random walk with hitting probability If I am at $0$, I can go up by $1$ with probability $q$, and go down by $1$ with probability $(1-q)$, what is the probability I hit $10$ at some time?
Let the probability I hit $i$ at some time be $p_i$, then,


*

*$p_2=p_1^2$ since I hit $1$ and then hit another $1$

*$p_1=q+(1-q)p_2$ since I condition on the first step


To solve this, I got two roots $p_1=1$ and $p_1=\frac{q}{1-q}$. Suppose $q<0.5$.
My question is: why we through away $p_1=1$? I want some rigorous explanation. 


*

*Of course, intuitively, if $q=0$ it never go up, so we choose $p_1=\frac{q}{1-q}$.

*It seems expectation($q*1+(1-q)*(-1)$) is $<0$ is not sufficient to claim $p_1<1$, right?


Thanks!

You can think of this as a grid where we go up with probability $q$. We will hit 10 if there are 10 more ups than downs. That is, P(hit 10) $= \sum_{n \geq 10} \sum_{k=10}^{n} {n \choose k} q^{k} (1-q)^{k-10}$.
