3 Door Logic Problem

[tsg]

If you're talking about the problem stated in the first post of this thread (which you linked to so I assume you are), I wholeheartedly agree that the 1/3 answer is correct.

I wasn't. I was talking about the coin problem linked to in the post I referenced and the subsequent discussion.

However, the parent at a daughter's graduation is a different animal, which I'll explain in the next post.

It isn't. Perform the coin experiment I posted eariler and you will see.

 

[CurtC]

It isn't. Perform the coin experiment I posted eariler and you will see.

I certainly agree with your analysis of the coin experiment. Were you discussing the graduation event too?

Yes. Knowing that the girl is the eldest child gives you more information with which to determine the possibilities.

It's not that you know that she is the eldest - the probability for that problem would still be 50% even if that tidbit had been left out. The fact that you know a specific kid is a girl gives you more information, regardless of which is older.

 

[tsg]

I certainly agree with your analysis of the coin experiment. Were you discussing the graduation event too?

Yes. It is identical to the coins problem. Substitute "girl" for "heads", "boy" for "tails" and "eldest child" for "first coin". There is no difference.

It's not that you know that she is the eldest - the probability for that problem would still be 50% even if that tidbit had been left out. The fact that you know a specific kid is a girl gives you more information, regardless of which is older.

But you don't know anything about a specific kid in the case when you are not told she is the eldest. All you know is that the mother has at least one girl. That the girl is with her doesn't eliminate any possibilities where the other child is a boy.

Let's throw some numbers at it just to see what happens:

A city of 1000 mothers with two children each.
500 of those will have a girl as the older child.
500 of those will have a boy as the older child.

Of the 500 with a girl as the older, 250 of those will have a girl as the younger child, and 250 will have a boy as the younger child.

Of the 500 with a boy as the older, 250 of those will have a girl as the younger child, and 250 will have a boy as the younger child.

250 mothers have two girls, 500 have a girl and a boy, 250 have two boys.

You meet a mother of two children that has a girl with her. What are the odds the other child is a girl? Of the 1000 mothers with two children, 750 of them have at least one girl. That she has chosen to bring the girl with her doesn't change this number[1]. Of those 750, only 250 of them have two girls. The chance her other child is a girl is 250 out of 750 or 1/3.

Now she tells you the girl she is with is her older child. What are the odds the other child is a girl? Of the 1000 mothers with two children, 500 of them have a girl as the oldest child. Of those 500, 250 have a girl as the younger child as well. The chance the other child is also a girl is 250 out of 500 or 1/2.

ETA: sentence at [1].

 

[tsg]

If the older / younger thing bothers you, let's go back to Wobble's puzzle with the same numbers:

1000 mothers who each have two children.
250 of them will have two girls.
500 of them will have a girl and a boy.
250 of them will have two boys.

The mothers who are able to are instructed to bring a girl child to the palace for a reward. You meet one of them on the road. What are the chances she has two girls? 750 mothers have at least one girl and thus qualify for the reward. 250 of those have two girls. 250 out of 750 is 1/3. The other 500 have only one girl. 500 out of 750 is 2/3. She is twice as likely to have only one girl as two.

 

[Wobble]

Here is my answer to the puzzle I posted earlier. I'm expecting people to disagree though:

Mad King Thud of Ugland is concerned that there are not enough girls in his kingdom. To encourage parents to have more girls, he decides to give a prize to each family that has at least one daughter. In order to claim the prize, a daughter from the family must go to the palace, where the King will give her 100 Uggets. The King decrees that if the family has more than one daughter, they must choose one at random to send to the palace.

While travelling in Ugland, you meet a mother and daughter who are travelling to the palace to collect the prize. She mentions to you that she has two children, and you ask her whether the child she has with her is the eldest. She says yes.

So what is the chance that the other child is also a girl?

The chance that the other child is also a girl is 1/3.

I agree with tsg's post about the chance being 1/3 if you just meet the mother, and know she has two children. Knowing that the girl with her is the eldest child does not make any difference.

