Problem 154 Solution

Let p_i be the probability of ruin given a bankroll of $i.

p₁ = .4 + .6*p₂
p₂ = .4*p₁ + .6*p₃
p₃ = .4*p₂ + .6*p₄
p₄ = .4*p₃ + .6*p₅
p₅ = .4*p₄ + .6*p₆
.
.
.

Now sum the equations.

Sum for i=1 to infinity of p_i = .4 + .4*(sum for i=1 to infinity of p_i) + .6*(sum for i=2 to infinity of p_i)

Sum for i=1 to infinity of p_i = .4 + (sum for i=1 to infinity of p_i) -.6*p₁)

.6*p₁=.4

p₁=.4/.6 = 2/3

At this point you can use substitution to find that p_i=(2/3)ⁱ.

So p₁₀=(2/3)¹⁰=~0.017342

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

Michael Shackleford, ASA - April 5, 2000