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From: levy@princeton.UUCP
Newsgroups: net.math
Subject: Re: Another counterintuitive problem in probabilities
Message-ID: <9@princeton.UUCP>
Date: Tue, 23-Aug-83 12:04:57 EDT
Article-I.D.: princeto.9
Posted: Tue Aug 23 12:04:57 1983
Date-Received: Wed, 24-Aug-83 01:55:54 EDT
References: <303@ihuxq.UUCP>
Organization: Princeton University
Lines: 21


OK, all you guys who said you'd choose the first number.  This gives you
*exactly* 10% chance of being right, and I can assure you you can do better
than that.  In particular your strategy gets worse and worse when you
have more numbers, whereas the optimal one does not.  Think again!

For those who say the question is unanswerable in the absence of more info,
here's why not:  The problem does not depend on the actual values of the
numbers, only on their ordering.  So we might as well assume they are 1
through 10, *as long as the strategy does not make use of this fact*.  In
other words, a strategy is a decision procedure (say deterministic) which
after the i-th number has been drawer tells you whether to declare it's
the biggest or not, *based only on the ordering of the first *i* numbers,
and not on their values*.  There are clearly only finitely many such
strategies, so one (or more) must be optimal.

Now of course if you have more information your chances are better!  If
you are competing agains someone who can only count up to 100, and s/he
draws that number in the first attempt, you win with 100% certainty...

			-Silvio Levy