3 Door Logic Problem

[tsg]

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

If you meet a mother of two with her daughter, she is twice as likely to have a boy as another girl. This has been shown. Knowing the age of the girl doesn't add any information.

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%

You're counting your girl-girl combination twice. Only one girl of the pair can be of graduating age.

 

[Oldpossum]

If you meet a mother of two with her daughter, she is twice as likely to have a boy as another girl. This has been shown. Knowing the age of the girl doesn't add any information.

(ETA Oldpossum Insert) Correct.

You're counting your girl-girl combination twice. Only one girl of the pair can be of graduating age.

This assumption is not Incorrect. In this case, the mother of two girls, is twice as likely to be present at the graduation of a Daughter, as the mother of a Boy and a girl. so the girl-girl combination needs to be counted twice, hence bringing the odds back up to 1 in 2 from the base 1 in 4.

ie a Mother of a girl and a boy, only gets to attend one Daughter graduation ceremony, (but there are two combinations, so base odds of this event is 1 in 2) where the mother of two girls, (a 1 in 4 base odds) will get to attend two ceremonies, thereby doubling the odds of encountering this woman at such a ceremony, thus bringing the final odds of this scenario back up to 1 in 2

Now if the Sex of the child graduating was not specified, only the sex of the child present, then your scenario of having double the chance of meeting a woman with a girl and boy, compared to meeting a woman with two girls, works at the odds you specified.

Specifying the sex of the graduating child, is akin to specifying that the first coin tossed is heads, it limits how further odds are calculated, by eliminating some posibilities.

 

[Wobble]

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.

The problem with this is that it is not the same as the original problem. You have made the assumption that person A would always give you exactly this information, provided it was true. I am saying that this assumption is not justified, and the answer you give depends on what assumption you make about person A's behaviour.

Let me ask you another question:

If person A had said "Yes, the second coin I tossed was a head.", what is the probability now that they have two heads?

 

[tsg]

The problem with this is that it is not the same as the original problem. You have made the assumption that person A would always give you exactly this information, provided it was true. I am saying that this assumption is not justified, and the answer you give depends on what assumption you make about person A's behaviour.

Without any further information, we can't make any assumptions about A's behavior. We can only count the number of times it is possible for him to answer that way truthfully.

Let me ask you another question:

If person A had said "Yes, the second coin I tossed was a head.", what is the probability now that they have two heads?

Read strictly, 1/2. The important point is that you know which coin is a head rather than just that one of them is. There is an argument to be made that he wouldn't have specified the second one if the first one was a head, but in a strict probability problem we can't make that assumption.

 

[tsg]

This assumption is not Incorrect. In this case, the mother of two girls, is twice as likely to be present at the graduation of a Daughter, as the mother of a Boy and a girl. so the girl-girl combination needs to be counted twice, hence bringing the odds back up to 1 in 2 from the base 1 in 4.

ie a Mother of a girl and a boy, only gets to attend one Daughter graduation ceremony, (but there are two combinations, so base odds of this event is 1 in 2) where the mother of two girls, (a 1 in 4 base odds) will get to attend two ceremonies, thereby doubling the odds of encountering this woman at such a ceremony, thus bringing the final odds of this scenario back up to 1 in 2

But there are twice as many mothers of a girl and a boy as mothers of two girls.

Now if the Sex of the child graduating was not specified, only the sex of the child present, then your scenario of having double the chance of meeting a woman with a girl and boy, compared to meeting a woman with two girls, works at the odds you specified.

Specifying the sex of the graduating child, is akin to specifying that the first coin tossed is heads, it limits how further odds are calculated, by eliminating some posibilities.

I have to think about this one some more, because there is definitely something wrong here, but I can't put my finger on it. What's bothering me is that the GG combinations are exclusive, ie if one girl is of graduating age, the other can't be, and it's only showing up on one side of the calculation because the boys are being eliminated based on gender. I am also not convinced that reconstructing the population as CurtC did accurately reflects the distribution of that population.

 

[CurtC]

I guess we're ready for another scenario, tsg. You meet an old long lost friend at the grocery store. This friend has a daughter with her. In conversation, you find out she now has two children.

Given this info, what's the probability that the other is a girl?

1/2

 

[CurtC]

Sorry getting back to this, but we simulposted.

But there are twice as many mothers of a girl and a boy as mothers of two girls.

True.

But, the mothers of two girls are represented in this sample at double the rate of the mixed parents.

 

[Wobble]

Read strictly, 1/2. The important point is that you know which coin is a head rather than just that one of them is. There is an argument to be made that he wouldn't have specified the second one if the first one was a head, but in a strict probability problem we can't make that assumption.

