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Re: the 2-children puzzle:
On April 20 2012 07:38 el_dawg wrote:Show nested quote +On April 20 2012 05:54 el_dawg wrote: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.
+ Show Spoiler +
First of all, the monty hall problem DOES NOT work if the reveal was random, if a goat is revealed.
Proof by decision tree:
Assume car behind 1, goats behind 2 and 3. P = player, H = host. All scenarios are known to be equally likely because both the player's choice and the host's subsequent choice are random:
P1H2 = stick to win
P1H3 = stick to win
P2H1 = either impossible to win or you win automatically depending upon unstated rule (host revealed car)
P2H2 = switch to win
P3H1 = impossible to win / automatic win (host revealed car)
P3H2 = switch to win
As you can see, in this scenario seeing a goat revealed is just as likely when you've picked the other goat as it is when you picked the car, so it's a complete toss-up.
Secondly, the 2-child problem. Assuming the odds of a child being born male or female are in fact 50/50:
Here's the WRONG way to work it out:
Four equally likely possibilities: MM, MF, FM and FF
Seeing a male child (or being told one exists) eliminates FF, leaving three equally likely possibilities.
In only one of these three are there two male children.
Therefore the odds are 33% male, 66% female.
Again, this answer is wrong.
Proof that it is wrong:
Let's say you have 1000 friends, each with 2 children you've never seen. Could you make a nice profit by betting even money the other child is female each time the first child you see or are told about is male?
Immediately you can see that the answer depends upon how often, out of the 1000 iterations, you are shown a male child first. Remember: the odds of a child being male or female are 50%. Thus out of your 1000 friends, 250ish will have MM, 250ish MF, 250ish FM and 250ish FF.
If the first child you see is random each time, then you will see boys first at all the MM houses, girls first at all the FF houses, and boys first roughly half the time at the MF/FM houses. So you'll see about 500 boys and 500 girls. And since we assume the genders of siblings are independent, we know that roughly half the boys we saw will have sisters and half of them brothers, and the same for the girls we saw. So no, we can't make money by betting 'F' when we see 'M'
Only if your chances of seeing a boy first are skewed somehow can we deduce different odds.
For instance, what if we the first child we see is a boy 750 times and a girl 250? The odds against that happening by chance if there are 250MM,250MF,250FM and 250FF are astronomical - we would have to assume people weren't showing us girls unless they had no choice. In which case yes, we could make money by betting, since in roughly 500 of the 750 times we are shown a boy, the other is a girl - and we would make even more money betting that the other is a girl if we are shown a girl, since it's true every time.
So if we see a boy AND know we would only have seen or been told about a girl as a last resort, the odds are 33 and 66. Ot
herwise they're 50/50
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On April 20 2012 01:22 TanGeng wrote:
Teams of three players are invited to engage in a guessing game and told the rules of the game ahead of the time and allowed time to come up with a strategy.
On the set, there are three soundproof rooms. Above each room, there are two lights, one blue, one red. The rooms are windowed such that someone inside can only see the lights above other the rooms and nothing else. Inside the room, there are three buttons, a red button, a blue button, and a white button labeled pass. At the start of the game, each player will be led into a separate room, and one of the two lights will be turned on above each rooms. The light that is turned on is determined by a flip of a fair coin independently for all three rooms. The game ends when all players have pressed a button inside their respective rooms.
The team is adjudged to win if at least of them pressed the button of the matching color of the light turn on above their room and none of them pressed the button of the non-matching color. A press of the white pass button counts as neither matching nor non-matching. In all other cases, the team loses.
You lead a team of three. Do you come up with a strategy to maximize your chances of winning? What is it?
at least _____ of them?
Wtf. 1? 2? 3? 0?
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On April 20 2012 03:46 kdgns wrote:Show nested quote +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.
Look at the original post, either he edited it or you changed his quote, because what B says differs between the 2 versions.
