lew@ihuxr.UUCP (Lew Mammel, Jr.) (12/02/83)
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