Subject: Re: An infinite flippage of coin
Date: Wed, 12 May 2000 22:24:57 -0400
From: Alex Bogomolny
I suspect that both your answers are wrong. The first one one is similar to the famous Achilles and Tortoise paradox. The fact that every try has a finite probability does not guaranty a win even after an infinite number of steps. Your second solution shows why. The total probability of a win after an infinite number of steps is the sum of a series which might have happpened to be less than 1.
Your second answer has a numerical flaw. Assume that the probability of a win on the nth try is pn and let qn = 1 - pn. (pn = 2-n, but this is not important.)
We can write
p1 + q1 =
p1 + q1(p2 + q2) =
p1 + q1p2 + q1q2(p3 + q3) =
p1 + q1p2 + q1q2p3 + q1q2q3(p4 + q4) =
p1 + q1p2 + q1q2p3 + q1q2q3p4 + q1q2q3q4p5 + ...
The sum of the first n terms is the probability of a win after n tries. The sum of the series is naturally 1! Try to find an error in your calculations.
The answer to your question is thus that the probability of a win after an infinite number of steps is 1. However, note that this does not guarantee you a win. Events that occur with probability 1 are not bound to happen.
For example, probability that a random number on the unit interval is irrational is 1. However, when picking a random number from that interval, one can get a rational number, although probability of this event is 0.
All the best,