Having troubles solving this one.
Consider an ant walking along the positive integers. At $n$ the ant moves to $n+1$ with probability $1/3$ and to $n−1$ with probability $2/3$. If the ant reaches $0$ it stops walking. Starting from $n=3$ what is the probability that the ant reaches $n=5$ before reaching $n=0$?
So far, I've found the following: The expected distance the ant travels with each move is: $1/3(+1) + 2/3(-1)= -(1/3)$.
Thus, if the ant starts at $n=3$, it is expected to take the ant $1/3(x) = 3 (x=9)$ moves to make it to zero.
How to further develop it to the solution?
Hint: We can draw a states diagram. Let $p_n$ be the probability that the ant reaches $n=5$ before $n=0$ starting at $n.$
First note that $p_5 = 1$ and $p_0 = 0.$Let's look at $p_4.$ There's a $\frac{1}{3}$ chance we go to $p_3$ and $\frac{2}{3}$ chance we go to $p_5.$ So $$p_4 = \frac{1}{3}p_3 + \frac{2}{3}p_5 = \frac{1}{3}p_3+\frac{2}{3}.$$
We write these equations for $p_1$ through $p_4.$ Then, solve for $p_3.$