|
On April 20 2012 04:52 ishboh wrote:Show nested quote +On April 20 2012 03:59 Orome wrote:On April 20 2012 03:37 Crownlol wrote:On April 20 2012 02:33 XiGua wrote:I am betting that nobody will solve this without cheating. I present to you, THE WARDEN The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell. "No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead. "Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you" What is the strategy they come up with so that they can be free? Ok, I'm not sure, but here ya go: + Show Spoiler +The group designates a single person the "counter". This person is the only one who will reset switch A to the bottom position. A starts in the down position. If it is your first visit to the room, you move switch A to up. If switch A is in the up position, you don't touch it, you just move B (also, if you have been there before you move B). If the counter finds the switch in the up position, he moves it down and adds +1 to his count. When the counter's count reaches 22, he frees himself and all his co-prisoners! your version doesn't account for the fact that the switch could be up to begin with^^ all he has to add to ensure that his solution is correct is add a couple conditions: 1) they decide who the counter is beforehand. 2) he doesn't start his count until after the first time he adjusts the switch
Yep, that's what I had implied.
|
On April 21 2012 05:11 nq07 wrote:Show nested quote +On April 21 2012 05:07 LastPrime wrote:On April 21 2012 03:07 TanGeng wrote:A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange? + Show Spoiler +
The probability is 2/3.
Consider: You reach into one random pocket and pull out a ping pong ball. What is the probability that the other ball in the pocket is the same color?
+ Show Spoiler +Since the wording of the problem says random pocket, the answer is 1/2. If it said random marble, the answer 2/3 would have been correct. + Show Spoiler +But the problem implies that after reaching into a random pocket, you also randomly select one of the two balls in that pocket (otherwise, how would you end up with a ball)? So in fact the problem is implying random ball, in which case the 2/3 interpretation would seem to be correct?
+ Show Spoiler +I would still argue that 1/2 is correct. The first ping pong ball pulled out being orange is stated as a given condition, not as a random event.
On April 21 2012 05:23 TanGeng wrote:Show nested quote +On April 21 2012 05:07 LastPrime wrote:On April 21 2012 03:07 TanGeng wrote:A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange? + Show Spoiler +
The probability is 2/3.
Consider: You reach into one random pocket and pull out a ping pong ball. What is the probability that the other ball in the pocket is the same color?
+ Show Spoiler +Since the wording of the problem says random pocket, the answer is 1/2. If it said random marble, the answer 2/3 would have been correct. The semantics doesn't work your way. BTW, who's pulling out random marbles?
+ Show Spoiler +Please elaborate on why you think the semantics don't work my way.
|
On April 21 2012 05:37 LastPrime wrote:Show nested quote +On April 21 2012 05:11 nq07 wrote:On April 21 2012 05:07 LastPrime wrote:On April 21 2012 03:07 TanGeng wrote:A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange? + Show Spoiler +
The probability is 2/3.
Consider: You reach into one random pocket and pull out a ping pong ball. What is the probability that the other ball in the pocket is the same color?
+ Show Spoiler +Since the wording of the problem says random pocket, the answer is 1/2. If it said random marble, the answer 2/3 would have been correct. + Show Spoiler +But the problem implies that after reaching into a random pocket, you also randomly select one of the two balls in that pocket (otherwise, how would you end up with a ball)? So in fact the problem is implying random ball, in which case the 2/3 interpretation would seem to be correct? + Show Spoiler +I would still argue that 1/2 is correct. The first ping pong ball pulled out being orange is stated as a given condition, not as a random event. Show nested quote +On April 21 2012 05:23 TanGeng wrote:On April 21 2012 05:07 LastPrime wrote:On April 21 2012 03:07 TanGeng wrote:A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange? + Show Spoiler +
The probability is 2/3.
Consider: You reach into one random pocket and pull out a ping pong ball. What is the probability that the other ball in the pocket is the same color?
+ Show Spoiler +Since the wording of the problem says random pocket, the answer is 1/2. If it said random marble, the answer 2/3 would have been correct. The semantics doesn't work your way. BTW, who's pulling out random marbles? + Show Spoiler +Please elaborate on why you think the semantics don't work my way.
+ Show Spoiler +Agreed. Answer is definitely 1/2 since they stated that you have already pulled out an orange ping pong ball. That means you are either in the pocket (lets call it 1) with 2 oranges or 1 orange and 1 white (lets call this 2). If you already chose 1 ping pong ball, the chances the other 1 is orange depends on which pocket you are in. Either you are in pocket 1 or pocket 2 which will give you a 50% chance of getting another orange.
|
Sanya12364 Posts
Beers.
At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar?
+ Show Spoiler +
The special is in effect equivalent to buy 2 beers get 3rd free with the caveat that you need to have an empty 3rd glass before you get that free one. The goal is therefore to drink 15 beers based on your initial haul of 10.
So you drink 10 + 3 + 1 and manage 14 beers with two empty glasses. For the 15th beer you grab a friend's full beer, drink it, and turn in the empty 3 glasses to give your friend his beer back.
|
Sanya12364 Posts
On April 21 2012 05:37 LastPrime wrote:Show nested quote +On April 21 2012 05:11 nq07 wrote:On April 21 2012 05:07 LastPrime wrote:On April 21 2012 03:07 TanGeng wrote:A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange? + Show Spoiler +
The probability is 2/3.
