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Path: utzoo!linus!decvax!cca!ima!ism780!jim
From: jim@ism780.UUCP (Jim Balter)
Newsgroups: net.math
Subject: Re: Solution to second coins problem
Message-ID: <25@ism780.UUCP>
Date: Wed, 24-Aug-83 19:47:00 EDT
Article-I.D.: ism780.25
Posted: Wed Aug 24 19:47:00 1983
Date-Received: Fri, 26-Aug-83 07:07:51 EDT
Lines: 24

NO! NO! NO! (much screaming and tearing of hair).  You have repeated the
same old mistake in a new context:

	Case 1 & 2 are equally likely, thus we have:

They are **not** equally likely!!!  The stipulation, that you find
a G coin when you select a bottom drawer at random, is twice as likely
to be satisfied when the S coin is in the top drawer as when it is in
the bottom drawer.  Therefore, given the stipulation, the odds that
the S coin was in fact in the top drawer is twice the odds that
it was in fact in the lower drawer (if you don't think this follows,
we need a formal linguist; I don't know how to formally demonstrate it).
So, the answer is (1/3 * 100%) + (2/3 * 50%) = 2/3 (!!!!).

Selecting the bottom drawer every time is just another way of randomly
selecting a drawer if the method of placing the coins is random.
So the two answers *must* be identical.  Substitute "the drawer manufactured
earlier" for "the top drawer" and "the drawer manufactured earlier" for
"the bottom drawer" and then do the analysis.  "But I don't know which was
manufactured first", you say?  Exactly; it doesn't matter.

Jim Balter (decvax!yale-co!ima!jim), Interactive Systems Corp

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