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From: chris@umcp-cs.UUCP
Newsgroups: net.math
Subject: Solution to second coins problem
Message-ID: <2124@umcp-cs.UUCP>
Date: Wed, 24-Aug-83 08:36:30 EDT
Article-I.D.: umcp-cs.2124
Posted: Wed Aug 24 08:36:30 1983
Date-Received: Wed, 24-Aug-83 14:26:17 EDT
Organization: Univ. of Maryland, Computer Science Dept.
Lines: 40

Everyone (including me) seems to think, "oh, that's easy, it's the
same as the original problem."  It's not.  The next thought is "well,
then, it's 50%."  Wrong again....

It turns out that the probability of a gold coin in the upper drawer
is 75%!  This amazing conclusion is reached as follows:

case 1: silver coin of G/S pair is in lower drawer.
	In this case, in order to get a gold coin when opening the lower
	drawer, you have to get the G/G pair.  Thus the upper drawer
	contains a gold coin.  This means that in case 1, you have a
	100% chance of a gold coin.

case 2: silver coin of G/S pair is in upper drawer.
	In case 2, you have an equal chance of picking the G/G pair or
	the G/S pair.  If you pick the G/G pair, you get a gold coin.
	If you pick the G/S pair, you get a silver coin.  So in case
	2, you have a 50% chance.

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

 [chance of case 1]  [chance of gold]  [total]
	0.5	 x	1.0	=	 0.50
plus
 [chance of case 2]  [chance of gold]
	0.5	 x	0.5	=	 0.25
					 ----
					 0.75

or 75% chance.

Judging by the number of wrong answers, this problem is substantially
more difficult than the preceding one.  (I'd certainly like to
think so, having twice come up with the wrong answer!)

Chris
-- 
In-Real-Life: Chris Torek, Univ of MD Comp Sci
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