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Path: utzoo!linus!decvax!harpo!eagle!alice!rabbit!ark
From: ark@rabbit.UUCP
Newsgroups: net.math,net.misc,net.rec.bridge
Subject: simple (?) statistics problem solved
Message-ID: <1814@rabbit.UUCP>
Date: Wed, 17-Aug-83 23:44:00 EDT
Article-I.D.: rabbit.1814
Posted: Wed Aug 17 23:44:00 1983
Date-Received: Thu, 18-Aug-83 11:08:46 EDT
Organization: Bell Labs, Murray Hill
Lines: 60

First, let me confess a small imprecision:  it isn't a
statistics problem but rather a probability problem.
Review:  there are three cabinets, each with two drawers.
One contains two gold coins, another contains two silver
coins, and the third contains one gold and one silver.
You walk up to a cabinet and open a drawer.  It contains
a gold coin.  What is the probability that the coin in the
other drawer will also be gold?

I have gotten about 80 responses so far.  Almost none were
from bridge players, and all but about five said that the
answer is 1/2.  They reason as follows:  I can rule out having
chosen the cabinet with two silver coins, and the other two
cabinets are equally likely, so I'll see another gold coin
half the time.

This answer, popular as it is, is wrong.  Here's the right one.

There are six drawers.  Three contain silver coins, and the other
three contain gold coins.  Before I have chosen anything, my probability
of choosing each drawer is 1/6.  Once I've seen a gold coin, I know
I didn't choose any of the three drawers with silver coins in them,
but my probability of having chosen each of the other drawers must
be equal, as I haven't learned anything beyond having ruled out the
three with silver coins.

Once I've seen the gold coin, then, the probability is 1/3 that I've
chosen any particular one of the drawers with a gold coin.

1/3 of the time, then, I've chosen the drawer in the cabinet with the
gold and silver coins.  In that case, the probability of seeing another gold is 0.
1/3 of the time, I've chosen one of the drawers in the cabinet with the
two golds.  In that case the probability of my seeing another gold is 1.
The same reasoning applies to the OTHER drawer in the cabinet with two
golds.  Thus, my overall probability of seeing another gold is

	(1/3)*0 + (1/3)*1 + (1/3)*1 or 2/3.

Another way to look at it is this.  Suppose I do the experiment six
times and choose a different (first) drawer each time.  Three times,
I'll get a silver on the first try and the initial conditions won't hold.
The remaining times I'll see a gold again 2/3 of the time.

The reason I asked about bridge players is that this is similar to
the following bridge problem.  You want to pick up this suit:

		Dummy
		x x x x

		You
		A K 10 x x

You cash the Ace and the Queen drops offside.  Do you assume the Queen
was singleton or that LHO was dealt QJ doubleton?

The rule that applies here is called the Principle of Restricted Choice.
It is normaly correct to assume that LHO's choice was restricted --
that is, that LHO was playing a singleton Q -- rather than that LHO chose
to exercise a choice in a particular way (playing Q from QJ).
If LHO had QJ, after all, he might have played the J.