## A Frog Puzzle

I stumbled upon a TED-Ed video with a frog puzzle:

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 of the species produces the antidote. The male and female frogs occur in equal numbers and look identical. There is no way to distinguish between them except that the male has a distinctive croak. To your left you spot a frog on a tree stump. You hear a croak from a clearing in the opposite direction, where you see two frogs. You can’t tell which one made the sound. You feel yourself starting to lose consciousness, and you realize that you only have time to run in one direction. Which way should you go: to the clearing and lick both frogs or to the tree stump and lick the stump frog?

My first thought was that male frogs croak to attract female frogs. That means the second frog in the clearing is probably an already-attracted female. The fact that the stump frog is not moving means it is male. I was wrong. This puzzle didn’t assume any knowledge of biology. The puzzle assumes that each frog’s gender is independent from other frogs. Thus this puzzle is similar to two-children puzzles that I wrote so much about. I not only blogged about this, but also wrote a paper: *Martin Gardner’s Mistake*.

As in two-children puzzles, the solution depends on why the frog croaked. It is easy to make a reasonable model here. Suppose the male frog croaks with probability p. Now the puzzle can be solved.

Consider the stump frog before the croaking:

**It is a female with probability 1/2.**- It is a croaking male with probability p/2.
**It is a silent male with probability (1-p)/2.**

Consider the two frogs in the clearing before the croaking:

- Both are female with probability 1/4.
**One is a female and another is a croaking male with probability p/2.**- One is a female and another is a silent male with probability (1-p)/2.
- Both are silent males with probability (1-p)
^{2}/4. - Both are croaking males with probability p
^{2}/4. **One is a silent male and another is a croaking male with probability p(1-p)/2.**

The probabilities corresponding to our outcome—a non-croaking frog on the stump and one croaking frog in the clearing—are in bold. Given that the stump frog is silent, the probability that it’s a female is 1/(2-p). Simillarly, given that one clearing frog croaked, the probability that one of them is a female is 1/(2-p). The probabilities are the same: it doesn’t matter where you go for the antidote.

The TED-Ed’s puzzle makes the same mistake that is common in the two-children puzzles. I don’t want to repeat their incorrect solution. The TED-Ed’s frog puzzle is wrong.

(The calculation in the second to last paragraph was corrected on Nov 13, 2021.)

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## Nick:

Just thought you should know that your result for the probability of the stump frog being female given that it is silent “1/(1-p)” is greater than 1 for 0<=p1 we have: 1/(1-p) -> infinity. Clearly there is a mistake here.

2 June 2016, 1:16 pmI have done a similar calculation and my result for the same probability is 1/(2-p).

In addition to this I worked out that if we allow the two frogs may have emitted >1 croak then the probability that one is female is 2/(4-p).

Comparing these two results for 0<=p<=1 we see that the latter is always less that or equal to the former with equality only when p=0, each giving 1/2.

## Nick:

Apologies in the first line the limits of p should read <=p<1.

2 June 2016, 1:18 pm## Nick:

Try again: 0<=p<1

2 June 2016, 1:19 pm## Daryl Mcullough:

This is one of the conceptually difficult aspects of probability theory. In nonprobabilistic mathematics, you can use whatever information you have to solve a problem, and it doesn’t make any difference where you got that information. But how you got the information is relevant in computing probabilities.

If there were no croaking involved, and somebody just told you: “At least one of the frogs in the clearing is male”, then what you would do is to list all the possibilities that are consistent with you knowledge, and give them equal probability. So there are 6 possible assignments of male/female to the three frogs consistent with the information. Out of those 6, there are 4 assignments where the other frog in the clearing is female, and only 3 assignments where the frog on the stump is female.

But if the information came from listening for croaks, then that changes the situation, because the fact that the frog on the stump DIDN’T croak is information, as well. It makes it slightly more likely that that frog is female.

This is a little bit of an unfair puzzle, though, because the statement didn’t actually say under what circumstances a frog croaks. Maybe they never croak when they are alone. Maybe they never croak when they are sitting on a stump. Maybe they only croak when a female is nearby. You have to model the croaking somehow. You chose the model in which male frogs just croak at random.

Now, is there a croaking model that supports the original reasoning? I guess if you assume that frogs never croak when they are alone.

15 June 2016, 9:10 am## Bill:

If you add in the probability for the female frogs to croak independently at probability q, the probabilities of survival for both cases are still equal: (1-q)/(2-p-q).

12 July 2016, 11:52 am