= Solution
A random time $T$ is a <stopping time> when the event $\{T\leq n\}$ is determined by $X_0,\ldots,X_n$. The <Strong Markov property> says that, conditionally on $T<\infty$ and $X_T=x$, the process $(X_{T+k})_{k\geq0}$ is a fresh Markov chain started at $x$, independent of the history before $T$.
Use the state space $\{1,2,4\}\times\mathbb Z_{\geq0}$, with $(2,0)$ absorbing. Observe the chain only when it is at square 2. From $(2,k)$, the change $Y$ in wealth by the next return to square 2 has distribution
$$
\Pr(Y=-1)=\frac12,\qquad
\Pr(Y=1)=\frac18,\qquad
\Pr(Y=2)=\frac38.
$$
Indeed, heads lands on square 3 and returns to square 2 after losing £1. After tails reaches square 4, the remaining two or three moves give the other cases.
For $r=2/3$,
$$
\mathbb E[r^Y]
=\frac12r^{-1}+\frac18r+\frac38r^2=1.
$$
Thus $r^{M_j}$ is a <martingale> for the embedded wealth random walk $M_j$. Stopping when it first reaches $m-1$ or a large upper level and then letting that level tend to infinity gives
$$
\Pr_{(2,m)}(\text{ever hit }(2,m-1))=r=\boxed{\frac23}.
$$
The upper-bound contribution vanishes because $0<r<1$; equivalently, this is the smaller probability solution of the first-step equation.
Back to article page
