[NeilCJames]

Older/younger doesn't exactly enter into it, but some ordering is necessary to properly describe the set of probabilities. This can be arbitrary, but birth order makes the most intuitive sense. The set of probabilities of sex distribution in two-child families is [(B, B), (B, G), (G, B), (G, G)], for any arbitrary ordering, this cannot be reduced to [(B,B), (B,G), (G,G)] as you have done without adding a variable to double the probability of (B,G).

Assuming a coin-flip probability for boy/girl distribution, you get the 1/3 answer if we select for two-child families where at least one child is a boy: [(B,B), (B,G), (G,B)]. If we don't pre-select for having at least one boy, (i.e. if we select the family because we just met the father socially), the probability rises to 1/2, because we have two cases to consider, each with a 1/2 probability: [(B,B), (B,G)], and [(B,B), (G,B)].

[bad_user]

Yes, but this is wrong ... the sex of the second child is in no way connected to the sex of the first child.

He didn't ask ... what's the sex of the second child? No ... he asked ...

  p(boys = 2 | boy >= 1) == p(child = boy)

Choosing a distribution that involves both children in this case is wrong, hence my answer that ordering doesn't matter. 33% is just wrong.

[NeilCJames]

What I wrote is not wrong unless, when you wrote: "So given 2 children, there are only 3 possibilities ... boy, boy girl, girl boy, girl" you are asserting that in the absence of any information about the sex of either child, 2 boys is a 1/3 probability, instead of (1/2)*(1/2)= 1/4.