Here's the dangerous scenario that you've gotten yourself into:

You’re stranded in a rainforest, and you’ve eaten a poisonous mushroom. To save your life, you need an antidote excreted by a certain species of frog. Unfortunately, only the female frog produces the antidote. The male and female look identical, but the male frog has a distinctive croak.

As you begin to lose consciousness, you find yourself standing equidistant between two points. At one of them is a single frog. At the other are two frogs. You just heard the sound of the male croaking from the second point. You have time to get to one of those points and begin frantic frog licking. Which direction do you choose?

The answer isn't quite so obvious. Derek Abbott explains in this demonstration of conditional probability.

"With the frogs you only have 2 possibilities, male or female,..."

You have to be careful, as just because something can be expressed in terms of two possibilities doesn't mean they 50-50. You can describe any generic pair of MF creatures as either four equal probability possibilities, or three possibilities: MM, M&F, FF, but in the latter case the split is 25-50-25 without any further constraints.

"The whole situation is really only about 1 frog and whether or not it is a male or female."

It isn't about just one frog though. You have information about a group of two frogs, and don't know which of the two it applies to.

"Why is it not true that, knowing that one of the 2 frogs is a male, there is only a 50% chance that the other one is a female. "

If you knew one specific frog was male, then the chances of the second one being male vs. female is 50-50. The problem with situations like the riddle is that you don't know specifically which frog is the one you heard. Even if you could later identify the sex of the frogs, in the 2 male situation you will never resolve which one you heard. You're still adding information though, and it eliminates the possibility of FF. Even if ignoring the whole ordering/positioning thing, you go back to the 1:2:1 ratio of MM, M&F, FF that you start off with before you heard a male frog, eliminate the FF possibility, and still have a 1:2 ratio.

"I know it is beyond my grasp of the nuances of math but,"

With high school level algebra and a little effort to learn basic probability notation, you can learn Bayes' theorem, and then situations like this can be evaluated quite straightforwardly with a simple formula that can replace a hard to understand probability with ones that are often easier to understand.
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I'm sorry to beat on this. I know it is beyond my grasp of the nuances of math but, when you give the dice example you have 3 different numbers that can possibly come up on the die to make a combination that equals 4. 1, 2 and 3. With the frogs you only have 2 possibilities, male or female, not 3. If the 2 represents the male frog, then 1 and 3 can't. If you were to say that #1 represents the male frog and #3 represents the female frog then the only combinations that you could roll to equal 4 would be a 1 and a 3 or a 3 and a 1. The #2 doesn't have any representation because the male and female are already represented by 1 and 3. In the frog situation you only have 2 different things (a male and a female) that you are trying to make 3 different combinations with. Why is it not true that, knowing that one of the 2 frogs is a male, there is only a 50% chance that the other one is a female. The whole situation is really only about 1 frog and whether or not it is a male or female. If you go and lick both frogs, you will 100% lick one male frog. The only unknown is what the 2nd frog will be, and there is an equal chance that it will be a male or female, (50% chance).
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Frog#1 is either a male or a female (50-50), whether you draw it on the left, top, etc. The same applies to Frog#2.
When you see the two frogs together, those are the only four possible combinations of their genders.

Imagine rolling two dice lots of times and you want to keep track of how often they add up to 4. Will you roll a 1+3 combination as often as you roll a 2+2 combination? The only combinations are:
A. Die#1 = 1, Die#2 = 3;
B. Die#1 = 2, Die#2 = 2;
C. Die#1 = 3, Die#2 = 1.

Although each die has an equal chance of rolling any number, the 1+3 combination is twice as likely to occur than the 2+2 combination. With a couple of dice and an hour or so, you can easily test this.
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The simple answer is you can run a simulation or experiment and you will get the 2/3 probability.

But otherwise, the ability to distinguish between two things is rather important in math, including applications to physics. All that matters is there is some distinction between the two such that you can label them, it doesn't matter how you label them, or that there exist multiple ways to label them. And in the example you give, if you want top & bottom to not be redundant ways of separating them, as in there is in you separate the situation of the frogs being beside each other or above/below each other, then you end up with a situation where not every case is equally likely, and can get back the 2/3rd number with more careful effort.

The importance of labeling though can go down some deep rabbit holes, as things like the Axiom of Choice can be confusing to even some of the greatest mathematicians, and many pages can be written trying to describe its implications. In physics, you have a drastic difference between bosons and fermions because fermions can be separated by spin while bosons can't.
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The part I don't get about the 2 in 3 probability is that the 3 possibilities shown were 2 males, or, female on the left and male on the right, or, male on the left with female on the right. I don't get how female on the left or right would be considered 2 different possibilities as the net result is the same, 1 male and 1 female. Givem that logic, why couldn't we add 2 more possibilities and say female on bottom male on top, and male on bottom female on top, as if they were standing on each other. This would increase the odds to 4 out of 5, or 80%.
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