In the Monty Hall problem, you are confronted with three identical doors, one of which conceals a car while the other two conceal goats. You choose a door at random, number one say, but do not open it. Monty now opens a door he knows to conceal a goat. He then gives you the option of sticking or switching. What should you do to maximize your chances of winning the car.

As we are all by now aware, the correct answer is that you double your chances of winning by switching doors. Most people find this counterintuitive on the grounds that after Monty opens a door, only two, equally likely options remain. Thus, there is no advantage to be gained from switching.

Consider the following argument in defense of that view:

I claim that you have erred in initially assigning a probability of 1/3 to each of the three doors. This assignemnt was premised on the idea that for all you knew to the contrary, each of the three doors was equally likely to conceal the car. This did not adequately consider all of the information you had at your disposal. In particular, you knew that (1) You were going to choose door one and that (2) Monty would then open a goat-concealing door. These facts alter the way you should assign probabilities.

You see, at the moment you choose door one you know that either door two or door three will be eliminated. Regardless of which it is, you know that there will then be two equally likely doors remaining, one of which will be door one. It follows that door one never has a 1/3 probability of being correct. Rather, it had a 1/2 probability of being correct all along.

What do you think?

There are a number of ways to expose the flaw in the argument, but I especially like the following: If we accept this sort of reasoning, then regardless of the situation it will always be impossible to assign probabilities in a non-contradictory way.

Imagine that we have *n* equally probable hypotheses, H1, H2, …,Hn. Each of these hypotheses gets a probability of *1/n*. Imagine now that we are talking to an omniscient being. We ask Him, “Please point to a true hypothesis out of the collection H1 or H2, H1 or H3, H1 or H4, …, H1 or Hn.” After he does so, we could follow our previous reasoning to conclude that H1 now has a probability of 1/2.

What if we took the hypotheses three at a time? We ask the omnisicent one to name a true hypothesis out of H1 or (H2 or H3), H1 or (H2 or H4), and on we go allowing the hypotheses in the parentheses to cycle through all possible pairs of subscripts. Once a true triple is pointed out, we will conclude that H1 has probability 1/3.

By continuing in this way, we could justify a probability of 1/n for H1, for any integer *n* we choose (this might require altering our partition of the sample space into possible hypotheses). And it works the other way too! Just replace H1 with its negation and use the same trick. Partition the hypothesis “Not H1” into some suitable number of equiprobable possibilities and use our previous argument to drive its probability down as low as you wish. This has the effect of driving the probability of H1 as high as you wish.

Thus, accepting the argument presented at the beginning of this post would force us to conclude that the whole idea of a meaningful calculus for probabilities is a pipe dream, and that the whole subject is a bottomless pit of contradictions and black holes. Then again, after pondering the Monty Hall problem for a while that seems like an all too reasonable conclusion!

Incidentally, I encountered this argument in a paper by Ruth Weintraub, entitled “A Paradox of Confirmation.” It was published in the philosophy journal *Erkenntnis* in 1988. It is part of the considerable professional philosophical literature on the Monty Hall problem (techinically, Weintraub was discussing the three prisoners problem, which is isomorphic to the MHP.) Wading through this literature has lately been a major project of mine.