't a EUREKA solution at all
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On September 19 2006 15:14 Rekrul wrote:Show nested quote +On September 19 2006 12:57 LTT wrote:#2 + Show Spoiler +We want a set of questions where there are three unique answers that are independant of the watches settings.
Consider the questions: Is the Car Behind door 1 or 2? and Is the Car Behind door 2 or 3?
C = car, G = goat. Doors are presented in order from 1 to 3
Case:
CGG Yes, No
GCG Yes, Yes
GGC No, Yes
Now, for our questions to ask the watch, use the following:
Is the answer to the question "Is the car behind door 1 or 2?" blue?
Is the answer to the question "Is the car behind door 2 or 3?" blue?
Blue = Yes, Yellow = No
cgg Blue, Yellow
gcg Blue, Blue
ggc Yellow, Blue
Blue = No, Yellow = Yes
cgg Blue, Yellow
gcg Yellow, Yellow
ggc Yellow, Blue
We get the same results irregardless of the truth values of the watch. If Blue, Yellow pick door 1, If the answers are the same pick door 2, If Yellow, Blue pick door 3. No. This still leaves you with a 50/50 cause you won't know if it's door one or three considering the fact that you don't know which light means no or yes.
Rek, look at the cases. You don't need to know which means yes or no because the cases are the exact same for each situation. The only thing that is different is the blue, blue or yellow, yellow outcome, but we just check to see if the color is the same. The actual color doesn;t matter in that case.
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On September 19 2006 12:57 LTT wrote:#2 + Show Spoiler +We want a set of questions where there are three unique answers that are independant of the watches settings.
Consider the questions: Is the Car Behind door 1 or 2? and Is the Car Behind door 2 or 3?
C = car, G = goat. Doors are presented in order from 1 to 3
Case:
CGG Yes, No
GCG Yes, Yes
GGC No, Yes
Now, for our questions to ask the watch, use the following:
Is the answer to the question "Is the car behind door 1 or 2?" blue?
Is the answer to the question "Is the car behind door 2 or 3?" blue?
Blue = Yes, Yellow = No
cgg Blue, Yellow
gcg Blue, Blue
ggc Yellow, Blue
Blue = No, Yellow = Yes
cgg Blue, Yellow
gcg Yellow, Yellow
ggc Yellow, Blue
We get the same results irregardless of the truth values of the watch. If Blue, Yellow pick door 1, If the answers are the same pick door 2, If Yellow, Blue pick door 3.
If you ask it "Is the answer to this question blue?" And blue = no, how can it respond with a blue light? It's contradictory, no?
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On September 19 2006 15:23 BlackJack wrote:Show nested quote +On September 19 2006 12:57 LTT wrote:#2 + Show Spoiler +We want a set of questions where there are three unique answers that are independant of the watches settings.
Consider the questions: Is the Car Behind door 1 or 2? and Is the Car Behind door 2 or 3?
C = car, G = goat. Doors are presented in order from 1 to 3
Case:
CGG Yes, No
GCG Yes, Yes
GGC No, Yes
Now, for our questions to ask the watch, use the following:
Is the answer to the question "Is the car behind door 1 or 2?" blue?
Is the answer to the question "Is the car behind door 2 or 3?" blue?
Blue = Yes, Yellow = No
cgg Blue, Yellow
gcg Blue, Blue
ggc Yellow, Blue
Blue = No, Yellow = Yes
cgg Blue, Yellow
gcg Yellow, Yellow
ggc Yellow, Blue
We get the same results irregardless of the truth values of the watch. If Blue, Yellow pick door 1, If the answers are the same pick door 2, If Yellow, Blue pick door 3. If you ask it "Is the answer to this question blue?" And blue = no, how can it respond with a blue light? It's contradictory, no?
It isn't. You just have to start from the inside and work your way out.
Take the CGG case. Is the car behind door number 1 or 2? Yes. What is yes? Yellow. Is the answer blue? No. What is no? Blue. Done.
Edit: To clarify, it isn't asking "Is the answer to this question blue?" It's asking "Is the answer to this other question blue?"
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Blue = no.
You ask it "Is the answer blue?"
It flashes a blue light.
That's not a paradox to you? Telling you the answer is not a blue light by giving you a blue light?
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No, because the answer to the other question is yellow. You need to think of the question being referenced independantly of the referencial question.
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anyone solved 8?
i think i've got a solution...
