The Monty Hall Problem

My recent foray into math problems yesterday brought something interesting to mind as well. In fact, I was just discussing it at the office a couple of weeks ago, and it illustrates part of my fascination with probability. (Yes, the word "combinatorics" lies on the list of hobbies on my resumé.) Even if you don't have a background in math, this is still a pretty good exercise.

The Monty Hall problem is actually a rather well-known probability puzzle. It's named after the original host of a 1960s game show called Let's Make a Deal, although I assume that Mr. Hall never actually came up with the situation below on live television. Regardless of that, for the purposes of this puzzle, you just happen to be a contestant on his game show.

For your final prize round, you are shown three closed doors:


The game is simple: Choose one of the three doors, and whatever lies behind it will be yours to take home. One of the doors has a car behind it, and the other two have goats.

Needless to say, you don't want to have anything to do with the goats, but you do want to win that car.


That little graphic above doesn't necessarily illustrate the actual positions of the car and the goats, of course. It's just there to give you an idea of what the game is like. Hypothetically, your potential prizes could also be arranged like this:


Or they could be arranged like this:


You don't know, really. All that you know is that there are three doors in front of you. You don't know which one holds the car, and you don't know which one holds the goats.

So you choose one of the three doors. It doesn't matter which one you choose, really -- given the circumstances, your chances of winning the car are obviously one in three, or 1/3.


But... wait a minute. Your host, Monty Hall, now decides to make the game a little more interesting. As a token of his generosity, he opens one of the other two doors (the ones you haven't chosen) to reveal... a goat.


Now he gives you a choice: You can either stay with the current door that you've chosen, or you can switch your choice to the other door. You still want to win that car, so... should you switch doors?

...

...

...

This is a lot harder for most people to wrap their heads around, it seems.

The common perception is that the switch doesn't matter. When your friendly game show host revealed a door with a goat behind it, he effectively cut down your choices to two doors. Each door therefore has a one-in-two chance of holding the car, or 1/2, which means that you have a fifty percent chance of winning the car regardless of which door you choose. That's just cake, right there.

Except that that's a fallacy of thought. It turns out that that's not the right answer.

It turns out that the correct reasoning sources from the very beginning of the puzzle, right as you're about to choose a door for the first time. Any door that you choose will obviously have a 1/3 chance of winning the car to begin with:


So let's say that you choose a door, along with its attached one-out-of-three chance of winning the car.


When Monty Hall opens one of the other two doors to reveal a goat, that effectively drops your chances of finding a car behind that door to zero-out-of-three:


But given this logic, the chances of the car being behind the third (unselected) door must be... two out of three!


And therefore it is actually in your best interest to switch doors -- because you have twice as much chance of winning the car if you do so!

I've seen arguments erupt over this proposition, of course. It wouldn't be interesting if I didn't know a couple of people who stopped speaking to each other over this. I did mention it, after all - some people just find it difficult to wrap their heads around such an idea. ("The probabilities change just because somebody opens a door? That's preposterous!")

If you'd rather see a more physical proof of this, however, consider the fact that there are really only six ways by which the car and the goats could be arranged:


Yes, the goats are considered to be distinct from each other. We can call them Bluebell and Cornflower, if you like. Or Laurel and Hardy. Or Bonnie and Clyde. For the purpose of this blog post, however, we'll call them One and Two.

Each column represents the range of possibilities that you have whenever you choose a specific door. What that means is that, given that you choose one of the three doors to start with, you effectively have a 2/6 chance of choosing the car outright, a 2/6 chance of choosing Goat One, and a 2/6 chance of choosing Goat Two.

If you choose the door with the car from the onset (a 2/6 chance), Monty Hall can open any of the two remaining doors to reveal a goat. In this case, it would definitely not be in your best interest to switch -- because the other door has a goat as well.

But if you choose a door with a goat from the onset (a 4/6 chance in total; you can get either Goat One or Goat Two), Monty Hall opens the only one of the two remaining doors to reveal a goat. In this case, it would be best for you to switch, because the other remaining door holds the car.

In short, given the situation above, it's better for you to switch doors two out of three times. And that's despite any intuition that screams the feasibility of a fifty-fifty chance.




* The goat silhouette was sourced from http://www.fotosearch.com. The car image (a Ferrari Enzo, if you're curious) was sourced from http://www.seriouswheels.com. Neither image is used for purposes of promotion or financial gain. Don't sue me, or I'll write another mathematical proof for you to read. And we wouldn't want that, now would we?