The Puzzle

A gambler starts with \( i \) dollars. Each time he plays:

  • He wins $1 with probability \( p \)
  • He loses $1 with probability \( q = 1 - p \)

He continues playing until:

  • He reaches \( N \) dollars (goal), or
  • He reaches 0 dollars (ruin)

Question:

What is the probability that he reaches \( N \) before going broke?

Solution

Let \( P_i \) be the probability that the gambler reaches \( N \) starting from \( i \).

Case 1: \( p = q = \frac{1}{2} \) (Fair Game)

This is the symmetric case. Then:

\[ P_i = \frac{i}{N} \]

So the probability of success is simply proportional to how far along he is toward \( N \).

Case 2: \( p \ne \frac{1}{2} \) (Biased Game)

The general formula is:

\[ P_i = \frac{1 - \left( \frac{q}{p} \right)^i}{1 - \left( \frac{q}{p} \right)^N} \quad \text{for } p \ne q \]

This is derived using recurrence relations and boundary conditions \( P_0 = 0 \), \( P_N = 1 \).

Summary

Case Probability of reaching \( N \)
Fair game (\( p = \frac{1}{2} \)) \( \frac{i}{N} \)
Biased game (\( p \ne \frac{1}{2} \)) \( \frac{1 - (q/p)^i}{1 - (q/p)^N} \)

Reference