Additional information in conditional probability problem gives a different answer but this does not seem correct to me? Q1.) I tell you that I have two children and that one of them is a girl. What is the probability that I have two girls? Assume girls and boys equally likely to be born and that the gender of one child is independent of the other.
The answer is 1/3 which I understand and agree with.
It can be seen that of the possible combinations {BB,GB,BG,GG} the first combination is not possible. In the remaining three cases, there is only one in which there are two girls which gives the correct answer.
Q2.) I tell you that I have two children and that one of them is a girl. You knock on the front door and are greeted by a girl who you correctly deduce is my daughter. What is the probability that I have two girls? Assume girls and boys equally likely to be born and that the gender off one child is independent of the other. Compare and contrast the answers.
The claimed answer here is 0.5. That is on the basis that the girl at the door is taken to be #1 and therefore the only randomness remaining is the gender of the second child which is given in the question is 50%.
However this doesn't seem correct to me, how can you know that the girl is #1 and not #2? Seeing the girl does not seem to preclude {GB,BG,GG} so the answer should be the same as above?
Baz
 A: In the first case, you are quite correct. By asserting that one of your children is a girl, what you effectively said is that the older child is a girl, or the younger child is a girl, and you haven't ruled out the possibility that both children could be girls. Hence, the answer is, indeed, $\frac13.$
In the second case, however, exactly one of the children has answered the door, and that child is a daughter. The only question is whether the child who didn't answer the door is a girl or a boy. You don't know if it's the older child or the younger child who has answered the door, but exactly one of them has. If the older child answered the door, then it is impossible that the older child is a boy, so there are only two possibilities remaining. Likewise, if it was the younger child who answered the door, then it is impossible that the younger child is a boy, and so there are only two possibilities remaining. Regardless of which child answered the door, the answer is then $\frac12.$
A: Assuming that boys and girls in this household are equally likely to answer the door and that only one person ever answers the door at a time, we may choose the order of the children to be based on first seen.
In this sense, we can indeed claim that the girl that we saw at the door is "child #1" and the result follows.
However, this assumes that boys and girls are equally likely to answer the door.
Consider an opposite extreme where you know that our entire family is home at the time and our family follows some very strict religious customs that only a boy may answer the door unless no boys are present.  In this scenario, the fact that a girl answered the door directly implies that there were no boys available to answer the door, which in turn implies that I must have two daughters and no sons.
I would say then that the problem is poorly worded and there is not yet enough information to conclude that the probability is indeed $\frac{1}{2}$ without additional assumptions (even if the assumption seems valid).
A: I'll expand two of the answers that are already given. They are based on two different models: in one, the child answering the door is chosen based on age, and in the other by sex. And the answer depends on which model is chosen.


*

*Suppose that the elder child opens the door with probability $p$ and the younger with probability $1-p$. Let G@D denote the event that you see a girl at the door. We have:
$$
P\left(GG|G@D\right)=\frac{P\left(G@D|GG\right)P(GG)}{P\left(G@D\right)}=\frac{1/3}{1/3+p/3+\left(1-p\right)/3}=\frac{1}{2}.
$$

*Suppose that, if the family has a boy and a girl, the girl opens the door with probability $q$. Now,
$$
P\left(GG|G@D\right)=\frac{P\left(G@D|GG\right)P(GG)}{P\left(G@D\right)}=\frac{1/3}{1/3+q/3+q/3}=\frac{1}{1+2q}.
$$
So the answer is $1/2$ only if we pick the first model or if $q=1/2$ in the second model.
