Two-child probability paradox, a nuanced explanation You have a co-worker named Jill. You know Jill has two children, but know nothing more about them. Jill invites you and your family to a holiday party at her house. When you arrive, you knock on the door and a teenager answers. You say “you must be one of Jill’s kids,” to which he replies “yes, I’m Daniel.” At this precise moment in time, you can now calculate the probability that Jill’s other child is a boy as well: 1/3. This is because the sample space goes from {BB, GB, BG}, each having an equal likelihood of being true, to {BB, GB, BG, GG}.
You then ask Daniel “Are you the younger sibling?” Regardless of what Daniel answers, the odds of the other being a boy jumps up to ½ immediately upon receiving a response. If Daniel answers yes, you know that the sample space, with pairs ordered {elder sibling, younger sibling} goes from {BB, GB, BG} to {BB, GB}. If he answers no, {BB, GB, BG} becomes {BB, BG}.
But we know that he is going to answer in either the affirmative or the negative… yet we can’t pare down the sample space until an answer is given. Why is this? If we know that one or the other is 100% likely to occur (an answer or ‘yes’ or ‘no’), why can’t we say from the get-go that the likelihood of another boy is 50%?
 A: The jump to $\frac12$ does not happen; the probability starts out at $\frac12$.
Let's be precise about our assumptions here: we assume that there are two siblings, each independently equally likely to be a boy or a girl. We also assume that (independently of gender) each sibling is equally likely to come to the door.
In that case, we can figure out the probability of seeing what we see (a boy coming to the door) under each of the four possibilities for the children's genders:


*

*If BB, the probability is $1$.

*If BG, the probability is $\frac12$.

*If GB, the probability is $\frac12$.

*If GG, the probability is $0$.


So (by using Bayes's theorem) the probability that both children are boys is already
$$\frac{1}{1 + \frac12 + \frac12 + 0} = \frac12.$$ This stays at $\frac12$ no matter what the answer to your question is.
In other words, even though the sample space has gone down to $\{BB, BG, GB\}$ these are also no longer equally likely: BB is twice as likely, because in the cases BG and GB there was only a $\frac12$ chance for you to see a boy and not a girl.
The paradox comes about if you do something like ask Jill "Is one of your children a boy?" and she says "Yes". In that case, we can again figure out the probability of hearing what we hear (that one of the children is a boy) under each of the four possibilities for the children's genders:


*

*If BB, the probability is $1$.

*If BG, the probability is $1$.

*If GB, the probability is $1$.

*If GG, the probability is $0$.


So the probability that both children are boys in this case is
$$\frac{1}{1+1+1+0} = \frac13.$$