To see this, imagine you start with 1000 families, and look at the different possibilities:

Families going to palace:

BG - 250
GB - 250
GG - 250

Now consider which families have sent the eldest daughter:

BG - 0
GB - 250
GG - 125 (The King ordered them to send a daughter at random, so the other 125 sent the youngest daughter.)

Since you know that this family sent the eldest daughter, you know that there is 1/3 chance she has a sister.

And is this the same as the coins question?

No. The coins puzzle was not well enough defined to answer. The point of this example was to get rid of the psychology aspect so it can be properly answered, and tested.

 

[tsg]

Families going to palace:

BG - 250
GB - 250
GG - 250

Now consider which families have sent the eldest daughter:

BG - 0
GB - 250
GG - 125 (The King ordered them to send a daughter at random, so the other 125 sent the youngest daughter.)

[ignore] Bah, I misread.

ETA: You're original puzzle never specified they would choose between the older or younger daughter with equal probability.

 

[CurtC]

You meet a mother of two children that has a girl with her. What are the odds the other child is a girl? Of the 1000 mothers with two children, 750 of them have at least one girl. That she has chosen to bring the girl with her doesn't change this number[1]. Of those 750, only 250 of them have two girls. The chance her other child is a girl is 250 out of 750 or 1/3.

There are two very different formulations of the problem that have been bandied about, so let's be specific about which one we're talking about.

In Wobble's example of the kingdom where mothers with a daughter must travel with a daughter, I see that we all agree that the probability of the other kid being a girl is 1/3, regardless of whether we know which is the older one. I think we agree, right?

In his other example (described in this post), where you meet a parent at an all-girl's graduation, and learn that this parent has two children. I am certain that in this case, the odds of the other kid being a girl are 1/2, again regardless of whether we know whether the graduating daughter is the oldest. Do you agree or disagree? I am quite certain of this and would bet any amount of money on it.

 

[Wobble]

[ignore] Bah, I misread.

ETA: You're original puzzle never specified they would choose between the older or younger daughter with equal probability.

No it didn't, though that was what I meant by "choose one at random".

This kind of shows the problem with the original question. I was not sure of the answer, and imagined trying to test it by running trials. But that brought me to the psychology aspect: if you want to make repeated tests, then person A must follow some kind of rule about when to give extra information, and what information to give. And the conclusions you can draw from their statement clearly depend on what rule they are following.

My example was supposed to be equivalent to the coin problem when person A is following the rule where (supposing they are asked if they have any heads) they reply:

TT - no
HT - first coin head
TH - second coin head
HH - 'first coin head' 50% of the time, 'second coin head' 50% of the time

Of course they could be using a completely different rule, which is why I think the original coin question is unanswerable.

 

[tsg]

No it didn't, though that was what I meant by "choose one at random".

It's no doubt a semantic quibble, but an important one. If you are going to eliminate half the population of mothers with two girls on the assumption that half of them would have brought the younger, you have to know that for a fact. Otherwise you can't eliminate them from the set.

This kind of shows the problem with the original question. I was not sure of the answer, and imagined trying to test it by running trials. But that brought me to the psychology aspect: if you want to make repeated tests, then person A must follow some kind of rule about when to give extra information, and what information to give. And the conclusions you can draw from their statement clearly depend on what rule they are following.

If you don't know the rule they are following, you can't make any assumptions on what their motivations were for making the statement.

My example was supposed to be equivalent to the coin problem when person A is following the rule where (supposing they are asked if they have any heads) they reply:

TT - no
HT - first coin head
TH - second coin head
HH - 'first coin head' 50% of the time, 'second coin head' 50% of the time

Of course they could be using a completely different rule, which is why I think the original coin question is unanswerable.

