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From: thomson@utcsrgv.UUCP (Brian Thomson)
Newsgroups: net.math,net.misc,net.rec.bridge
Subject: Re: simple (?) statistics problem solved
Message-ID: <2019@utcsrgv.UUCP>
Date: Fri, 19-Aug-83 15:03:05 EDT
Article-I.D.: utcsrgv.2019
Posted: Fri Aug 19 15:03:05 1983
Date-Received: Fri, 19-Aug-83 17:17:43 EDT
References: <1814@rabbit.UUCP>, <1287@tektronix.UUCP>
Organization: CSRG, University of Toronto
Lines: 20

No, the probability of a silver coin really IS 1/3.
Perhaps the following explanation will clear this up:

After finding one gold coin, you can indeed eliminate the
cabinet containing two silver coins.  Let's (figuratively)
toss that cabinet out the window.  We are left with two
cabinets and four drawers.  Three drawers contain gold coins,
one drawer contains a silver coin.

We have already opened one drawer and found a gold coin.  That
means there are three drawers remaining, two with gold coins and
one with a silver coin.  We have NO WAY of knowing which of those
three drawers is on the other side of our selected cabinet.
It could be any one of the three remaining drawers, so the probability
of it being the drawer with the silver coin is 1 in 3.

Does this help?
-- 
			Brian Thomson,	    CSRG Univ. of Toronto
			{linus,ihnp4,uw-beaver,floyd,utzoo}!utcsrgv!thomson