Okay, now imagine you have been playing this game for a few rounds with the same person. Each round person A tosses two coins, and you ask whether there were any heads:

Round 1:

Person A says "Yes, the first coin is a head."

What is the chance the other coin is also a head?

Round 2:

Person A says "Nope."

Round 3:

Person A says "Yes, the second coin is a head."

What is the chance the other coin is also a head?

 

[tsg]

I guess we're ready for another scenario, tsg. You meet an old long lost friend at the grocery store. This friend has a daughter with her. In conversation, you find out she now has two children.

Given this info, what's the probability that the other is a girl?

We've been through this. It's 1/3. She's twice as likely to have a brother as a sister.

Of families with two children:

25% will be BB
50% will be BG
25% will be GG

Knowing one child is a girl only eliminates the possibility of two boys. Of the remaining possibilities, BG outnumbers GG 2 to 1.

ETA:
This is different from the graduation problem. If the answer to that problem is indeed 1/2 (and I am not yet ready to concede this, although it looks like it might be so), it's because the parent in question is required to bring a daughter with him. There is no chance she will bring a boy, and we know that. There is nothing in this scenario that says she can't bring a boy, only that she happened not to. All we know is that she has at least one daughter, nothing more.

 

[tsg]

Okay, now imagine you have been playing this game for a few rounds with the same person. Each round person A tosses two coins, and you ask whether there were any heads:

Round 1:

Person A says "Yes, the first coin is a head."

What is the chance the other coin is also a head?

Round 2:

Person A says "Nope."

Round 3:

Person A says "Yes, the second coin is a head."

What is the chance the other coin is also a head?

This is a different scenario than the original problem. Now you can begin to question why he chose to tell you it was the second coin rather than the first one.

But in the original problem, there are two possibilities that satisfy his statement, HH and HT. TT and TH do not. Of those two possibilities, which are equally likely, HH occurs once. Therefore the chances his other coin is a head is 1/2.

 

[Wobble]

This is a different scenario than the original problem. Now you can begin to question why he chose to tell you it was the second coin rather than the first one.

But in the original problem, there are two possibilities that satisfy his statement, HH and HT. TT and TH do not. Of those two possibilities, which are equally likely, HH occurs once. Therefore the chances his other coin is a head is 1/2.

So simply repeating the game makes it different? After the first round you are happy to tell me the answer but if I then play the game again that changes the answer you gave?

If you cannot repeat the game then you cannot test it, which was what I said earlier. And if you can't test it how can you say it has an answer at all?

 

[Wobble]

I guess we're ready for another scenario, tsg. You meet an old long lost friend at the grocery store. This friend has a daughter with her. In conversation, you find out she now has two children.

Given this info, what's the probability that the other is a girl?

Assuming that she is equally likely to have either child with her, 1/2.

 

[tsg]

So simply repeating the game makes it different?

Possibly. You get more information in your scenario. But the original problem never said anything about doing it more than once. And the scenario in the original problem is very specific: he tells you that the first coin is a head. This is all you have to go on. Speculating about what he would do if you repeated the game, while possibly interesting, doesn't affect the problem we are given.

After the first round you are happy to tell me the answer but if I then play the game again that changes the answer you gave?

If you cannot repeat the game then you cannot test it, which was what I said earlier. And if you can't test it how can you say it has an answer at all?

Because it does. When a person tells you he has two coins, one of which is a head, there are three equally likely possibilities that satisfy that statement, one of which is true. If he tells you a specific coin is a head, there are two equally likely possibilities that satisfy that statement, one of which is true. Trying to read any motivation on the part of the person making the statement is making assumptions that aren't justified. If you know that he wouldn't specify which coin was heads up if they both were, then you can eliminate the possibilities when they are both heads-up. But we don't know that, therefore we can't.

 

[CurtC]

I guess we're ready for another scenario, tsg. You meet an old long lost friend at the grocery store. This friend has a daughter with her. In conversation, you find out she now has two children.

Given this info, what's the probability that the other is a girl?

We've been through this. It's 1/3. She's twice as likely to have a brother as a sister.

Nope, it's 1/2.

Here is the key difference between this puzzle, and the classic puzzle for which the answer is 1/3: in the classic puzzle, you're asking whether either of the kids is a girl, so the probabilities of those two are mixed together. In my puzzle here, and in the graduation example, we know that a specific child is a girl, so the chance that the other kid is also a girl is 1/2.

