Consider (simple) random walk on $\mathbb{Z}^2$ started at the origin. The probability that the walk returns to the origin before hitting $(0,1)$ is $1/2$.

To see this, let $a(x)$ be the potential kernel for random walk on $\mathbb{Z}^2$. Then $2a(x)$ counts the expected number of visits to the origin by a random walk started at the origin before the first time it hits $x$. Let $p$ denote the probability that the walk hits $x$ before returning to the origin.

Hence,

\begin{equation*} 2a(x) = (1-p)+p(1-p)+p^2(1-p)+\cdots = \frac{1}{1-p}, \end{equation*}

and consequently,

\begin{equation*} p = 1-\frac{1}{2a(x)}. \end{equation*}

On the other hand, one can also obtain that

\begin{equation*} a(x) = \frac{1}{(2\pi)^2}\int_{[-\pi,\pi]^2}\frac{1-\exp(-ix\cdot\theta)}{1-\phi(\theta)}d\theta, \end{equation*}

where $\phi(\theta)$ is the characteristic function of the random walk. In particular, one can work out that $a(0,1) = 1$, and hence that the probability we are looking for is $1/2$. With some more work, one can also work out what the answer would be for, say, $(3,2)$ in place of $(0,1)$ (it turns out to be $(16\pi+3)/(16\pi+6)$).

But the expression for $(0,1)$ suggests that there is some sort of underlying symmetry. Is there a simple argument for why this is the case?

I'm aware that this problem can be rephrased in terms of electrical networks, and that there appears to be a simple solution involving symmetry in that setting, but I'd prefer a simple solution involving symmetry without these techniques, if possible.