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Sanya12364 Posts
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 +
Since the Warden says prisoners are chosen at random rather than at whim, it's best to just wait for a long time without a counting strategy. The first person to visit the room N time with N being sufficiently large (such as 40) should give a good probability of going free.
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On April 20 2012 01:22 TanGeng wrote:
Teams of three players are invited to engage in a guessing game and told the rules of the game ahead of the time and allowed time to come up with a strategy.
On the set, there are three soundproof rooms. Above each room, there are two lights, one blue, one red. The rooms are windowed such that someone inside can only see the lights above other the rooms and nothing else. Inside the room, there are three buttons, a red button, a blue button, and a white button labeled pass. At the start of the game, each player will be led into a separate room, and one of the two lights will be turned on above each rooms. The light that is turned on is determined by a flip of a fair coin independently for all three rooms. The game ends when all players have pressed a button inside their respective rooms.
The team is adjudged to win if at least one of them pressed the button of the matching color of the light turn on above their room and none of them pressed the button of the non-matching color. A press of the white pass button counts as neither matching nor non-matching. In all other cases, the team loses.
You lead a team of three. Do you come up with a strategy to maximize your chances of winning? What is it?
+ Show Spoiler + If you see both other rooms have the same color, than guess the other color. Otherwise pass.
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Sanya12364 Posts
On April 20 2012 01:31 Bahamuth wrote:Show nested quote +On April 20 2012 01:22 TanGeng wrote:
Teams of three players are invited to engage in a guessing game and told the rules of the game ahead of the time and allowed time to come up with a strategy.
On the set, there are three soundproof rooms. Above each room, there are two lights, one blue, one red. The rooms are windowed such that someone inside can only see the lights above other the rooms and nothing else. Inside the room, there are three buttons, a red button, a blue button, and a white button labeled pass. At the start of the game, each player will be led into a separate room, and one of the two lights will be turned on above each rooms. The light that is turned on is determined by a flip of a fair coin independently for all three rooms. The game ends when all players have pressed a button inside their respective rooms.
The team is adjudged to win if at least of them pressed the button of the matching color of the light turn on above their room and none of them pressed the button of the non-matching color. A press of the white pass button counts as neither matching nor non-matching. In all other cases, the team loses.
You lead a team of three. Do you come up with a strategy to maximize your chances of winning? What is it? Just to clarify, I'm guessing players don't have to press their buttons simultaneously, and that other players can see what button the others pressed? Am I allowed to use some notion of time? As in, the strategy is to do something under a certain circumstance. If no player does anything for a time x, proceed to the next step?
Players don't know if other team members have pressed buttons unless they are being led out of the room in which case the game is already over. You may have a concept of time.
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To the OP of the boy girl thing: It is 50%.
BB,BG,GB,GG accounts for order (BG and GB are the same but in a different order).
If you want to do it this way, then you must say that the boy you met at the door is either: Boy 1 being case of B_
or he is Boy 2 being case of _B.
Either way, there is one case of second child being boy and one case of second child being a girl.
If he is boy 1, then: BB and BG are you only option.
If he is boy 2, then: BB and GB are you only option.
GB and BG can't be in the same case because the boy is in two different spots (think of it as the one you met vs the one you didn't meet).
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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 +
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are unknown at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
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On April 20 2012 10:35 TanGeng 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 +
Since the Warden says prisoners are chosen at random rather than at whim, it's best to just wait for a long time without a counting strategy. The first person to visit the room N time with N being sufficiently large (such as 40) should give a good probability of going free.
+ Show Spoiler +
Would I be on the right track if I treated the 2 switches as a 2 bit memory cell, and allow only a certain amount of prisoners to change certain states?
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On April 18 2012 23:02 kochanfe wrote:
You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?
Second Problem: You need exactly four gallons. How do you do it?
There are two solutions for both problems, and they are each related.
+ Show Spoiler +#1 -- the 5 + 2 approach.
Start by filling the 5 gallon jug.
Pour 3 gallons into the 3 gallon jug (filling it) leaving 2 gallons in the 5 gallon jug.
Dump the 3 gallon jug.
