Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Path: utzoo!linus!decvax!cca!ima!ism780!jim
From: jim@ism780.UUCP (Jim Balter)
Newsgroups: net.math
Subject: Re(2): Statistics problem solved
Message-ID: <21@ism780.UUCP>
Date: Tue, 23-Aug-83 12:36:00 EDT
Article-I.D.: ism780.21
Posted: Tue Aug 23 12:36:00 1983
Date-Received: Wed, 24-Aug-83 10:26:52 EDT
Lines: 66

I find it incredible that so many bright people can miss such a simple,
straightforward problem.  Maybe it isn't so incredible; bright programmers
are excellent at coming up with clever rationalizations to protect their
egos from the pain of being in error.

    I was always taught that past probabilities can not be used to determine
    the probability of a future occurence.

I doubt that you were taught that, although you appear to have learned it,
most likely by using flipped coins as your only example.
The effect of past occurrences upon the probability of future occurrences
is the basis of induction, and enables us to think at all.  The proper
concept is that the probability of the occurrence of one event is of no
bearing upon the probability of occurrence of an *independent* event.  As an
illustration, I play Russian Roulette and get a blank chamber.
I spin the barrel and fire again; what are my odds?  What would my odds
be if I didn't spin the barrel?  In the first case the two events are
independent, but not in the second.

				 If the drawer contains a gold piece we know
    that the cabinet does not contain two silver pieces.  At this stage we
    have two possible cabinets.  One that contains a gold piece and a silver
    piece, and another that contains two gold pieces.  One of the drawers
    containing a gold piece is open so the remaining drawers contain a gold
    piece, a gold piece and a silver piece.  Thus the probability for finding
    a gold piece when we open another drawer is 2/3.

Wrong.  At this point you claim to be ignoring the constraint that you had to
open the drawer in the same cabinet (see below).  So, there are three
cabinets, and you don't know the contents of any drawer other than the open
one.  The odds of opening a random drawer and finding a gold coin in this
situation is of course 2/5.

    But we have another constraint.  Since
    we are trying to determine the probability that the
    OTHER DRAWER OF THE SAME CABINET contains a gold piece

Yes that is the constraint.  Since we don't know which coins are in which
drawers other than the g/g s/s g/s distribution given, being constrained
to select the drawer in the same cabinet gives the same results
(probability-wise) as being constrained to not pick a drawer in the s/s
cabinet (because all permutations with each constraint satisfy the other
constraint), which as you showed above is 2/3.

   we have to change or [sic] model so that the drawer
   that we opened on the selected cabinet is open in BOTH POSSIBLE CABINETS.

This is nonsensical, but it does seem to satisfy the need to arrive at the
wrong answer.  Please, quit trying to imagine how many teeth there are in the
horse's mouth, and try the experiment (but be sure you follow the original
statement of the problem).

As for intuition, why would anyone intuitively expect a woman who has
already borne one boy to have the same chance for two boys after her
second child as a woman who has as yet borne no children?

And for the latest permutation of the problem with selecting the bottom
drawer first instead of a random drawer, why would that possibly matter,
any more than the stipulation that you pick the drawer which was manufactured
first?  The problem made no claim at all about the effect of top
or bottom drawers upon the distribution of the coins in the drawers.
For all you know, the damn drawers are side by side.  Sigh.

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

--------