You keep wanting to break it down by which is older or younger, and that is one way to specify a certain kid, but it's just as valid to specify a certain kid according to which one is with the parent and which one is still at home. If we know that one specific kid is a girl, the chance that the other is also a girl is 1/2.

 

[tsg]

Okay, here's where I'm at.

In both the graduation problem and the Mad King problem, the chances of the other child also being a girl, I'm almost convinced, are 1/2. But that's because the person we meet is required to bring a girl. The possibilities of them choosing to bring a boy with them are eliminated by this constraint, and therefore half the BG possibilities are eliminated. To equate it with the coin analogy, it would be the same as one coin having a head on both sides. Half the HT combinations are eliminated because there is only one coin that can come up tails.

Meeting a woman with a daughter, on the other hand, is not limited by this constraint. Why the daughter is with her, we are not told. There is no known requirement that it be a daughter rather than a son. She just happens to have the daughter with her. In this case, the fact the daughter is with her is identical to being told she has at least one daughter, nothing more. Without knowing more than that, we can't eliminate half the BG possibilities like we can in the other two problems and the problem where we know the daughter is the oldest child. The chances she has another daughter are 1/3.

 

[CurtC]

Meeting a woman with a daughter, on the other hand, is not limited by this constraint. Why the daughter is with her, we are not told. There is no known requirement that it be a daughter rather than a son. She just happens to have the daughter with her. In this case, the fact the daughter is with her is identical to being told she has at least one daughter, nothing more.

That's where you are going astray - it's not identical. Saying she has at least one daughter (out of two kids) is equivalent to saying that she does not have two boys. Seeing that the particular kid with her is a girl is not the same thing.

Without knowing more than that, we can't eliminate half the BG possibilities like we can in the other two problems and the problem where we know the daughter is the oldest child. The chances she has another daughter are 1/3.

Again, it doesn't matter which is oldest or youngest. We know that the specific kid with her is a girl, therefore the other kid being a girl has a 1/2 probability.

 

[Wobble]

Possibly. You get more information in your scenario. But the original problem never said anything about doing it more than once. And the scenario in the original problem is very specific: he tells you that the first coin is a head. This is all you have to go on. Speculating about what he would do if you repeated the game, while possibly interesting, doesn't affect the problem we are given.

So if in the original problem had said that the game would be repeated, you would have answered differently? What would your answer be in that case? And if it is different, how can merely knowing the game might be played more than once give you more information?

Also do you agree that there is no way to test the original problem?

 

[Wobble]

That's where you are going astray - it's not identical. Saying she has at least one daughter (out of two kids) is equivalent to saying that she does not have two boys. Seeing that the particular kid with her is a girl is not the same thing.

Again, it doesn't matter which is oldest or youngest. We know that the specific kid with her is a girl, therefore the other kid being a girl has a 1/2 probability.

The equivalent using coins is to have a friend toss two coins, then pick one at random (and yes tsg I mean that there is an equal chance of picking either coin) and hand it to you. What is the chance the other coin has the same side showing?

This could easily be tested.

 

[tsg]

That's where you are going astray - it's not identical. Saying she has at least one daughter (out of two kids) is equivalent to saying that she does not have two boys.

And the remaining choices are she has a boy and a girl, and two girls. A boy and a girl is twice as likely.

Seeing that the particular kid with her is a girl is not the same thing.

We know nothing of that particular kid except that it is a girl. That the girl is standing with her doesn't give us any more information than that she has at least one girl.

Again, it doesn't matter which is oldest or youngest. We know that the specific kid with her is a girl, therefore the other kid being a girl has a 1/2 probability.

No, it's not.

 

[Skeptic Ginger]

Everyone has a different way of looking at this and perhaps one of them is the same as mine, I haven't read through the thread. However, after trying to find a simple explanation in logic this is my explanation:

What changes are the odds of the un-picked door. Everything else is the same. When you remove one of the two wrong choices you essentially affect the odds of the remaining un-picked door.

Were you to remove one of the two un-picked doors randomly, and not disclose the result, there would be no impact on the odds of the remaining door vs the chosen door. But by selecting one of the two remaining doors to remove, you now have changed the odds of the unselected door.

Another way to look at it is when you pick one of the three doors, the odds are equal and the decision is made from the direction of selecting. When the host eliminates one of the two remaining doors he/she changes the odds but in the opposite direction, from the direction of the odds of the choices. The odds are affected by changing the choices regardless of the fact the odds were the same at the beginning when the first door was chosen.

The host's actions increased the odds of the un-picked door. The first chosen door was out of the equation when the un-picked door's odds were increased.