Let us consider a directed graph $\Gamma=(V,E,s,t)$ ($V$ set of vertices, $E$ set of edges, $s,t: E \rightarrow V$ are the "source" and "target" maps).

Assume that $\Gamma$ is **bi-regular**, that is there are two integers $d_1 \geq 1$ and $d_2 \geq 1$ such that $|s^{-1}(v)|=d_1$ and $|t^{-1}(v)|=d_2$ for every $v \in D$. Assume that $\Gamma$ **weakly connected** (there is an undirected path relying any two vertices).
And finally assume that $\Gamma$ has **no forward-closed finite subset** (i.e. if $S$ is a finite subset of $V$, there is an edge in $E$ with source in $S$ and target not in $S$). In particular $\Gamma$ is infinite.

Consider the standard discrete-time forward random walk on $\Gamma$, starting at $x \in V$: at time $0$ you are at $x$ with probability $1$; for any $n \geq 0$, at time $n+1$, conditional to being at $y$ at time $n$, your odds of being at any of the $d_1$ forward-neighbors of $y$ (i.e. $t(e)$ for $e \in E$, $s(e)=y$) is $1/d_1$.

Let $p^n_{x,y}$ be the probability of being at $y$ at time $n$.

Is it true that for every $y$, $p_n(x,y) \rightarrow 0$ as $n \rightarrow \infty$?

Here are some remarks. This question is related to Fedja's beautiful answer to this question. Fedja proves the result when $\Gamma$ is an undirected regular graph (seen as an undirected graph by replacing each undirected edge by two directed edges going both way). Unfortunately, I have not been able to extend his argument to my directed case.

The hypothesis that $\Gamma$ has **no forward-closed finite subset** is certainly necessary: If $S$ was such a subset, and $x \in S$, then you would be sure to stay in the finite set $S$ forever, so $\sum_{y \in S} {p^n_{x,y}} = 1$ and one of these $p^n_{x,y}$ at least can not tend to $0$.

The hypothesis that $|s^{-1}(v)|=d_1$ (or at least $s^{-1}(v)$ finite) for all $v$ is necessary to define the random walk, but that $|t^{-1}(v)|=d_2$ (or at least is finite) for all $v$ is also necessary for the theorem to be true. Without it, consider the graph with $V=\mathbb Z$, and for every $a \in \mathbb Z$, there is one directed edge from $a$ to $a+1$ and one directed edge from $a$ to $0$ (called the "speedy return" edge). Thus $|s^{-1}(a)|=2$ for every $a \in \mathbb Z$, but $|t^{-1}(0)|=\infty$. The probability $p^n_{0,0}$ is $\geq 1/2$ since every path that ends up with the "speedy return edge" goes from $0$ to $0$.