>>14018465Assume that all females in the scenario had equal croak distribution for the duration before we make our choice, i.e. p : N to R where p(x) is chance of x croaks. Let f be real in [0,1], and let p(0) = 1-f, p(1) = f, and p(n) = 0 for n>1. In other words each female had equal croaking chance in the time span. Assuming this allows us to show that licking the pair is always at least as good as the lone frog, regardless of croak chance.
To find the chance that the once-croaking pair has a male, we can first compute the chance that the pair is FF. We can use Bayes to find that Pr(FF | 1croak) = (1-f)/(8-4f). Hence the probability of success when licking the pair is Pr(MF or FM | 1croak) = 1-Pr(FF | 1croak) = (7-3f)/(8-4f).
Chance of success when licking lone frog is Pr(male | 0croak) = 1/(2-f), by Bayes, as other anons have observed. Now, suppose (7-3f)/(8-4f) >= 1/(2-f). Basic algebra shows that this holds iff f <= 1, and therefore picking the pair is always an equally good or better choice under the assumption of IID female croaks.