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Calgary25980 Posts
On September 19 2006 14:18 rpf289 wrote:Show nested quote +
4. A man is in a rowing boat floating on a lake, holding a brick in his hands. He throws the brick over the side of the boat. The brick sinks quickly. Does the water level in the lake go up or down? Explain.
+ Show Spoiler +The water level of the lake does not go up or down. The brick going into the water does not displace any more water than it did when being supported by the canoe.
This has been solved like 9 on pages 2-4. Your solution is wrong.
I'm also discouraged by the people who can't do simple probabilities that you learn in grade 10.
You flip two coins, here are the probable outcomes:
(HH, HT, TH, TT).
So for that guy who keeps saying that it's 25% same and 50% other (which doesn't even make sense in itself because that doesn't sume to 1), please stop.
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how can i do a fucking spoiler? 8[
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got it
okay concerning riddle 8
+ Show Spoiler +
just eat the pills out of the boxes or whatever
ignore the three you have in your hand and put them somewhere else
so after a certain amount of days, in your boxes will be 1 pill left
either a or be no matter
that makes 4 pills you have left
cut each pill excaktly into a half
eat 1 half of each pill
do the same thing the next day
voila
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On September 19 2006 11:44 Dametri wrote:Show nested quote +On September 19 2006 11:35 -_- wrote:I'VE SOLVED 2 ANSWER IN SPOILER AND AS MUCH EXPLANATION AS I CAN FIT IN 10 MINS + Show Spoiler +
Q1 Are there exactly 2 goats in 1/2
Q2 Is there exactly 1 goat in 2/3
If you get a Blue Blue lights (or yellow yellow irrelevant) they can either mean yes or no. If they mean Yes Yes then the car must be 3. If it's a no no then there must be 1 or 0 goats in 1/2 and 2 or 0 goats in 2/3. If 0 in 1/2 there must be 2 in 2/3 so there must be 1 in 1/2. Impossible. If there's 1 goat in 1/2 it must be either goat 1 or 2. Regardless of the goat there then must be 1 in 2/3.
SO THERE CAN'T BE NO NO IT MUST BE YES YES
Now for BY or YB (doesn't matter)
If it's a yes to 1 and a no to 2 there must be 2 in 1/2. So there must be a 2 in 2/3 SO THERE CAN'T BE A YES NO. It must be no yes.
If it's a no to 1 then there must be 0 or 1 goats in 1/2. If it's 0, there's 2 in 2/3 and thus 1 in 1/2. THERE CAN'T BE 0 thusly. If there's 1 it must be either 1/2. If it's 2, then 3 must be the other goat, and we have 2 and 3 being goats thus 2 goats in 2/3 and impossible. If it's a 1, then the other one can't be 2 so the other goat must be 3 and thus car is 2
A bit might be off because of fast typing in thinking but that's it. I'm so proud! + Show Spoiler +sorry.. there can be a no no (car in 1, goat in 2, goat in 3 - 1 goat in 1/2, 2 goats in 2/3)
Nope. I am very sure of my solution. (heh not really but you didn't prove me wrong)
If there is one goat in 1/2 and 2 in 2/3 then the answer to one is no and the answer to 2 is no. The no no situation I described. So there there must be 1/2 goat in one or 0 goats in 1/2. If there's 0 goats, then then there must be 2 in 2/3 which means there's 1 in 1/2 and thus impossible. If there's 1, it can be either 1 or 2. If it's 1, HERE'S WHERE I LOST YOU then there CAN ONLY BE 1 goat in 2/3. IF there is only one goat in 2/3 THEN THE ANSWER TO SECOND QUESTION MUST BE YES. So the whole thing is impossible. There is no NO NO (wrong wrong)
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Basically I'm saying you cannot have it flash blue or yellow twice and be wrong twice.
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wtf hotbid, your riddle is messed up lol... maybe you didn't think about the answer, but it really will make no difference what side you pick, no matter what conditions you consider, because they will affect each random possiblity equaly in probability.
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dear -_-,
i think your's is the nicest answer but wrong:
car in door 1
Q1 Are there exactly 2 goats in 1/2 : NO
Q2 Is there exactly 1 goat in 2/3 : NO
say blue blue
car in door 2
Q1 Are there exactly 2 goats in 1/2 : NO
Q2 Is there exactly 1 goat in 2/3 : YES
say blue yellow
car in door 3
Q1 Are there exactly 2 goats in 1/2 : YES
Q2 Is there exactly 1 goat in 2/3 : YES
say blue blue
so it can't tell the diffrence between 1 and 3
Even if i'm wrong you've only got 2 light combos which can at most identify 2 doors, what if it's behind the 3rd?
