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Suppose we start at State $1$, directly to the right of a State $0$. What's the probability that we ever get to State $0$, if we take n steps to the right with probability p, 1 step to the left with probability q, and stay forever at State i (the current state) with probability r.

I know how to construct the solutions to this question but I don't know the why. I would appreciate a basic, intuitive, and logical way to get these equations.

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Let $P(n)$ be the probability that you get to $0$ from $n$. If you are at $k$ (as long as $k \gt 0$), you go to $k-1$ with probability $q$, to $k+n$ with probability $p$ and get frozen with probability $r=1-p-q$ Then write the recurrence $P(k)=pP(k+n)+qP(k-1)$ because you either go left (probability $q$) to $k-1$ or go right $n$ steps (probability $p$) to $k+n$ and the chance of success in each direction is the product of the chance you went that way times the chance you succeed when you get there.

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    @WuschelbeutelKartoffelhuhn: p vs q fixed, two places. You would have $P(2)=P(1)^2$ in this walk if $p=0$, because starting from $2$ you have to move left twice in succession to get to $0$, but I don't think that is what you are asking.2012-09-25