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From: lew@ihuxr.UUCP
Newsgroups: net.math
Subject: Chris Torek vs. Thomas Bayes
Message-ID: <578@ihuxr.UUCP>
Date: Wed, 24-Aug-83 17:47:27 EDT
Article-I.D.: ihuxr.578
Posted: Wed Aug 24 17:47:27 1983
Date-Received: Fri, 26-Aug-83 02:24:16 EDT
Organization: BTL Naperville, Il.
Lines: 19

In Chris Torek's problem, the probability that a gold coin is in the bottom
drawer of the second cabinet, given that we found a gold coin in the
bottom drawer of the chosen cabinet is 2/3, not 1/2. Chris has turned
the problem into a 3-stage decision instead of a 2-stage decision, but
this is no problem for Bayes' theorem! We let the events A and B be the
"gold in bottom of 2" and "silver in bottom of 2" then A1,A2,A3,B1,B2,B3
are the events "A & we chose cabinet 1", "A & we chose cabinet 2", etc.

Now we have P(A1) = P(A2) = ... = P(B3) = 1/6. Letting E be "found gold
in bottom of chosen cabinet", we have:

	P(E|A1) = 0	P(E|B1) = 0
	P(E|A2) = 1	P(E|B2) = 0
	P(E|A3) = 1	P(E|B3) = 1

... so Bayes' theorem gives P(A2|E) = P(A3|E) = P(B3|E) = 1/3
and P(A|E) = P(3|E) = 2/3.

		Lew Mammel, Jr. ihuxr!lew