The Game Show Problem

How about a little Math, Scarecrow?

The best ones look simple but are counter-intuitive. And I will admit, I didn't get this. Even after reading this good explanation (ain't the Internet grand?) I reluctantly came around.

I saw this in the movie "21," a decent (3.675 starts) if imperfect flick. They gave it enough time that I felt they must have had good backing, but I couldn't get it. It's "The Game Show Problem" and our beloved protagonist, Ben Campbell, solves it correctly to ingratiate himself with his professor at MIT (played by Kevin Spacey) and eventually secure his spot on the school's unofficial intermural blackjack squad.

Here it is, identical to its appearance in 21:

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?

I watched the movie twice and gave the screen the angled-head-quizzical-dog look both times. I majored in Math (probability was not my thing, and I left school to pursue a music career) but it seems like third grade rules apply, that once the door is opened, you have a 50-50 shot with either door. I err in good company; many of the PhD commenters make the same claim.

Writing a little java program to test it empirically, it becomes obvious. Sticking with Door #1, you will win only if it is in there (33%). The host will not show you the car, so switching gives you the known best choice of the other two.

Counterintuitive. Clever. I would not have made the Blackjack team.

UPDATE: I just ran the computer program 1,000,000 times (boy, my finger is tired!) and sticking with door 1 wins 33.3134%, switching 66.6866%

Posted by jk at August 14, 2008 6:42 PM