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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.

Daniel
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  • Not sure what exactly you mean by 'frame of reference' or 'medium' ... but take a look here: https://math.stackexchange.com/questions/2149812/the-two-daughter-problem/2150036#2150036 maybe this is along the lines of what you're thinking? – Bram28 Apr 15 '20 at 00:49
  • Not quite because Mr. Smith knows the gender of both of his children. Although we can take the scenario where he knows one of his children is a boy but doesn't know the other's gender. Is this still analogous to my second scenario listed in the original post? – Daniel Apr 15 '20 at 01:04
  • Yes, that would be analogous. Or: I ran into one of Mr. Smith children and it was a boy. Then I learn Mr.Smith has a second child. So: I know that Mr.Smith has two children, at elast one of which is a boy ... in that case the probability of both being boys is $\frac{1}{2}$. But there are other scenarios where I also 'know that one of Mr. Smith's children is a boy, and yet now the probability of both being boys is $\frac{1}{3}$. That's what I tried to describe in that post. – Bram28 Apr 15 '20 at 01:06
  • I see. Then, is the ted-ed frog problem analogous to situation 2, wherein we know there is a male, then learn that there is a second child/frog? – Daniel Apr 16 '20 at 02:00
  • The way you phrase that is ambuiguous: "we know there is a male" .... how do you know there is a male? If you are looking at a specific frog and know for that specific frog it is a male, and then someone holds up a second frog for which you don;t kow the gender, then it's back to $\frac{1}{2}$ that both are males. But that is not the Ted-frog scnario: there you see two frogs, and know (given the call) that at least one is male. Now the probability both are male is $\frac{1}{3}$. – Bram28 Apr 16 '20 at 11:37

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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.

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.

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.

JeffJo
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  • The paragraph where you claim that I'm wrong in the doctor problem, either there is a misunderstanding or you're wrong. This is a very [link]well-known statistics problem(https://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Bayesian_analysis). I don't know why you believe you can only assume he picks one with equal probability. I clearly state that he looks at both and states at least one of them is a boy, which I juxtapose in the next problem by saying he only looks at one of the babies (which I argued is 1/2) – Daniel Apr 16 '20 at 01:54
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    It is indeed a well-known problem. It goes back to 1889, but the modern version was introduced by Martin Gardner in the April, 1959 issue of Scientific American. And he did say the answer was 1/3. But in October, he withdrew that answer because, as he stated it, the problem is ambiguous. The answer is 1/3 if, and only if, you ask the doctor "Is at least one a boy?" The answer is 1/2 if the doctor could have said "at least one is a girl" when there is one of each. Which he can do in your version. – JeffJo Apr 18 '20 at 01:29
  • There are 4 possibilities in the Doctor Problem: The doctor looks at ... (A) ... two boys and says "At least one male." (B) ... two girls and says "At least one female." (C) .... one of each and says "at least one male." (D) ... one of each and says "at least one female." The prior probabilities for cases A and B are 25%. If the prior probability for case C is P, then the prior probability for D is 50%-P. You assumed P=50%, which means that the answer for the identical problem that swaps boys for girls is 1. That's a paradox. If we aren't told how the doctor decides, we must assume P=25% – JeffJo May 20 '20 at 13:08