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My answer is by far the nicest, but alas I have seen my folly. I Admit... DEFEAT.
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hey, post the answers on the first post because certain people have trouble understanding which one is correct...
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On September 19 2006 15:14 Rekrul wrote:Show nested quote +On September 19 2006 12:57 LTT wrote:#2 + Show Spoiler +We want a set of questions where there are three unique answers that are independant of the watches settings.
Consider the questions: Is the Car Behind door 1 or 2? and Is the Car Behind door 2 or 3?
C = car, G = goat. Doors are presented in order from 1 to 3
Case:
CGG Yes, No
GCG Yes, Yes
GGC No, Yes
Now, for our questions to ask the watch, use the following:
Is the answer to the question "Is the car behind door 1 or 2?" blue?
Is the answer to the question "Is the car behind door 2 or 3?" blue?
Blue = Yes, Yellow = No
cgg Blue, Yellow
gcg Blue, Blue
ggc Yellow, Blue
Blue = No, Yellow = Yes
cgg Blue, Yellow
gcg Yellow, Yellow
ggc Yellow, Blue
We get the same results irregardless of the truth values of the watch. If Blue, Yellow pick door 1, If the answers are the same pick door 2, If Yellow, Blue pick door 3. No. This still leaves you with a 50/50 cause you won't know if it's door one or three considering the fact that you don't know which light means no or yes.
This answer is wrong. You don't know which color means yes. The only way to determine which door it is according to your answer is if the car is in the middle door and you get the same light twice.
"""We get the same results irregardless of the truth values of the watch. If Blue, Yellow pick door 1, If the answers are the same pick door 2, If Yellow, Blue pick door 3."""
Since when did Blue and Yellow mean yes and no.. Lol, Switch the values of blue and yellow and your answers are all wrong. The Blue Blue or Yellow Yellow works tho. It would be in the middle.
Edit: If it tells you if the answer is blue, and it says blue it might mean no. If it tells you blue for the 2nd it might mean yes leaving door 3 open, so I'm actually wrong above.. I'm just thinking and thinking and its not making sense PLZ someone explain the answer to me if im wrong or someone at least agree with me haha im really tired.
has anyone heard this b4? this riddle?
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On September 19 2006 18:05 JoopeyDoop wrote:Show nested quote +On September 19 2006 15:14 Rekrul wrote:On September 19 2006 12:57 LTT wrote:#2 + Show Spoiler +We want a set of questions where there are three unique answers that are independant of the watches settings.
Consider the questions: Is the Car Behind door 1 or 2? and Is the Car Behind door 2 or 3?
C = car, G = goat. Doors are presented in order from 1 to 3
Case:
CGG Yes, No
GCG Yes, Yes
GGC No, Yes
Now, for our questions to ask the watch, use the following:
Is the answer to the question "Is the car behind door 1 or 2?" blue?
Is the answer to the question "Is the car behind door 2 or 3?" blue?
Blue = Yes, Yellow = No
cgg Blue, Yellow
gcg Blue, Blue
ggc Yellow, Blue
Blue = No, Yellow = Yes
cgg Blue, Yellow
gcg Yellow, Yellow
ggc Yellow, Blue
We get the same results irregardless of the truth values of the watch. If Blue, Yellow pick door 1, If the answers are the same pick door 2, If Yellow, Blue pick door 3. No. This still leaves you with a 50/50 cause you won't know if it's door one or three considering the fact that you don't know which light means no or yes. This answer is wrong. You don't know which color means yes. The only way to determine which door it is according to your answer is if the car is in the middle door and you get the same light twice. """We get the same results irregardless of the truth values of the watch. If Blue, Yellow pick door 1, If the answers are the same pick door 2, If Yellow, Blue pick door 3.""" Since when did Blue and Yellow mean yes and no.. Lol, Switch the values of blue and yellow and your answers are all wrong. The Blue Blue or Yellow Yellow works tho. It would be in the middle.
You need to read my post again. I covered both cases. Either blue means yes and yellow means no, or yellow means yes and blue means no. That's two cases. The car can be behind door number 1, door number 2, or door number 3. That's three possibilities for each of the two possible truth values for the watch responses. Here's the chart I originally posted:
Blue = Yes, Yellow = No
cgg Blue, Yellow
gcg Blue, Blue
ggc Yellow, Blue
Blue = No, Yellow = Yes
cgg Blue, Yellow
gcg Yellow, Yellow
ggc Yellow, Blue
Read it as follows:
Possible Truth Values
Door orientation, Watches answer to question one, watches answer to question two.
