Consider a random walk on $\mathbb{Z}$ starting from $i >0$. With probability $p$ it moves to the nearest neighbor on the left, with the same probability it moves to the nearest neighbor on the right, with probability $q$ the walker dies. Naturally $2p+q=1$.
I want to estimate the probability that the walke reaches $0$ before reaching $N>i$ or before dying. I call this probability $P(i, N)$ and I try to calculate it. This probability solves the following equation, $$ P(i, N) = \, \, p \,P(i-1, N) + p\, P(i+1, N), $$ with boundary conditions $P(0, N) = 0$ and $P(L, N) = 1$. Is it right? What is the expression for $P(i,N)$?
Recall that every sequence $(x_n)$ such that $x_n=ax_{n+1}+bx_{n-1}$ for every $n$ in an interval of integers is a linear combination of the sequences $(r^n)$ and $(s^n)$ where $r$ and $s$ solve the identity $u=au^2+b$ (that is, if the roots are different, otherwise a modification is necessary).
Here, $P(i,N)=Ar^i+Bs^i$ for every $0\leqslant i\leqslant N$, where $r$ and $s$ solve $u=pu^2+p$ hence $$r=\frac{1+\sqrt{1-4p^2}}{2p},\qquad s=\frac{1-\sqrt{1-4p^2}}{2p}.$$ The boundary conditions are $P(0,N)=1$ and $P(N,N)=0$ hence $A+B=1$ and $Ar^N+Bs^N=0$. Finally, $$ P(i,N)=\frac{s^Nr^i-r^Ns^i}{s^N-r^N}. $$