= Last-excursion direction law at a transient junction
{title2=$\Pr(\text{escape through branch }j)=a_j/\sum_i a_i$}
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 $\sum_{m\geq0}(1-a)^m a_j=a_j/a$. When the branches have outward random-walk drift, the final infinite excursion tends to the corresponding end almost surely.
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