Gambler's Ruin

Medium3–4 hrsRandom WalkGambler's RuinAbsorption

Overview

Gambler's Ruin is the canonical risk-of-ruin problem. The key insight: for a FAIR game, your probability of winning is just the fraction of the total range you start at. This means: if you start with $1 trying to reach$ 100 in a fair game, you have only a 1% chance. The lesson for trading: even with zero edge, if your opponent has more capital, you will almost certainly be driven to ruin before winning. This is why bankroll management matters.

How to Recognize

→Start at position x, absorbing barriers at 0 and N
→'What is P(reaching N before 0)?'
→Asymmetric coin flip walked between two walls
→Risk of ruin with a fixed bankroll

Step-by-Step Approach

1.For fair coin (p=1/2): P(reach N | start x) = x/N — beautiful and simple
2.For biased coin (p≠1/2): use ratio r = q/p. P = (1−r^x)/(1−r^N)
3.Expected duration: E[T | start x, fair] = x(N−x)
4.Martingale method: find a martingale, use Optional Stopping to extract the result

Key Formulas

P(\text{reach } N \mid x, \text{fair}) = \frac{x}{N}

P(\text{reach } N \mid x, \text{biased}) = \frac{1-(q/p)^x}{1-(q/p)^N}

E[T \mid x, p=\tfrac{1}{2}] = x(N-x)

E[T \mid x, p \neq \tfrac{1}{2}] = \frac{x}{q-p} - \frac{N}{q-p}\cdot\frac{1-(q/p)^x}{1-(q/p)^N}

Worked Examples

0Quick Example

Fair game, start at

50, ruin at

0, stop at

100. P(reach

100) = 50/100 = 50%. E[duration] = 50·50 = 2500 bets.

Problem

You have $100, trying to reach$ 200 before going broke. The casino has a 51% edge on each bet. What is P(ruin)?

Solution

p = 0.49 (your win prob), q = 0.51 (loss prob), r = q/p = 51/49.

P(reach N=200 | start x=100) = (1−r^100)/(1−r^200).

r^100 = (51/49)^100. Use ln: 100·ln(51/49) = 100·ln(1.0408) ≈ 100·0.04 = 4. So r^100 ≈ e^4 ≈ 54.6.

r^200 ≈ e^8 ≈ 2981.

P(reach 200) = (1−54.6)/(1−2981) = (−53.6)/(−2980) ≈ 0.018 = 1.8%.

P(ruin) = 1 − 1.8% = 98.2%.

Answer

$P(\text{ruin}) \approx 98.2\%$ . A tiny 1% house edge with equal bankrolls makes ruin almost certain. This is why casinos are profitable even with near-50/50 games.

Common Mistakes

!
Using x/N for biased games. The formula x/N only holds when p = q = 1/2.
!
Forgetting that 'ruin' means reaching 0, so P(ruin) = 1 − P(reaching N).
!
Not using log to compute (q/p)^k for large k. Direct computation overflows; use k·ln(q/p).
!
Confusing expected DURATION (steps until game ends) with expected OUTCOME.

Practice Problems

Click "Show Answer" to reveal

You start with $1, trying to reach$ 10 before going broke. Fair game. What is P(success)?

Hint: Apply x/N formula.

A drunk man starts at lamp post 5 on a street from 0 to 10. He staggers left or right with equal probability. What is P(he reaches his home at 10 before falling in the river at 0)?

Hint: Fair random walk: P(reach N | start x) = x/N.

In a game with p=0.6, q=0.4 (biased in your favor), you start with $10 trying to reach$ 20. What is P(winning)?

Hint: r = q/p = 0.4/0.6 = 2/3. Formula: (1−r^10)/(1−r^20).

Back to Study Plan

Kelly Criterion & Bet Sizing Brownian Motion & Hitting Times