# 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:

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