Last-excursion direction law at a transient junction ($\mathrm{Pr}(\text{escape through branch }j)=a_j/{\sum }_i{a}_i$)

At a junction, let $a_j$ be the probability that one excursion leaves through branch $j$ and never returns. If $a=\sum a_j>0$, repeated returns restart the same probabilities by the Strong Markov property. Thus the probability of $m$ returns is $(1-a{)}^m$, and eventual escape through $j$ is ${\sum }_{m\ge 0}(1-a{)}^m{a}_j=a_j/a$. When the branches have outward random-walk drift, the final infinite excursion tends to the corresponding end almost surely.