# The win rate of three player rotation battle

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:

• You need to show what effort you have put in, and where you are stuck before appropriate help can be given. May 18, 2022 at 8:53
• Clarification: suppose player A wins round (A over B), player B wins round 2 (B over C), and player C wins round 3 (A over C). Does this count as "two in a row," since player A did not play the second game? May 18, 2022 at 15:52
• @AaronMontgomery, A wins in this case, it can be equivalent to the score system, one point is added for a win, and the scores are cleared when lost. May 19, 2022 at 0:31

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.

If the game ends when a particular player wins the game at the first opportunity, then the answers are (in the order they can win) $$B: \dfrac38,\;\; C: \dfrac18,\;\; A: \dfrac{15}{128}$$

But if the game continues to infinity, it becomes complex, as it does not seem to be a G.P. for each player. However, you can try if you can find a pattern, else simulation is the only way.

Using binary digits, with $$1$$ denoting a win, and $$0$$ a loss, for trying to find a pattern, I can give a hint, eg for $$C$$ winning two in a row only on the second opportunity, possible routes are

$$ab|bc|ca|ab|bc|ca$$
$$01|01|01|01|01|1x$$
$$\color{red}{10}|01|01|01|01|1x$$
$$\color{red}{10|10}|01|01|01|1x$$
$$\color{red}{10|10|10}|01|01|1x$$
$$\color{red}{10|10|10|10}|01|1x$$

The idea is that two $$1's$$ (wins) should only come once.

Looking backwards, we shall always have a block $$x110$$, and continue as $$1010..$$ except that at any stage the sequence can be changed to $$0101...$$, but once changed to $$0101..$$ it can't be changed back to $$1010...$$

You can also have a look at this