For the classical gambler's ruin: $p(i) \equiv p$, $\Delta_i = 1$, absorbing boundaries at 0 and $N$
This simplifies to the fundamental recursion:
Assume $P_i = \lambda^i$ and substitute:
Solving: $\lambda = \frac{1 \pm \sqrt{1-4pq}}{2p} = \frac{1 \pm |p-q|}{2p}$
Case 1: $p \neq \frac{1}{2}$ (distinct roots)
Applying $P_0 = 1$ and $P_N = 0$:
Case 2: $p = \frac{1}{2}$ (repeated root)
Linear ruin probability, quadratic expected time.
Exponential growth in ruin probability as casino edge increases.
Surprising result: ruin probability unchanged, but variance and time differ dramatically.
Jump size modification completely changes the game dynamics.
Counterintuitive: coefficient modification can turn losing games into winners!
Continuous limit of master equation yields barrier option pricing.
Coefficient modification models evolutionary forces in finite populations.
For the classical case with $p(i) \equiv p$ and $\Delta_i = 1$:
Problem: Start at 0, stop when hitting +k
Boundary: T(k) = 0, boundedness as i → -∞
Homogeneous solution: $T_h(i) = A + Br^i$ with $r = \frac{q}{p}$
Particular solution: Try $T_p(i) = \alpha i$
Substituting: $\alpha(p-q) = -1 \Rightarrow \alpha = -\frac{1}{p-q}$
General solution: $T(i) = A + Br^i - \frac{i}{p-q}$
If $p > q$: boundedness forces $B = 0$
From $T(k) = 0$: $A = \frac{k}{p-q}$
Problem: Start at 0, stop when hitting ±k
Boundary: T(-k) = T(+k) = 0
Shift coordinates: $j = i + k$, so we have interval [0, 2k]
Start at $j = k$, boundaries at 0 and 2k
Solve on finite interval: Same method but with two boundary conditions
The transition from biased to fair game is smooth:
This can be verified using L'Hôpital's rule or by expanding around $p = 1/2$:
Let $\epsilon = p - 1/2$, so $p = 1/2 + \epsilon$ and $q = 1/2 - \epsilon$:
As $\epsilon \to 0$:
Substituting back recovers the linear formula.