The Monty Hall problem – The numbers don’t lie

I’ve filed this in science as it’s basically a mathematical conundrum. I was chatting about this subtle problem with R last night. She’s a self-confessed numbers phobic but interested in how the world works. I worship at the altar of Phi and am perfectly willing to believe a logical mathematical prediction even if it flies in the face of a more immediate intuition. The Monty Hall problem is called after the presenter of a 60s/70s game show called “Let’s Make a Deal”. It’s a weaker version of the 3 prisoners problem and is generally stated as

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

We’re also asked to assume that the host is honest, the show isn’t rigged, we have no prior knowledge of the appearance of the goat and a whole bunch of other “butterfly wings” that could influence the result in any way. The answer is that it IS to your advantage to switch doors. For example if you pick door A and the host shows you a goat behind door C your first pick was made with a probability of 1 in 3 of correctly identifying the one door with the car behind it from 3 equally advantageous probabilities. The mistake that most people (almost everyone) seems to make is that they then misunderstand the probabilistic basis for switching. I also don’t particularly like the Goat-1,Goat-2 explanation presented by some including professional smart person Marilyn Vos Savant.
When A was picked I had a 1/3 chance of finding a car but I had a 2/3 chance of finding a goat. I suppose it depends on how you feel about goats but I’m indifferent and would prefer a car. The resale value is generally higher, unless it’s an Alfa of course! Changing to door B doesn’t have a 2/3 chance of finding a goat. There’s only one goat left. Damn, I’ve halved my chances of finding a goat. So let’s let the probability that switching to door B is a good by P(SwitchB).
P(SwitchB) = P(A was a goat) = 1 - (1/3) = 2/3
The probability that switching is a good idea is DIRECTLY affected by the probability the first choice was a goat. You can’t ignore the past and must treat the problem as a continuation of the same game. This in itself makes sense but for several reasons our brains find it difficult to combine temporal and logical context and we get a bit confused.