Henry, I think I have finally figured this out. Let's use your notation. Before any frog croaked, the possibilities for the frogs sitting on the stump are:
And these are all equal possibilities. If we observed 80 stumps, we would find 20 with two males, 40 with one male and one female, and 20 with two females.
Now if we wait long enough before each stump to hear one croak. we will find that the 20 with two males will have 10 croaks from Frog 1 and 10 from frog 2. So now our distribution of stumps where we heard a croak will be
Frog1 Frog2 Count
M m 10
m M 10
M f 20
f M 20
f f 0
There is a 50% (30/60) chance we will pick the frog that croaked (5 from Mm, 5 from mM, 10 from Mf, and 10 from fM), and it will be male.
There is a 50% chance (30/60) we will pick a non-croaker (5 from Mm, 5 from mM, 10 from Mf, and 10 from fM), but of those 30 picks, only the 5 from Mm and 5 from mM will be males
So, in total, we will have picked a male in 40/60 cases, 67%.