Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Path: utzoo!linus!decvax!cca!ima!johnl
From: johnl@ima.UUCP
Newsgroups: net.math
Subject: Re: simple (?) statistics problem solved - (nf)
Message-ID: <378@ima.UUCP>
Date: Fri, 19-Aug-83 18:42:55 EDT
Article-I.D.: ima.378
Posted: Fri Aug 19 18:42:55 1983
Date-Received: Sat, 20-Aug-83 09:05:24 EDT
Lines: 35

#R:rabbit:-181400:ima:16700002:000:1583
ima!johnl    Aug 19 17:44:00 1983

***** ima:general / ism780!jim /  6:48 am  Aug 19, 1983
All you bozos insist on counting cabinets.  Why not count coins?
There are three gold coins, equally likely to be selected.
There are two cabinets, not equally likely to be selected
(*given* that a gold coin will be seen in the drawer).
tektronix!richl talks about ordered pairs (S,S), (G,S), and (G,G),
eliminates the first, and then claims 50/50.
Since he did realize the pairs are ordered, why didn't he mention them
all: (S,S), (S,S), (S,G), (G,S), (G,G), and (G,G)?  Eliminating
the first three leaves 2/3.

And for the people who eliminate the (S,S) cabinet and notice they have
two left; why don't they eliminate half of the (S,G) cabinet?
Because they don't realize that "eliminating a cabinet"
corresponds to the quantitative assignment of a probability of 0 to the
selection of that cabinet, just as the (G,S) cabinet has a 1/3 probability
of selection, and the (G,G) cabinet has a 2/3 probability of selection,
*given* that a gold coin is selected ("Gee, but what if I open the
drawer and there is a silver coin?"  Then you are in the wrong universe,
bozo).

Here is an interesting view of the problem:
Open all the drawers, and drop into each a piece of paper stating the
contents of the other drawer in the same cabinet.  Now, take out the
three drawers with gold coins, shuffle them, and pick one at random.
What are the odds that it contains a slip of paper saying "the other
drawer contains a gold coin"?  Do you still say 1/2?

I worry about a society built upon the logic of these minds.

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