Notice the questions are crafted in such a way that the watch responses are unique to each door, but the same regardless of which color means yes or no.
If you still don't follow, I can write out every case.
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On September 19 2006 08:31 Pacifist wrote:Show nested quote +On September 19 2006 08:24 Ethenielle wrote:On September 19 2006 08:05 GrandInquisitor wrote:
2. You have a very nice, shiny watch. But this is no ordinary watch. This watch can answer two yes or no questions 100% accurately per day. You ask, and either a blue or yellow light will flash. Unfortunately, you don't know which light means yes and which light means no, and you can never find out because there's a 50-50 chance of the lights switching every night.
You happen to be on a game show. The rules are simple. There are three doors. Behind two are goats. Behind one is a shiny red car. Obviously, you want the car. You can choose any door, after asking the watch two questions.
What are the two yes/no questions you can ask so that you will have a 100% guaranteed chance of getting the car? + Show Spoiler +I'm not entirely sure if it's allowed, but you could just ask "are there goats behind these two doors?", if you get a yes, then it's the third door, if you get a no you can ask "is there a goat behind this door(= one of the two doors you chose)", if no then it's the car, if yes then it's the other. If you can't ask about two doors at once, though.. + Show Spoiler +You don't know which light means yes and which light means no.
[spoiler] 1. Ask is there goat in door 1, 2? If you get blue, assume it means no. 2. Ask if there is goat in door 3. If you get blue, that means car is door 3. If you get yellow, then that means blue = yes. Thus, goat is in door 3.[spoiler]
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lol im havin trouble followin, i dont want you to spell it for me but if u did id love u
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1. Ask is there goat in door 1, 2? If you get blue, assume it means no. 2. Ask if there is goat in door 3. If you get blue, that means car is door 3. If you get yellow, then that means blue = yes. Thus, goat is in door 3.
Um.. If you ask the first question, then you already will know that the answer to the second question is the opposite color that the watch flashed the first time... I don't think you've solved it with 1 question.
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On September 19 2006 18:21 kdog3682 wrote:Show nested quote +On September 19 2006 08:31 Pacifist wrote:On September 19 2006 08:24 Ethenielle wrote:On September 19 2006 08:05 GrandInquisitor wrote:
2. You have a very nice, shiny watch. But this is no ordinary watch. This watch can answer two yes or no questions 100% accurately per day. You ask, and either a blue or yellow light will flash. Unfortunately, you don't know which light means yes and which light means no, and you can never find out because there's a 50-50 chance of the lights switching every night.
You happen to be on a game show. The rules are simple. There are three doors. Behind two are goats. Behind one is a shiny red car. Obviously, you want the car. You can choose any door, after asking the watch two questions.
What are the two yes/no questions you can ask so that you will have a 100% guaranteed chance of getting the car? + Show Spoiler +I'm not entirely sure if it's allowed, but you could just ask "are there goats behind these two doors?", if you get a yes, then it's the third door, if you get a no you can ask "is there a goat behind this door(= one of the two doors you chose)", if no then it's the car, if yes then it's the other. If you can't ask about two doors at once, though.. + Show Spoiler +You don't know which light means yes and which light means no. [spoiler] 1. Ask is there goat in door 1, 2? If you get blue, assume it means no. 2. Ask if there is goat in door 3. If you get blue, that means car is door 3. If you get yellow, then that means blue = yes. Thus, goat is in door 3.[spoiler]
dude how can you assume lol you have no idea if its yes or no... if its telling you yes and its blue ur *#&$ed lol
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Looks like this simplifies even better than I thought. The second case isn't blue, blue and yellow, yellow. It is always blue, blue! I guess I was rushing my logic before. ^_^;
This is a breakdown of every case. I start out with the entire question and begin with the inner piece. I then work my way out until I've answered the original question. Think of it as substitution.
Case 1: Door 1 = Car, Door 2 = Goat, Door 3 = Goat
-Subcase 1: Blue = Yes, Yellow = No
Is (the answer to the question "Is the car behind door 1 or 2?" ) blue?
*Is the car behind door 1 or 2?" -> Yes
*Yes -> Blue
*(the answer to the question "Is the car behind door 1 or 2?" ) = blue
*Is (blue) blue? -> Yes
*Yes -> Blue
Is (the answer to the question "Is the car behind door 2 or 3?" ) blue?
