Solved – The Frog Riddle – Conditional Probabilities

I saw this riddle doing the rounds on the internet:

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!

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.

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