= Solution
Until the first visit to zero, the positive-side chain is a nearest-neighbour random walk with upward probability $2/3$ and downward probability $1/3$. For an <integer> $m>1$, let $h_i^{(m)}=\Pr_i(T_0<T_m)$, $0\leq i\leq m$. The <Markov property> gives
$$
h_i^{(m)}=\frac13h_{i-1}^{(m)}+\frac23h_{i+1}^{(m)},\qquad h_0^{(m)}=1,\quad h_m^{(m)}=0.
$$
The characteristic roots of this recurrence are $1$ and $1/2$. Solving its two boundary equations gives the <biased gambler's ruin probability>
$$
h_i^{(m)}=\frac{2^{-i}-2^{-m}}{1-2^{-m}}.
$$
A finite path that reaches zero has a finite maximum before doing so, and is counted by $\{T_0<T_m\}$ for every sufficiently large $m$. Conversely each such event implies a visit to zero. Therefore these events increase to $\{T_0<\infty\}$, and <continuity> of probability gives
$$
\boxed{\Pr_1(T_0<\infty)=\lim_{m\to\infty}h_1^{(m)}=\frac12.}
$$
More generally the hitting probability from $i>0$ is $2^{-i}$. The finite-boundary derivation avoids choosing an unjustified solution of the infinite recurrence.
Back to article page