Consider: You reach into one random pocket and pull out a ping pong ball. What is the probability that the other ball in the pocket is the same color?
+ Show Spoiler +Since the wording of the problem says random pocket, the answer is 1/2. If it said random marble, the answer 2/3 would have been correct. + Show Spoiler +But the problem implies that after reaching into a random pocket, you also randomly select one of the two balls in that pocket (otherwise, how would you end up with a ball)? So in fact the problem is implying random ball, in which case the 2/3 interpretation would seem to be correct? + Show Spoiler +I would still argue that 1/2 is correct. The first ping pong ball pulled out being orange is stated as a given condition, not as a random event. Show nested quote +On April 21 2012 05:23 TanGeng wrote:On April 21 2012 05:07 LastPrime wrote:On April 21 2012 03:07 TanGeng wrote:A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange? + Show Spoiler +
The probability is 2/3.
Consider: You reach into one random pocket and pull out a ping pong ball. What is the probability that the other ball in the pocket is the same color?
+ Show Spoiler +Since the wording of the problem says random pocket, the answer is 1/2. If it said random marble, the answer 2/3 would have been correct. The semantics doesn't work your way. BTW, who's pulling out random marbles? + Show Spoiler +Please elaborate on why you think the semantics don't work my way.
+ Show Spoiler +
The one orange ball that is pull out is the given event. The given event is NOT that you picked a pocket with at least one orange ball.
Based on the given, we count three possibilities, and two chances to be in the pocket with two orange balls and one chance to be in the pocket with the pocket with one of each. The probability is clearly 2/3.
|
On April 21 2012 06:46 TanGeng wrote:
Beers.
At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar?
+ Show Spoiler +
We don't have enough information to answer this question.
If you and your friends are male, the group would be able to drink 14 beers split between them, plus any empties you can snatch from other tables while they're not looking.
If female, however much you want.
|
On April 21 2012 06:46 TanGeng wrote:
Beers.
At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar?
+ Show Spoiler +
|
On April 21 2012 06:46 TanGeng wrote:
Beers.
At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar?
+ Show Spoiler +You can get 15 out of it. Buy 10 and you can drink 9 of them, getting 3 new.
Drink the 3 new ones and get a new-new one.
Then borrow a beer, drink these 3 (the new-new, the borrowed and the one from the original 10) and return the last free beer to the person you borrowed from.
|
On April 21 2012 07:03 GaiaCaT wrote:Show nested quote +On April 21 2012 06:46 TanGeng wrote:
Beers.
At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar? + Show Spoiler +You can get 15 out of it. Buy 10 and you can drink 9 of them, getting 3 new.
Drink the 3 new ones and get a new-new one.
Then borrow a beer, drink these 3 (the new-new, the borrowed and the one from the original 10) and return the last free beer to the person you borrowed from.
But you still get + Show Spoiler +, the question doesn't say how many the bar loses.
|
On April 21 2012 06:46 TanGeng wrote:Beers. At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar? + Show Spoiler +
The special is in effect equivalent to buy 2 beers get 3rd free with the caveat that you need to have an empty 3rd glass before you get that free one. The goal is therefore to drink 15 beers based on your initial haul of 10.
So you drink 10 + 3 + 1 and manage 14 beers with two empty glasses. For the 15th beer you grab a friend's full beer, drink it, and turn in the empty 3 glasses to give your friend his beer back.
Your answer assumes that you have 11 beers, not 10. If you can borrow 1 beer from another patron, why not collect empty glasses from them as well or panhandle outside for more money? The whole "riddle" falls apart when you introduce borrowing.
|
On April 21 2012 07:23 Zealotdriver wrote:Show nested quote +On April 21 2012 06:46 TanGeng wrote:Beers. At a local bar, the Weekly Friday Special is a free glass of beer for every three empty finished glasses. You're with some friends and spent all of your money to order 10 glasses of beer. How many beers will you be able to drink at the bar? + Show Spoiler +
The special is in effect equivalent to buy 2 beers get 3rd free with the caveat that you need to have an empty 3rd glass before you get that free one. The goal is therefore to drink 15 beers based on your initial haul of 10.
So you drink 10 + 3 + 1 and manage 14 beers with two empty glasses. For the 15th beer you grab a friend's full beer, drink it, and turn in the empty 3 glasses to give your friend his beer back.
Your answer assumes that you have 11 beers, not 10. If you can borrow 1 beer from another patron, why not collect empty glasses from them as well or panhandle outside for more money? The whole "riddle" falls apart when you introduce borrowing.
The point is that you return to the friend exactly what was borrowed so that he is not affected in any way. Collecting empty glasses or panhandling outside for more money would not be analogous.
|
United States7483 Posts
On April 21 2012 05:14 radscorpion9 wrote:Show nested quote +On April 21 2012 03:52 Whitewing wrote:
If a chicken and a half can lay an egg and a half in a day and a half, how many eggs could six chickens lay in six days? Here's what I got: + Show Spoiler +
6 days is 1.5*4. So 1.5 chickens will lay 1.5*4 eggs in 1.5*4 days.
