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The hitting probability of a set is one on that set and transition-harmonic elsewhere by the Markov property. Finite-horizon hitting probabilities begin at the boundary indicator and increase through the same positive transition recursion. Any nonnegative solution dominates every iterate; monotone passage to infinite time shows that it dominates the hitting probability. This proves minimality without requiring almost-sure hitting.

Write $h_i={\mathbb{P}}_i(H^A<\infty )$. For $i\in A$, $h_i=1$. For $i\notin A$, conditioning on the first step and using the Markov property gives $h_i={\sum }_j{p}_{ij}{h}_j$. This establishes the required equations for the hitting probability.

To prove the hitting probability is the minimal nonnegative harmonic extension, let $h_i^{(n)}={\mathbb{P}}_i(H^A\le n)$. Then $h^{(0)}=1_A$ andIf $g$ is any nonnegative solution of the same boundary equations, $g\ge h^{(0)}$. Positivity of the transition probabilities implies by induction $g\ge h^{(n)}$ for every $n$. The events $\{H^A\le n\}$ increase to $\{H^A<\infty \}$, so $h^{(n)}\uparrow h$ and $g\ge h$. Therefore $h$ is the minimal nonnegative solution.

For the tournament won by two consecutive victories, a transient pair $ij$, with $i\ne j$ and third player $k$, moves with equal probabilities to the absorbing state $jj$ or the transient pair $jk$. Thus the two transient cycles arewith probability $1/2$ of absorption at each step. A full cycle without absorption has probability $1/8$, so absorption occurs almost surely.

Starting from $AC$, the successive absorption winners are $C,B,A$, with first-cycle probabilities $1/2,1/4,1/8$. Summing the geometric repetitions givesStarting from $BC$, the order is $C,A,B$, giving $(2/7,1/7,4/7)$.

Because the first game is between $A$ and $B$, the initial ordered pair of winners is $AA,AC,BB,BC$, each with probability $1/4$. Combining the immediate wins with the transient hitting probabilities,Their sum is one, consistent with almost-sure absorption.

Put $D_t=B_t-B_t^{\prime }$. By independence of the two Brownian motions, $D_t/\sqrt{2}$ is a standard Brownian motion: its independent increments have the normal distribution with variance equal to elapsed time. The stopping time $\tau$ is the first time $D$ hits $-1$ or $3$.

To justify finiteness without presupposing the desired hitting probability, each unit-time increment of $D$ has the normal distribution $N(0,2)$. An increment of absolute value greater than $4$ forces exit from $[-1,3]$ if the path was inside at its start. These independent increments give $\mathbb{P}(\tau >n)\le (1-p{)}^n$, where $p=\mathbb{P}(|N(0,2)|>4)>0$. Hence $\tau <\infty$ almost surely.

The stopped martingale $D_{t\wedge \tau }$ lies in $[-1,3]$. The optional stopping theorem at $t\wedge \tau$ gives $\mathbb{E}{D}_{t\wedge \tau }=0$, and the bounded convergence theorem gives $\mathbb{E}{D}_{\tau }=0$. Writing $q=\mathbb{P}(D_{\tau }=-1)$, path continuity gives $0=-q+3(1-q)$. Therefore, as in Brownian exit from an interval,

Let $h(i)$ be the probability of reaching state $3$ before state $0$. The boundary values are $h(0)=0$ and $h(3)=1$. First-step analysis at states $1$ and $2$ givesTherefore $h(1)=h(1)/4+1/4$, so the requested hitting probability is