*Is the car behind door 2 or 3?" -> No
*No -> Yellow
*(the answer to the question "Is the car behind door 2 or 3?" ) = yellow
*Is (yellow) blue? -> No
*No -> Yellow
-Subcase 2: Blue = No, Yellow = Yes
Is (the answer to the question "Is the car behind door 1 or 2?" ) blue?
*Is the car behind door 1 or 2?" -> Yes
*Yes -> Yellow
*(the answer to the question "Is the car behind door 1 or 2?" ) = yellow
*Is (yellow) blue? -> No
*No -> Blue
Is (the answer to the question "Is the car behind door 2 or 3?" ) blue?
*Is the car behind door 2 or 3?" -> No
*No -> Blue
*(the answer to the question "Is the car behind door 2 or 3?" ) = blue
*Is (blue) blue? -> Yes
*Yes -> Yellow
Conclusion: For both possibilities of the truth values for blue/yellow, the response to the questions will be Blue for #1 and Yellow for #2.
Case 2: Door 1 = Goat, Door 2 = Car, Door 3 = Goat
-Subcase 1: Blue = Yes, Yellow = No
Is (the answer to the question "Is the car behind door 1 or 2?" ) blue?
*Is the car behind door 1 or 2?" -> Yes
*Yes -> Blue
*(the answer to the question "Is the car behind door 1 or 2?" ) = blue
*Is (blue) blue? -> Yes
*Yes -> Blue
Is (the answer to the question "Is the car behind door 2 or 3?" ) blue?
*Is the car behind door 2 or 3?" -> Yes
*Yes -> Blue
*(the answer to the question "Is the car behind door 2 or 3?" ) = blue
*Is (blue) blue? -> Yes
*Yes -> Blue
-Subcase 2: Blue = No, Yellow = Yes
Is (the answer to the question "Is the car behind door 1 or 2?" ) blue?
*Is the car behind door 1 or 2?" -> Yes
*Yes -> Yellow
*(the answer to the question "Is the car behind door 1 or 2?" ) = yellow
*Is (yellow) blue? -> No
*No -> Blue
Is (the answer to the question "Is the car behind door 2 or 3?" ) blue?
*Is the car behind door 2 or 3?" -> Yes
*Yes -> Yellow
*(the answer to the question "Is the car behind door 2 or 3?" ) = yellow
*Is (yellow) blue? -> No
*No -> Blue
Conclusion: For both possibilities of the truth values for blue/yellow, the response to the questions will be Blue for #1 and Blue for #2.
Case 3: Door 1 = Goat, Door 2 = Goat, Door 3 = Car
-Subcase 1: Blue = Yes, Yellow = No
Is (the answer to the question "Is the car behind door 1 or 2?" ) blue?
*Is the car behind door 1 or 2?" -> No
*No -> Yellow
*(the answer to the question "Is the car behind door 1 or 2?" ) = yellow
*Is (yellow) blue? -> No
*No -> yellow
Is (the answer to the question "Is the car behind door 2 or 3?" ) blue?
*Is the car behind door 2 or 3?" -> Yes
*Yes -> Blue
*(the answer to the question "Is the car behind door 2 or 3?" ) = blue
*Is (blue) blue? -> Yes
*Yes -> Blue
-Subcase 2: Blue = No, Yellow = Yes
Is (the answer to the question "Is the car behind door 1 or 2?" ) blue?
*Is the car behind door 1 or 2?" -> No
*No -> Blue
*(the answer to the question "Is the car behind door 1 or 2?" ) = blue
*Is (blue) blue? -> Yes
*Yes -> yellow
Is (the answer to the question "Is the car behind door 2 or 3?" ) blue?
*Is the car behind door 2 or 3?" -> Yes
*Yes -> Yellow
*(the answer to the question "Is the car behind door 2 or 3?" ) = yellow
*Is (yellow) blue? -> No
*No -> Blue
Conclusion: For both possibilities of the truth values for blue/yellow, the response to the questions will be Yellow for #1 and Blue for #2.
Final Conclusion: If the answers are #1 Blue #2 Yellow, pick the first door. If the answers are #1 Blue #2 Blue, pick the second door. If the answers are #1 Yellow #2 Blue, pick the third door.
Edit: Damn smilies.
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they're good riddles, only some of the answers are too long. short answers would be better...anyone else got any more?
