On how to pick goats from cars: The Monty Hall Problem

Back in my days at Palantir Technologies, I distinctly remember being befuddled by the Monty Hall problem. And I also remember distinctly being an utter skeptic when it came to the actual solution. I've since come around. Here's the succinct version:

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 No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Now you'd think it wouldn't matter whether or not you switched, after all, you already made the choice, right? Probability holds constant, regardless of whether someone knows about it, right?

Wrong! You should always switch doors. And there's a simple intuitive explanation. While you made a choice initially with no information, the HOST always has to reveal one of the doors that has a goat behind it, and he essentially has more information than you do, which also affects the outcome.

There are two scenarios here:

  • 1 out of 3 times, you'll select a door with the car. In this case, the host will have a choice between two doors that both have goats. In this case, if you switch, you'll lose every time. If you don't switch, you win.
  • 2 out of 3 times, you'll select a door with a goat. In this case, the host will ALWAYS open the other door with a goat. In this case, if you switch, you win every time. If you don't switch, you always lose.

So there you have it. Switching will cause you to win 2 out of 3 times.

Interestingly, the intuitive thing to do is to not switch. Out of 228 subjects in one study, only 13% chose to switch according to the wikipedia article, which is excellent by the way.

And if you still don't believe it, here's some monte carlo simulations by Antonio Cangiano to prove it. Code proves everything!