# Gambler’s Ruin

## Problem

Player \(M\) has $\(1\), and Player \(N\) has $\(2\). Each play gives one of the players $\(1\) from the other. Player \(M\) is enough better than Player \(N\) that he wins \(\frac{2}{3}\) of the plays. They play until one is bankrupt. What is the chance that Player \(M\) wins?

## Solution

Let \(M\) and \(N\) represent victories by Player \(M\) and Player \(N\), respectively. Then, consider the various sequences that result in Player \(M\)’s victory:

\[\text{Case 0: } M M\] \[\text{Case 1: } M N M M\] \[\text{Case 2: } M N M N M M\] \[\text{\ldots}\] \[\text{Case \textit{n}: } (M N)^n M M\]

From the above, we can draw the following insights regarding sequences that result in Player \(M\)’s victory:

Player \(N\) must not win the \(1^{st}\) play

Player \(M\) winning \(2\) plays in a row results in a victory

Player \(N\) must not win \(2\) players in a row

These insights lead to the following convergent geometric series representing Player \(M\)’s probability of victory:

\[P(M\; Victory) = \sum_{i=0}^{\infty} \left[
\left( \frac{2}{3} \right) \left( \frac{1}{3}\right) \right] ^n\left(\frac{2}{3}\right)^2 = \frac{4}{7}\]