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simple answer for #2 + Show Spoiler + question 1 "is the car behind door 1 red?" question 2 "is the card behind door 2 red?" if the watch flashes a color it means that the car is behind that door. if the car isn't behind that door, the question doesn't make sense and therefore, the watch wont flash anything. if the watch doesnt flash something on both doors, it means the car is behind door 3.
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looking over LLT's post... @_@. trying to make sense of it.
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United States24673 Posts
This thread is another example of how people post before reading what other people have said...often completely contradicting what has already been proven....so many lazy ppl ._. I mean look at how many people gave the wrong answer for the brick boat and lake problem after it was solved...
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Didn't read all posts .. though no1 is pretty easy ... u burn string 1 from both sides and string 2 from one side at the same time. After the 1st string is burned u burn 2nd string from 2nd side. GG no re
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Korea (South)17174 Posts
ltt, in your original post you put " "Is the car behind door 1 or 2?" blue? " with the parenthesis fucked so i missread, i didn't read your long post but i assume it is also just a very complicated way to do the solution
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can someone put all the answers in one post?
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Aotearoa39261 Posts
Your plane crashed in the middle of an unknown jungle and everyone except you was killed, nothing remained in the plane as it burnt to a crisp. You walk for a while, and find a path... you follow it. After about 30 minutes of walking you reach a fork, with a man standing on the far left, and another of the far right of the fork. There is a sign in the middle which reads "Eternal bliss awaits those who venture down one path. Those who venture down the other will be tortured and killed. These two men will help you, they will answer one, and only one question which you ask them. One of these men will always tell you the truth, the other will always lie."
What question do you ask to find out the correct path? Goodluck :D
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Lots of people don't seem to understand, so I'm going to explain in detail about #2:
+ Show Spoiler +Q1: If I ask is there a car behind 1 will you light in blue? Q2: If I ask is there a car behind 3 will you light in blue? This way the answers will always be the same no matter which color means yes and which no. Car at 1 - Blue, Yellow Blue = Yes, Yellow = No Q1: The car is at 1, so the answer to "Is there a car behind 1" is yes - Blue To the full question it has to answer yes, because it will light Blue, so in the end the answer will be yes - Blue Q2: The car is not at 3, so the answer to "Is there a car behind 3" is no - Yellow To the full question it has to answer no, because it will light Yellow, so in the end the answer will be no - Yellow Blue = No, Yellow = Yes Q1: The car is at 1, so the answer to "Is there a car behind 1" is yes - Yellow To the full question it has to answer no, because it will light Yellow, so in the end the answer will be no - Blue Q2: The car is not at 3, so the answer to "Is there a car behind 3" is no - Blue To the full question it has to answer yes, because it will light Blue, so in the end the answer will be yes - Yellow Car at 2 - Yellow, Yellow Blue = Yes, Yellow = No The car is not at 1, so the answer to "Is there a car behind 1" is no - Yellow To the full question it has to answer no, because it will light Yellow, so in the end the answer will be no - Yellow The car is not at 3, so the answer to "Is there a car behind 3" is no - Yellow To the full question it has to answer no, because it will light Yellow, so in the end the answer will be no - Yellow Blue = No, Yellow = Yes Q1: The car is not at 1, so the answer to "Is there a car behind 1" is no - Blue To the full question it has to answer no, because it will light Blue, so in the end the answer will be no - Yellow Q2: The car is not at 3, so the answer to "Is there a car behind 3" is no - Blue To the full question it has to answer yes, because it will light Blue, so in the end the answer will be yes - Yellow Car at 3 - Yellow, Blue Blue = Yes, Yellow = No Q1: The car is not at 1, so the answer to "Is there a car behind 1" is no - Yellow To the full question it has to answer no, because it will light Yellow, so in the end the answer will be no - Yellow Q2: The car is at 3, so the answer to "Is there a car behind 3" is yes - Blue To the full question it has to answer yes, because it will light Blue, so in the end the answer will be yes - Blue Blue = No, Yellow = Yes Q1: The car is not at 1, so the answer to "Is there a car behind 1" is no - Blue To the full question it has to answer yes, because it will light Blue, so in the end the answer will be yes - Yellow Q2: The car is at 3, so the answer to "Is there a car behind 3" is yes - Yellow To the full question it has to answer no, because it will light Yellow, so in the end the answer will be no - Blue ... For those wondering what would Blue, Blue mean: there are cars behind both door 1 and door 3 Ofc, it there are lots of possible 2 question combinations to solve the riddle, they could be for cars at 1/2 or 2/3 or goats behind 2 door combinations or yellow instead of blue, it doesn't matter.
