Talk:Monty Hall problem/Analysis/Population of hosts

The Morgan paper is even wrong in the condition that the player has chosen door 1 and the host has opened door 3.

The reason is that Morgan's analysis fails to take account of the information given by the door which has been opened!

To start with an easy example:

Suppose we have a population of hosts half of whom have Morgan's q=1 (they always open door 3 if they can) and half of whom have q=0 (they never open door 3 unless they must).

Suppose we consider two games one with each type of host. We then consider only those cases where door 3 has been opened.

Subject to that condition, it is twice as likely (2/3 chance) that we have a host with q=1 where the probability of winning is 1/2 as it is that we will have a host with q=0, where the probability of winning is 1. Thus the probability of winning over the two games is:  3/2 * 1/2 + 1/3 *1 = 2/3  Martin Hogbin (talk) 08:45, 14 April 2009 (UTC)