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On April 20 2012 03:37 Crownlol wrote:Show nested quote +On April 20 2012 02:33 XiGua wrote:I am betting that nobody will solve this without cheating. I present to you, THE WARDEN The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell. "No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead. "Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you" What is the strategy they come up with so that they can be free? 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!
what if the warden picks the counter to go 22 times in a row?
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On April 19 2012 08:44 Nehsb wrote:Show nested quote +On April 19 2012 07:54 keplersfolly wrote:I'm not a regular on this forum so apologies if I'm quoting wrong :D Solution discussion for Nehsb and Miicro: + Show Spoiler +I'm impressed :D Was that off the top of your heads or had you heard it before? It took me and my colleagues a while to work it out from a position of total ignorance. Nehsb is more correct though as he doesn't assume the starting state of the levers and he hints towards fully understanding the issue (but falls a little short). Basically there are two related facets:-
You MUST flip a lever exactly once + There are 2 levers.
What this effectively means is that the prisoners decide that the left lever is the "noop" lever. So the counter only cares about the right lever and you follow the logic you both described, so if a non-counter enters the room and the right lever is flipped to indicate someone needs counting then they throw away their flip on the left lever. You can think of this as changing the puzzle conditions to be having just one lever but flipping it is optional. Still, congrats if you worked this out from scratch! Ninja Edit: Nehsb's solution accounts for the unknown starting state - good job! Miicro's solution accounts for the fact there are two levers - good job! I'd heard it before. There's also a really hard version that I'm curious if anyone has an easier solution to: Show nested quote +20 prisoners are in jail who are all gathered in the main hall and addressed by the warden who explains the following. "In 1 hour you will all be placed in solitary confinement and never see each other again. You will individually be escorted by the guards to the "Lever Room" in which the are a pair of levers that have no function except being able to be flipped either Up or Down. You can flip any number of levers, including all of them or none of them. We make no guarantees about the order you will be called to the Lever Room in, and we may take any of you there more than once. If any of you prisoners tell the guard "We have all been to the lever room" and this is true you will all be released, otherwise you will all be shot."You do not know the starting position of the levers. You are allowed to write down a strategy and give it to all of the prisoners, but the strategy each prisoner gets must be the same; you are not allowed to "pick a leader" or anything like that. The changes are in bold: - Still 2 levers - You can now flip any number of levers - Every prisoner must have the same strategy - You do not know the starting states of the levers The problem gets much easier if you do know the starting state of the levers.
A hint:
+ Show Spoiler +Instead of choosing a leader as in the earlier puzzles, use the levers to split the people into two about equally-sized groups.
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what if the warden picks the counter to go 22 times in a row?
+ Show Spoiler +His count is still at "one", and the A switch is still in the down position. He just diddles the B switch 21 times.
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On April 20 2012 03:36 Taekwon wrote:
A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!"
Boy B said: "Neither A nor B ate it."
Boy C said: "I swear on my momma's grave I didn't eat it!"
Girl D said: "C is telling the truth!"
One of these children is lying. Who is it?
+ Show Spoiler +A can not be lying, if she did, B would also be lying
C can not be lying, if he did D would also be lying
D can not be lying, if she did, C would be lying
B could be lying, if he was the one to eat it, A C and D would still be telling the truth.
or it could be the boy that says he had his cake eaten that's lying, seriously what kid goes around saving cake for later, he probably ate his cake and have it too.
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On April 20 2012 02:33 XiGua wrote:I am betting that nobody will solve this without cheating. I present to you, THE WARDEN The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell. "No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead. "Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you" What is the strategy they come up with so that they can be free?
I didn't see it in the thread earlier, so here's my try:
+ Show Spoiler +
Since there's no way to store the information of how many people have been in the room with the levers themselves, one of the prisoners needs to act as counter. The procedure is simple:
Simplified solution assuming switch 1 is down to begin with: If one of the prisoners (except the counter) enters the room for the first time, they flip switch 1 up. If switch 1 is already flipped up, they instead flip switch 2. If any of them are brought in again after the first time, they also flip switch 2.
