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From: levy@princeton.UUCP
Newsgroups: net.math,net.misc,net.rec.bridge
Subject: Re: simple (?) statistics problem solved
Message-ID: <4@princeton.UUCP>
Date: Sat, 20-Aug-83 22:37:59 EDT
Article-I.D.: princeto.4
Posted: Sat Aug 20 22:37:59 1983
Date-Received: Sun, 21-Aug-83 07:19:42 EDT
References: <908@utcsstat.UUCP>
Organization: Princeton University
Lines: 31

Laura -
It is not true that the "initial state" of the problem is one drawer open.
It should be "one drawer open at random, but such that it contains a gold
coin".  Then it is not true that the two relevant cabinets are equivalent,
since the one with two golds is more likely to have been picked up in the
first place.  
It's a bit like saying "One out of every five people born every day are
Chinese, except in China -- where every one of them is".  If you pick a
human being at random, and you verify that s/he is a Chinese, what are the
chances that your picked you human in China?  Very high indeed.  (In particular
the chance that this person's father is Chinese is close to one, while the
chance that an arbitrary person's father is Chinese is only close to 1/5.)
     Why should you choose drawers rather than cabinets for your sample
space?  The answer to that one is simple.  Just a choice of cabinets does
not determine the outcome of the event.  You need more information, i.e.
you need to subdivide your sample space further. 
     Finally, as many people pointed out, the SS cabinet is totally 
irrelevant.
     I suggest that you actually do the experiment (you may substitute dimes
and nickels for gold and silver coins respectively, with approximately the
same effect).  *This is the experiment implied by the statement of the
problem:*
           i=0;
           while  (i < BIG) {
                     choose drawer at random;
                     if (coin = gold) {
                              look at other drawer;
                              if (other coin = gold) j = j+1;  }  }
                       -- Silvio Levy
           probability = j/BIG.
iiate of