Now 1.5*4 = 6 eggs
But there are 6 chickens, which is equivalent to 1.5*4. So multiplying 4 "groups" of 1.5 chickens by the output of each chicken group for 6 days, we get:
4*6 = 24 eggs! Cool puzzle, I hope that worked
You are correct.
|
United States7483 Posts
On April 21 2012 05:32 TheJizWiz wrote:Show nested quote +On April 21 2012 03:52 Whitewing wrote:
If a chicken and a half can lay an egg and a half in a day and a half, how many eggs could six chickens lay in six days? + Show Spoiler +I don't think half a chicken is able to lay eggs. So it's just 1 chicken that lays 1.5 eggs in 1.5 days. Which basically means 1 chicken lays 1 egg per day.
Based on this 6 chicken would lay 6 eggs in a day and in six days that would be 36 eggs.
No, the riddle explicitly states that a chicken and a half can indeed lay one egg and a half in a day and a half. Your answer is incorrect.
|
United States7483 Posts
Pure word riddle for you: good luck!
You are trapped in a perfectly cylindrical room with perfectly smooth walls, no friction at all. There is a light at the top of the room so that you can see, but it is a good 10 feet above your head. Your only possessions inside this room are a mirror and a table. The walls are sufficiently strong that you have no hope of damaging them directly.
Escape the room.
Hint:+ Show Spoiler +The solution is purely word play, do not try to logic your way out of the room.
|
On April 21 2012 07:53 Whitewing wrote:Pure word riddle for you: good luck! You are trapped in a perfectly cylindrical room with perfectly smooth walls, no friction at all. There is a light at the top of the room so that you can see, but it is a good 10 feet above your head. Your only possessions inside this room are a mirror and a table. The walls are sufficiently strong that you have no hope of damaging them directly. Escape the room. Hint: + Show Spoiler +The solution is purely word play, do not try to logic your way out of the room.
+ Show Spoiler +
A solution from a simpler puzzle should work here - we don't need to use the mirror or the light.
1: Bang your hand on the table until it's sore.
2: Use the saw to cut the table in half.
3: Put the two halves together on the wall to make a whole.
4: Climb out the hole.
(optionally)
5: Yell until you're hoarse.
6: Ride off into the sunset.
|
Sanya12364 Posts
Marble Probability
You have 100 marbles and two empty marble jars. To start you place 1 marble in each. Then you take time putting the marbles one at a time into either of the two jars. For each marble, the probability that the marble goes into a jar is equal to the fraction of number of marbles that the jar holds of the number of marbles in the two jars (example jar A has 2 marbles, jar B has 5 marbles; the probability is 2/7 and 5/7 respectively.)
You're also running a gambling ring. What is your betting line for the number of marbles in the jar with fewer marbles?
|
|
|
United States7483 Posts
On April 21 2012 08:22 Aragnis wrote:Show nested quote +On April 21 2012 07:53 Whitewing wrote:Pure word riddle for you: good luck! You are trapped in a perfectly cylindrical room with perfectly smooth walls, no friction at all. There is a light at the top of the room so that you can see, but it is a good 10 feet above your head. Your only possessions inside this room are a mirror and a table. The walls are sufficiently strong that you have no hope of damaging them directly. Escape the room. Hint: + Show Spoiler +The solution is purely word play, do not try to logic your way out of the room. + Show Spoiler +
A solution from a simpler puzzle should work here - we don't need to use the mirror or the light.
1: Bang your hand on the table until it's sore.
2: Use the saw to cut the table in half.
3: Put the two halves together on the wall to make a whole.
4: Climb out the hole.
(optionally)
5: Yell until you're hoarse.
6: Ride off into the sunset.
+ Show Spoiler +
|
Sanya12364 Posts
Blind Man's Deck
There was once a very wise blind man, and the town philosopher was jealous of the man's popularity. One day, he took a full deck of 52 cards, turned all the Aces, Kings, and Queens face up, shuffled the deck well and handed the deck to the blind man. He challenged the wise man to split the cards into piles such that the number of cards facing up in all of the piles was equal.
Without pausing for a beat, the wise man counted off a few cards, split the decks into piles, and completed the challenge in a matter of seconds. How did the wise man do it?
|
On April 21 2012 08:31 TanGeng wrote:
Marble Probability
You have 100 marbles and two empty marble jars. To start you place 1 marble in each. Then you take time putting the marbles one at a time into either of the two jars. For each marble, the probability that the marble goes into a jar is equal to the fraction of number of marbles that the jar holds of the number of marbles in the two jars (example jar A has 2 marbles, jar B has 5 marbles; the probability is 2/7 and 5/7 respectively.)
You're also running a gambling ring. What is your betting line for the number of marbles in the jar with fewer marbles?
+ Show Spoiler +
|
|
|
|
|
|
EPT![[image loading]](http://images.wikia.com/wowwiki/images/a/a4/Warden_artwork.jpg)
