Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.1exp 11/4/83; site ihuxv.UUCP
Path: utzoo!linus!decvax!harpo!floyd!clyde!ihnp4!ihuxv!hon
From: hon@ihuxv.UUCP (Herb Norton)
Newsgroups: net.math
Subject: Re: simple statistics question
Message-ID: <589@ihuxv.UUCP>
Date: Wed, 30-Nov-83 13:29:55 EST
Article-I.D.: ihuxv.589
Posted: Wed Nov 30 13:29:55 1983
Date-Received: Fri, 2-Dec-83 00:12:02 EST
References: <130@hou3c.UUCP>
Organization: AT&T Bell Labs, Naperville, Il
Lines: 10

I have seen the solution to this type of problem given as:
if you have observed an the same outcome in N trials the
probability of not observing that outcome in the next trial
is 1/(N+1).  This is intuitively resonable since if you do
observe  the different outcome on the N+1st trial then you
have observed it once in N+1 trials which is just the probablity
you assigned.  If you don't observe a different outcome, it is
still what you expect since that the different outcome had
low probability.
				Herb Norton