Rock-paper-scissors probabilities: draws and the round of a final winner

Specify the random model first

Assume every player independently chooses rock, paper, or scissors with probability $1/3$, independently on each round. Then $n$ labeled players have $3^n$ equally likely hand patterns.

For two players, six of the nine patterns give a winner and three are draws. Thus the probability of a decisive round is $2/3$ and the draw probability is $1/3$.

For more than two players, we use the standard group rule: a round is decisive when exactly two hand types appear. Everyone showing the winning type wins that round. If all players show the same type, or all three types appear, the round is a draw.

This distinction matters: a decisive round can leave more than one winning player.

Problem 1: the draw probability for n players

Problem. Find the probability that a single round with $n\ge2$ players is a draw.

Count the complement. Choose the two types appearing in the round in $\binom32=3$ ways. For a chosen pair, there are $2^n$ patterns, but the two patterns with every player showing the same type must be excluded. Hence the number of decisive patterns is $3(2^n-2)$.

These cases do not overlap because each contains exactly its chosen pair of types. Therefore

$$\boxed{P(\text{draw})=1-\frac{3(2^n-2)}{3^n}=\frac{3^{n-1}-2^n+2}{3^{n-1}}.}$$

At $n=2$ and $n=3$, this gives $1/3$. For larger groups, draws become more common; counting all-three-type draws directly would be less convenient than using the complement.

Problem 2: one final winner after exactly n rounds

Three players repeatedly play under the following elimination rule. In a decisive round, all losing players leave and the winners continue. A draw leaves the player set unchanged. The process ends when exactly one player remains. Count every round, including draws.

Problem. Find the probability that the final winner is determined on round $n$.

For three players, a round has these probabilities:

• Exactly one winner, reducing the group from three to one: $1/3$.
• Exactly two winners, reducing the group from three to two: $1/3$.
• A draw, leaving three players: $1/3$.

For example, a unique winner is specified by choosing the player in three ways and that player's hand in three ways; the other two must both show the losing hand. Thus there are nine of the $27$ hand patterns.

With two players, a draw has probability $1/3$, and a unique winner has probability $2/3$.

First solution: split the possible elimination paths

Three players directly to one

There must be $n-1$ draws followed by a unique winner:

$$P(3\to1\text{ on round }n)=\left(\frac13\right)^{n-1}\frac13=\frac1{3^n}.$$

Three players to two, then to one

Suppose the reduction to two occurs on round $k$, where $1\le k\le n-1$. There are $k-1$ initial draws, the reduction with probability $1/3$, then $n-k-1$ two-player draws and a final decisive round with probability $2/3$.

For each $k$, the probability is

$$\left(\frac13\right)^{k-1}\frac13\left(\frac13\right)^{n-k-1}\frac23=\frac2{3^n}.$$

There are $n-1$ choices of $k$. Adding both paths gives

$$\boxed{P(T=n)=\frac1{3^n}+\frac{2(n-1)}{3^n}=\frac{2n-1}{3^n},\qquad n\ge1.}$$

For $n=1$, the second path is impossible and its term is zero, so the same final formula still works.

Second solution: track the unfinished states

Let $p_n$ be the probability that three players remain after $n$ rounds and $q_n$ the probability that two remain. Initially $p_0=1,q_0=0$. The transition probabilities give

$$p_{n+1}=\frac13p_n,\qquad q_{n+1}=\frac13p_n+\frac13q_n.$$

The diagram displays only the unfinished states. The omitted probability corresponds to a finished game, not to a missing possible outcome.

The first recurrence gives $p_n=3^{-n}$. Multiply the second by $3^{n+1}$:

$$3^{n+1}q_{n+1}=1+3^nq_n.$$

Since $q_0=0$, this yields $q_n=n/3^n$. A game ending on round $n$ must have been unfinished after round $n-1$, so

$$P(T=n)=\frac13p_{n-1}+\frac23q_{n-1}=\boxed{\frac{2n-1}{3^n}.}$$

Using time zero makes the argument work uniformly at $n=1$ without a separate path case.

What to watch for

Distinguish a decisive round from a unique final winner. Include all possible group-size paths and all possible times of an intermediate reduction. When draw probabilities differ between group sizes, the path sum may become a genuine geometric sum rather than a constant term repeated $n-1$ times.

Related: probability recurrence relations and telescoping and geometric-factor sums.