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Path: utzoo!linus!security!genrad!decvax!cca!ima!ism780!jim
From: jim@ism780.UUCP (Jim Balter)
Newsgroups: net.math
Subject: Re: Why is it 2/3?????
Message-ID: <24@ism780.UUCP>
Date: Wed, 24-Aug-83 11:16:00 EDT
Article-I.D.: ism780.24
Posted: Wed Aug 24 11:16:00 1983
Date-Received: Fri, 26-Aug-83 05:03:16 EDT
Lines: 99

You are not reading the solutions carefully.

    this answer seems to rest on one or both of the following
    assumptions:

   1)  Any of the three remaining drawers (having tossed out the SS cabinet) 
       can be chosen.

No such assumption was ever made, although it happens to turn out that
the results are the same if you do choose any of the other drawers not
in the S/S cabinet, instead of just the one in the same cabinet, because all
of the permutations of those drawers satisfy the stipulations in the problem,
so picking a drawer other than the one in the same cabinet is just equivalent
to picking the one in the same cabinet given a different random arrangement
of the drawers.  If you find my language too convoluted, please just ignore
it instead of using it as a further grounds for claims of errors in the
analysis.

   2)  The gold coin you know about was *not* necessarily the first one
       you chose.

No such assumption has ever been made, and I don't understand how you could
have gotten this impression.  Certainly the coin known about is the coin
chosen.

    The problem with the 2/3 answer is that once we have that gold coin in our
    grubby little paws, we have reduced the problem to two cabinets (not three
    remaining drawers).

You only say this because you have a mental image in which cabinets play
a bigger role than drawers.  You don't know which cabinet (among those
containing G and G or G and S) you chose; nor do you know which coin
(among the *three* (G, G, S) with a G in the other drawer of the same
cabinet (the given)) you are going to find in the second drawer.  Imagine the
coins are attached by long strings, instead of sharing cabinets.  I put them
all in another room, and bring out a gold one.  You know that the S/S
combo is out.  You have a G in your hand.  Which of the remaining three coins
is tied to the string?  Suppose that, just before you pull on the string to
retrieve the coin, my accomplice in the other room randomly reattaches it to
one of the three.  Does that affect the probabilities?

    If the problem had asked about the probability before
    we knew we had one gold coin, or if we were free to choose from the three
    remaining drawers, the answer would be different.

The probability of what?  Of finding a gold coin in the first drawer?
That is 1/2, but surely you don't mean that.  Of finding a gold coin in
the second drawer if you do in fact find a gold coin in the first drawer?
That is the problem as stated.  If you would try to formalize what you
mean here, you might get a hint of the imprecise thinking which is misleading
you.
Many people seem to be having a problem with subjunctive situations.
You are to determine the probability after reading the problem; the problem
describes a situation, in which there is a drawer open containing a gold coin.
If you are the subject of the problem, or you are just reading about it,
the situation is the same.
Also, you talk about the three remaining drawers.  Imagine
that you are in the room with the cabinets.  Which of the drawers are those
three?  Are there only two cabinets?  The notion of throwing out the cabinet
with 2 silver coins merely indicates that the odds of *having selected it
given that one of its drawers contains a gold coin* is (2-2)/3,
just as the odds of having selected the cabinet with 1 silver coin is (2-1)/3,
and the odds of having selected the cabinet with 0 silver coins is (2-0)/3
(the answer!).

   However, by choosing a gold coin, we know:

   1) The silver cabinet is out of the question.

Hmmm, yes, 0/3, yes, that sounds right.  But, this non-quantitative language
suggests you might be about to make a mistake.

   2) The one remaining drawer we must open to determine which cabinet 
      we have chosen must contain either a gold coin or a silver coin.

Hmmm, yes, but are those two possibilities, a gold coin or a silver coin,
equally likely?  Why do you insist on ignoring this question when it has
been brought up repeatedly?  We are looking for a coin which shares a cabinet
with a gold coin.  There are three such coins; two are gold.

    The probability associated with picking a gold coin in the first place
    is *not* part of the original question as stated.  It is a *given* that
    the gold coin is chosen.

You are certainly right.  Couldn't agree with you more.  Never said otherwise.

I am amazed that so many people, who never would have questioned the answer
given in the back of the textbook, will so thoroughly question the same
answer when given on the net.  I think that is a move in the right direction,
but perhaps a trip back to the textbook might help in this case.

Someone wrote a program, ran it, and posted the results, which were
much closer to 2/3 than 1/2.  If you can come up with a program consistent
with the problem that gives a different result, please post it.
Otherwise, please just contemplate quietly.

Jim Balter (decvax!yale-co!ima!jim), Interactive Systems Corp

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