# Two Gamblers Play Until One is Bankrupt: Chance That the Better Player Wins

**Problem**

Player *M* has Php1, and Player *N* has Php2. Each play gives one the players Php1 from the other. Player *M* is enough better than player *N* that he wins 2/3 of the plays. They play until one is bankrupt. What is the chance that Player *M* wins?

A. 3/4 | C. 4/7 |

B. 5/7 | D. 2/3 |

**Answer Key**

**Solution**

*W*= win

*L*= lose

Player *M* wins:

$P = WW + WLWW + WLWLWW + WLWLWLWW + \ldots$

$P = W^2 + (WL)W^2 + (WL)^2 W^2 + (WL)^3 W^2 + \ldots$

$P = \left( \frac{2}{3} \right)^2 + \left( \frac{2}{3} \cdot \frac{1}{3} \right) \left( \frac{2}{3} \right)^2 + \left( \frac{2}{3} \cdot \frac{1}{3} \right)^2 \left( \frac{2}{3} \right)^2 + \left( \frac{2}{3} \cdot \frac{1}{3} \right)^3 \left( \frac{2}{3} \right)^2 + \ldots$

Sum of Infinite Geometric Progression

$P = \dfrac{a_1}{1 - r} = \dfrac{\left( \frac{2}{3} \right)^2}{1 - \left( \frac{2}{3} \cdot \frac{1}{3} \right)}$

$P = 4/7$ ← Answer: [ C ]