As an undergraduate, I was taught the concept of the “physicist’s proof”, a sort of silly concept that was a professor’s attempt to get us students to think a little harder about some problems. Here, I give you a “physicist’s proof” of the famous Monty Hall problem, which (to me!) is easier to think about than the typical Bayesian approach.

The Monty Hall problem, which was developed on a TV game show, goes something like this (if you already know the Monty Hall problem, you can skip the paragraphs in italics):

*Suppose you are a contestant on a game show where there are three doors and a car behind one of them. You must select the correct door to win the car.*

*You therefore select one of the three doors. Now, the host of the show, who knows where the car is, opens a different door for you and shows you that there is no car behind that door.*

*There are two remaining unopened doors — the one you have chosen and one other. Now, before you find out whether or not you have guessed correctly, the host gives you the option to change your selection from the door you initially chose to the other remaining unopened door.*

*Should you switch or should you remain with you initial selection?*

When I first heard this problem, I remember thinking, like most people, that there was a 50/50 chance of the car being behind either door. However, there is a way to convince yourself that this is not so. This is by taking the limit of a large number of doors. I’ll explain what I mean in a second, but let me just emphasize that taking limits is a common and important technique that physicists must master to think about problems in general.

In Einstein’s book, Relativity, he describes using this kind of thinking to point out absurd consequences of Galilean relativity. Einstein imagined himself running away from a clock at the speed of light: in this scenario, the light from the clock would be matching his pace and he would therefore observe the hands of the clock remaining stationary and time standing still. Were he able to run just a little bit faster than the light emanating from the clock, he would see the hands of the clock start to wind backwards. This would enable him to violate causality! However, Einstein held causality to be a dearer physical truth than Newton’s laws. Special relativity was Einstein’s answer to this contradiction, a conclusion he reached by considering a physical limit.

Now, let us return to the Monty Hall problem. And this time, instead of three doors, let’s think about the limit of, say, a million doors. In this scenario, suppose that you have to choose one door from one million doors instead of just three. For the sake of argument, suppose you select door number 999,983. The host, who knows where the car is, opens all the other doors, apart from door number 17. Should you stick to door 999,983 or should you switch to door 17?

Let’s think about this for a second — there are two scenarios. Either you were correct on your first guess and the car is behind door 999,983 or you were incorrect on your first guess and the car is behind door 17. When you initially made your selection, the chance of you having made the right guess was 1/1,000,000! The probability of you having chosen the right door is almost zero! If you had chosen any other door apart from door 17, you would have been faced with the same option: the door you chose vs. door 17. And there are 999,999 doors for you to select and not win the car. In some sense, by opening all the other doors, the host is basically telling you that the car is behind door 17 (there is a 99.9999% chance!).

To me, at least, the million door scenario demonstrates quite emphatically that switching from your initial choice is more logical. For some reason, the three door case appears to be more psychologically challenging, and the probabilities are not as obvious. Taking the appropriate limit of the Monty Hall problem is therefore (at least to me) much more intuitive!

Especially for those who are soon to take the physics GRE — remember to take appropriate limits, this will often eliminate at least a couple answers!

**For completeness, I show below the more rigorous Bayesian method for the three-door case:**

Bayes theorem says that:

For the sake of argument, suppose that you select door 3. The host then shows you that there is no car behind door 2. The calculation goes something like this. Below, “car3” translates to “the car was behind door 3” and “opened2” translates to “the host opened door 2”

The probabilities in the numerator are easy to obtain and . However, the is a little harder to calculate. It helps to enumerate all the scenarios. Given that you have chosen door three, if the car is behind door 1, then the probability that the host opens door two is 1. Given that you have chosen door three and are correct, the probability of the host choosing door 2 is 1/2. Obviously, the probability of the car being behind door 2 is zero. Therefore, considering that all doors have a 1/3 possibility of having the car behind them at the outset, the denominator becomes:

and hence:

.

Likewise, the probability that the car is behind door 1 is:

which can similarly be calculated:

.

It is a bizarre answer, but Bayesian results often are.