= Solution
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 $h_i=\sum_jp_{ij}h_j$. 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 $h^{(0)}=\mathbf1_A$ and
$$
h_i^{(n+1)}=
\begin{cases}
1,&i\in A,\\
\sum_jp_{ij}h_j^{(n)},&i\notin A.
\end{cases}
$$
If $g$ 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 \b[$h$ 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 are
$$
AB\longrightarrow BC\longrightarrow CA\longrightarrow AB,\qquad
AC\longrightarrow CB\longrightarrow BA\longrightarrow AC,
$$
with 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 gives
$$
\mathbb P(\text{winner}=A,B,C\mid AC)=(1/7,2/7,4/7).
$$
Starting 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>,
$$
\boxed{\mathbb P(A\text{ wins})=\frac5{14},\qquad
\mathbb P(B\text{ wins})=\frac5{14},\qquad
\mathbb P(C\text{ wins})=\frac27.}
$$
Their sum is one, consistent with almost-sure absorption.
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