Let $F_2$ be the free group on two elements, generated by $\{a, b\}$. We perform a random walk on $F_2$, starting at the identity element $e$ and uniformly at random selecting one of $\{a,a^{-1},b,b^{-1}\}$ to append to the current walk. So e.g. there's a 1/16 probability your second step will be $\{ba^{-1}\}$, and 1/4 probability your second step will be to $\{e\}$.
My question is: what is the probability we hit $a$ before we hit $b$?
My approach has been the following. Let $x_n$ be the probability we hit $a$ before hitting $b$ when starting at a word of length $n$, the first symbol of which is neither $a$ nor $b$ (ie it can only start with $a^{-1}$ or $b^{-1}$). It's not hard to see this is well defined. We are seeking $x_0$. By conditioning on possible next steps it's not hard to get the equations:
$x_0 = 1/4 + (1/2)x_1$
$x_{n+1} = (1/4)x_n + (3/4)x_{n+2}$
Unfortunately I don't think those alone are enough to infer the value of $x_0$, which makes me think there's an extra insight I'm missing. Any ideas?
(This is taken from 123Ga in Weighing the Odds by Williams, which I've been working through in self study)
By symmetry, the probability of hitting $a$ before we hit $b$ is the same as the probability of hitting $b$ before we hit $a$. So, what is the probability of never hitting either element?