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From: larry@grkermit.UUCP (Larry Kolodney)
Newsgroups: net.math,net.misc,net.rec.bridge
Subject: Re: simple (?) statistics problem solved
Message-ID: <597@grkermit.UUCP>
Date: Fri, 19-Aug-83 12:57:07 EDT
Article-I.D.: grkermit.597
Posted: Fri Aug 19 12:57:07 1983
Date-Received: Fri, 19-Aug-83 21:27:32 EDT
References: <1814@rabbit.UUCP> <1287@tektronix.UUCP>
Organization: GenRad Inc., Concord, MA
Lines: 24

From richl@tektronix

Put another way, if you put me in the above described situation,
with one gold coin in hand and one drawer left to open, I can assure
you that I would be correct 50% of the time by just saying, "Yep,
the other one must be gold".

~~~~~~~~~~~~~
No you wouldn't.  If it were true that the coins was picked randomly,
then the odds that the coin came from the gold gold cabinet is still
2/3.  

Think about this.  Let say the I present you with the situation where I
hold one gold coin in my hand and there is one open drawer, but I have
fiendishly decided to always choose the gold gold drawer.  In that
case, if you knew that, you would have to say the prob. was 1/1 for the
gold gold drawer.  If you know that the coin was chosen randomly, you
have to still say 2/3.  Only if it were chosen with a method that gave
each drawer an equal chance rather than each coin would you be able to
say 1/2.
-- 
Larry Kolodney (The Devil's Advocate)
{linus decvax}!genrad!grkermit!larry
(ARPA)  rms.g.lkk@mit-ai