The original question is answerable. Probability is about estimating the likelihood of an event with certain given information. If I have flipped a coin but not looked at it, what is the probability it is heads? I don't know what the outcome is but I know that there is a 50% chance it is heads, assuming the canonical fair coin. Half the time I perform this experiment the coin will turn out to be heads and the other half it will turn out to be tails. It is certainly one or the other - it can't be half heads and half tails. But I don't know which. If I look at the coin and see it is indeed heads-up, then the probability that the coin landed heads up is now 100%, because I have more information that I can use to eliminate the possible outcomes where it came up tails. If we don't know the motivations behind person A for answering "yes the first one was a head" we can't eliminate any of the possibilities where the second coin was also a head. Exploring the motivations of A is a psychological discussion based on what most people in his position would be more likely to do. Not that it isn't a valid discussion, but it isn't the point of the puzzle. What is the point of the puzzle is that knowing the first coin was a head eliminates the possibilities that the first coin was a tail, where knowing only that at least one coin is a head does not.

 

[Wobble]

The original question is answerable. Probability is about estimating the likelihood of an event with certain given information. If I have flipped a coin but not looked at it, what is the probability it is heads? I don't know what the outcome is but I know that there is a 50% chance it is heads, assuming the canonical fair coin. Half the time I perform this experiment the coin will turn out to be heads and the other half it will turn out to be tails. It is certainly one or the other - it can't be half heads and half tails. But I don't know which. If I look at the coin and see it is indeed heads-up, then the probability that the coin landed heads up is now 100%, because I have more information that I can use to eliminate the possible outcomes where it came up tails. If we don't know the motivations behind person A for answering "yes the first one was a head" we can't eliminate any of the possibilities where the second coin was also a head. Exploring the motivations of A is a psychological discussion based on what most people in his position would be more likely to do. Not that it isn't a valid discussion, but it isn't the point of the puzzle. What is the point of the puzzle is that knowing the first coin was a head eliminates the possibilities that the first coin was a tail, where knowing only that at least one coin is a head does not.

The problem I have with the original puzzle is that I cannot see any way to test it, even in theory. And in that case, how can it be said to have an answer? How can you work out the probability of something unless you can at least imagine it happening more than once?

 

[tsg]

In Wobble's example of the kingdom where mothers with a daughter must travel with a daughter, I see that we all agree that the probability of the other kid being a girl is 1/3, regardless of whether we know which is the older one. I think we agree, right?

Given the stipulation that half the mothers with two daughters will leave the older one home, yes I agree.

In his other example (described in this post), where you meet a parent at an all-girl's graduation, and learn that this parent has two children. I am certain that in this case, the odds of the other kid being a girl are 1/2, again regardless of whether we know whether the graduating daughter is the oldest. Do you agree or disagree?

I disagree. In order for this to be true, having a boy and a girl would have to be equally as likely as having having two girls, which isn't the case. With no other information to eliminate half the mothers with a boy and a girl, the chance of having two girls is 1/3.

 

[CurtC]

Great, so now we know exactly what we disagree about. Specifically what I disagree with in your last paragraph, is that the mothers with two daughters go to twice as many graduations as mothers with mixed kids, so they are over-represented in the population we're selecting from.

I think the easiest way to look at it is to divide the population of parents who have two kids, one of graduating age, into these four equally likely groups:

Sex of graduating age kid / Sex of non-graduating kid
B / B : 25%
B / G : 25%
G / B : 25%
G / G : 25%

The fact that this parent is at a girl's graduation lets us rule out the first two, therefore the chance of the non-graduating kid being a girl is 1/2.

Can you find anything wrong with my analysis?

 

[tsg]

The problem I have with the original puzzle is that I cannot see any way to test it, even in theory. And in that case, how can it be said to have an answer? How can you work out the probability of something unless you can at least imagine it happening more than once?

Toss a pair of coins 100 times. Count the number of times that you get A) two heads, B) one head and one tail, C) two tails, D) at least one head (sum of A and B), and E) the first coin is a head.

A should be approximately 25.
B should be approximately 50.
C should be approximately 25.
D should be approximately 75.
E should be approximately 50.

The chances that the other coin is a head when you only know that at least one coin is a head will be A / D or 25 / 75 = 1/3.

The chances that the other coin is a head when you know the first coin is a head will be A / E or 25 / 50 = 1/2.

 

[tsg]

Great, so now we know exactly what we disagree about. Specifically what I disagree with in your last paragraph, is that the mothers with two daughters go to twice as many graduations as mothers with mixed kids, so they are over-represented in the population we're selecting from.

