Gambler’s Ruin


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?


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:

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

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