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Plexa, exact this riddle was solved and explained by Day in other thread about puzzles few months back. But good one for those who never heard it.
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Aotearoa39261 Posts
On September 20 2006 03:18 One Page Memory wrote:
Plexa, exact this riddle was solved and explained by Day in other thread about puzzles few months back. But good one for those who never heard it. lol i never saw it!
i think i was studying for exams then.. i cant remember but i remember not being at tl for ages =( hard times... hard times....
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I understood what hotbid's logic was. If the two coins are identical, then their flaw would be identical too, so if either coin is pre-determined to land on one specific side 51% of the time, then the other one will too.
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Did anyone got #4 Boat/Brick correct yet? Because i don't think anyone really did since it is a trick question.
It comes down to volume vs mass density or weight. and since there are many kind of bricks and we don't know the volume and mass density of the brick in question. We cannot answer the question. It could be those new light ass clay brick or a gold brick. So really the best thing is to find out what is the brick's dimension and how much does it weigh. Until then, all answer is based on assumptions.
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For the boat one:
i couldn't quite see what the answer was from the thread but i heard Archimede's name mentioned and that immediately rang alarm bells in my head cos this is the wrong application for this principle...
"Archimede's principle states that the weight of a floating object is equal to the weight of the water displaced by the object. For an object to be in equilibrium, the upward force of buoyancy must be equal the downward force of gravity (weight). Also, the center of gravity and the center of the underwater volume (center of buoyancy) must be vertically aligned."
however, archimede's law does not apply when something (e.g. the brick) is submerged fully.
who cares if the brick is equal to the WEIGHT of water it displaces? When the brick is submerged it's VOLUME is the important point (so long as it's average density is greater than the waters - thus keeping it submergered).
-----------------------
Consider this example: submerge one 1-cubic cm marble weighing 10 tonnes into some water, the displacement of water is not 10 cubic metres (i.e. 10 tones of water), it is 1 cubic cm (i.e. ~1 gram).
Now consider thes two extensions:
A) putting the 10-tonne marble into a large empty vessel (weighing v tonnes), it will cause the apparent weight of the vessel to increase to 10+v tonnes, thus causing a huge increase in displacement.
B) putting a normal glass marble (weighing a coupe of grams) into the boat. It will not change the boat's apparent weight by a noticable amount and so will not change the displacement by a noticable amount.
-------------------
To answer the question correctly you need to know the following single detail:
What is the brick's average density?
if the answer is greater than the density of water... then the boat displaces more than the brick (so the level of the water falls).
this whole question is works on the assumption that this last 'if' is correct (bricks are generally quite heavy after all).
archimede's principle only applies if you sink both the boat AND the brick.
I hope this makes sense when read, Please point out if this is wrong.
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GrandInquisitor
New York City13113 Posts
Resonate: note what it says. The brick sinks quickly - therefore its average density >> that of water.
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Braavos36374 Posts
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On September 20 2006 06:32 GrandInquisitor wrote:
Resonate: note what it says. The brick sinks quickly - therefore its average density >> that of water.
ok fair cop.
But it doesn't change the point of my post. I won't bother editing it or it'll get confusing
On September 20 2006 07:07 Hot_Bid wrote:
ZOMG MAGIC BRICK
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I read Hot Bid's two quarter's post, and the responses, and just had to post this.
Hot Bid is right.
It's to your advantage if you pick "same" rather than "different." Why? Examine:
Say that the probability that the coin lands heads is 'p'. Therefore, the probability that the coin lands tails is (1-p). We see that
Probability of same = p^2 + (1 - p)^2
Probability of different = 2*p*(1-p)
where p is between 0 and 1, inclusive.
Doing some algebra (completing the square), we see that
probaiblity of same = 2(p-1/2)^2 + 1/2
probability of different = -2(p-1/2)^2 + 1/2
which implies that
4(p-1/2)^2 >=0 (trivial inequaility, x^2 is always >= 0)
2(p-1/2)^2 >= -2(p-1/2)^2
2(p-1/2)^2 + 1/2 >= -2(p-1/2)^2 + 1/2
probability of same >= probability of different
with equality occuring when p = 1/2.
In short, this means, that picking same will always give you better or equal odds than picking different. (The only time when the odds are equal is when p = 1/2.)
QED
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Hey, do anyone have the link to where [9]day solves and explains that question PLexa posted? I'm very curious about it, I won't sleep before I find it out. . .
