"So you're stranded in a huge rainforest and you've eaten a poisonous mushroom. To save your life, you need the antidote excreted by a certain species of frog. Unfortunately, only the female of the species produces the antidote. And to make matters worse, the male and female occur in equal numbers and look identical with no way for you to tell them apart, except that the male has a distinctive croak. And it may just be your lucky day. To your left, you've spotted a frog on a tree stump. But before you start running to it, you're startled by the croak of a male frog coming from a clearing in the opposite direction. There you see two frogs, but you can't tell which one made the sound. You feel yourself starting to lose consciousness and realize you only have time to go in one direction before you collapse. What are your chances of survival if you head for the clearing and lick both of the frogs there? What about if you go to the tree stump? Which way should you go?"

Run to the clearing or run to the stump (or die trying to solve this riddle)?

A horrible conditional probability example: This TED-Ed riddle is a horrible example of conditional probability because it leaves many critical assumptions open to interpretation. What about biology and the fact that a male frog croaks to attact a female frog? Apparently we must ignore that. Does the female frog croak at all? Not in real life! So, keep as much as possible a variable and solve. So, just state that male frogs croak with a probability of 'p' and females croak with the probability of 'q' and solve for that (you can always set q=0 after the fact)!

It appears very clear that TED-Ed intended for the distinctive croak of a male frog from the clearing to simply (only) inform us that there is a male frog in the clearing. However, what TED-Ed did not plan for is that it also tells us that there is no croaking male on the tree stump! So it matters HOW you learned that there was a male frog in the clearing! Did an 'oracle' tell you there was a male frog in the clearing (does not affect tree stump). Or did you hear only a croak coming from the clearing (also affects the tree stump)? It matters. See 'oracle mode' in simulation far below.

Also, the "Dig Deeper" lesson section for the frog riddle on the TED-Ed site points to the Monty Hall problem (more info from wiki), which confirms further that TED-Ed intended to present the 'oracle' problem.

Monty Hall problem: You are on a game show and there are three boxes. Inside only one box are the keys to a brand new car! You are asked to select a box. After your box selection, the game show host reveals an empty box and asks if you would like to stick with your original selection, or switch to the one remaining box. What should you do?

For most people, the answer -- that switching results in a 2/3 chance to win the car -- appears impossible. But a little thought makes it obvious. For the 1/3 of the time that you initially picked the box with the car keys, the host reveals one losing box, you switch to the one other losing box, and you lose. But for the 2/3 of the time that you initially picked a losing box, the host reveals the other losing box, you switch to the box with the car keys, and you win!

Croak probability matters!

To clearly 'see' that the probability of a male frog croaking matters a LOT, consider what happens when the probability of a male croaking is 100%. Then any 'silent' frog MUST be a female frog! And the fact that you heard a single male distinctive croak coming from the clearing means that the other two (silent) frogs (one in clearing; one on tree stump) MUST both be female frogs!

The response from TED-Ed was from support, and because that person was not a mathematician, I will not quote that person directly, but they defended the frog video, effectively stating 'on the stump there is no male croak, but likewise, there is also no female croak, so that means 1/2 is the right answer'.

Answer: The probability of finding a female frog in the clearing, given that you heard a distinctive male croak coming from the clearing (and no other croaks), is (1-q)/(2-p-q), where 'p' is the probability of a male frog croaking and 'q' is the probability of a female frog croaking. With p=0.5 and q=0, the probability is 2/3.

Answer: The probability of finding a female frog on the tree stump, given that you heard a distinctive male croak coming from the clearing (and no other croaks), is (1-q)/(2-p-q), where 'p' is the probability of a male croaking, and 'q' is the probability of a female frog croaking. With p=0.5 and q=0, the probability is 2/3.

Answer: No matter what the probability of a male frog croaking or female frog croaking is, both ways result in the same probability of finding a female frog!

But, given that you about to "lose consciousness" and "collapse", it is better to not waste time -- and therefore, you should lick just ONE frog, not two, so...

Your best option is go to to the tree stump!

Another way of thinking about conditional probability is that it in effect creates a new sample space 'B', which is a subset of the original sample space -- and then what is the probability of 'A' happening within that new sample space 'B'.

Now, what is the probability only one of the rolls was a 5, if the sum of both dice equals 9? P(B) is P(if the sum of both dice equals 9), and P(A) is P(what is the probability only one of the rolls was a 5). Solving, we get a P(B) of 4/36, and a P(A∩B) of 2/36 for a final answer is then (2/36)/(4/36), or 50%. The key to understanding the different answer is realizing that the condition creates a new sample space.

Something similar is happening with the frog riddle. The key to answering the frog riddle is correctly finding the new sample space 'B'.

P(male or female): We must know the probability of a male vs female frog. The original riddle states 'equal numbers', which we will translate as 'equally likely' -- even though it is not precisely that (see a later section for a more detailed look at this).

P(male croak): We will use the probability of a male frog croaking of 'p' -- keeping it a variable! This way, after the fact, you can set 'p' to anything.

