Consider a random walk on a line of integers. Suppose we start from the state $x$. Then, the probability of jumping from $x$ to $x+1$ is $p(x, x+1)=p$, and the probability of jumping from $x$ to $x-1$ is $p(x, x-1)=1-p$. Furthermore, suppose each jump is independent with each other. Now, I want to calculate the probability of starting from $1$ and eventually reaching $0$, namely $p(V_0<\infty |x_0=1)$, where $x_0$ is our initial state and $V_0$ denotes the minimum time of reaching $0$: $V_0=\min\{n: x_n=0\}$. The correct answer is just $(1-p)/p$, but I don't understand and get something different.
In my calculation, it should be $$p(V_0<\infty |x_0=1)=(1-p)+p(1-p)^2+p^2(1-p)^3+\dots=\frac{1-p}{1-p(1-p)},$$ where $p^2(1-p)^3$ is the probability of jumping from $1$ to $2$ to $3$ and then jumping back from $3$ to $2$ to $1$ to $0$.
Could anyone explain a little bit why $p(V_0<\infty |x_0=1)$ should just be $(1-p)/p$? I am also wondering why my computation does not work. Thanks in advance!
Edit: have just realised there is a problem with this, see below.
Let $q$ be the required probability.
There is a probability $1-p$ that after one step you will be at $0$.
There is a probability $p$ that after one step you will be at $2$. Then you reach $0$ in a finite number of steps if and only if you
The probability of the latter is by definition $q$. The probability of the former is also $q$, because it is clear by symmetry that the probability of reaching $1$ in a finite number of steps, starting at $2$, is the same as the probability of reaching $0$ in a finite number of steps, starting at $1$. Therefore $$q=1-p+pq^2\ ;$$ solving the quadratic we get $$q=1\quad\hbox{or}\quad q=\frac{1-p}p\ .$$ Now if $p<\frac12$ then the second solution is greater than $1$ and so we must choose the first, $q=1$. If $p=\frac12$ then both solutions are $1$. If $p>\frac12$ then there is always a greater chance of moving right than left, so it seems plausible (can anyone justify it more carefully?) that $q<1$, and so $q=\frac{1-p}p$.
Your calculation is wrong because you have only considered the possibility of going straight up and back. As in your example: $\def\temp{\raise1pt\hbox{-}}1\temp2\temp3\temp2\temp1\temp0$. But you have not considered paths like $1\temp2\temp3\temp2\temp1\temp2\temp1\temp2\temp3\temp2\temp3\temp2\temp1\temp0$.