# A simple explanation of the Monty Hall problem

You are participating in a game show. There are 3 doors in front of you; 2 are empty and 1 contains a prize. You are asked to pick a door. You do so, but you don’t open it yet. The game show host—let’s call him Monty Hall—now opens another door; not the one you picked. You see that it is empty. You are asked if you want to stick with your choice or switch to the remaining closed door? What should you do if you want to win the prize?

This is the famous Monty Hall problem. I first heard of this problem in high school while reading The Curious Case of the Dog in the Night-Time. I don’t remember any of the rest of the book. But I remember this problem and talking about it with my friends, my parents, my sister and constantly infuriating everyone.

It seems like you have to pick between a door that has a prize and one that is empty. So the probability should be 1/2 of winning. So it seems like it shouldn’t matter if you stick to your choice or switch.

But another argument goes like this: when you first picked a door, it was more likely that you picked an empty door. If you picked an empty door, then Monty has to open the other remaining empty door. Thus, the door that he did not open contains the prize. Since with probability 2/3 you picked an empty door at the beginning, if you switch, then you will win with probability 2/3.

What’s going on?

The key to resolving the confusion behind the problem, is to realize that the best strategy depends on what knowledge Monty had when he picked his door. Let’s consider two possible scenarios:

(1) Monty knows which door contains the prize, and therefore opens only a door that doesn’t contain the prize.

(2) Monty does not know which door contains the prize and he opens one of the remaining doors and at random. Even though you see that it is empty, Monty could’ve opened the door with the prize.

These two cases have different answers! In Case 1, you will win with probability 2/3 if you switch. While in Case 2, you will win with probability 1/2 if you switch. Let’s see how.

Case 1. Monty Hall knows where the prize is and opens the door that’s empty:

Let us label the doors A,B and C. Let door A contain the prize: this assumption doesn’t matter, because what we call A, B or C doesn’t matter. I indicate the door containing the prize with P.

Consider many, many copies of your universe, and you are playing this game in all of these different universes.

In 1/3rd of the universes—Universe 1 in the picture–you pick door A; in 1/3rd of the universes—Universe 2—you pick door B; and in the remaining 1/3rd —Universe 3—you pick door C. In the picture, the door that you pick originally is indicated in green. (Note that each Universe 1,2 or 3 contains many sub-universes).

Thus, in 2/3rds of the universes—Universes 2&3—you have picked the door without the prize. In these Universes, Monty has to open the remaining empty door. We indicate the door Monty opens by blue.

Thus, in Universes 2&3, Monty, by making his choice, has given away information about where the prize is. Only in Universe 1 is he free to open either of the two doors, and he doesn’t give away any information.

The case where Monty knows which door contains the prize. P indicates the prize. Green indicates the door you originally picked. Blue indicates the door picked by Monty. A,B,C are the door labels.

It is clear from the figure that you should switch doors in Universes 2 & 3. Only in Universe 1 should you stick to your original choice.

Thus, in 2/3rds of the Universe if you switch, you will win the prize.

And since you should always behave as though you are in the most likely universe, you should switch. And you will win with probability 2/3.

Case 2. Monty Hall does not know which door contains the prize:

Again, please see the figure below.

In 2/3rds of the universes, you have picked the empty door (indicated by green). In the remaining 1/3rd you picked the door with the prize.

In the universes where you picked the door with the prize, Monty will always open an empty door (the door he picks is indicated in blue). So far, so good. This is Universe 1 in the picture.

But, in the universes where you picked an empty door, among the doors that remain there is one empty door and one door with the prize. These are Universes 2&3 in the picture.

Now, because Monty does not know which door contains the prize, in half of the universes he will open the door with the prize. This is Universe 3 in the picture. And in the other half he will open the empty door, which is Universe 2.

But we know that we are not in Universe 3 because we see that Monty opens an empty door.

Thus, we must be either in Universe 1 or 2. It is clear that the if we are in Universe 2, we should switch and if we are in Universe 1, we should stick to our choice.

In this case Monty does not know where the prize is. Again, green indicates the door you picked. Blue indicates the door picked by Monty. P indicates the door with the prize. Notice that in Universe 3, Monty opens the door with the prize, therefore we can’t be in Universe 3.

But the number of sub-universes in both Universe 1 & 2 are the same!

Thus, we are equally likely to win the prize irrespective of whether we stick with our original or switch! Therefore in this case, the probability of winning is 1/2 either way.

The Monty Hall problem beautifully illustrates how the processes behind the evidence that we see is crucial in deciding how we go forward.

Acknowledgements: Kenny Easwaran pointed out the difference between the two cases when answering a question I asked at a physics department colloquium. I think that’s when the Monty Hall really clicked for me. Also, reading Eliezer Yudkowsky  really clarified some notions of probability relevant to this problem for me.