For a randomly selected family with two children, there are four possible boy/girl combinations:
B B,
B G,
G B,
G G
In the first case we are told that the older child is a boy. This leaves only two cases:
B B,
B G
Therefore, there is a 50% chance the second child is a boy.
In the second case, we are told only that [at least] one child is a boy. This leaves three possibilities:
B B,
B G,
G B
Therefore, the probability that both children are boys is 1/3.
Enumerating possible states of the world like this is the fundamental insight you need to have to be able to understand these types of problems - but it does take a while to get used to!
But you are assuming that each of those three possibilities has equal probability; can you explain the rationale for that? (it is clear why BB, BG, GB, and GG have equal probability in the unrestricted case, but less clear why BB, BG, and GB have equal probability in this restricted case)
Besides, this is just a rephrasing of the original article's argument, and doesn't counter mine at all. I am open to the possibility that there is a flaw in my argument, but where is it?
So you say: "it is clear why BB, BG, GB, and GG have equal probability in the unrestricted case"
The restricted case is just the unrestricted case + one additional bit of information, that is, you're told that GG is not an option. This eliminates GG from the unrestricted case, but says nothing more about the probabilities of the other options. So the probabilities stay equal, although they now equal 1/3 each (if you eliminate options, the remaining options all become more likely).
What you're missing is this: The statement "the older child is a boy" has more information than the statement "one of the children is a boy". The first statement allows you to eliminate two options (GB and GG), while the second statement only allows you to eliminate one option (GG).
The "older" part is not fundamental to the problem. Equally, the statement "the taller child is a boy" has more information than "one of the children is a boy". The problem with this is that probabilities for height are not so friendly like the 50/50 probabilities for birth order (e.g., boys are likely to be taller than girls, older children are taller than younger, etc), which introduces unnecessary complexities to a logic problem. So that's why birth order is used for these types of puzzles.
In the first case we are told that the older child is a boy. This leaves only two cases: B B, B G Therefore, there is a 50% chance the second child is a boy.
In the second case, we are told only that [at least] one child is a boy. This leaves three possibilities: B B, B G, G B Therefore, the probability that both children are boys is 1/3.
Enumerating possible states of the world like this is the fundamental insight you need to have to be able to understand these types of problems - but it does take a while to get used to!