Solution: Gambler's Ruin for a Biased Random Walk

Exercise: Gambler's Ruin for a Biased Random Walk

Part 1

$M_{n+1} = \rho^{S_n + X_{n+1}} = M_n\cdot \rho^{X_{n+1}}$ . Since $X_{n+1}$ is independent of $\mathcal{F}_n$ :

\mathbb{E}[M_{n+1} \mid \mathcal{F}_n] = M_n\cdot\mathbb{E}[\rho^{X_{n+1}}] = M_n\cdot(p\rho + q\rho^{-1}) = M_n\cdot (p\cdot q/p + q\cdot p/q) = M_n\cdot(q + p) = M_n.

Part 2

On $\{n \le \tau\}$ , $S_n \in [0, N]$ , so $M_n = \rho^{S_n} \in [\min(1, \rho^N), \max(1, \rho^N)]$ . Hence $(M_{n \wedge \tau})$ is bounded — OST's condition (2) holds.

Part 3

Apply OST: $\mathbb{E}[M_\tau] = M_0 = \rho^k$ . Now $S_\tau \in \{0, N\}$ , so:

\mathbb{E}[M_\tau] = \mathbb{P}(S_\tau = N)\cdot \rho^N + \mathbb{P}(S_\tau = 0)\cdot \rho^0 = \mathbb{P}(S_\tau = N)\rho^N + (1 - \mathbb{P}(S_\tau = N)).

Setting equal to $\rho^k$ and solving:

\mathbb{P}(S_\tau = N) = \frac{\rho^k - 1}{\rho^N - 1}, \qquad \mathbb{P}(S_\tau = 0) = 1 - \mathbb{P}(S_\tau = N) = \frac{\rho^N - \rho^k}{\rho^N - 1}.

Part 4

For $(p, k, N) = (0.49, 50, 100)$ : $\rho = 0.51/0.49 \approx 1.0408$ .

\mathbb{P}(S_\tau = 100) = \frac{1.0408^{50} - 1}{1.0408^{100} - 1} \approx \frac{6.568}{49.11} \approx 0.119.

For $(p, k, N) = (0.45, 50, 100)$ : $\rho = 0.55/0.45 \approx 1.2222$ .

\mathbb{P}(S_\tau = 100) = \frac{1.2222^{50} - 1}{1.2222^{100} - 1} \approx \frac{13{,}780}{1.898\cdot 10^8} \approx 7.3\cdot 10^{-5}.

Interpretation. A 1% house edge (

p = 0.49

) already drops success probability to 12%. A 5% house edge (

p = 0.45

) brings it to essentially zero. Casino house edges of 1–5% ensure that the gambler's probability of reaching a goal is exponentially small in the size of the goal relative to the capital. This is why casinos are robustly profitable despite each individual game being nearly-fair; the ruin probability compounds over the number of rounds played.

Takeaways

Gambler's ruin is an OST computation. Find a martingale that takes distinct values at each boundary, apply OST, solve the two-equation system.
Tiny edges compound to huge ruin probabilities. A 1% house edge produces an exponentially-small probability of doubling your stake, and a 5% edge makes doubling essentially impossible.
The key step is finding the right exponential martingale. $\rho^{S_n}$ works only because the specific value $\rho = q/p$ makes $\mathbb{E}[\rho^{X}] = 1$ , cancelling the bias.
Gambler's-ruin is isomorphic to knockout-option pricing. The probability of "hitting the upper barrier" is exactly the undiscounted price of a digital knock-in option — the barrier-option machinery is gambler's ruin with discounting.