# New Two Children problem

The well known "Two Children problem" is answered in " In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?"

M. Smith says: I have two children and at least one is a girl named Jane .

What are the chances that both children are girls if the probability that a girl is named Jane is p ?

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I do not see how this is different from the original. Where does the name come in? As far as I can tell, the probability remains $\frac{1}{3}$ – Alex Becker Apr 18 '11 at 2:03
I suspect the idea is that if the name Jane is somehow "uncommon", the odds of having a single girl given that she is called Jane "feel" somewhat lower. Of course, the name of the girls are not independent (if you happen to name your first born Jane, the odds that your second born will be called Jane are 0, even if she is a girl, since you are not going to call both your kids by the same first name). – Arturo Magidin Apr 18 '11 at 2:19
Maybe the problem is not well-phrased, but I think the idea is that the probability of boy-girl with girl named Jane is $\frac{p}{3}$, and the probability of girl-girl with at least one girl named Jane is meant to be $\frac{2p-p^2}{3}$ (since both not named jane is given by $(1-p)(1-p) = 1-2p+p^2$. Since $p(2-p)\gt p$, girl-girl is more likely than boy-girl (or than girl-boy). But I could be falling into one of the many common errors in trying to compute probability off the cuff. – Arturo Magidin Apr 18 '11 at 2:39
This issue was discussed in math.stackexchange.com/questions/4400/… – Ross Millikan Apr 18 '11 at 2:58

To evaluate this probability we would need to know the conditional probabilities of M. Smith saying this, given the various cases. Since this information is not available, we can only speculate. As usual in such questions (but not in real life), we assume that the a priori probabilities are $1/4$ that both children would be girls, $1/2$ that one is a girl and one a boy, $1/4$ that both are boys. Here are three scenarios:

A) M. Smith is choosing at random one of his children, and telling you that child's name. There is nothing special about the name "Jane".

B) Without knowing anything about Smith's family, you asked him "How many children do you have? Is one named Jane?"

C) Without knowing anything about Smith's family, you asked him "How many children do you have? Tell me the name of one of your girls, if you have any." There is nothing special about the name "Jane".

In case (A), the probability that both are girls is 1/2 (because in the boy-girl case he might have chosen to tell you about a boy).

In case (B), the probability is approximately 1/2, because the probability of a Jane in a two-girl family is approximately twice the probability of a Jane in a one-girl family (this will depend on Smith's naming practices, but in typical models it will be exactly twice).

In case (C), the probability that both are girls is 1/3, because the conditional probability of the given response is the same in the boy-girl case as in the girl-girl case.

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Okay, maybe I'm about to make one of the mistakes so common with this type of problem and mess up royally, but then I'll learn something from people correcting me.

I think the idea/intention of the problem is that if the name Jane is rare, then it's more likely that one child is named Jane if you have two girls than if you have a girl and a boy (though the problem is possibly badly phrased, presumably boys are never called Jane, and we allow the possibility that a parent might, by choosing names at random, name both her girls "Jane" by happenstance).

I would try to compute as follows: since at least one child is a girl, we have three equally likely possibilities, each with probability $\frac{1}{3}$: boy-girl, girl-boy, or girl-girl. At this point, the "standard" problem shows that the odds of both being girls is $\frac{1}{3}$.

But if the names of the girls are random and independent, and Jane has probability $p$, you have the following possibilities:

• Boy-Girl, and girl is named Jane is $\frac{1}{3}\times p = \frac{p}{3}$;
• Boy-Girl, and girl is not named Jane: $\frac{1-p}{3}$;
• Girl-Boy, and girl is named Jane is $\frac{p}{3}$;
• Girl-Boy, and girl is not named Jane is $\frac{1-p}{3}$;
• Girl-Girl, and neither girl is named Jane: $\frac{1}{3}(1-p)(1-p) = \frac{1-2p+p^2}{3}$.
• Girl-Girl, and at least one girl is named Jane: $\frac{2p-p^2}{3}$.

So: since the events in which there is at least one girl named Jane have total probability $$\frac{4p-p^2}{3}$$ and the probability of both children being girls with at least one named Jane is $\frac{2p-p^2}{3}$, then the the chances that both children are girls given that at least one is a girl and is named Jane is $$\frac{\quad\frac{2p-p^2}{3}\quad}{\quad\frac{4p-p^2}{3}\quad} = \frac{2-p}{4-p}.$$ If girls are always named Jane ($p=1$), then this gives $\frac{1}{3}$, the probability that both children are girls.

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I agree with this. The other limit is also enlightening: if very few girls are named Jane $(p\ll 1)$, the probability that both are girls goes to $\frac{1}{2}$. This comes out of the boy born on Tuesday link as well. math.stackexchange.com/questions/4400/… – Ross Millikan Apr 18 '11 at 3:01
@Ross: Well, glad I don't have to wipe egg off my face. Also: if the probability is $p=\frac{1}{7}$ (as in the "born on a Tuesday" question), the answer also gives $\frac{13}{27}$, same as the answer in the previous question. – Arturo Magidin Apr 18 '11 at 5:12