Gambler’s Ruin: Probability that M Wins

Two players, M and N, play until one is ruined. M starts with $1, N starts with $2, so the total capital is $3.

Goal: Find P(M reaches $3 before $0)

1) State Diagram

Let the state be M’s current amount of money. The absorbing states are: $0 (M is ruined) and $3 (M wins).

p = 2/3 q = 1/3 p = 2/3 q = 1/3 q p $0 M ruined $1 start $2 intermediate $3 M wins

2) Final Answer

4/7
≈ 0.5714 ≈ 57.14%
Start State
M begins at $1.
Winning State
M wins if he reaches $3.
Losing State
M loses if he reaches $0.

3) Solve Using State Probabilities

Let:

From State $1
P₁ = (2/3)·P₂ + (1/3)·0

From $1, M either:

  • wins the next game with probability 2/3, moving to $2
  • loses the next game with probability 1/3, moving to $0 and losing immediately
So: P₁ = (2/3)P₂
From State $2
P₂ = (2/3)·1 + (1/3)·P₁

From $2, M either:

  • wins the next game with probability 2/3, moving to $3 and winning immediately
  • loses the next game with probability 1/3, moving back to $1
Substitute and Solve
P₂ = 2/3 + (1/3)P₁
P₁ = (2/3)P₂ = (2/3)[2/3 + (1/3)P₁]
P₁ = 4/9 + 2P₁/9
7P₁/9 = 4/9
P₁ = 4/7

4) One-Game Simulation

Click the button to simulate one random path of the game starting from M = $1.

$0
M ruined
$1
start
$2
middle
$3
M wins
Ready. Current state: $1

Since p = 2/3 is larger than q = 1/3, M is favored, which is why the winning probability is above 50%.