Subject: | Monty Hall problem |
---|---|

Date: | Thu, 05 Mar 1998 08:59:37 -0700 |

From: | TERRY PASCAL |

Alex,

I've been reading the Monty Hall problem and would like to post the answer to demonstrate where most people math is incorrect.

Given 2 doors, one with a prize and one without, which should you pick?

I think you'd agree that odds are 50-50 you'll be right. The fact that I may have had 3 doors initially and opened one, or 1000 doors and opened 998, does not change the current problem. There is still only a 50/50 chance of winning.

Because the probability of the problem can be expressed in three different ways.

P1 - 1/3 chance prize is door #1, 2/3 chance it is not P2 - 1/3 chance prize is door #2, 2/3 chance it is not P3 - 1/3 chance prize is door #3, 2/3 chance it is not

Most of the logic I've seen has the following flaw. As soon as the contestant picks door #1, they immediately assume that P2 and P3 are not valid. Well, they are. So they follow only P1. Regardless of which door I pick, all 3 are still valid. It is only when Monty opens a door that things change. By opening door #3 we now have

P3 - 0/0 chance prize is door #3, 3/3 chance it is not.

Now here everyone says, "Hey by P1, then the 2/3 chance all goes to door #2' But then why not, "By P2, then the 2/3 chance all goes to door #1"? Both are incorrect. When Monty picks #3, the whole problem is changed to:

P1 - 1/2 chance prize is door #1, 1/2 chance it is not P2 - 1/2 chance prize is door #2, 1/2 chance it is not P3 - 0 chance prize is door #3, 1 chance it is not

Look at it another way.

What's being said is that if I pick #1 and monty shows #3, then 2/3 of the time #2 will win. So If I had picked #2 to start with and monty opens #3, then the odds go 2/3 to #1. Why would they change? What if I don't have to tell monty? What if I could write it on a secret ballot?

Terry Pascal

Copyright © 1996-2018 Alexander Bogomolny

71409454