Triangular communication network

A communication network is made of nodes conected with wire. The net sends packets in such a way that if one packet is located in an internal node $$x$$ (internal node is the one connected to more than one node), it chooses randomly the output node. The probability of going out through the node $$y$$ connected to $$x$$ is equal to $$p_{xy}$$, such that $$\sum_{y} p_{xy}=1$$. When the packet reaches an external node $$X$$, it remains there. $$p_X$$ denotes the probability of going to an external node, when we are connected to it.

We are thinking about calculating the probability $$P(xX)$$. That's to say, the probability that being the packet in an internal node $$x$$ it finishes in the external node $$X$$.

Solve the problem for a triangular net with nodes $$a$$,$$b$$,$$c$$ conected to the external nodes $$A$$,$$B$$,$$C$$,respectively. Solve it using linear equations that verify the different probabilities $$P(xY)$$.

If I want to the calculate probability $$P(aA)$$, I think that I have to solve a linear equation system with $$P(aA)$$, $$P(bA)$$ and $$P(cA)$$.

$$P(aA)= P_A +P_A P_{ac}P_{ca} + P_A P_{ab}P_{ba}$$. I don't know if I have to add the probabilites $$P_A P_{ac}P_{cb}P_{ba}$$ and $$P_A P_{ab}P_{bc}P_{ca}$$. Also if the packet goes from $$a$$ to $$c$$, instead of returning to $$a$$, $$(P_{ac}P_{ca})$$, it could go from $$c$$ to $$C$$.

To answer your last question directly, yes you have to consider all complicated possibilities, like $$abcbcbcbabacbaA$$ etc. That's why that is not the way to go.
$$P(aA) = P_A + P_{ab} P(bA) + P_{ac} P(cA)$$
Why? Because from $$a$$, either you go to $$A$$ and win, or you go to $$b$$ in which case you now have prob $$P(bA)$$ of winning, or you go to $$c$$ and now have prob $$P(cA)$$ of winning. There are $$3$$ unknowns, $$P(aA), P(bA), P(cA)$$ and you need to find $$3$$ equations (one of them given above) and then solve. That accounts for $$P(xA)$$. Similarly you need other equation-systems for $$P(xB)$$ and $$P(xC)$$.