The Frog Riddle - Conditional Probabilities I saw this riddle doing the rounds on the internet: https://ed.ted.com/lessons/can-you-solve-the-frog-riddle-derek-abbott
In summary; There is a population of frogs with male:female occurring in 50:50 ratio. There are two patches of ground near you, one containing a single frog, the other containing two frogs. Your survival depends on you finding a female frog in one of these two patches, but you only get to make one attempt. You cannot tell which frogs are which in advance, except that you know that one of the frogs in the patch with two frogs in is male.
The answer given to the riddle is that the odds of the single frog being female is 50%, but the odds of one of the two frogs being female is 2/3 (67%). The explanation being that there are four possible combinations of male female pairs, one is excluded because we know one frog is male, hence 2/3 combinations where we find a female frog in the pair and 1/3 where we don't.
The probabilities just seem wrong to me; can anyone clarify the reason why this is the case?
I suspect that there is a subtly in the framing of the question that I'm missing.
As i read the problem, we have a choice of two options, both of which are simply a 50:50 chance of whether a single frog is male or female.  Not knowing which frog in the pair is definitely male should have no effect on the probability of the other. 
If I am wrong I really want to understand why!
 A: Let's look at the pair of frogs. Male frogs are identified by croaking in the video. 
As explained in the video, before we hear any croaking, there are 4 equally likely outcomes given 2 frogs:


*

*Frog 1 is Male, Frog 2 is Male

*Frog 1 is Female, Frog 2 is Male

*Frog 1 is Male, Frog 2 is Female

*Frog 1 is Female, Frog 2 is Female


Making the assumptions about males and females occurring equally and independently, our sample space is $\{(M,M),(F,M),(M,F),(F,F)\}$, and we have probability $1/4$ for each element.
Now, once we hear the croak coming from this pair, we know that at least one frog is male. Thus the event $(F,F)$ is impossible. We then have a new, reduced sample space induced by this condition: $\{(M,M),(F,M),(M,F)\}$. Each remaining possibility is still equally likely, and the probability of all the events added together must be $1$. So the probability of each of these three events in the new sample space must be $1/3$.
The only event that ends badly for us is $(M,M)$, so there is a $2/3$ chance of survival.

More formally, the definition of conditional probability says:
$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
So if $A$ is the event that at least one female is present and $B$ is the event that at least one male is present, we have:
\begin{align}P(\text{F given at least 1 M}) &= \frac{P(\text{F and at least 1 male})}{P(\text{at least 1 M})}\\
&= \frac{P(\text{1 M and 1 F})}{P(\text{1 M or 2 M})} \\
&= \frac{P[(M,F),(F,M)]}{P[(M,M),(F,M),(M,F)]} \\
&= \frac{1/2}{3/4} = 2/3 \end{align}
This is really the same procedure we reasoned through as above.
A: Since the math is already laid out I'll try to provide some intuition.  The issue is that knowing that at least one frog is male is different from knowing that any particular frog is male.  The former case carries less information and this effectively increases our chances over the latter situation.
Call the frogs left and right, and suppose we are told that the right frog is male.  Then we have eliminated two possible events from the sample space: the event where both frogs are female and the event where the left frog is male and the right frog is female.  Now the probability truly is one half and it doesn't matter which one we choose.  The exact same argument is true if we learn that the left frog is male.
But if we are told only that at least one frog is male, which is what happens when we hear the croak, then we cannot eliminate the event that the left frog is male and the right frog is female.  We can only eliminate the event that both are female, which makes the event that at least one is female more likely than the previous setting.
I think the reason why this is confusing is that we naturally think learning that at least one is male should make us disinclined to choose the pair of frogs.  It is true that this information makes it less probable that at least one is female, but recognize also that there was a full three quarters chance of at least one female before we learned anything at all.  It's the ambiguity of the information we receive which makes it so we should still prefer the two frogs over the one.
A: Your intuition is correct in this case. As the problem is stated your odds of survival are 50%. The video incorrectly states the problem space based on the information we have and therefore comes to an incorrect conclusion. The correct problem space contains 8 conditions and is as follows. 
We have two frogs on a log, and one of them has croaked what are our possibilities? (M designates male, F designates female and c designates croaked, first position is left, second position is right)
[
  [Mc, M], 
  [M, Mc],
  [Mc, F], 
  [M, Fc], (X No Male croak) 
  [Fc, M], (X No Male croak)
  [F, Mc], 
  [Fc, F], (X No Male croak)
  [F, Fc], (X No Male croak)
]