Pour the remaining 2 gallons from the 5 gallon jug into the 3 gallon jug.
Fill the 5 gallon jug.
Grats, you now have 7 gallons.
Pour 1 gallon from the 5 gallon jug into the 3 gallon jug.
(Dump or ignore the 3 gallon jug.)
Grats, you now have 4 gallons (in the 5 gallon jug).
#2 -- the 4 + 3 approach.
Start by filling the 3 gallon jug.
Pour all 3 gallons into the 5 gallon jug.
Again fill the 3 gallon jug.
Pour 2 gallons into the 5 gallon jug (filling it) leaving 1 gallon in the 3 gallon jug.
Dump the 5 gallon jug.
Pour the remaining 1 gallon from the 3 gallon jug into the 5 gallon jug.
Fill the 3 gallon jug.
Pour the 3 gallons into the 5 gallon jug.
Grats, you now have 4 gallons (in the 5 gallon jug).
Fill the 3 gallon jug.
Grats, you now have 7 gallons.
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[QUOTE]On April 20 2012 10:51 drew-chan wrote:
[QUOTE]On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.
I present to you,
THE WARDEN
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.
"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.
"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"
What is the strategy they come up with so that they can be free?
[/QUOTE]
+ Show Spoiler +
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
[QUOTE]
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory.
+ Show Spoiler +i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years.
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United States7483 Posts
On April 20 2012 11:30 whatwhatanut wrote:Show nested quote +On April 20 2012 10:51 drew-chan 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 +
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory. + Show Spoiler +i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years.
+ Show Spoiler + Solution is more or less the same, simply have one of the two switches be the one the counter uses, and have people who aren't going to modify it flip the other switch. Say there's one on the left and one on the right, have the counter only count when the left switch is up, and have him flip it down when he goes in. Everyone else only flips the right switch, unless the left switch is down, then they flip it up. They only ever flip the left switch up twice, never more than that, even if it is down.
Personally, I prefer the method of "wait until I've gone 50 times." Assuming the selection is truly random and that the probability of choosing any given prisoner is approximately the same as choosing any other prisoner, the odds of having been chosen 50 times before every other person has gone at least once is absurdly small.
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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
Yeah....no
+ Show Spoiler +Like others have said, you're wrong and the correct answer is 50%.
Before answering the door there are 4 possible outcomes:
Male | Male
Male | Female
Female | Male
Female | Female
Once you open the door to reveal a male child, you know that there is at least one male child. Therefore there are only two remaining outcomes left:
Male | Male
Male | Female
I have no idea why you thought Female | Male was an additional outcome, as it would only be an outcome if the one opening the door was female.
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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 D ate it."
Boy C said: "I swear on my momma's grave I didn't eat it!"
Girl D said: "C is telling the truth!"
One of these children is lying. Who is it?
Alice said: "I didn't eat it!"
Brad said: "Neither Alice nor Dena ate it."
Cory said: "I swear on my momma's grave I didn't eat it!"
Dena said: "Cory is telling the truth!"
[For reference, I changed their names so it would be easier to talk about]
+ Show Spoiler [Solution] +Alice
Brad
Cory
Dena
Well let's work through it. For starters we know that one of these children are lying. That means, since no one mentions Brad at all, he can't be the one who ate it.
Alice
Cory
Dena
Now, Cory claims he didn't eat it, and Dena says that Cory is telling the truth. If Cory did eat it, both of them would be lying, and since we know only one person lied, that can't be true. Which means that Cory did not eat it either.
Alice
Dena
Finally Alice claims she did not eat it, and Brad claims that neither Alice or Dena ate it. If Alice really did eat it, then both Brad and Alice would be lying, so Alice cannot be the one who ate it.
Dena
The only suspect left is Dena. Brad claims that neither Alice nor Dena ate the cake. However Dena was the one who ate the cake, and so Brad is lying.
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On April 20 2012 12:55 killa_robot wrote:Show nested quote +On April 20 2012 03:36 Taekwon wrote:
A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!"
