Thursday, July 24, 2008

Monty Hall Problem

This was shown in the movie 21. I got the answer wrong, and had a vigorous debate with a friend who was watching the show with me. It turned out she had taken a Math class on this in her undergrad. Bugger. She kept repeating - "conditional probability!".

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. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?


Ray said...

Mathematically, you should change your choice. since previously it was 1/3 chance, and now by choosing again, its 1/2 chance.

But keeping your choice constant is a choice in itself, so really, what's the difference? By choosing to stick with your original choice, its like asking you to pick again, and you pick the same one.


bubblene said...

i remember the answer for this!
it's always to switch your choices...! (though i haven't quite fully comprehended why)..

Eileen Chew said...

Oh, it's not saying, pick again out of the two doors. The clever lady friend is right - the answer is, always switch from your original choice as it always gives you a greater prob to win.

akikonomu said...

Marilyn vos Savant is your friend. I believe she wrote about the Monty Hall problem a few decades ago.