I'm aware this question has been asked many times - however, I feel I have a different take on the topic.
First, let's take a classic scenario. Here, a doctor has 2 babies. He checks both of their genders and tells the nurse "at least one of them is male". Here, the odds babies are both boys is clearly 1/3.
Next, let's spin this scenario. Here, the doctor still has 2 babies. He checks only one of their genders (you are unsure which one) and tells you "at least one of them is male". Here, the odds the babies are both boys is clearly 1/2. (unless I'm wrong here)
Finally, let me pose my question: does the frame of reference from which you are acquiring the information impact the statistics? In other words, if the medium from which you are acquiring the information is only aware of one of the two binary options (as in scenario 2), does that change the statistics?
More relevantly, in the TED-ed frog problem, is the scenario closer to my scenario 2? The medium by which you obtain the information can only know one of their sexes for certain, in my opinion.
On a side note, I also feel as if the riddle is confuddled if you group the three frogs into one triplet. For the sake of this post, though, we will not take this into consideration.
First, TED-ed’s answer is wrong. It’s equally likely that there is a female frog in either direction. The short explanation is that, if there was a male frog on the stump, it could have croaked first. Eliminating the cases where that could have happened reduces the chances that it was male.
I point that out, because it isolates the error made in this entire class of problems. It is incorrect to base the answer on the combinations that could produce the information. You need to also eliminate the cases that could, but wouldn’t. A good example is the first doctor case you mentioned. You said the answer was clearly 1/3, but that is also wrong. It is 1/2.
Think about it: there is a 1/4 chance of BB, in which case the doctor will tell you there is a boy. There is a similar 1/4 chance of GG, in which case the doctor will tell you there is a girl. But in the 1/2 chance that there is one of each, why do you assume the doctor will always tell you there is a boy? You can only assume he picks either one with a 50% probability, in which case the answer is 1/2.
The information you have (the “frame of reference”) does not impact “the statistics” of the cases that can exist in the absence of that information. But it can affect what you know can’t exist with it. And how you obtained the information can affect it even more. The question you need to ask yourself is “Are there situations where this information is true, but I would have learned something else?” TED-ed’s frog problem is a good example.
+++++
Here's the solution to TED-Ed's riddle. I'll use the event M2:F to mean two males in the clearing and a female on the stump, and M1F1:M to mean one of each in the clearing and a male on the stump. The event C means the first croak comes from the clearing.
Pr(M2:M)=1/8, Pr(M2:M & C)=(1/8)*(2/3)=1/12
Pr(M2:F)=1/8, Pr(M2:M & C)=1/8
Pr(M1F1:M)=1/4, Pr(M1F1:M & C)=(1/4)*(1/2)=1/8
Pr(M1F1:F)=1/4, Pr(M1F1:F & C)=1/4
Pr(C) = sum of all these = 1/12+1/8+1/8+1/4 = 7/12
Pr(M1F1|C) = [Pr(M1F1:M & C)+Pr(M1F1:F & C)]/Pr(C) = [1/8+1/4]/(7/12) = 9/14
Pr(F|C) = [Pr(M2:F & C)+Pr(M1F1 & C:F)]/(Pr(C) = [1/8+1/4]/(7/12) = 9/14
+++++
[Edit] I finally noticed the edit to the original post.
Saying "I'm ignoring any situation where my answer might be wrong" does not make for a convincing argument that it is right. The riddle is not "confuddled" [sic] by considering all of the elements that are part of the problem as stated. It is addressed properly. There are three frogs in the problem.