Whenever the counter enters the room and finds switch 1 flipped up, he flips it back down. Otherwise he flips switch 2.
Switch 2 is completely useless here, it's simply used because the prisoners have to flip one switch. All the information is stored in switch 1.
If the prisoners knew that switch 1 was flipped down to begin with, this would be the end of the puzzle. The counter would simply announce that everyone had been in the room after he sees switch 1 flipped up for the 22nd time.
Because he doesn't know however, we need to account for the fact that the switch might have been up to begin with. I've been thinking about an elegant solution to this, but I don't think there is one. The prisoners simply have to take the longer way around.
The idea is that you can account for the possible mistake of 1 by doubling the amount of counts. If each of the non-counting prisoners flips the switch up the first two times they enter the room and the counter waits for switch 1 being up 44 times, then one of those times being a potential error doesn't matter. There would still be 43 real counts remaining, which could only have come from all the 22 non-counters (21 * 2 + 1 * 1).
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On April 20 2012 03:37 Crownlol wrote:Show nested quote +On April 20 2012 02:33 XiGua wrote:I am betting that nobody will solve this without cheating. I present to you, THE WARDEN The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell. "No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead. "Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you" What is the strategy they come up with so that they can be free? 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^^
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Riddle:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)
+ Show Spoiler +33%Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%. For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probabilityProbability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33 Edit + Update: This question turned out a lot more ambiguous than I had intended. The problem is in how one interprets "a male child answers." In case 1, a random child is chosen and happened to be male. In case 2, a male child is chosen if at all possible. Imagining a community of 100 houses with 2 children each, 25 BB, 50 BG, and 25 GG. In the first case, the 25 BB and 50% of the 50 BG houses are considered (since the chance of a boy answering the door in a BG house is 50%), leaving 25 BB and 25 BG with the probability of the unknown child being a boy equal to 25/50=50%. In the second case, a boy answers at all 50 of the BG houses which combined with the 25 BB houses makes 75 with 25/75=33% of the houses having the unknown child being a boy. I have honestly always interpreted the question as the second case, but it seems like most people interpret it as the first case. The hint "The answer is not 50%" suggests that it is not case 1, but I should definitely have added some additional language to be more specific.
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On April 20 2012 03:59 Orome wrote:Show nested quote +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
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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
How does he know that he's the first one to adjust the switch? What if one of the other prisoners originally saw it in the down position and flipped it up?
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On April 20 2012 04:41 el_dawg wrote:Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%) + Show Spoiler +33%Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%. For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probabilityI have to disagree with this one. Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33
+ Show Spoiler +
You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.
The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.
In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.
In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.
This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!
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On April 20 2012 05:19 Aim Here wrote:Show nested quote +On April 20 2012 04:41 el_dawg wrote:Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%) + Show Spoiler +33%Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%. For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probabilityI have to disagree with this one. Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33 + Show Spoiler +
You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.
The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.
In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.
In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.
This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!
+ Show Spoiler +So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement.
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On April 20 2012 05:29 Vega62a wrote:Show nested quote +On April 20 2012 05:19 Aim Here wrote:On April 20 2012 04:41 el_dawg wrote:Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%) + Show Spoiler +33%Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%. For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probabilityI have to disagree with this one. Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33 + Show Spoiler +
You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.
The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.
In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.
In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.
This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!
+ Show Spoiler +So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement.
+ Show Spoiler +Yeah the point is that you're supposed to introduce the idea that one is a male, what is the probability of the other being male?
Then you introduce the fact that you visited the house and opened the door - what is the probability now? (Since it changes to 50% upon opening the door).
A real good mind fuck.
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On April 20 2012 05:29 Vega62a wrote:Show nested quote +On April 20 2012 05:19 Aim Here wrote:On April 20 2012 04:41 el_dawg wrote:Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%) + Show Spoiler +33%Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%. For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probabilityI have to disagree with this one. Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33 + Show Spoiler +
You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.
