3 Door Logic Problem

[Wobble]

Okay I've been thinking about this some more, and I think I have worked out what was bothering me about it. Apparently the original question involved sons and daughters, so here is my own example:

You have gone to see your daughter graduate from an all girls college. While at the graduation you start chatting to one of the other parents. She mentions that she has two children. Given that one is obviously a girl, what is the chance that the other is also a girl?

Then you ask if she has been to a graduation before. "No", she tells you, "it's my eldest who is graduating today." Now that you know her eldest child is a girl, what is the chance the other child is also a girl?

And finally, is this the same problem as the coins and as the original question?

 

[tsg]

Okay I've been thinking about this some more, and I think I have worked out what was bothering me about it. Apparently the original question involved sons and daughters, so here is my own example:

You have gone to see your daughter graduate from an all girls college. While at the graduation you start chatting to one of the other parents. She mentions that she has two children. Given that one is obviously a girl, what is the chance that the other is also a girl?

Assuming each child has an equal probability of being a boy or a girl, the probability that both children are girls is 1/3.

Of BB, BG, GB, and GG, only BB is ruled out.

Then you ask if she has been to a graduation before. "No", she tells you, "it's my eldest who is graduating today." Now that you know her eldest child is a girl, what is the chance the other child is also a girl?

1/2. Now BB and BG are ruled out.

And finally, is this the same problem as the coins and as the original question?

It is identical to the coins problem. It's similar to Monty Hall (the strict version) in that it has to do with how much information you are being given.

 

[Wobble]

Damn, now I come to look at this again I think I screwed my example up. Since parents who have two daughters are likely to attend twice as many graduations (at that college), than those with one daughter, there should be twice as many there as expected. So if all the parents there had exactly two children, one would expect half of them to have two girls, and half to have one boy and one girl.

Guess I will have to think some more.

 

[Lamuella]

Damn, now I come to look at this again I think I screwed my example up. Since parents who have two daughters are likely to attend twice as many graduations (at that college), than those with one daughter, there should be twice as many there as expected. So if all the parents there had exactly two children, one would expect half of them to have two girls, and half to have one boy and one girl.

Guess I will have to think some more.

I really don't see how that follows, and I really don't see what that would have to do with the puzzle.

You're not looking at this woman as a representative sample of the group, you're looking at her as an individual. The only variables that matter are the information you know about her children. All you know about her children is that she has at least one girl, or that her eldest is a girl.

 

[tsg]

Damn, now I come to look at this again I think I screwed my example up. Since parents who have two daughters are likely to attend twice as many graduations (at that college), than those with one daughter, there should be twice as many there as expected. So if all the parents there had exactly two children, one would expect half of them to have two girls, and half to have one boy and one girl.

Assuming a constant class size from year to year, the parents of two daughters will go to twice as many graduations as the parents of one daughter and one son. That doesn't mean the number of parents of two daughters will be double at each graduation, because there will be twice as many parents who go to one graduation at different times (excluding twins or any other event that would have two siblings in the same class).

I hope that was clear.

 

[Wobble]

Assuming a constant class size from year to year, the parents of two daughters will go to twice as many graduations as the parents of one daughter and one son. That doesn't mean the number of parents of two daughters will be double at each graduation, because there will be twice as many parents who go to one graduation at different times (excluding twins or any other event that would have two siblings in the same class).

I hope that was clear.

It's not clear to me I'm afraid.

Imagine if every girl in the city went to this college, and they all graduated from it. You don't know the ages of any of the girls, so you can't predict which will be at the graduation. As far as you know, every girl has an equal chance of being there. So if a family has two girls, it has twice the chance of being at the graduation. (Ignoring twins.)

What is wrong with that?

 

[CurtC]

I think you two are saying the same thing, and I agree with both of you - the parent with two girls is twice as likely to be at that graduation, so the other kid will be a boy 50% of the time. You can list all the options and look at the proportion to find probabilities, only when each of those listed options has equal chance of happening.

