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:

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  • $\begingroup$ You need to show what effort you have put in, and where you are stuck before appropriate help can be given. $\endgroup$ May 18, 2022 at 8:53
  • 1
    $\begingroup$ 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? $\endgroup$ May 18, 2022 at 15:52
  • $\begingroup$ @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. $\endgroup$
    – Aster
    May 19, 2022 at 0:31

2 Answers 2


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


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


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