Boy B said: "Neither A nor D ate it."
Boy C said: "I swear on my momma's grave I didn't eat it!"
Girl D said: "C is telling the truth!"
One of these children is lying. Who is it? Alice said: "I didn't eat it!" Brad said: "Neither Alice nor Dena ate it." Cory said: "I swear on my momma's grave I didn't eat it!" Dena said: "Cory is telling the truth!" [For reference, I changed their names so it would be easier to talk about] + Show Spoiler [Solution] +Alice
Brad
Cory
Dena
Well let's work through it. For starters we know that one of these children are lying. That means, since no one mentions Brad at all, he can't be the one who ate it.
Alice
Cory
Dena
Now, Cory claims he didn't eat it, and Dena says that Cory is telling the truth. If Cory did eat it, both of them would be lying, and since we know only one person lied, that can't be true. Which means that Cory did not eat it either.
Alice
Dena
Finally Alice claims she did not eat it, and Brad claims that neither Alice or Dena ate it. If Alice really did eat it, then both Brad and Alice would be lying, so Alice cannot be the one who ate it.
Dena
The only suspect left is Dena. Brad claims that neither Alice nor Dena ate the cake. However Dena was the one who ate the cake, and so Brad is lying.
Easiest way: If D is lying, so is C. If D is telling the truth, so is C. Both C and D must be telling the truth.
If A is lying, both A an B are lying. Not possible. A is telling truth. Only B is lying because D must have ate it.
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Yes, my apologies, I made a typing error - the second version is correct
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Re: all the 2 children replies.
Ya, I should have written it as "A boy answers the door if at all possible" since the way it is written now is ambiguous as to whether a boy just happened to open the door or whether the question is controlling that event. In the Monty hall example, "the host opens a door and reveals a goat" is always assumed (or explicitly stated) to be not random. As Umpteen mentioned, you get different answers if the information was revealed randomly or not.
I really like the 2 part version of the children question that was suggested.
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On April 20 2012 13:32 el_dawg wrote:
Re: all the 2 children replies.
Ya, I should have written it as "A boy answers the door if at all possible" since the way it is written now is ambiguous as to whether a boy just happened to open the door or whether the question is controlling that event. In the Monty hall example, "the host opens a door and reveals a goat" is always assumed (or explicitly stated) to be not random. As Umpteen mentioned, you get different answers if the information was revealed randomly or not.
I really like the 2 part version of the children question that was suggested.
If that's the case it's still 50% as soon as you find out a boy answers the door lol.
+ Show Spoiler [Original Riddle] +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 +So at first there are four possiblities:
Male | Male
Male | Female
Female | Male
Female | Female
Where the left column is who answer the door, and the right column is the unseen child. However you change the rule to say if a boy can answer it, he will. So, if a female answers the door, it's because there is no chance of a male answering it .Which means there are only three possibilities to begin with now..
Male | Male
Male | Female
Female | Female
The thing is, in the question itself, you already determine that the person to answer the door is male. This eliminates the last outcome leaving only the two outcomes with males answering the door:
Male | Male
Male | Female
Which means there's a 50/50 chance the other child is male.
The question would actually be a thinking question if you had instead asked:
What are the chances of a Male child answering the door?
Before telling us who answers the door, in which case the odds would be 66% since there are only 3 possible scenarios and in two of them a male answers it.
Alternatively, you could not tell us the gender of who answer the door at all, and instead say this:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, and you know that a male child will always answer the door first if possible, but you don't know the gender of either of them. Upon arriving, one of the children answers the door, but you can't tell if they are a boy or girl. What is the % chance that the other child is a male. (hint: it is not 50%)
By not telling them who answers the door, you have set it up so each of the three possibilities are still possible, so there's a 66% chance the second child is female, and a 33% chance they are male.
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On April 20 2012 13:20 Release wrote:Show nested quote +On April 20 2012 12:55 killa_robot wrote:On April 20 2012 03:36 Taekwon wrote:
A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!"
