Let's consider an asymmetric Random Walk on $Z$, with transition probabilities $p\_{i, i+1}=p$, $~~p\_{i, i+1}=q$, $\forall i \in \mathcal{Z}$, $p+q=1$ and $p>q$.


I am interested in the probability of the first hitting time of a bareer in $i=n$, assuming that the walk started from $i=0$ at time $t=0$. Is there an explicit formula for it? How does it depend on $n$? 



asymmetric random walk, hitting time probability
The explicit formula is: $P[N\_m=n]=(m/n)P[S\_n=m]$, where $P[N\_m=n]$ is the
probability the position $m$ is hit after exactly $n$ steps, 
$S\_n = X\_1+X\_2+\dots X\_n$ and $P[S\_n=m]$ is the probablity after $n$ steps
the path to be at the position $m$. This last is well-known and is given
by


$$P[S\_n=m]=\frac{n!}{[(n+m)/2]![(n-m)/2]!}p^{(n+m)/2}q^{(n-m)/2}.$$


This is true when $n$ and $m$ have the same parity, else the probability
is zero. Additionally $m$ is positive and $n$ is greater than or equal to $m$.
A similar expression can be found for negative $m$.