Each case is equally likely based on the information that we have, when we eliminate the conditions given the knowledge that a male frog has croaked. We find that there are 4 outcomes to expect. Left male frog croaked next to a right male frog that was silent. Right male frog croaked next to a left male frog that was silent. Or there was a croaking male frog paired with a single female frog in either direction. For an intuitive way to understand this, the two male frogs are twice as likely to croak than the single male frog paired with a female, so we have to weight it appropriately.
You could also divide the search space by croaking frog (C) and non croaking frog (N). Since the croaking frog is 100% a male, you can eliminate it from your search since it has no chance of helping you survive. While the author intended to create a "monty hall problem" they inadvertently created a "boy or girl paradox". 
The following questions yield different results:
Given that there is a male what is the likelihood the other is female?
Given that a male frog croaked what is the likelihood the other is female?
I know more information in the second case
https://en.wikipedia.org/wiki/Monty_Hall_problem
https://en.wikipedia.org/wiki/Boy_or_Girl_paradox
A: A clearer answer to this, since the previous was too long and not easy to understand.

The possible outcomes are different, although I used same letters. To make clear the sample space, I will describe the possible outcomes
M M -->  "The male is on the left" - "A random male on the right"
M F --> "The male is on the left" - "A random female on the right"
M M --> "The male is on the right" - "A random male on the left"
M F --> "The male is on the right" - "A random female on the left"
A: The problem I have with this problem, is that the solution seems to be using different rules for what it considers a possible result for the two frogs being male and female, and male and male.
The F/M pair, and the M/F pair, are different because we don't know whether the first frog or the second frog is male, so F/M and M/F are two separate possibilities, even though the result still amounts to "one female frog, one male frog". 
But the M/M pair is only considered one possible result, even though the same logic should apply: we don't know which frog is the one that made the croaking sound, so either frog could be the one we heard, and the other one could still be male, it just didn't happen to croak.    
A: Not knowing anything: $\{(M,M), (M,F), (F,M), (F,F)\}$.
Three pairs with at least one female out of four possible combinations: $3/4$ or $75\%$
Knowing the first one is male: $\{(M,M), (M,F)\}$. One pair with at least one female out of two possible combinations: $1/2$ or $50\%$
Knowing that there is at least one male: $\{(M,M), (M,F), (F,M)\}$. Two pairs with at least one female out of three possible combinations: $2/3$ or $67\%$
A: The correct answer is given above by tomciopp and the video is incorrect. I want to elaborate on their answer with a little more to diagnose the video's mistake and give an intuitive illustration or two.
The Video's Mistake
In the video, they conditionalize on the statement "at least one frog is male." However, the information we receive is "there was exactly one croak in the clearing." This statement implies that there is at least one male frog in the clearing, but it is not logically equivalent to it. If exactly one croaked, then at least one is male; however, "if at least one is male, exactly one croaked" is false.
If we conditionalize on the information we receive (instead of a statement which it merely implies) tomciopp's analysis is correct.
An intuitive illustration
Here's an intuitive case to help understand the difference. Suppose I flip two coins and exactly one of them is a penny. Then I ask you "what's the probability that the non-penny is a heads?" The answer is 0.5. It doesn't matter what the penny did. If I add a piece of information like "All the pennies I have ever flipped were heads" that doesn't change one whit your expectation about non-pennies.
