"A family has two children, at least one of whom is a boy; what is the probability that both are boys?"

and

"A family has two children, at least one of whom is a boy named Bartholomew; what is the probability that both are boys?"

have different answers.

Well, I could be wrong; these matters are notoriously slippery. But let me reconstruct my understanding (it's been a few years since I taught probability) and then you can tell me if you buy it. I'll start by taking probability theory out of the problem and turning the two questions into demographic questions: "Of those families that have two children, at least one of whom is a boy, what fraction are two-boy families?" "Of those families that have two children, at least one of whom is a boy named Bartholomew, what fraction are two-boy families?" Among four million two-child families, we'll have three million that have at least one boy, and one million of those families are two-boy families, so the answer to the first question is 1 million divided by 3 million, or 1/3. For the second question, let's assume (for definiteness) that one child in a thousand is named Bartholomew. So, of the two million families with one boy and one girl, there are 2000 families with a boy named Bartholomew, and of the one million families with two boys, there are 1000 two-boy families in which the older boy is named Bartholomew and 1000 two-boy famlies in which the younger boy is named Bartholomew. (There may also be 1 family in which both boys are named Bartholomew --- cf. the Dr. Seuss poem "Too Many Daves" --- but they are too few to skew the demographics by much.) That makes 2000 two-boy families in which at least one of the children is a boy named Bartholomew. So the answer to the second question is 2000 divided by 2000+2000, or 1/2. If you think my reasoning is wrong (not just dependent on unstated assumptions that might shift the answers slightly from 1/3 and 1/2, but seriously wrong), please let me know where my mistake is. Jim Propp