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Path: utzoo!linus!decvax!harpo!seismo!hao!woods
From: woods@hao.UUCP (Greg Woods)
Newsgroups: net.math,net.games
Subject: Re: Party game of chance
Message-ID: <846@hao.UUCP>
Date: Sun, 12-Feb-84 17:35:49 EST
Article-I.D.: hao.846
Posted: Sun Feb 12 17:35:49 1984
Date-Received: Wed, 15-Feb-84 06:09:09 EST
References: <211@heurikon.UUCP>
Organization: High Altitude Obs./NCAR, Boulder CO
Lines: 30


  This is a common fallacy. Just because there are 365 "chances" does *not*
mean the probability of winning is 1. When your "chance" total hit 365 exactly,
that means you have a 50% chance, not a 100% chance, of winning. If you are
going to play that game for money, you better have enough people so that
your "chances" are a lot higher than 365! 
  This is one of those things where it is easier to calculate the probability
of *losing* and subtract it from one. As you can see, the chance of winning
does not become 100% until you have more than 365 people, which is what your
intuition should tell you. The chance of losing is number of outcomes leading
to a loss divided by total possible outcomes. The total possible outcomes is
easy: if you have N people, it is 365^N (ignoring leap-years for simplicity's
sake. It doesn't change the result significantly). The number of possible ways
to lose is a little more tricky, but not too bad. The first person can have
any of 365 birthdays. The second, if you are to lose, can have any of the
remaining 364. The third, 363, and so on. Therefore the number of ways to lose
is 365*364*....*(365-N+1). This can be expressed mathematically as

     365!/(365-N)! where "!" means factorial, fact(n) { n=0 ? return(1) : 
     return(n*fact(n-1)); }

     Therefore, your chance of losing is (365!/(365-N)!)/(365^N). This correctly
predicts your chance of losing to be 1 if N=1, and nearly 0 if N=365. 
The formula requires defining -1! if N>365, but surely you can see that it
really is impossible to lose if you have 366 or more people! :-)

			 GREG
-- 
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