3 Door Logic Problem

[tsg]

OK, yet another scenario:
You run into an old friend at the store. She tells you she has two teenaged children. What are the chances of her having two girls if the odds of her having a girl or a boy are 50/50?

1/4

As you talk and walk around the store, she mentions something about getting a training bra. As boys do not as a rule wear training bras, you now know she has at least one daughter. To put it another way, you now know that she doesn't have two sons. What are the odds of her having two girls?

How old are girls usually when they need training bras? I've never had to buy one, but isn't it usually towards the younger end of the teenage spectrum? Wouldn't that make it more likely she's talking about her younger child if they are both teenagers?

 

[tsg]

and another one:

you are at a vast social gathering that is attended by an entirely random cross section of the population. The only requirement is that you must be married and attend with your spouse.

The MC asks everyone to stand for a fun game.

The MC tells everyone with more than two children to sit down

The MC tells everyone with less than two children to sit down. The only people remaining are those with two children.

Of the people still standing, what are the chances of them having two daughters, assuming a 50/50 boy/girl split?

1/4 of them will have two girls.

The MC then tells everyone who doesn't have a daughter to sit down. The only people still standing are people with at least one daughter.

Of the people still standing, what are the chances of them having two daughters, assuming a 50/50 boy/girl split?

1/3 of the standing people will have two girls.

The MC then tells everyone whose first child was a boy to sit down. The only people still standing are those whose first child was a girl.

Of the people still standing, what are the chances of them having two daughters, assuming a 50/50 boy/girl split?

1/2 of the standing people will have two girls.

 

[CurtC]

1/3 of the standing people will have two girls.

The MC then tells everyone whose first child was a boy to sit down.

1/2 of the standing people will have two girls.

So you see what happened here? The people with two girls were standing the whole time, but in the larger group, they represented 1/3 of it, then when some of the people sat down, now the two-girl group is 1/2 of it. The fact that there are now fewer total people standing has increased the proportion, or another way to look at it, the chance of having two girls.

From your previous comments, I think you believe that it was the age being nailed down that caused the difference, but with this example I think I might get through.

What if, instead of telling everyone whose first child was a boy to sit down, he had told everyone whose kid who alphabetically came first was a boy to sit down? There will be the same proportion of people sit down in this case, the two-girl couples will still all be standing, so their proportion will now go up, just the same, to 1/2. It can't still be 1/3, because some of the people have now sat down. And we haven't found anything out about birth order!

In fact, you can come up with any number of questions to ask at that point, to select one specific kid, regardless of age/birth order, and still the same proportion would sit down. That's because when you know the sex of one specific kid, the chance of the other being a boy is 1/2. No age or birth order information required.

 

[tsg]

So you see what happened here? The people with two girls were standing the whole time, but in the larger group, they represented 1/3 of it, then when some of the people sat down, now the two-girl group is 1/2 of it. The fact that there are now fewer total people standing has increased the proportion, or another way to look at it, the chance of having two girls.

From your previous comments, I think you believe that it was the age being nailed down that caused the difference, but with this example I think I might get through.

No. It was that it eliminated a known quantity of the possibilities.

What if, instead of telling everyone whose first child was a boy to sit down, he had told everyone whose kid who alphabetically came first was a boy to sit down? There will be the same proportion of people sit down in this case, the two-girl couples will still all be standing, so their proportion will now go up, just the same, to 1/2. It can't still be 1/3, because some of the people have now sat down. And we haven't found anything out about birth order!

But you have eliminated a known quantity of the possibilities. The parents do not have a choice which child they are going to choose to represent their children. They are forced into a specific action and the results of that action are quantifiable.

In fact, you can come up with any number of questions to ask at that point, to select one specific kid, regardless of age/birth order, and still the same proportion would sit down. That's because when you know the sex of one specific kid, the chance of the other being a boy is 1/2. No age or birth order information required.

Yes. There are many ways to eliminate a known quantity of the possibilities. That the child happens to be with the parent isn't one of them.

Nobody was claiming that knowing which child was older was the only way to do it.

 

[CurtC]

So are you saying, that since you don't know the parent's motivations for picking which child to bring to the store, that the answer is not (necessarily) 1/2?

I can accept that. In that case, however, it would be equally wrong to assign a probability of 1/3. Doing so would imply that you know something about the parent's motivation.

The closest you could get would be to say "Since I don't know what motivated my old friend to discriminate based on sex which kid to bring along to the store, we cannot assess the probability of the other kid's being a girl. However, if the pick was purely random or was a criteria not based on the child's sex, the chance of the other kid being a girl is 1/2."

Saying that the chance is 1/3 implies that you have more information than this.

 

[tsg]

So are you saying, that since you don't know the parent's motivations for picking which child to bring to the store, that the answer is not (necessarily) 1/2?

I can accept that. In that case, however, it would be equally wrong to assign a probability of 1/3. Doing so would imply that you know something about the parent's motivation.

The closest you could get would be to say "Since I don't know what motivated my old friend to discriminate based on sex which kid to bring along to the store, we cannot assess the probability of the other kid's being a girl. However, if the pick was purely random or was a criteria not based on the child's sex, the chance of the other kid being a girl is 1/2."

Saying that the chance is 1/3 implies that you have more information than this.

No, it's precisely because you don't have more information that you can't eliminate the possibilities that make it 1/2. All you know about that woman is that she has at least one daughter. Having a son and a daughter is twice as likely as two daughters. She is twice as likely to have a son as another daughter, based on the information we have.

Pick a card out of a shuffled deck without looking. What are the chances that card is the Ace of Spades? 1 in 52. Player B then takes 26 cards out of the deck without looking. What are the chances you have the Ace of Spades? Still 1 in 52. Player B then turns over the 26 cards he picked and the Ace of Spades isn't among them. What are the chances you have the Ace of Spades now? 1 in 26. If Player B instead takes 26 cards out of the deck telling you he is deliberately not going to take the Ace of Spades, and shows them to you, what are your chances of having the Ace of Spades? 1 in 52, because none of those cards had a chance of being the Ace of Spades, so no new information is gained by seeing them. We already knew there were 26 cards that weren't the Ace of Spades in the deck, seeing them doesn't add anything new to the equation, where his picking cards without looking does once we know the Ace of Spades isn't among them.

It's about what you know and that is precisely the point behind Monty Hall.

 

[Lamuella]

How old are girls usually when they need training bras? I've never had to buy one, but isn't it usually towards the younger end of the teenage spectrum? Wouldn't that make it more likely she's talking about her younger child if they are both teenagers?

All right, let me rephrase:

She mentions that she is on the PTA at Wolfingham Ladies College, a girls school in the area that has students between the ages of 11 and 18. She's not a teacher, and in fact works as a dental hygenist. From this you know that she must be the parent of at least one girl.

 

[tsg]

All right, let me rephrase:

She mentions that she is on the PTA at Wolfingham Ladies College, a girls school in the area that has students between the ages of 11 and 18. She's not a teacher, and in fact works as a dental hygenist. From this you know that she must be the parent of at least one girl.

If all I'm supposed to gather from this is that she's the parent of at least one girl, the chances her other child is a girl is 1/3. If there's something else I'm supposed to know, I'm missing it.