Probability, Philosophy, and Monty Hall

I'm not sure that it is possible to come up with something original every weekday, and today is one of those days where my mind is elsewhere. So here is something I've plagiarized word-for-word from a photocopy tacked to a wall in the mathematics department at UCT (the original author is Dr. J Ritchie, a senior lecturer on the Philosophy of Science here at UCT). I thought the ending fit nicely with some of the things I said yesterday with regard to consequentialism and finitism.

The Monty Hall Problem

Here's a puzzle you might have heard before. Imagine you are taking part in a game show. The host, Monty Hall, has three doors in front of him. Behind one there is a car and behind the other two a goat. At the end of the game you will have chosen one of the doors and you'll win whatever is inside. You want to win the car.

You start by choosing one door at random. Monty looks behind the other two doors and opens one of them to reveal a goat. He now offers you the chance to swap your door for the one he didn't open. What should you do?

Many people argue like this. There are only two possibilities: the car is either behind your door or Monty's, each is equally likely, so it doesn't matter which door you choose. But that's a mistake. There are three possibilities at the beginning of the game, all of which we assume are equally likely. Either you have chosen the door with the car or you've chosen the door with goat 1 or you've chosen the door with goat 2. If you swap in the first case, you'll win a goat. If you swap in the other two cases, then you'll win the car. Hence you're twice as likely to win if you swap, so you should swap.

That's a simple problem in probability theory but let's now think about it in a different way. Imagine instead of winning a car that Monty promises to shower you in riches, if you win. But if you lose, Monty will shoot you. Of course, you might not want to play that game. But tough, Monty has kidnapped all your family and threatened to kill them all and you unless you play. The game proceeds as before. You choose a door, Monty opens one of his doors to reveal a goat (which now represents your imminent death) and asks if you want to swap. What should you do?

We've ratcheted up the drama a bit, you might think, but the logic of the case remains the same. You are more likely to win if you swap. So you should swap. Let's say you swap and it turns out the door you are left with contains a goat; so Monty shoots you. In what sense, then, was it the right decision to swap?

The obvious response is that it is the right thing to do because, if you play the game a lot, you will win twice as often (on average) as you lose. Probabilities in other words tell us about long-run frequencies. But this kind of game you can't play a lot. Once you lose, you are in a very serious way out of the game for good. In fact since relative frequencies only converge on probabilities in the infinite long run and in the long-run we're all dead, is there any good reason ever to choose the more probable option? Now we've moved from a simple problem in probability theory to a hard problem in philosophy... But I've run out of space to offer you any solutions.

(By the way, you are invited to send possible solutions to jack.ritchie@uct.ac.za)