Consider a simple asymmetric random walk $S_n$ which goes up with probability $p$ and down with $1-p$.
For $b<x<a$ let $$r(x) = P( S_n\text{ hits }a \text{ before }b |S_0 = x). $$ This probability satisfies the recurrence relation $r(x)=p \times r(x+1)+(1-p) \times r(x-1)$ which has the solution
$$ r(x)=c_1+c_2 \left(\frac{1-p}{p} \right)^x $$ but what are the boundary conditions? I don't quite see what these might be.
Intuitively this probability should approach $1$ if $b$ approaches $-\infty$ and $0$ if $a$ approaches $\infty$ but even if this is correct I am unable to devise an expression using this fact.
Any hint is much appreciated.
Well, if $S_0 = a$ then $P(S_n \text{ hits A before B}) = 1$.
On the other hand, if $S_0 = b$ then $P(S_n \text{ hits A before B}) = 0$.
These are the only boundaries. The equation is about $r(x\in [a,b])$, $a$ and $b$ are fixed.