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From: mcewan@uiucdcs.UUCP (mcewan )
Newsgroups: net.math
Subject: Re: Let's Make a Deal - (nf)
Message-ID: <3102@uiucdcs.UUCP>
Date: Tue, 4-Oct-83 23:30:23 EDT
Article-I.D.: uiucdcs.3102
Posted: Tue Oct  4 23:30:23 1983
Date-Received: Fri, 7-Oct-83 05:18:47 EDT
Lines: 39

#R:mi-cec:-16500:uiucdcs:28200022:000:1133
uiucdcs!mcewan    Oct  4 15:45:00 1983

*sigh* I'll try again.

Wolog, assume that you chose door #1. Let W(i) be the event that door i
is the winner, and let S(i) be the event that Monty shows you door i.
Then
	P(S(1)) = 0
	P(S(2) | W(1)) = 1/2	P(S(3) | W(1)) = 1/2
	P(S(2) | W(2)) = 0	P(S(3) | W(2)) = 1
	P(S(2) | W(3)) = 1	P(S(3) | W(3)) = 0
	a priori P(W(i)) = 1/3 for i=1,2,3

Applying Bayes formula

                                       P(S(2)|W(3))P(W(3))
P(W(3) | S(2)) = -------------------------------------------------------------
                   P(S(2)|W(1))P(W(1))+P(S(2)|W(2))P(W(2))+P(S(2)|W(3))P(W(3))

                       1*1/3
               = ---------------------
                  1/2*1/3+0*1/3+1*1/3

                    1/3
               = ---------
                    1/2

               = 2/3

Similarly, P(W(3) | S(2)) = 2/3.

The trick is that no matter what door you chose, Monty is going to show you
a losing door. Since this is independent of the door you chose, being shown
a losing door gives you NO information about your door, and cannot affect
the probability of that door being the winner.

				Scott McEwan
				uiucdcs!mcewan