Muddle related to “boy-or-girl paradox” Suppose there is a species of frog such that just one clue permits you to sex its individuals: the fact that only males can croak, although they almost never do. If you collect two individuals and happen to observe one of them croaking, then you can be certain the croaker is male, but since you have no information about the other frog you will presumably say that it is equally likely to be male or female. (Actually you do have a little information: the frog is slightly likelier to be female since it had a chance to croak and didn’t, but let’s assume the error is negligible.)
This much seems fairly uncontroversial, and we can re-formulate our little conclusion in the statement that, if you have two frogs and you observe one of them croaking, there is probability $\frac{1}{2}$ of there being a female in the pair.
But what if you collect two specimens and then, while your back is turned, you hear a single croak come from one of the frogs? What has changed in this case is only that you cannot now point to one of the frogs and say, “This one is certainly male and the other one I have no information about.” Nevertheless there is still a frog (the croaker) who is definitely known to be male and another (the noncroaker) about whom nothing is known. Consequently the probability of a female in the pair is still $\frac{1}{2}$.
LATER: If we insist on distinguishing the frogs by a visible criterion (left versus right rather than the more natural croaker versus noncroaker) then we have the following argument: $(1)$ if the croaker is on the left then the frog on the right is equally likely to be male or female; and $(2)$ if the croaker is on the right than the frog on the left is equally likely to be male or female; therefore $(3)$ whichever frog is the croaker, the other one is equally likely to be male or female. To be explicit, we have the following equally likely alternatives:
$1)$ croaker on the left, noncroaking male on the right
$2)$ croaker on the right, noncroaking male on the left
$3)$ croaker on the left, female on the right
$4)$ croaker on the right, female on the left
TO RESUME: Now suppose there is another species of frog such that just one clue permits you to sex its individuals: fights are almost unknown in this species, but when a fight does occur there is always a male combattant. You collect two individuals of this species and happen to observe a fight. You now know that the frogs are not both females, but what is the probability that one of them is female? The argument of the previous cases doesn’t seem to go through. It might be argued that, since what you learn from the fight is exactly the fact that the frogs are not both female, the remaining possibilities (two males/a male on the left and a female on the right/a male on the right and a female on the left) should be considered equally probable, so that in this case the probability of a female in the pair is $\frac{2}{3}$.
Can these conclusions be right? And if not, where does the reasoning go wrong? Thanks if you can help.
Peace,
Sean
 A: Have you seen the Wikipedia article on the boy or girl paradox?

Nevertheless there is still a frog (the croaker) who is definitely known to be male and another (the noncroaker) about whom nothing is known. Consequently the probability of a female in the pair is still 1/2.

This is incorrect; the probability of a female in the pair is $\frac{2}{3}$ as in your second argument, because as in your second argument all you've learned is that of the four possibilities MM, MF, FM, FF (which started out equally likely) you've excluded the last one.
Intuitively you might say that the issue is that there are two possible croakers, and in the first possibility MM you don't know which of the two males is "the" croaker. The $\frac{1}{2}$ comes in some sense from assigning the MM case double the weight it should have.
A: Assume a male frog is observed to croak with probability c and a female frog is observed to croak with probability zero. Then the probability of a female in the pair, given that exactly one frog is heard to croak, is 1/(2-c). Now assume a pair of frogs is observed to fight with probability f if a male is in the pair and with probability zero if no male is in the pair. Then the probability of a female in the pair, given that a fight is observed, is 2/3.
The take-home lesson, I now think, is that the first task in solving a probability puzzle should always be to turn the puzzle into a sensible question, no matter how many unwarranted assumptions you need to make along the way. Do NOT try to “honour” the ambiguity of vague information by treating it “with a light hand” and hoping that the inferences you draw from it are valid on any of its possible interpretations: that way leads to paradox.