The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.
In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.
In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.
This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!
+ Show Spoiler +So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement.
+ Show Spoiler +Hmm, I didn't think the child answering the door made a difference. I will have to rethink this a bit.
If we play a game where I flip 2 coins (re-flipping both if both are tails) and show you that one is heads while hiding the other, you would guess that the hidden coin has a 66% chance of being tails, right?
I really like this question, but the hardest part is finding a way that asks it correctly.
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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?
+ Show Spoiler +
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
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On April 20 2012 06:51 TheAngryZergling wrote:Show nested quote +On April 20 2012 02:33 XiGua wrote:I am betting that nobody will solve this without cheating. I present to you, THE WARDEN The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell. "No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead. "Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you" What is the strategy they come up with so that they can be free? + Show Spoiler +
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
+ Show Spoiler +Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.
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On April 20 2012 05:54 el_dawg wrote:Show nested quote +On April 20 2012 05:29 Vega62a wrote:On April 20 2012 05:19 Aim Here wrote:On April 20 2012 04:41 el_dawg wrote:Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%) + Show Spoiler +33%Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%. For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probabilityI have to disagree with this one. Probability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33 + Show Spoiler +
You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.
The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.
In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.
In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.
This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!
+ Show Spoiler +So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement. + Show Spoiler +Hmm, I didn't think the child answering the door made a difference. I will have to rethink this a bit.
If we play a game where I flip 2 coins (re-flipping both if both are tails) and show you that one is heads while hiding the other, you would guess that the hidden coin has a 66% chance of being tails, right?
I really like this question, but the hardest part is finding a way that asks it correctly.
+ Show Spoiler +I think what is missing from my question is whether or not the child opening the door was random. For example, in the Monty hall problem, it is understood that the host will never open the door with the car. So this question needs something to say that not only did a boy happen to answer the door, but a boy will always answer the door.
Edit: the monty hall problem still works if the reveal was random (assuming a goat was revealed), I'm just using it as an example where the revealed info is usually chosen by the asker who knows the status of all the doors/ coins/ children.
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On April 20 2012 04:41 el_dawg wrote:Riddle: You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%) + Show Spoiler +33%Before you knock on the door, there are 4 equally possible situations. First child - Second child Boy - Boy Boy - Girl Girl - Boy Girl - Girl Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only: Boy - Boy Boy - Girl Girl - Boy Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%. For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probabilityProbability(A given B)=P(A and B)/P(B) P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33
What's funny is that even if your answer is mathematically incorrect. The answer is not 50% even if you consider the chance of having a girl or a boy to be 50/50.
Can you guess why??
+ Show Spoiler +The two child could be true twins, making the probably of having a second boy in the house slighty higher
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United States2474 Posts
On April 20 2012 07:06 Excludos wrote:Show nested quote +On April 20 2012 06:51 TheAngryZergling 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? + Show Spoiler +
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
+ Show Spoiler +Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.
+ Show Spoiler +The same solution works, you just count to 43 or whatever (I'm too lazy to figure out the exact number) and make sure everyone flips the counting lever twice
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On April 20 2012 09:03 crate wrote:Show nested quote +On April 20 2012 07:06 Excludos wrote:On April 20 2012 06:51 TheAngryZergling 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? + Show Spoiler +
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
+ Show Spoiler +Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room. + Show Spoiler +The same solution works, you just count to 43 or whatever (I'm too lazy to figure out the exact number) and make sure everyone flips the counting lever twice
crate's solution is true, just to make sure
+ Show Spoiler +just make sure everybody flips the switch down twice, so at some point (too lazy to figure the number too) it still insures everybody has AT LEAST flipped the switch once, which is the only requirement
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EPT![[image loading]](http://images.wikia.com/wowwiki/images/a/a4/Warden_artwork.jpg)
