= Solution
One excursion from zero escapes forever on the positive branch with probability $a_+=\tfrac12(1-\tfrac12)=1/4$, and on the negative branch with probability $a_-=\tfrac12(1-\tfrac13)=1/3$. The remaining probability is the return probability $r=5/12$. By the <last-excursion direction law at a transient junction>, the eventual positive escape probability is
$$
\boxed{\sum_{m=0}^\infty r^m a_+=\frac{1/4}{1-5/12}=\frac37.}
$$
Equivalently, if $p$ denotes the desired probability, the <Strong Markov property> gives $p=1/4+(5/12)p$. The negative escape probability is $4/7$, and their sum is one, as independently established in part (b).
Back to article page
