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Gambler's ruin is the absorption problem for a fortune that changes by one unit per game. Let $S_n\in \{0,1,\dots ,N\}$, with independent wins of Probability $p\in (0,1)$ and losses of Probability $q=1-p$. Stop when the fortune reaches zero or the target $N$. Zero is ruin; reaching $N$ is success.

The success Probability $h_i={\mathbb{P}}_i(T_N<T_0)$ has boundary values $h_0=0$, $h_N=1$ and satisfies $h_i=p{h}_{i+1}+q{h}_{i-1}$. Solving this second-order recurrence givesFinite-interval exit times have geometric tails, so absorption occurs almost surely and ruin has Probability $1-h_i$. The fair case is linear because the fortune itself is harmonic; the biased case uses the harmonic exponential ${\rho }^i$.

For a fair game with initial fortune three and target ten, success has Probability $3/10$. The Expected duration of symmetric gambler's ruin is $i(N-i)$, giving 21 steps in this example. The same local first-step method produces expected durations, discounted absorption probabilities and conditional durations; their different boundary conditions encode different questions about the same paths.

Write $u_k$ for the quantity denoted ${\xi }_k$ in the question, to distinguish it from the random-walk increments. Absorption at zero at time zero gives $u_0=1$, and starting at $N$ gives $u_N=0$. For $1\le k\le N-1$, condition on the first step and use independence of the remaining increments. That step contributes one factor of $s$, so the discounted gambler's ruin hitting probability satisfiesThe absorption time is finite almost surely: in each block of $N$ steps there is a probability bounded below by a positive constant of reaching a boundary by successive steps in one direction.

Trying $u_k=r^k$ gives $ps{r}^2-r+qs=0$, with rootsThey are positive and distinct because $4pq\le 1$ and $0<s<1$. Hence the general solution is $A{r}_{+}^k+B{r}_{-}^k$. Enforcing both boundary values givesThis includes the unbiased walk without a special limiting case, since the discount keeps the roots distinct. For uniqueness, a difference of two solutions has zero boundary values and its largest absolute value $M$ satisfies $M\le sM$ if attained in the interior. Since $s<1$, $M=0$.