# very simple conditional probability question [closed]

I know this isn't a research question, so it might get voted off, but here goes:

You know that a couple has two children. You go to the couple's house and one of their children, a young boy, opens the door. What is the probability that the couple's other child is a girl?

If you list all possibilities for the sexes of two children, BB, BG, GB, GG, you see that 2 of the 3 pairs that have B (for boy) in them also have a girl, so the answer one could argue is 2/3.

On the other hand, one could argue that the answer is 1/2, since the probability that any one child is a girl is 1/2, and intuitively (?) should be independent of the gender of its siblings.

Some background to possibly justify posting it here: the question was asked at an interview for an actuarial/insurance type position, and the interviewer was the answer was 2/3, whereas my friend who was being interviewed (and has a masters in math) thought the answer was 1/2, even after the interviewer explained his logic. My friend felt that the interviewer wasn't taking into account the fact that it is not equally likely that a boy will open the door in the BB versus the BG combination, and one has to take into account that fact. I have no idea which is the correct answer, both sound somewhat convincing to me (I have a Ph.D. in math, but I won't mention from where in an effort to avoid embarrassing my degree granting institution!). Anyways, any help would be appreciated and I apologize if this is too simple a question for this forum.

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## closed as off topic by Qiaochu Yuan, Robin Chapman, Andy Putman, Yemon Choi, Pete L. ClarkAug 18 '10 at 22:02

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This question would probably be fine at math.stackexchange.com – Jon Bannon Aug 18 '10 at 19:21
oh, sorry, I didn't know of that site - that looks like a more appropriate site. Is there an easy way to move it over there? – Not Bayes Aug 18 '10 at 19:26
okay, done. closing it here would be fine with me. – Not Bayes Aug 18 '10 at 19:45
A town has 100 houses each with 2 kids. 25 BB, 50 BG, 25 GG. You knock on each door and a random child answers. You'd get 50 boys answering, 25 from a BB house and 25 from a BG hose. So the chance of (non answering child=B | answering child was a B)=1/2. – Aaron Meyerowitz Aug 19 '10 at 9:18
Okay, I finally figured it out. Tom LaGatta (and maybe also Yemon) believe that "a boy opens the door" is meant strictly to rule out the possibility of GG in a Monty Hall-esque way, i.e., without changing the relative probabilities of BG, GB and BB. KalEl, Aaron, Dominic and I all believe that "a boy opens the door" means that of the two children, one is chosen with equal probability to open the door, and it happens that this one is male. The question is ambiguous as to which of these two is intended. – JBL Aug 21 '10 at 15:05

We will assume all the obvious implicit assumptions (eg. random child being boy of girl is 50/50, boys and girls open the door uniformly, etc.).

If you had a slightly different question, i.e. if you asked the couple if they have at least one boy, and the answer is yes, then the chance of the other one being a girl is 2/3. Intuitively, the probability is not 1/2 because in this case the answer depends on both the children, i.e. it is a function of both of them considered together.

However, if you asked the couple to pick a child at random, then she/he bears no information about the other child, and consequently his/her gender does not give you any information about the sibling.

Your case is the second case, where the child opening the door is selected at random, and she happens to be female. This does not bear any information regarding the other child.

Mathematically,

$P(Other\ is\ B|G\ opens\ door) = P(BG|G\ opens\ door) =$ $\frac{P(BG\ and\ G\ opens)}{P(GG\ and\ G\ opens\ door) + P(BG\ and\ G\ opens\ door)} = \frac{1/2*1/2}{1/4+1/2*1/2} = 1/2$

(Note here, that $P(BG\ and\ G\ opens)=P(G\ opens|BG)*P(BG)=1/2*1/2$.)

However as a digression, a twist in the question can be brought about - if you take probabilities for a boy and girl to be different for opening the door.

Eg. suppose boy opens with probability $p$, girl with $q=1-p$, in a family with BG.

Then $P(Other\ is\ B|G\ opened\ door) = P(BG)/P(G\ opened\ door) =$

$\frac{P(BG\ and\ G\ opens)}{P(GG\ and\ G\ opens\ door) + P(BG\ and\ G\ opens\ door)} = \frac{1/2*q }{1/4+1/2*q} = \frac{2q}{2q+1}$.

This means $q = 0 \implies P(BG|G\ opens)=0$. That makes sense, since girls don't open the door if there is a boy, so definitely the other one is girl too.

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I don't agree with this interpretation of the question. Surely what was intended by the question is the following: given that you have seen a boy open the door, and given that you know there are two children, what is the probability that the other child is a girl? That interpretation seems to lead back to the 2/3 answer. – Yemon Choi Aug 18 '10 at 20:49
The error in that reasoning is that the events $\{BG\}$ and $\{G~opens~door\}$ are not independent, hence the probability of both of them occurring does not factor as the product of each individual probability. – Tom LaGatta Aug 18 '10 at 21:58
@Yemon, I don't see how it leads back to 2/3. @Tom, I skipped the step $P(BG\ and\ G opens)=P(G\ opens|BG)*P(BG)=q*1/2$, which is $1/2*1/2$ with $q=1/2$ ($q$ is probability of G openning the door for a family with BG, as defined in answer). – KalEl Aug 19 '10 at 12:09
@KalEl: I've had a more serious think about the problem and would like (at least for now!) to retract the comment I made above. If you edit your answer then I think I can reverse my downvote (as it stands, too much time has elapsed for me to do so). – Yemon Choi Aug 21 '10 at 17:24
Edited... but feel free to downvote again if you don't agree with the solution! – KalEl Aug 24 '10 at 10:59

Let me make it even more confusing. Suppose we let our probability space be the possible genders of the first and second child AND which of the two children came to the door, i.e. BG1 means that the first child is male, the second child is female and it was the first child who answered the door (Let's suppose that which child answers the door is independent of gender and each child is equally likely to answer it).

Then we condition on the subset: BG1, BB1, BB2, GB2

Then half of these situations have the second child as a boy.

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