Let $\tau=T_a\wedge T_b$. The walk exits the finite interval $[a,b]$ almost surely, and the stopped martingale $\phi(S_{n\wedge\tau})$ is bounded between $\lambda^b$ and $\lambda^a$. The <optional stopping theorem> and <bounded convergence theorem> give
$$
1=\phi(0)=\mathbb E[\phi(S_\tau)]
=\lambda^a\mathbb P(T_a<T_b)
+\lambda^b\mathbb P(T_b<T_a).
$$
Solving for the first probability gives the <biased gambler's ruin probability>
$$
\boxed{\mathbb P(T_a<T_b)
=\frac{\lambda^b-1}{\lambda^b-\lambda^a}
=\frac{\phi(b)-\phi(0)}{\phi(b)-\phi(a)}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 201 / 1 / b / Solution

For a walk that steps right with probability $p>1/2$, put $\lambda=(1-p)/p$. If $a<0<b$ and $T_x$ is the first hitting time of $x$, then the martingale $\lambda^{S_n}$ and the <optional stopping theorem> give
$$
\mathbb P(T_a<T_b)
=\frac{\lambda^b-1}{\lambda^b-\lambda^a}.
$$
Letting the opposite boundary tend to infinity gives
$$
\mathbb P(T_a<\infty)=\lambda^{-a},
\qquad
\mathbb P(T_b<\infty)=1.
$$

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