I agree but set B is not a uniformly chosen subset of A in this case. That is the core of the trick. The rule for choosing B is intuitively uniform but actually slightly favours families with a girl and a boy over those with two boys.
Yeah, what matters is the contents of the initial set of families over which we determine probability.
"A man has two children, and one is a son born on a Tuesday. What is the probability that the other child is also a son?"
If the man is randomly chosen from the set of all families the answer is 1/2.
If the man is randomly chosen from the set of all families with a son born on a Tuesday then the answer is 13/27.
The reason for the difference is that a boy/girl family has a 1/7 chance that the boy was born on a Tuesday whereas the boy/boy family has only a 13/49 chance.
> I have two children. One or more is a son. Exactly one of them was born on a Tuesday.
I'm not sure that's what you are supposed to infer.
Looking at your earlier statement:
> I have two children. Here is some information about one of them: son, born on Tuesday.
There are two ways to interpret this.
(1) I am a man pulled at random from the set of [families with two children of indeterminate gender]. Here is some information about one of them: son, born on Tuesday.
(2) I am a man pulled at random from the set of [families with two children of indeterminate gender, one of whom was born on a Tuesday]. Here is some information about one of them: son, born on Tuesday.
We're not selecting from the same initial set in each case - set (2) is more restrictive. A difference in probability is maybe not surprising.
Still, I totally agree with you that it's a bit of a jump to conclude that the man is referring to scenario (2) in which the birthday information is used to narrow the initial set while the gender information is used to determine the probability. Just seems like a trick question to me.
To expand on the numbers a bit more... In scenario (2) the possible combinations are:
(Combo A) G G (0/49 at least one boy TB)
(Combo B) B G (7/49 at least one boy TB)
(Combo C) G B (7/49 at least one boy TB)
(Combo D) B B (13/49 at least one boy TB)
If the man has one boy, then combo A does not apply and the probabilty he has two boys must be 13 / (7 + 7 + 13) == 13/27.