I think your scenario is equivalent to the formulation that I have used in the past, where you happen to meet up with an old acquaintance at the grocery store. She has a girl with her, and tells you that she has two children. What's the chance that the other is a boy? You know it's 50%, even though you don't know whether the kid with her is the older or younger sibling.

 

[tsg]

It's not clear to me I'm afraid.

Imagine if every girl in the city went to this college, and they all graduated from it. You don't know the ages of any of the girls, so you can't predict which will be at the graduation. As far as you know, every girl has an equal chance of being there. So if a family has two girls, it has twice the chance of being at the graduation. (Ignoring twins.)

What is wrong with that?

You're confusing your probability space. Choosing a parent in the city at random, a parent of two girls is twice as likely to be at this year's graduation than a parent of one girl. But, once at the graduation, a parent picked at random is twice as likely to be the parent of one girl than the parent of two girls. That 1/3 of the parents in the city will also be at another graduation (past or future) doesn't mean there will be more of them at this one.

 

[tsg]

I think you two are saying the same thing, and I agree with both of you - the parent with two girls is twice as likely to be at that graduation, so the other kid will be a boy 50% of the time. You can list all the options and look at the proportion to find probabilities, only when each of those listed options has equal chance of happening.

I think your scenario is equivalent to the formulation that I have used in the past, where you happen to meet up with an old acquaintance at the grocery store. She has a girl with her, and tells you that she has two children. What's the chance that the other is a boy? You know it's 50%, even though you don't know whether the kid with her is the older or younger sibling.

This is incorrect for reasons previously stated.

 

[CurtC]

This is incorrect for reasons previously stated.

I guess I didn't understand those reasons previously stated. Could you clarify them? I believe that if you see a parent of two children at a daughter's graduation ceremony, there is a 50% chance that the other kid is a boy. If you think otherwise, please explain.

 

[Wobble]

I am not sure whether tsg is right or not about the graduation thing, but it's not really relevent to the actual problem. So I have made another example, which hopefully avoids the problem and actually phrases the question the way I want:

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?

And is this the same as the coins question?

 

[Wobble]

I think you two are saying the same thing, and I agree with both of you - the parent with two girls is twice as likely to be at that graduation, so the other kid will be a boy 50% of the time. You can list all the options and look at the proportion to find probabilities, only when each of those listed options has equal chance of happening.

I think your scenario is equivalent to the formulation that I have used in the past, where you happen to meet up with an old acquaintance at the grocery store. She has a girl with her, and tells you that she has two children. What's the chance that the other is a boy? You know it's 50%, even though you don't know whether the kid with her is the older or younger sibling.

I agree with this.

 

[Rasmus]

I am not sure whether tsg is right or not about the graduation thing, but it's not really relevent to the actual problem. So I have made another example, which hopefully avoids the problem and actually phrases the question the way I want:

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?

Any family with two children is equally likely to have

- two girls
- a younger brother and an older sister
- a younger sister and an older brother
- two boys

Now, any family with an older sibling that is a girl is either case one or case two.

So: 50%.

(If you didn't know it was the older child, it could have also been case 3, and thus a 1/3 chance that the other kid would also be a girl.)

And is this the same as the coins question?[/QUOTE]

Yes, I think so.

 

[Lamuella]

I guess I didn't understand those reasons previously stated. Could you clarify them? I believe that if you see a parent of two children at a daughter's graduation ceremony, there is a 50% chance that the other kid is a boy. If you think otherwise, please explain.

Explanation:

Let's assume that having a boy has a probability of exactly 50%.

Thus, when you have two children, the odds of having two girls is 25%; the odds of having a boy, then a girl, is 25%; the odds of having a girl, then a boy is 25%; and the odds of having two boys is 25%.

We know that this woman did not have two boys. This is all we know about her. Thus the chances of her having had two girls is now 33%, the chances of her having had a boy then a girl is 33%, and the chances of her having had a girl then a boy is 33%

 

[Wobble]

Explanation:

Let's assume that having a boy has a probability of exactly 50%.

