, 2017

Solution:

First we compute probability that the mouse returns to $0$ before being eaten. Then probability that it is at $0$ in $2n$ minutes without being eaten is given by $\frac{1}{3^{2n}} \binom{2n}{n}$. Therefore, the overall expectation is given by
$$\begin{aligned} & \sum_{n \geq 1} \binom{2n}{n} 9^{-n} = -1 + \sum_{n \geq 0} \binom{2n}{n} 9^{-n} \\ = & -1 + \frac{1}{\sqrt{1-4/9}} = -1 + \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}-5}{5} \end{aligned}$$
where we use the well known fact that
$$\sum_{n \geq 0} \binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}}$$
for $x=\frac{1}{9}$.