Connection probability on oriented percolation does not depend on root

There is a long time that I have not updated my blog. This time, I would like to talk about once again a question in Alibaba competition (2021) as I did not solve it that day. 

Question: On a general connected graph $G=(V, E)$, every bond contains two oriented edges. Then we pose an oriented percolation model on it. We choose a vertex as root, then try to prove that the probability to have oriented path from every vertex to root is independent to the choice of root.

It is natural to think of a bijection, but it is not easy. During the exam, I only treat the very simple case that $G$ is a tree, which is very easy. However, it is still useful because we will see an argument to combine the two.

We denote by $h(a) = \mathbb P_p[G \to a]$ the connection probability. It suffices to prove that $h(a) = h(b)$ for every $a \sim b$. We prove it by induction on the number of bonds $\vert E \vert$. Then the basic case is the tree and we have already proved it and we need to finish the induction step.

For every configuration $\omega$, we define $$Pivot(\omega, a, \overrightarrow{ba}) := \{v \in V: v \to a \text{ when } \omega(\overrightarrow{ba}) = 1, v \nrightarrow a \text{ otherwise}\}.$$

Then we treat two cases:

Case 1: $\omega \in \{G \to a\}, Pivot(\omega, a, \overrightarrow{ba}) = \emptyset$. For this case, $\omega \setminus \overrightarrow{ba}$ can be thought as a connection configuration in $(V, E\setminus \{a,b\})$. Then by induction, for this case the probability does not depend on the choice of $a,b$ by adding an oriented edge and we can flip it.

Case 2: $\omega \in \{G \to a\}, Pivot(\omega, a, \overrightarrow{ba}) \neq \emptyset$. Then this implies that $\omega(\overrightarrow{ba}) = 1$ and $b \in Pivot(\omega, a, \overrightarrow{ba})$. We should add one more restriction that $a \nrightarrow b$, otherwise it is the common part for two. Then we flip the orientation  $\overrightarrow{ba}$, and one can check that 

$$\mathbb P_p[\omega \in \{G \to a\} \cap \{a \nrightarrow b\}, b \in Pivot(\omega, a, \overrightarrow{ba}) ] \\= \mathbb P_p[\omega \in \{G \to b\} \cap \{b \nrightarrow a\}, a \in Pivot(\omega, b, \overrightarrow{ab}) ].$$

With these arguments, we finish the step of induction and establish the result for the general graph $G$.