Let $\mathbb{S}$ be countable and $X_n:\mathbb{S}^\mathbb{N} \rightarrow \mathbb{S}$ be the coordinate map, i.e., $X_n(\omega) = \omega_n$. Let $p:\mathbb{S}\times\mathbb{S} \rightarrow [0,1]$ be an irreducible transitional probability so that $X_n$ is a Markov chain and let $\pi\in (0,\infty)$ be a reversible measure with respect to $p$. Let $T^x = \inf \{n \ge 1: X_n = x\}$.
Question:
In this case, is there a relation between $P^x(T^y<\infty)$ and $P^y (T^x<\infty)$?
Some thought process:
I initially thought that it would satisfy $\pi(x) P^x(T^y<\infty) = \pi(y) P^y(T^x<\infty)$, since if I took a path $x=x_0,x_1,...,x_n = y$ for $x_1,...,x_{n-1} \ne y$ and computed $$ \pi(x) \prod_{i=1}^n p(x_{i-1},x_i) = \pi(y) \prod_{i=1}^n p(x_i,x_{i-1}) $$ But then I realized that $x_i = x$ is possible for some $1 \le i <n$ so that $x=x_0,x_1,...,x_n=y$ is a path from $x$ to the first time hitting $y$, while the reverse path $y=x_n,x_{n-1},...,x_0=x$ is not a path from $y$ to the first time hitting $x$.
A comment that I write here for lack of reputation: Note that the event $(T^x < \infty)$ means " do we ever hit x" and does not really break your argument. The only thing that changes is that you now have a bijection of paths with different hitting times but since you sum over all times anyway it does not matter.