Tuesday, 18 November 2008

The Monty Hall Problem

一條幾得意的數學題.......

You are on a game show on television. On this game show the idea is to win a car as a prize. The game show host shows you three doors. He says that there is a car behind one of the doors and there are goats behind the other two doors. He asks you to pick a door. You pick a door but the door is not opened. Then the game show host opens one of the doors you didn't pick to show a goat (because he knows what is behind the doors). Then he says that you have one final chance to change your mind before the doors are opened and you get a car or a goat. So he asks you if you want to change your mind and pick the other unopened door instead. what should you do? (From "The curious incident of the dog in the night-time" )

答案在
http://mathworld.wolfram.com/MontyHallProblem.html

6 comments:

Eric Ting said...

it will increase the probability to win no matter switch or not...ma.

Pomme d'amour said...

The sentence "he knows what is behind the doors" is crucial. If the host doesn't know it, then the answer will be different.

Angel Lee said...

Can you tell me more?

Pomme d'amour said...
This comment has been removed by the author.
Pomme d'amour said...

Let us label the door with the car as C and the other two doors as A, B.

Let us denote by (A,B) if the player has chosen door A and the host has chosen door B. So we have six possible situations in each game:

(A,B), (A,C), (B,C), (B,A), (C,A), (C,B).

Note: If the host doesn't know where the car is, it may happen that the host has actually opened a door with the car in some games.

Now, the probability of occurrence of each of the above situation is the same (as no particular one is in favor).

GIVEN the host has opened a door with a goat, then only the following situations are consistent with this given condition:

(A,B), (B,A), (C,A), (C,B).

Among these, (C,A) and (C,B) are the situations where the player has chosen the car. Hence, among four equally probable situations, two of them correspond to a win and therefore under the above given condition the probability that the player has chosen the car is 1/2. In other words, there is no advantage to switch or not to switch the choice.

In the language of mathematics, the above stuff is just what one calls "conditional probability" and in fact there is a simple formula for calculating it.

Angel Lee said...

In Math, I am weak at ProB. hehe, i do not have any patience to think of it.