Friday, April 11, 2008

Let's Make a Deal

There are three doors. Behind two of the doors, there is a goat. Behind one of the doors is a new car. You pick a door, then the host will open one of the remaining two doors to reveal a goat. He then asks you - do you want to switch to the other door?

What should you do? Does it matter? What do you think the odds are of success (winning the car) if you stay, and if you switch?

You can run a number of simulations HERE to test your ideas. If you're unfamiliar with the scenario, I'd like you to leave your guess as to the results in the comments here. To get a good sample, run 20 or 30 rounds both switching and not switching before you ask it for an explanation.

3 comments:

Anonymous said...

I remember reading about this somewhere. I remember that it seemed to go counter to reason. It shouldn't matter because my chances are 50/50 after I see the goat, right?

Unless I don't discount the goat-door. But then aren't my chances still 1 in 3?

Is this related to having babies? It's supposed to be 50/50 chance every time, but if I already have two boys (biology aside) does my chance to have another boy increase?

-Dave said...

Knowing the Lange and Pastrell families, I honestly wonder if there is some non-random biological factor at work that influences boy/girl outcomes.

If you're not talking biological or hand-of-God involvement, then the fact that you've had 2 or 50 boys in a row has no impact statistically on the odds of the next child being a boy or a girl. It's very unlikely to flip a coin as heads 51 times, but just because you've hit 50 in a row doesn't mean the odds of hitting #51 are suddenly 4 in 10 Quadrillion. Your odds of hitting #51, given that you've hit #50, are still just 1 in 2.

-Dave said...

I waited long enough:

The answer is that if you don't switch, you will lose 2/3 of the time. If you do switch, you win 2/3 of the time.

It looks like a 50/50 choice, but the opening a door to reveal a goat changes the game. Let's call the goats Goat A and Goat B.

If you initially select the door with the car, the host will reveal either Goat A or Goat B. If you switch, you will lose. If you stay, you will win.

If you initially select the door with Goat A, then the host must open the door for Goat B. If you switch, you win. If you stay, you lose.

If you initially select the door with Goat B, the host must reveal the door for Goat A. If you switch, you win. If you stay, you lose.

Since each of these three scenarios really is random, they should all occur with roughly the saem frequency. Therefore, switching wil win 2/3 of the time. The only way to win by staying is if you choose the car originally, which only happens 1/3 of the time.