Transient random walk on 3-color 3-regular tree Suppose that $T=(V,E)$ is a 3-regular tree with root $0$. Suppose that $0$ is colored green. All other vertices are colored blue, red or green, such that each vertex has exactly one neighbour of each color. Multiple colorings are possible. Just pick one.

Let $R=(R(n))_{n \geq 0}$ be a random walk on $V$ that starts at the root, so $R(0)=0$. At each step, it steps to the (nearest) neighbour colored $x\in\{\text{blue},\text{red},\text{green}\}$ with probability $p_x$. Assume that $p_{\text{blue}},p_{\text{red}},p_{\text{green}}>0$ and $p_{\text{blue}}+p_{\text{red}}+p_{\text{green}}=1$. It is known that $R$ is transient. For each subtree induced by vertices L, M, R (see figure) I wish to compute the probability that $R$ "escapes to infinity" in that specific subtree. This problem is trivial if $p_{\text{blue}},p_{\text{red}},p_{\text{green}}=\frac{1}{3}$: for each subtree induced by L, M, R the walk $R$ escapes in it with probability $\frac{1}{3}$ by symmetry. I have tried to use similar symmetry arguments in the general case without succes. Can anyone help me with the general case? 
 A: Your process can be thought of as a random walk on the letters $r, g$ and $b$. I will represent $R_{n}$ as a string such as $ggbbgb$. Denote $|R_n|$ as the number of symbols in such a string.
Your question is, I believe: what are the probabilities that the first "letter" in the limiting word is $r, g$ or $b$?
We'll need the first visit random variable $T(x) = \min_{k \geq 1} \{R_{k} = x \}$ and the generating functions $S_{i, j} = E[\lambda^{T(j)} I_{|R| = 2} | R_0 = i]$ for $i, j \in \{r, g, b \}$. In words, $S_{i, j}$ is the generating function for the first visit to the nearest node of color $j$, starting at color $i$. I'm being a bit lazy and using translational symmetry, i.e the fact that every vertex of a given color looks the same. It follows from a first-step anaylsis that we have nine equations
\begin{align}
S_{i, j} = \lambda(p_{j} + p_{j + 1} S_{j + 1, i} S_{i, j} +  p_{j + 2} S_{j + 2, i} S_{i, j}) \tag{1} 
\end{align}
where I'm indexing the permutations $r, g, b$ by $1, 2, 3$. Next, I'll define the return generating function $S_{\text{self}}(i) = E[\lambda^{T(i)} | R_0 = i]$, i.e. the generating function for the first return to the current word. We have
$$
S_{\text{self}}(i) = \lambda(p_{j} S_{j, i} + p_{j+1} S_{j+1, i} + p_{j+2} S_{j+2, i}) \tag{2}
$$
Finally, we need the escape probability. It will be easier to calculate the probability of not escaping. The generating function for not escaping from node $i$ is the generating function for the first visit to the green root node or to itself. Hence,
$$
S_{\text{not escape}}(i) = \lambda(p_g + p_b S_{b, i} + p_r S_{r, i})
$$
So the probability of escaping is
$$
p_{\text{escape}}^{(i)} = 1 - (p_g + p_b P_{b, i} + p_r P_{r, i}) \tag{3}
$$
where I have introduced the notation $P_{i, j} = S_{i, j}(\lambda = 1)$. Putting this all together, the probability of escaping from any individual node $P_{\text{escape}}^{(i)}$ is
\begin{align}
P_{\text{escape}}^{(i)} = \frac{P_{g, i} p_{\text{escape}}^{(i)}}{1 - P_{\text{self}}^{(i)}}
\end{align}
It remains to solve Eq. (1) numerically in order to find Eqs. (2) and (3). One can check that $S_{i, j} = (3 - \sqrt{9 - 8 \lambda^2})/4\lambda$ in the symmetric case, which leads to $P_{\text{escape}}^{(i)} = 1/3$ as expected.
