Variant of boy or girl paradox Here is an interesting question my friend brought up:

You are invited to a family party. A Boy opens the door for you. There
  are two children there. What is the probability that a boy opens door for
  you next time?

This is my solution:
$$P(\text{boy second time | boy first time}) = \frac{P(\text{boy both time)}}{P(\text{boy first time)}}$$
$$=\frac{1/3+1/3*1/2*1/2+1/3*1/2*1/2}{1/3+1/3*1/2+1/3*1/2}=3/4$$
The $1/3$ comes from the fact that given we have one boy already so the combination can only be {boy, girl}, {girl, boy}, {boy, boy}. The 1/2 comes from for each time there is a 1/2 probability that a boy will open the door if there is one boy one girl.
Is my calculation correct? I feel not confident about the 1/3 argument.
Edited: @Rolf proposed another point that actually the probability of two boys and one boy/one girl should be 1/2, not 1/3 each.
Edited again: Actually there are two approaches, as shown below. The key is to whether to assume the prior or not in the calculation.

A boy opened the door the first time around, but we are told there are two children (and we are assuming the boy is one of the two). Assuming there is an equal chance the other child is a boy or girl, that means there is also an equal chance for the household to be a two-boy household or a boy-girl household. So, the sample space is BB or BG with equal likelihood. Hence, a boy opening the door at any time is 
$$\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4}$$
So ... @James ... you got the right answer ... but the wrong method. In fact, you made two mistakes.  The first mistake was the same I originally made (see link to my original answer below) that with a boy opening the door there would be a $\frac{1}{3}$ chance of there being two boys and a $\frac{2}{3}$ chance of there being one boy. No, that is really just $\frac{1}{2}$ and $\frac{1}{2}$. The second mistake is that you used that new sample space, that is based on the fact that a boy opened the door the first time, to calculate the probability of a boy opening the door the first time! No, if you calculate it this way, you should make no assumptions at all and thus you have a sample space of BB, BG, GB, and GG. .. this is what Robert did in his answer.
Original (incorrect) answer
