Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.1exp 11/4/83; site ihuxr.UUCP
Path: utzoo!linus!philabs!cmcl2!floyd!clyde!burl!hou3c!hocda!houxm!ihnp4!ihuxr!lew
From: lew@ihuxr.UUCP (Lew Mammel, Jr.)
Newsgroups: net.math
Subject: Surviving discrete events
Message-ID: <794@ihuxr.UUCP>
Date: Thu, 1-Dec-83 18:05:49 EST
Article-I.D.: ihuxr.794
Posted: Thu Dec  1 18:05:49 1983
Date-Received: Sat, 3-Dec-83 12:38:29 EST
Organization: AT&T Bell Labs, Naperville, Il
Lines: 40

If p is the probability of surviving a discrete event, the probability
of surviving n such events is p^n. If f(p) represents our a priori
estimation of the distribution of p over the interval 0 to 1, then
our expectation of surviving 1 event is:

	<p> = int 0 to 1 of p * f(p) * dp

... and our expectation of surviving n events is (of course):

	< p^n > = int 0 to 1 of p^n * f(p) * dp

Our revised estimation of f(p) after surviving n events is given
by Bayes' Theorem (what else):

	f'(p,n) = p^n * f(p) / <p^n>

If f(p) = 1 (uniform distribution) we get:

	f'(p,n) = (n+1) * p^n

... and our expectation of surviving the n+1st event, given that we
survived the first n, is:

	<p'> = int 0 to 1 of p * f'(p,n) * dp = 1 - 1/(n+2)

To get <p'> = 1 - 1/(n+1) as Herb Norton suggested, we have to set
f(p) = 1/p. This is problematical since it is unnormalizable and
gives a zero expectation of surviving the first event. It also allows
us to assign a finite probability that we are certain to die in
any event when we have just survived one. Anyway, it works out formally.

Actually, a slight rephrasing of Herb's answer conforms with the
uniform distribution model. We can say that the probability of dying
in the nth event, given that we have survived n-1 events, is 1/(n+1).

Note that f(p) = 1 is modeled by choosing an urn from a uniform distribution
over p, the fraction of red balls in the urn, and then choosing balls
from that urn. Of course, "surviving" means picking a red ball.

	Lew Mammel, Jr. ihnp4!ihuxr!lew