# Can you Solve TED’s Frog Riddle? Can TED?

Critics argue that it’s not.

For those of you who want to see the problem laid out in detail, you can watch the video here and read the comments below to understand the debate.

For the rest of you, I’ll lay out the basics in a simplified way.

There are two tree stumps, A and BStump A holds one frogStumb B holds two frogsFrogs can be male (M) or female (F) each with 50% probabilityYou want to choose the stump with the highest probability of holding a female frogWith no other information, one would obviously choose stump B since it contains two frogs rather than one.

However, before you can choose, you hear a croak from stump B, which is indicative of a male frog.

You now know that there is at least one male frog on stump B.

Which stump do you choose, and what is the probability that that stump contains a female frog?What makes this problem especially interesting is that there is controversy over how to correctly solve it, and what its solution actually is.

Specifically, there are two lines of very sensible reasoning that produce conflicting answers.

One is presented by TED in the video (you can listen to that here or continue reading for a summary) and one presented by critics in the comment section.

Answer 1: TED’s SolutionThe idea behind TED’s solution is that you begin by considering the initial sample space before you hear a frog croak.

The sample space is the set of possibilities for the genders of the frogs on stump B.

These possibilities are (MM), (MF), (FM), and (FF) where letters M and F indicate male and female.

Each possibility is equally likely.

Sample space of frogs.

Image from TED-Ed’s video “Can you solve the frog riddle? — Derek Abbott”After hearing a croak, we update the sample space using the new information we’ve gained.

That is, after hearing a croak, we know that stump B cannot have two female frogs because female frogs don’t croak.

Therefore, we eliminate (FF) from the sample space.

We now have the remaining possibilities: (MM), (MF), and (FM).

Two out of these three possibilities contain a female frog so the probability of a female frog being on stump two is 2 out of 3, or 66.

7%.

Obviously, the probability of a female frog on stump A is just 50% since we know nothing about it.

Therefore, we choose stump B.

Answer 2: Critics’ SolutionTED’s solution makes sense.

However, it also contradicts some simple intuition.

Let’s say we consider Stump B with two frogs that can be male or female.

Once we hear a croak, we know that there is at least one male frog.

We don’t know which one is male but it doesn’t matter.

There is now one frog which we know is male and one frog which we really don’t know anything about.

Because the frogs are independent, the probability that the second frog is female is still 1/2, or 50%, and so the probability that Stump B contains a female frog is 50%.

Therefore, it makes no difference whether you choose Stump A or Stump B.

This is the argument that dominates the comment section of YouTube.

Users have come up with a number of ways to explain the reasoning behind why the probability is 50% either way.

Some of these explanations are just as compelling as TED’s explanation.

Answer 3: Bayes’ SolutionWe now have two different solutions that both seem correct, but provide conflicting answers.

How can we tell for sure which one is really true?.A good approach, though not as fun, is to reduce the problem to something transparent and objective.

To accomplish this, we translate the problem into math-friendly notation.

If you’re not familiar with probability and don’t want to get bogged down in formulas, feel free to skip to the section “Settling the Debate”.

For the rest of you, I recommend pulling out your notes from probability and working it out yourself.

The first steps, as always, are to define variables for different events and define the quantity we want to determine.

This quantity is ‘the probability of there being two males on stump B given that a croak was observed’, written P(M₂|C ).

Right off the bat, we can use Bayes’ theorem to transform our desired probability into a combination of values that are easier to find.

For those that are new to Bayes’ Rule (glad you’re still reading!), a definition might be marginally helpful.

An article from the Corporate Finance Institute states that Bayes’ Rule¹…is a mathematical formula used to determine the conditional probability of events… based on prior knowledge of the conditions that might be relevant to the eventThe most important thing to understand about Bayes’ Rule is that it allows us to invert the conditional probability P(M₂|C) to P(C|M₂) as shown below.

Notice I said this form is easier to solve for, not easy.

Finding P(C), the probability of observing croaking, requires using the law of total probability and a few other steps.

These are shown below.

An important note is that P(C) is the unconditional probability of observing croaking.

People often want to write P(C)=1, since it is known that a croak was heard, but that is not correct as P(C) represents the probability of hearing croaking in general.

Also, it is assumed that the probability of one frog croaking is independent of the other frog croaking.

This is a common assumption in probability and seems plausible for this problem.

There we have it.

P(C)=Pc – 0.

25Pc ².

Simple, right?.But the obvious question is “what is Pc?” We defined it above as the probability of observing croaking from a single frog in the given time period.

We’ll discuss its implications in a minute.

First, let’s put P(M₂|C) into its final form.

If you’re confused about where the values of P(M₂) or P(C|M₂) come from, look back at how we solved for Pc.

P(M₂) and P(C|M₂) we’re solved for along the way.

Settling the DebateWe have now completed our objective of solving for P(M₂|C)!.However, this answer is not a number like 66.

7%.

It’s an equation.

This doesn’t seem right, but the problem is not actually solvable in the way that it is presented.

Getting a numeric solution requires one to make a significant assumption: what value of Pc to use!.For this reason, the two incompatible solutions are actually both correct depending on how likely it is that one observes a croak from a single male frog.

The graph below demonstrates the range of values of P(M₂|C) for different values of Pc (between 0 and 1).

Graph of P(M₂|C) on the Y-axis versus Pc on the X-axis.

As Pc approaches 0, (that is when the probability of observing a croak from a male frog is extremely rare) the probability of two male frogs approaches 50%.

As Pc approaches 1, (that is when the probability of observing a croak from a male frog is extremely common) the probability of two male frogs approaches 33%.

This corresponds to TED’s solution, though they frame the problem as the probability of at least one female, which is 1–0.

33=0.

67.

As the actual value of Pc is inevitably somewhere between 0 and 1 exclusive, the probability of at least one female at stump B will be higher than 50% and so stump B is the better choice.

Why does the probability of hearing a croak affect anything?.Well, if croaking is really rare, we’re about twice as likely to observe it when there are two males as compared to when there is only one.

When croaking is completely common, however, the probability of observing croaking from a set of two males is approximately the same as from a single male.

How likely it is to observe croaking when there are two males affects how likely there are to be two males when croaking is observed.

This relationship can feel confusing but the interaction can be seen in our solution for the equation of P(C|M₂) above.

If you’ve never heard of Bayes’ theorem before today, or if this problem has convinced you that you need a refresher, check out this video from Khan Academy.

Final Note: Further Questionable AssumptionsTED and YouTube are not the only ones guilty of making unstated assumptions.

In the solution above, I made the assumption that it would not be possible to hear or distinguish two croaks in the given time period.

I lumped cases of one croak and two croaks into the one general category of hearing a croak.

This makes sense in the context of the problem but is probably not realistic.

If we remove this assumption and specify that in the given case only a single croak was heard, we get the following equation for P(M₂|C), where C specifically refers to the event of hearing a single croak.

This function also gives a value of 0.

5 at Pc=0, but decreases faster than the first function to a value of 0 for Pc=1 (if you always hear a croak when there is a male frog then you will never hear just one croak when there are two male frogs.

) As the actual value of Pc is inevitably somewhere between 0 and 1 exclusive, the probability of two males at stump B will be lower than 50%, and thus the probability of at least one female will be greater than 50%.

Therefore stump B is the better option.

Additionally, because this function decreases faster than the first function, it gives a higher probability of there being a female frog on stump B for all values of Pc>0.

[1] Corporate Finance Institute, What is the Bayes’ Theorem?.(2018), corporatefinanceinstitute.

com.