Frog probability

[Aractus]

Okay, so there is a special species of frog. It is impossible for you to tell the difference between the males and females, however male frogs croak and females do not. You hear a croak and then see two frogs sitting together. What is the probability that one of them is a female? Have a look at the TED-Ed video and see what you think:

https://www.youtube.com/watch?v=cpwSGsb-rTs

Now have a look at MindYourDecisions/Presh Talwalkar clip which comes up with a different answer:

https://www.youtube.com/watch?v=go3xtDdsNQM

So my question to you is which of the above is correct?

My answer ...

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Neither, but TED-Ed is closer. Let's start with the premise that before hearing a croak there are FOUR possible pairs of frogs:

1. Boy, Boy
2. Boy, Girl
3. Girl, Boy
4. Girl, Girl

Once we hear a croak then possibility 4 is eliminated, however we also have to consider the further probabilities that happen.

There is an equal possibility of starting from these conditions:
1. Boy, Boy
2. Boy, Girl
3. Girl, Boy

The Boy-Boy pair can either:
1. Both croak
2. Frog1 croaks
3. Frog2 croaks
4. Neither croaks

And the other two pairs each have these two possibilities:
1. Boy frog croaks
2. Boy frog doesn't croak

So now we can see that there are four possible outcomes for the boy-boy pair. However there is a big unknown in this calculation: what is the probability that a male frog would croak in the time that you have been there? So let's put in a few values and see what happens. First let's imagine Pr(.1) - that is a 10%chance that a male frog would croak:

1. Both croak ....Pr(.1*.1) = .01
2. Frog1 croaks ....Pr(.1*.9) = .09
3. Frog2 croaks ....Pr(.1*.9) = .09
4. Neither croaks ....Pr(.9*.9) = 0.81

As you can see the probabilities add up nicely to 100.

Pair 2
1. Boy frog croaks ....Pr(.1) = 0.1
2. Boy frog doesn't croak ....Pr(.9) = 0.9

Pair 3
1. Boy frog croaks ....Pr(.1) = 0.1
2. Boy frog doesn't croak ....Pr(.9) = 0.9

Now we'll add them together, and exclude the events leading to failure and we get:

Boy-Boy ....Pr(.18/.38) = 47.4%
Pair with a girl ....Pr(.2/.38) = 52.6%

Right now let's see what would happen if there was a 90% chance of a male frog croaking:

Boy-Boy pair:
1. Both croak ....Pr(.9*.9) = .81
2. Frog1 croaks ....Pr(.9*.1) = .09
3. Frog2 croaks ....Pr(.9*.1) = .09
4. Neither croaks ....Pr(.1*.1) = 0.01

As you can see for the Boy-Boy pair it does not change the probability at all - it remains 18% probable that one of the frogs in the pair croaks.

Pair 2
1. Boy frog croaks ....Pr(.9) = 0.9
2. Boy frog doesn't croak ....Pr(.1) = 0.1

Pair 3
1. Boy frog croaks ....Pr(.9) = 0.9
2. Boy frog doesn't croak ....Pr(.1) = 0.1

Now we'll add them together, and exclude the events leading to failure and we get:

Boy-Boy ....Pr(.18/1.98) = 9.09%
Pair with a girl ....Pr(1.8/1.98) = 90.9%

And for completion let's consider if the chance was 50% that a male frog would croak in the time you were there:

Boy-Boy pair:
1. Both croak ....Pr(.5*.5) = .25
2. Frog1 croaks ....Pr(.5*.5) = .25
3. Frog2 croaks ....Pr(.5*.5) = .25
4. Neither croaks ....Pr(.5*.5) = .25

Pair 2
1. Boy frog croaks ....Pr(.5) = 0.5
2. Boy frog doesn't croak ....Pr(.5) = 0.5

Pair 3
1. Boy frog croaks ....Pr(.5) = 0.5
2. Boy frog doesn't croak ....Pr(.5) = 0.5

Now this calculation looks more like TED-Ed:

Boy-Boy ....Pr(.5/1.5) = 33.3%
Pair with a girl ....Pr(1/1.5) = 66.7%

Talwalkar's criticism would be true if the four stating conditions were equally likely, however we can show that they are not. The probability is dependant on the population spread at the start of the sample: for example let's say you had 400 male frogs, and 400 females, and paired them up randomly. The probability of male-male pairs remains the same before and after you hear croaking coming from some of your 400 pairs, and if you wait long enough eventually all the pairs with at least one male will croak. The fact that you've heard a croak doesn't suddenly double the number of available male-male pairs to hear in the set of 400 pairs - there should still be about 100 pairs.

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