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United States20661 Posts
+ Show Spoiler +
Ask either man "what would the other man say the correct path is?"
Liar knows other man would tell you the right path, so he would tell you the wrong path.
Honest man knows other man would tell you the wrong path, so he will tell you the wrong path.
Either way, you know not to go down whatever path they tell you to, and pick the other one.
^response to Plexa's question. I'm not sure how Day did it, but it should be similar.
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I just about figured it out myself, or well, yeah similar to it aswell :D This one was better explained, easier to understand/explain though. Thanks!
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-__- don't write QED after that kind of proof
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#5 is the most trivial case for a much more general problem.
See http://www.cs.huji.ac.il/~dolev/pubs/cheating.pdf for a serious paper which describes that case and many more complicated (great to catch any friends of yours who might have heard the original).
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On September 19 2006 20:16 IIICodeIIIIIII wrote:simple answer for #2 + Show Spoiler + question 1 "is the car behind door 1 red?" question 2 "is the card behind door 2 red?" if the watch flashes a color it means that the car is behind that door. if the car isn't behind that door, the question doesn't make sense and therefore, the watch wont flash anything. if the watch doesnt flash something on both doors, it means the car is behind door 3.
sounds nice, but doesn't work
if the premise is wrong, the answer will always be yes (thus a light will flash)
if i ask you if birds are blue when the sun is green, the answer is yes. it's a rule of propositional logic. (correct me if i'm wrong)
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wow snowbird why did u revive this
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there was a link in the other riddle thread, and i didn't know this thread before, and i think it's a good one
also i wanted to clarify that mistake (correctly i hope)
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All the answers for #6 are crazy complicated. It's pretty simple.
Put 6 marbles on each side. One side will have the heavier ball. Keep those 6 marbles, discard the rest.
Put 3 marbles on each side. One side will have the heavier ball. Keep those 3, discard the rest
Of the 3 you have, one is heavier. Randomly put 1 on each side, and hold on to the 3rd. If one side goes down, that's the heavier one. If they are even, you're holding the heavier one.
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On January 26 2007 09:59 Haemonculus wrote:
All the answers for #6 are crazy complicated. It's pretty simple.
Put 6 marbles on each side. One side will have the heavier ball. Keep those 6 marbles, discard the rest.
Put 3 marbles on each side. One side will have the heavier ball. Keep those 3, discard the rest
Of the 3 you have, one is heavier. Randomly put 1 on each side, and hold on to the 3rd. If one side goes down, that's the heavier one. If they are even, you're holding the heavier one.
your answer does not include: "Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light."
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On January 26 2007 10:28 snowbird wrote:Show nested quote +On January 26 2007 09:59 Haemonculus wrote:
All the answers for #6 are crazy complicated. It's pretty simple.
Put 6 marbles on each side. One side will have the heavier ball. Keep those 6 marbles, discard the rest.
Put 3 marbles on each side. One side will have the heavier ball. Keep those 3, discard the rest
Of the 3 you have, one is heavier. Randomly put 1 on each side, and hold on to the 3rd. If one side goes down, that's the heavier one. If they are even, you're holding the heavier one. your answer does not include: "Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light."
LOL. My bad. Didn't even read the whole question. gg me.
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As for number 4 the brick in the lake one. The water level stays the same because it was boyant in the boat which is holding its weight in the first place. So throwing it in the lake wont change the water level at all.
Don't know if this has been said. old thread.
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#1
Hm i think i can measure 45min with just 1 string.
as a preparation for the measuring:
I burn the string and start walking with constant speed(and counting steps) and i walk till its burnt out.
now the measuring starts: i multiply the number of steps with 0,75 and start walking back the same speed, and when i reach the new number of steps - 45mins have passed. you do have to walk the same speed all the way but i think humans are capable of that.
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Russian Federation106 Posts
I got answer for the 3rd one: you gotta play with one switch , turn it up and turn it down until the bulb in another room would burn off .
After the light is the other room stoped turning on and off you switch on the 2nd switch and go the the other room:
bulb that is on connected to the 2nd switch ,
bulb that is hot and and off is poor bulb connected to the 1st switch
and bulb that is off and cold connected to the 3rd switch j/k
You can just hold the 1st switch on for some time , than turn it off , switch the 2nd one , go to the other room ... than read the explanation above.
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i was gonna say wtf u making another one of these for then saw the date
riddles are cool but i sux at them =(
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EPT![[image loading]](http://www.soaringpanda.com/gelblog/archives/orly.jpeg)
