Find the Lady?

On the way back from Barley someone (who probably wishes to remain anonymous) referred to the conundrum concerning the three cups with a pea hidden underneath one. The stooge is asked to guess where the pea is. The one in the know then takes away one of the other cups (which is empty) and invites the stooge to change her selection. “Go on, you only had a one in three chance of a correct guess and that still remains for your selection, but the other remaining cup has a 50/50 chance of hiding the pea, so it’s sensible to change your choice, and increase your chances from 1 in 3 to 1 in 2 ~ you know it makes sense.”

I said that this was obviously incorrect (was I that polite?) and that I would post a rational explanation at the earliest opportunity as to why there is no advantage to either changing or sticking.

First thoughts were to present all the possible scenarios and then to show all the outcomes, and work out what that meant as probabilities. However, just a little more thought confirmed my initial view that the main flaw was in declaring that the initial 1 in 3 chance hadn’t changed, so the other remaining cup is a better bet. If the remaining unchosen cup has a 1 in 2 chance of being loaded, and the chosen cup has a 1 in 3 chance, is there, therefore, a 1 in 6 chance that the elusive pea has gone to Heaven? Obviously not; with 2 cups and the pea being under one of them, they both have a 50/50 chance of holding the pea.

Just liken it to the Grand National, where you’ve backed a horse at 100:1. After the final fence (no.14, which is jumped twice, making it the 30th of 30 on the second lap) and there’s only yours and one other horse left running; are your chances still 100:1? Again, obviously not!

Which only goes to show that it is safe to play poker with a maths teacher.