P(female croak): We will use the probability of a female frog croaking of 'q' -- keeping it a variable! This way, after the fact, you can set 'q' to anything -- including zero, which eliminates female croaking.

one and only one male croak: The original TED-Ed video makes it clear in its subsequent narration that "startled by the croak of a male frog" means hearing ONE and only one male croaking, which eliminates multiple males croaking at the same time.

and no other croaking: The riddle narration makes it clear that you are looking at the frog on the stump when you are started by a distinctive male croak from the clearing. So you know there was no croak coming from the stump. As for the clearing, the riddle says "which one made the sound", which means a single sound was heard.

Format: Each item is "(stump frog) — (clearing frog pair)". Notice that each column is one stump frog option, and the 16 'frog pair' entries within all columns are identical.

The probability of an item, given 'M' males of which 'm' are croaking, and 'F' females of which 'f' are croaking is:


When p=0.5 and q=0.5 this greatly simplifies to just (1/2)^6, or 1/64.

So the answer for P(female on stump|B) is:

So the answer for P(female in clearing|B) is:

So ignoring female croaking (no impact a hearing the 'distinctive' male croak) is exactly the same as no female croaking. That makes sense. But then where do you do this? Only in the clearing formula? In the tree stump formula? Both? What a mess.

Tree stump probability tree

Tree Stump: For the tree stump, there is only one frog with possibilities (M,F). When you extend that to only one male frog croaking in the clearing and no female croaking, you get either (1) a female, or (2) a non-croaking male -- or a sample space of (F_*, M₀). Also, refer to the probability tree seen right. Therefore, the conditional probability of finding a female frog at the tree stump, given a male croak only from the clearing (and p=0.5, q=0) is (1/2)/(1/2+1/4), or 2/3. But TED-Ed used a sample space of (M,F), and calculated 1/2, the wrong answer!

The key mistake that TED-Ed made is not understanding that the probability of a male frog croaking affects both the clearing and the tree stump -- and simply (wrongly) presumed that F_* and M₀ are both equally likely.

The only way to get the 1/2 TED-Ed answer is if male frogs never croak (p=0)!

Clearing: For the clearing, there are two frogs with possibilities (FM,MF,MM). When you extend that to only one male frog croaking and no female croaking, you get (F_*M₁, M₁F_*, M₁M₀, M₀M₁). Therefore, the conditional probability of finding a female frog in the clearing, given a male croak only from the clearing (and p=0.5, q=0) is (1/8+1/8)/(1/8+1/8+1/16+1/16) = 2/3. But TED-Ed used a sample space of (FM,MF,MM), calculated 2/3, and got lucky getting the correct answer using the wrong math.

But the only way to get that result was to assume that female frogs croak with probability of 0.5 (but female frogs don't croak in the real world). That was a mistake and the wrong tactic to take.

Presh should have instead just calculated the probabilities either way and shown that they are equal!

Isn't it amazing that mathematicians that find fault with the TED-Ed riddle incorrectly translate "the male and female occur in equal numbers" to "equally likely". Both are very close to each other, but not the same thing! "Equal numbers" means that you start out with 'N' male frogs and 'N' female frogs!

With N female frogs and N male frogs, there is a N/(2N) = 1/2 probability that the first frog is a female. The second frog is known to be a male. So what about the third frog? Do the math and there is a 1/(2-1/N) probability that the third frog is a female, giving an ever so slight advantage to the third frog being a female! As N goes to infinity, 1/(2-1/N) becomes the expected 1/2 in the case of "equally likely".

TIP: A spreadsheet is your best friend. My steps: (1) create a working model in a spreadsheet, (2) derive the formula on paper, and (3) enter the formula into the spreadsheet and confirm that it and the model produce the same results for varied 'N'.

So the answer is: You should lick the third frog.

First answer is 1/2 (the only probability fork when a female does not croak is for the male croaking, or not) Second answer is also 1/2. (because you are excluding when they both don't croak or when they both do croak). Third answer is 3/4 (1/4 is when they both don't croak).

The single croak two frog paradox: Given the probability of a frog croaking is 0.5, what is the probability of hearing a single croak from (1) ONE frog, (2) TWO frogs. Surprisingly, the same! For one frog, it is obviously 1/2. For two frogs, it is 2/4 (01,10) of (00, 01, 10, 11), or the same 1/2. The key is realizing that with two frogs, one frog is croaking and the other frog MUST not croak!

▶Run the frog simulation - confirms that the answer is the same both ways

Warning: There is a danger that you make a wrong assumption and then code that wrong assumption into the simulation, which only further confirms your wrong assumption. If the simulation confirms your math calculations, your math calculation may be correct. But if the simulation refutes your math calculations, then you absolutely KNOW you have a problem (either in the math, or in the simulation).

• Can you solve the frog riddle? - TED-Ed video on YouTube
• TED-Ed Lesson - Full lesson on YouTube
  
  
• TED-Ed's Frog Riddle Is Wrong - YouTube video by Presh Talwalkar
• TED-Ed s Frog Riddle Is Not Entirely Correct - a blog post by Presh Talwalkar
• A Frog Puzzle - a post on Tanya Khovanova's Math Blog.
• TED-Ed open discussion - an open discussion on TED-Ed