The Monty Hall Problem

The Monty Hall problem is one of the most (in)famous little math puzzlers in the field of probability. I was first introduced to it via a Marilyn Vos Savant article my high school physics teacher handed out in class, but it was also featured in a scene from the movie 21.



In case the video ever disappears, I'll reproduce it here, from the Marilyn vos Savant article in question:

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 #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

According to the maths, it is most assuredly to your advantage to switch doors in this scenario.

I have never, ever, ever understood how it works. I've had incredibly smart friends attempt to explain it, I've tried to sit down and work it out with diagrams. For some reason, it came up in a discussion with the boy last night, and things became very...heated.


The Monty Hall problem is SERIOUS BUSINESS.

There's no reason, I argued, for the door you didn't pick to magically have the car behind it. I conceptualized the scenario as such: when Monty Hall opens the door with the goat, the doors reset and you now have a 50% chance of getting the car so you have no reason to switch (or not to switch). That is the "common sense," instinctual conceptualization of the problem that most people have. The boy, of course, was arguing the other point.

In a pique of frustration, I turned to the computer. Even if I didn't understand the math behind it, if you could simulate the Monty Hall scenario enough times, I would be willing to accept the math at face value. Empirical proof is empirical proof.

For your edification, here is the Monty Hall simulator I used. If it boggles your mind as much as it boggles mine, it's definitely very helpful. Not in explaining the math, so much, but more so in demonstrating why vos Savant is right.

Here is the Fancy Monty Hall simulator, which will run itself within parameters of your own defining (how many times, whether to keep or switch doors, etc.). A little window will pop up telling you to use Internet Explorer, but you can pretty much ignore that. It worked just fine for me in Chrome. However, if that one fails, this Monty Hall simulator will work. It takes a little bit to load, so don't worry if it takes a couple minutes to start up.

Lo and behold after 200 runs, the probabilities work out just as vos Savant theorized:

Keeping the door:
Wins: 36 cars (36%)
Losses: 64 goats (64%)

Changing the door:
Wins: 77 cars (77%)
Losses: 23 goats (23%)

What manner of witchcraft is this?!

The best explanation I can give myself is that by choosing a door—any one of the doors—you basically damn it to probably being a goat. If you have three doors to choose from, your chances of making the right choice are one out of three. Not particularly good odds. At this stage in the game, if someone were to ask you: "Do you think the car is behind that door?", you'd answer, "Probably not."

Monty opening the door doesn't change the fact that the odds when you initially chose were one out of three. The "reset" your common sense assumes happens, doesn't: the door you picked still probably doesn't have a car. When Monty opens the door with the goat, you now know two things: one door is definitely a goat, and one door is probably a goat. What's left, then, is the door that is least likely a goat and most likely a car.

(Of course, this is a situation in which Monty Hall ALWAYS opens a door to show a goat regardless of whether or not your choice is correct. Situations where he usually opens the door only under certain parameters to try and psych you out, and so on, can't be so rigorously calculated.)

It seems a bit mystical and woo-woo, but there you have it. Numbers, like hips, don't lie.