|Subject:||Solution to Monty Hall problem|
|Date:||Mon, 25 Sep 2000 12:24:01 +0100 (BST)|
This is regarding the Monty Hall problem. The problem admittedly befuddled me for some time but I now feel that I have found a way to look at it that will make it crystal clear to most people.
The main thing that confuses in the problem is that there are only 3 doors. Lets generalize the problem as follows:
There are n number of doors. One of them has a prize behind it and n-1 are empty. After I select one door the host opens n-2 empty doors and gives me the option to switch doors. Should I switch?
When n is 3 as in the original problem, things get confusing but now let n be a large number say 1000. So there are 999 empty doors.I select one of them and the host shows me 998 empty doors. Now it is clear that it is definitely advantageous for me to switch. There was only a 0.001 chance of me picking the correct door immediately so obviously one can't say that it is now a 50-50 chance. It is still 0.001 that I had picked the right door initially. So the chance of me winning if I now switch is 0.999 or (n-1)/n. Since this is the solution for the general problem of n doors, the answer for n=3 is (3-1)/2 or there is a 2/3 chance of winning if I switch my choice of doors.
Hoping for your comment Alex!.
Copyright © 1996-2018 Alexander Bogomolny