Some friends and I were arguing about the two following word problems, which are variants of the classic boy/girl paradox.
(i) "Mrs.Smith lives in a world where the eldest child is always male. She has two children, one older and one younger (although you do not know which is which). What is the probability that they are both male?"
(ii) "Mrs.Smith has two children, one older and one younger (although you do not know which is which). The eldest child is male. What is the probability that they are both male?" You can assume that being the eldest child is independent of being male in this problem.
My answer to (i) was $\frac{1}{3}$ and to (ii) $\frac{1}{2}$. My friends believe the answer to both is $\frac{1}{2}$. In what follows, I make an assumption throughout that one child has to be (at least slightly) older than the other since this didn't seem to be a point of disagreement between us. Here is my way of answering each question.
Question (i)
I make the background assumptions that there is always an eldest child and eldest $\implies$ male. Notice that the two background assumptions imply that at least one child in every pair is male. We can thus reformulate the problem as asking whether for the probability that both children are male given that at least one is male. Let $A = \text{"Both children are male."}$, $B = \text{"At least one child is male."}$. We wish to find $P(A|B)$. By Bayes's theorem, $$P(A|B) = \frac{P(B|A)P(A)}{P(B)}.$$
If both children are male, clearly at least one child is male. Hence, $P(B|A) = 1$. Since our background assumptions imply at least one child in every pair is male (and imply nothing stronger than that), the possible sex distributions are $\{MM,MF,FM\}$, so $P(A) = \frac{1}{3}$ and $P(B) = 1$. Therefore, $$P(A|B) = \frac{1 \cdot \frac{1}{3}}{1} = \frac{1}{3}.$$
Question (ii)
I make the background assumption that there is always an eldest child (who is not necessarily male in general). Let $A = \text{"Both children are male."}$, $B = \text{"The eldest child is male."}$. By Bayes's theorem, $$P(A|B) = \frac{P(B|A)P(A)}{P(B)}.$$
If both children are male, then clearly the eldest child is male, so $P(B|A) = 1$. The possible sex distributions are $\{MM,MF,FM,FF\}$, so $P(A) = \frac{1}{4}$. Since being the eldest child is independent of being male, $P(B) = \frac{1}{2}$. Therefore, $$P(A|B) = \frac{1 \cdot \frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}.$$
Are these two arguments and their results correct?
EDIT: I have a follow-up question that came up in the comments. Watch this video. They say the probability that at least one of the two frogs on the left is female is $\frac{2}{3}$. But consider the following statement, seemingly analogous to (i).
(i') "Mrs.Smith lives in a world where a frog that croaks is always male. She has two frogs, one that croaks and one that doesn't (although you do not know which is which). What is the probability that they are not both male?"
Then by analogy with (i), you get that the probability that both are male is $\frac{1}{2}$, so the probability that not both are male is also $\frac{1}{2}$.
It seems to me that if you believe (i) is correct, you should also believe the frog video is incorrect and that the answer is actually $\frac{1}{2}$. What goes wrong here (if anything)?