Codex @codex  0

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.