That they go to twice as many graduations doesn't mean there are twice as many of them at this one. That I buy a lottery ticket for two different drawings doesn't mean I have more chance of winning any single drawing than the people who only buy one.

I think the easiest way to look at it is to divide the population of parents who have two kids, one of graduating age, into these four equally likely groups:

Sex of graduating age kid / Sex of non-graduating kid
B / B : 25%
B / G : 25%
G / B : 25%
G / G : 25%

This assumes that being of graduating age is equally likely as not, which is not a reasonable assumption to make.

 

[CurtC]

That they go to twice as many graduations doesn't mean there are twice as many of them at this one.

Yes, I think it does, on average anyway. If a subset of the population attends twice as often on average, then on average they will be over-represented in this sample by 2x their proportion in the population.

This assumes that being of graduating age is equally likely as not, which is not a reasonable assumption to make.

Out of the population with two kids, one of which is of graduating age, half of the time the graduating kid will be the older, half the time it will be the younger.

 

[tsg]

Yes, I think it does, on average anyway. If a subset of the population attends twice as often on average, then on average they will be over-represented in this sample by 2x their proportion in the population.

It doesn't. The mothers of two girls will attend 2 graduations in their lifetimes. They won't all be at the same graduations. Most mothers of two girls will not be at a particular graduation. Assuming a population that remains relatively constant and evenly distributed, the number of girls of graduating age will also be relatively constant from graduation to graduation. By the definition of the problem, a girl of graduating age can only have one mother and a mother can only have at most one girl of graduating age. If 1/3 of the girls in the population have a sister, then 1/3 of the girls of graduating age will have a sister. The only way the mothers of two girls could be over-represented is if the number of girls of graduating age with sisters is over-represented.

Out of the population with two kids, one of which is of graduating age, half of the time the graduating kid will be the older, half the time it will be the younger.

This is not so. We have no information that tells us a child of graduating age is equally likely to have an older or younger sibling.

 

[CurtC]

This is not so. We have no information that tells us a child of graduating age is equally likely to have an older or younger sibling.

With the fairly simple assumption that the population of girl births and boy births has remained constant over a very few years, we can then say that with a child of graduating age with one sibling, it is equally likely that that sibling can be younger, or older. But this doesn't really matter anyway, because the whole age issue is a red herring.

In the population of mothers with two children, one of whom is a daughter at graduating age, half of them will have a two girls, half will have a girl and a boy. And I'm resisting the temptation to state it as an analagous scenario, until we get agreement on this one.

 

[Lamuella]

The problem I have with the original puzzle is that I cannot see any way to test it, even in theory.

I can.

get two friends. Give each two coins. Get Friend 1 to flip two coins one hundred times each and record the outcome each time it lands heads up on the first coin. Get friend 2 to record the outcome each time it lands heads up on at least one coin.

Get friend 1 to tally the number of double heads as a percentage of outcomes he recorded.

Get friend two to tally the number of double heads as a percentage of outcomes he recorded.

Compare.

 

[tsg]

In the population of mothers with two children, one of whom is a daughter at graduating age, half of them will have a two girls, half will have a girl and a boy. And I'm resisting the temptation to state it as an analagous scenario, until we get agreement on this one.

There is no way this can be true unless having two girls is as likely as a boy and a girl. It isn't.

 

[CurtC]

Really? It seems self-apparent to me, so I'll try to explain myself more methodically.

I'll divide up the population of parents who have two kids, one of graduating age, into these equally-likely groups:

grad boy, older girl 12.5%
grad boy, younger girl 12.5%
grad boy, older boy 12.5%
grad boy, younger boy 12.5%
grad girl, older girl 12.5%
grad girl, younger girl 12.5%
grad girl, older boy 12.5%
grad girl, younger boy 12.5%

Now, since we know that this parent has a girl who is graduating, it lets us eliminate the first four from the pool. Out of those that are left (and each is equally likely), half of them have a boy, and half have two girls.

Again, in the case of the meeting-a-parent-at-an-all-girls-graduation scenario, the chance of the other kid being a girl is 1/2.