Probability and Pancakes

9/18/18

Let B stand for "burnt" and U stand for "unburnt". Imagine we have 3 types of pancakes.

• Burnt on both sides (BB)
• Burnt on one side (BU or UB)
• Not burnt on either side (UU)

Suddenly, you are presented with a pancake that is burnt on just one side, but the other side is not revealed.

Question. What is the probability that the other side is burnt?

Some possible answers are 1/3 or 1/2, all depending on your assumptions. Another reasonable approach is to use conditional probability and get 2/3 as an answer.

Using conditional probability, we consider P(want to know | what we already know) (read "probability of what we want to know given what we already know"), and we have P(burnt down | burnt up) = P(burnt up AND burnt down) / P(burnt up).

The P(burnt up) = 1*P(BB) + .5*P(BU) = 1/3 + .5*(1/3) = .5. So putting it all together, P(burnt down | burnt up) = (1/3)/(1/2) = 2/3.

One could also write out the sample space of ways of getting two sides, S_original = {UU, BB, UB, BU}, and then trim it to only consider the sample space with a B on one side, S_trimmed = {BB, UB, BU}, and note that 2 out of the 3 possibilities have a B for the second side.

I put up a spreadsheet demonstrating this here.

Thanks for reading!