Boy B said: "Neither A nor D ate it."
Boy C said: "I swear on my momma's grave I didn't eat it!"
Girl D said: "C is telling the truth!"
One of these children is lying. Who is it? Alice said: "I didn't eat it!" Brad said: "Neither Alice nor Dena ate it." Cory said: "I swear on my momma's grave I didn't eat it!" Dena said: "Cory is telling the truth!" [For reference, I changed their names so it would be easier to talk about] + Show Spoiler [Solution] +Alice
Brad
Cory
Dena
Well let's work through it. For starters we know that one of these children are lying. That means, since no one mentions Brad at all, he can't be the one who ate it.
Alice
Cory
Dena
Now, Cory claims he didn't eat it, and Dena says that Cory is telling the truth. If Cory did eat it, both of them would be lying, and since we know only one person lied, that can't be true. Which means that Cory did not eat it either.
Alice
Dena
Finally Alice claims she did not eat it, and Brad claims that neither Alice or Dena ate it. If Alice really did eat it, then both Brad and Alice would be lying, so Alice cannot be the one who ate it.
Dena
The only suspect left is Dena. Brad claims that neither Alice nor Dena ate the cake. However Dena was the one who ate the cake, and so Brad is lying.
Easiest way: If D is lying, so is C. If D is telling the truth, so is C. Both C and D must be telling the truth. If A is lying, both A an B are lying. Not possible. A is telling truth. Only B is lying because D must have ate it.
As per the rules of the thread, please spoiler that Release
+ Show Spoiler +Regardless, the answer given is incorrect.
Though I appreciate Killa's nuance on the names.
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On April 20 2012 14:01 killa_robot wrote:Show nested quote +On April 20 2012 13:32 el_dawg wrote:
Re: all the 2 children replies.
Ya, I should have written it as "A boy answers the door if at all possible" since the way it is written now is ambiguous as to whether a boy just happened to open the door or whether the question is controlling that event. In the Monty hall example, "the host opens a door and reveals a goat" is always assumed (or explicitly stated) to be not random. As Umpteen mentioned, you get different answers if the information was revealed randomly or not.
I really like the 2 part version of the children question that was suggested. If that's the case it's still 50% as soon as you find out a boy answers the door lol. + Show Spoiler [Original Riddle] +Riddle:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%) So at first there are four possiblities:
Male | Male
Male | Female
Female | Male
Female | Female
Where the left column is who answer the door, and the right column is the unseen child. However you change the rule to say if a boy can answer it, he will. So, if a female answers the door, it's because there is no chance of a male answering it .Which means there are only three possibilities to begin with now..
Male | Male
Male | Female
Female | Female
The thing is, in the question itself, you already determine that the person to answer the door is male. This eliminates the last outcome leaving only the two outcomes with males answering the door:
Male | Male
Male | Female
Which means there's a 50/50 chance the other child is male.
The question would actually be a thinking question if you had instead asked:
What are the chances of a Male child answering the door?
Before telling us who answers the door, in which case the odds would be 66% since there are only 3 possible scenarios and in two of them a male answers it.
Alternatively, you could not tell us the gender of who answer the door at all, and instead say this:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, and you know that a male child will always answer the door first if possible, but you don't know the gender of either of them. Upon arriving, one of the children answers the door, but you can't tell if they are a boy or girl. What is the % chance that the other child is a male. (hint: it is not 50%)
By not telling them who answers the door, you have set it up so each of the three possibilities are still possible, so there's a 66% chance the second child is female, and a 33% chance they are male.
If we play a game where I flip 2 coins (re-flipping both if both are tails) and show you that one is heads while hiding the other, you would guess that the hidden coin has a 33% chance of being heads, right? Since the only way for me to show a heads while hiding a heads is if both were heads (33% chance), otherwise I will show a heads while hiding a tails (66% chance). That is sort of what I was trying to ask with the original question.
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EPT![[image loading]](http://images.wikia.com/wowwiki/images/a/a4/Warden_artwork.jpg)
