The Setup

Two players, M and N, play a series of games:

  • M starts with \( $1 \)
  • N starts with \( $2 \)
  • Total capital is \( $3 \)

At each step:

  • M wins $1 with probability \( p = \frac{2}{3} \)
  • M loses $1 with probability \( q = \frac{1}{3} \)

The game continues until one player is ruined.

Question:

What is the probability that M wins (i.e., reaches \( $3 \)) before being ruined?

The Gambler’s Ruin Formula

Let \( P(i) \) be the probability that M wins, starting with \( i \) dollars out of \( N = 3 \).

For a biased game \( (p \ne q) \), the probability of reaching \( N \) before 0 is:

\[ P(i) = \frac{1 - (q/p)^i}{1 - (q/p)^N} \]

Given:

  • \( i = 1 \)
  • \( N = 3 \)
  • \( p = \frac{2}{3} \), \( q = \frac{1}{3} \)
  • \( q/p = \frac{1/3}{2/3} = \frac{1}{2} \)

Plug In:

\[ P(1) = \frac{1 - (1/2)^1}{1 - (1/2)^3} = \frac{1 - 1/2}{1 - 1/8} = \frac{1/2}{7/8} = \boxed{\frac{4}{7}} \]

Final Answer

\[ \boxed{\frac{4}{7}} \]

So, the probability that M wins the game is 4/7.

Reference