Thus, when you have two children, the odds of having two girls is 25%; the odds of having a boy, then a girl, is 25%; the odds of having a girl, then a boy is 25%; and the odds of having two boys is 25%.

We know that this woman did not have two boys. This is all we know about her. Thus the chances of her having had two girls is now 33%, the chances of her having had a boy then a girl is 33%, and the chances of her having had a girl then a boy is 33%

But what if you look at it this way: There are two children, the one in the shop and the other one. So there are four possibilities:

shop | B B G G
____________________

other | B G B G

Since the one we have met is a girl, we are left with the last two, giving a 50% chance for either boy or girl at home.

On the other hand, if we meet the woman when she is alone, and ask whether she has any girls (and we know that she has two children), then we only eliminate one of the possibilities (the first one), so there is indeed a 33% chance that the other child is a girl.

 

[tsg]

I guess I didn't understand those reasons previously stated. Could you clarify them? I believe that if you see a parent of two children at a daughter's graduation ceremony, there is a 50% chance that the other kid is a boy. If you think otherwise, please explain.

Start here. The problems are identical.

There is a very simple way to test this. Toss two coins 100 times. Count the number of times there is at least one head. If you're right, the number of times there is two heads should be around half that number.

Here's how I predict it will happen.
~25 times you will get two heads.
~50 times you will get a head and a tail.
~25 times you will two tails.

Of the ~75 times there is at least one head, ~25 of them will be two heads, and ~50 times there is a head and a tail, making the probability 25/75 or 1/3 that the other coin is also a head.

 

[CurtC]

Now, any family with an older sibling that is a girl is either case one or case two.

So: 50%.

(If you didn't know it was the older child, it could have also been case 3, and thus a 1/3 chance that the other kid would also be a girl.)

I think Wobble included the statement about which is the older sibling, as a red herring. That can't make a difference in the probability. Even if you don't know which is the elder child, the probability is 50%. In that example, slicing the data sets by older/younger doesn't make sense. Instead you can divide the population of possible parents of two children, accompanied by one of them only, into these populations:

Boy with mom, Boy at home.
Boy with mom, Girl at home.
Girl with mom, Boy at home.
Girl with mom, Girl at home.

This should make it clearer - there is a girl with her mom, which rules out the first two equally likely possibilities, therefore the chance that the other is a boy is 50%, without any regard to which is the older one.

 

[CurtC]

Start here. The problems are identical.

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.

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

 

[CurtC]

Thus, when you have two children, the odds of having two girls is 25%; the odds of having a boy, then a girl, is 25%; the odds of having a girl, then a boy is 25%; and the odds of having two boys is 25%.

We know that this woman did not have two boys. This is all we know about her. Thus the chances of her having had two girls is now 33%, the chances of her having had a boy then a girl is 33%, and the chances of her having had a girl then a boy is 33%

I mentioned earlier about the problem with figuring odds by listing all the possible outcomes, and counting them. The problem is that to do that, each of those possibilities needs to be equally as likely. In the problem with meeting a parent at her daughter's graduation, that's not the case.

Sure, the parents of two boys will not be represented here, but the parents of two girls will be doubly represented. The subset of them in the general population is still 25%, but here at this graduation ceremony, there will be twice as many from this subset. Therefore the number of parents at the graduation with two girls is equal to the number with a boy and a girl, therefore 50%.

Another way to look at it is by slicing the population, not by age, but by which kid is of graduating age:

Kid of graduating age / kid not of graduating age
B / B : 25%
B / G : 25%
G / B : 25%
G / G : 25%

Since the parent you are talking to has a graduating daughter, the first two possibilities are ruled out, and among the equally likely last two, half have two girls, half have a boy and a girl.

 

[tsg]

I am not sure whether tsg is right or not about the graduation thing, but it's not really relevent to the actual problem. So I have made another example, which hopefully avoids the problem and actually phrases the question the way I want:

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?

1/2

And is this the same as the coins question?

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