Three players $A,B,C$ are playing a game.
Players play against each other in round, the order of battle is
$$AB \to BC \to CA \to AB \to \cdots$$
Players need to win two consecutive rounds to win the game.
The winning percentage of each round is as follows:
| Player | Left Win | Right Win |
|---|---|---|
| A vs B | 25% | 75% |
| B vs C | 50% | 50% |
| C vs A | 25% | 75% |
Asking the probability of winning the game for each player.
My attempt
My result using Monte Carlo is:
$$
\begin{aligned}
P_A &= 28.80\% \\
P_B &= 54.53\% \\
P_C &= 16.67\% \\
\end{aligned}
$$
let $p_{a,b,c}$ means the current score.
One point is added for a win, and the scores are cleared when lost.
Whoever reaches two points first wins.
Legal status seems to be these:
Best Answer
Definition: A quick win is one that occurs within the first three rounds. A slow win is one that occurs in round $4$ or later.
Notation: A sequence of letters involving $A, B, C$ indicates a sequence of winners. For instance $ABA$ means $A$ wins round one; $B$ wins round two; $A$ wins round three. Also something like $(ABC)^n$ means that there are $n$ repetitions of $ABC$ in the sequence of rounds. With this notation we will always take $n\ge 1$.
The initial sequence $ABC$ has probability $\frac14\cdot\frac12\cdot\frac14=\frac{1}{32}$. Consequently an initial sequence of the form $(ABC)^n$ has probability $\left(\frac{1}{32}\right)^n$.
Similarly, an initial sequence $(BCA)^n$ has probability $\left(\frac{9}{32}\right)^n$.
Wins for $A$: Quick wins for $A$ are $ABA$ and $ACA$. Slow wins for $A$ are of the forms $(ABC)^n ABA$ and $(BCA)^n A$.
Wins for $B$: The only quick win for $B$ is $BB$. Slow wins for $B$ are of the forms $(ABC)^n B$ and $(BCA)^n BB$
Wins for $C$: Quick wins for $C$ are $ACC$ and $BCC$. Slow wins for $C$ are of the forms $(ABC)^n AC$ and $(BCA)^n BCC$.
Probability that $A$ wins is: $$\begin{array}{ccccccc} & P(ABA) & + & P(ACA) & + & \displaystyle\sum_{n=1}^\infty P[(ABC)^n ABA] & + & \displaystyle\sum_{n=1}^\infty P[(BCA)^n A]\\ =& \frac14\cdot\frac12\cdot\frac34 &+& \frac14\cdot\frac12\cdot\frac34 &+& \displaystyle\sum_{n=1}^\infty \left[\left(\frac{1}{32}\right)^n\cdot \frac14\cdot \frac12 \cdot\frac34\right] &+& \displaystyle\sum_{n=1}^\infty \left[\left(\frac{9}{32}\right)^n\cdot \frac14\right]\\ =&\frac{3}{32} &+& \frac{3}{32} &+& \frac{1}{31}\cdot \frac{3}{32} &+& \frac{9}{23}\cdot\frac{1}{4} \\ =& \frac{6579}{22816} &\approx & 0.28835\end{array} $$ Note that going from the second to the third line of the preceding calculation, both summations are geometric series.
Similar calculations show that the probability that $B$ wins is given by $$\frac{1557}{2852}\approx 0.54593$$
And the probability that $C$ wins is given by $$\frac{3781}{22816}\approx 0.16572$$
I note that these values are quite similar to OP's Monte Carlo values.