In the case of the clearing, we have a croaker (=penny) and a non-croaker (=non-penny.) We also know that croakers are irrelevant to our decisions. 50% of non-croakers are female. The case is symmetric to coins. (This is sort of implied in the problem. Technically, if a frog hasn't croaked for a certain amount of time, I probably start to think it is more likely to be female, since males sometimes croak. In this case, however, we can fairly ignore this since each non-croaking frog has been observed the same about of time, they have the same increasing probability of being female.)
A Different Case
There is a species of toad that prevents mushroom poisoning. Only the females of this species have the necessary secretion to prevent mushroom poisoning. The females of this species do a funny dance whenever there is another female around, but not otherwise.
In this case, the pair is not dancing if and only if at least one is male. Thus, we can conditionalize on "at least one is male" and validly draw the conclusion there is a 2/3 probability of a female in a non-dancing pair.
A: The issue with considering Mm and mM separately is that there's a 50% chance of Mx and a 50% chance of Cy, where M is male and C is the croaker, but that's only if x is also male. If x is female, then the chances of the first frog being the croaker (i.e. Cy) doubles to 100%, which means the likelihood of that situation (i.e. MF) is also double (compared to Mm).
A: Before we hear any croaking, there are 4 equally likely outcomes given 2 frogs:
Frog 1 is Male, Frog 2 is Male
Frog 1 is Female, Frog 2 is Male
Frog 1 is Male, Frog 2 is Female
Frog 1 is Female, Frog 2 is Female
Making the assumptions about males and females occurring equally and independently, our sample space is  {(M,M),(F,M),(M,F),(F,F)}, and we have probability 1/4 for each element.
Once we hear the croak coming from this pair, we know that at least one frog is male. This male can equally likely be Frog 1 or Frog 2.
So there are 2 equally likely outcomes for the Frog 1:
Frog 1 is Male
Frog 1 is Random Frog
Making the assumptions about males and females occurring equally and independently, the Random Frog is equally likely to be a Random Male or a Random Female.
P(Frog 1 is Random Male given Frog 1 is Random Frog)=P(Frog 1 is Random Female given Frog 1 is Random Frog)=1/2
P(Frog 1 is Random Male and Frog 1 is Random Frog)=P(Frog 1 is Random Frog)P(Frog 1 is Random Male given Frog 1 is Random Frog)=(1/2)(1/2)=1/4
P(Frog 1 is Random Female and Frog 1 is Random Frog)=P(Frog 1 is Random Frog)P(Frog 1 is Random Female given Frog 1 is Random Frog)=(1/2)(1/2)=1/4
So there are 3 possible outcomes for the Frog 1:
Frog 1 is Male
Frog 1 is Random Male
Frog 1 is Random Female
and probabilities are:
P(Frog 1 is Male)=1/2
P(Frog 1 is Random Male)=1/4
P(Frog 1 is Random Female)=1/4
Now, for each possible outcome for Frog 1, there are 2 possible outcomes for the Frog 2:
Frog 2 is Male
Frog 2 is Random Frog
For each possible outcome for Frog 1, the Random Frog is equally likely to be a Random Male or a Random Female.
So, for each possible outcome for Frog 1, there are 3 possible outcomes for the Frog 2:
Frog 2 is Male
Frog 2 is Random Male
Frog 2 is Random Female
P(Frog 2 is Male given Frog 1 is Male)=0
P(Frog 2 is Male given Frog 1 is Random Male)=1
P(Frog 2 is Male given Frog 1 is Random Female)=1
P(Frog 2 is Random Male given Frog 1 is Male)=1/2
P(Frog 2 is Random Male given Frog 1 is Random Male)=0
P(Frog 2 is Random Male given Frog 1 is Random Female)=0
P(Frog 2 is Random Female given Frog 1 is Male)=1/2
P(Frog 2 is Random Female given Frog 1 is Random Male)=0
P(Frog 2 is Random Female given Frog 1 is Random Female)=0
P(Frog 2 is Random Male and Frog 1 is Male)=P(Frog 1 is Male)P(Frog 2 is Random Male given Frog 1 is Male)=(1/2)(1/2)=1/4
P(Frog 2 is Random Female and Frog 1 is Male)=P(Frog 1 is Male)P(Frog 2 is Random Female given Frog 1 is Male)=(1/2)(1/2)=1/4
P(Frog 2 is Male and Frog 1 is Random Male)=P(Frog 1 is Random Male)*P(Frog 2 is Male given Frog 1 is Random Male)=(1/4)*1=1/4
P(Frog 2 is Male and Frog 1 is Random Female)=P(Frog 1 is Random Female)*P(Frog 2 is Male given Frog 1 is Random Female)=(1/4)*1=1/4
So, our sample space is {(Male,Random Male),(Male,Random Female),(Random Male,Male),(Random Female,Male)}, and we have probability 1/4 for each element.
P(F given at least 1 M)=P(F and at least 1 male)/P(at least 1 M)=P(1 M and 1 F)/P(1 M or 2 M)=P[(Male,Random Female),(Random Female,Male)]/P[(Male,Random Male),(Male,Random Female),(Random Male,Male),(Random Female,Male)]=(1/2)/(4/4)=1/2
