Problem 155 Solution

Let p_i be the probability that the first flip with bankroll of $i will be heads.

p_i given eventual ruin = (p_i&pr(ruin at $i+1))/pr(ruin at $i)=

.6*(.4/.6)¹¹/(.4/.6)¹⁰ (see solution to problem 154 for probability of ruin) =

.6*(.4/.6) = 0.4

So the probability that any given flip will be heads is 0.4, thus the expected total flips that are heads is also 0.4 .

This problem is from the March 2000 issue of The Actuary, published by the Society of Actuaries.

Michael Shackleford, ASA - April 5, 2000