Generalizing Rock, Paper, Scissors game?

Question

Problem of playing rock-paper-scissors with $n$ elements and keeping it balanced

How to extend the rock-paper-scissors game to more than just $3$ elemetns, and keep it balanced?

In a balanced game, optimal strategy would be to pick every choice with equal probability.

Is there a general rule that can be used to efficiently evaluate a winner between any two distinct options?

For example, there is one generalization to five elements, rock-paper-scissors-lizard-spock:

Answer 1

Stating the problem

Let $G(n)$ be a rock-paper-scissors game with $n\in\mathbb N$ elements.

Let those $n$ elements be the fist $n$ nonnegative integers $\{0,1,2,\dots,n-1\}=N_n$.

For every $n_0,n_1\in N$, if $n_0$ beats $n_1$, then $n_0$ is a "counter" to $n_1$, and $n_1$ is a "anticounter" to $n_0$. Counters and anticounters of $n_0$ are together called "interactions" of $n_0$.

We want to establish a rule for counters and anticounters of every $n_0\in N_n$, such that:

• $n_0$ has an equal number of counters and anticounters - interactions are balanced.

• $n_0$'s interactions in $G(n)$ should be kept in $G(n+1)$.

Solution to the problem

The first property requires $n=2k+1,k\in\mathbb N$ to be odd.

If we start observing first few cases of $G(n)$, you'll notice that to satisfy the second property, $n_0=m$ needs to be a counter to $m+1,m-2,m+3,m-4,\dots$

This is finally equivalent to the following rule: The winner between $n_0,n_1\in N_n$ is:

• if $n_0,n_1$ have the same parity (both odd or both even), then winner is: $\max\{n_0,n_1\}$

• else, if $n_0,n_1$ have $\ne$ parity, then the winner is: $\min\{n_0,n_1\}$

This rule works for defining counters and anticounters for every element.

This rule allows a balanced rock-paper-scissors to be played with any number of odd elements, and to easily add/remove interactions to extend or reduce the game by $2$ elements.

For example, $n=3,5$ cases and with the classic labels to elements $0,1,2,3,4$ :

Generalization to countable sets

Notice that the general observed rule does not depend on $n$.

Thus if we let $n\to\infty$, we can play rock-paper-scissors on $\mathbb N_0 = \mathbb N \cup \{0\}$.

We can also include negative numbers and extend it to the whole integers $\mathbb Z$, as the "parity" and "$\lt$" are still defined on integers.

We can also play it on any countable set. We establish a bijection from some countable set $X$, such as rationals $X=\mathbb Q$ for example, to either $\mathbb N_0$ or $\mathbb Z$, and call it $B(x)$.

Then the winner between $x_1,x_2\in X$ is:

$$=\begin{cases} \displaystyle \max\{B(x_1), B(x_2)\}, & B(x_1)\bmod2=B(x_2)\bmod2\\ \displaystyle \min\{B(x_1), B(x_2)\}, & B(x_1)\bmod2\ne B(x_2)\bmod2 \end{cases}$$

I believe this rule is the most "balanced" way to extend rock-paper-scissors to any finite and countable sets.

Additionally, by symmetry, we can invert the rule and keep all the properties. That is, invert the conditions for min and max calculations.

Answer 2

For any odd number $n$, consider a game with outcomes in the range $0$ to $n-1$. Outcome $x$ beats outcome $y$ iff $(y-x) \bmod{n} < \frac{n}{2}$. In other words, each outcome beats the next half of the outcomes, and loses to the other half.

Answer 3

This answer is meant as an addition to the answers of Vepir and Thomas. I was wondering in how many ways we could generalize rock-paper-scissors. Here my findings.

We can represent the rules of the game with a directed graph like the one in the question. An arc from A to B means A beats B. For every pair of vertices A and B we need to have either A → B or B → A. Such a graph is called a tournament. To keep the game balanced every vertex needs to have the same indegree as outdegree. A tournament with that property is called a regular tournament. Note that a regular tournament needs to have an odd number of vertices.

In the table below we can see the number of (nonisomorphic) regular tournaments of a given vertex count. We can introduce extra requirements to prefer one regular tournament over an other:

• We can require the graph to be vertex-transitive. This adds a nice symmetry so that, loosely speaking, every vertex is the same. Note that a vertex-transitive tournament is automatically regular.
• We can require that we have to be able to build the tournament by starting at $n=1$, and then adding two vertices and multiple arcs at each step, while at each step having a regular tournament. This is similar to Vepir's "$n_0$'s interactions in $G(n)$ should be kept in $G(n+1)$". I'll call this propery extended.
• We can combine these two properties: We start at $n=1$, add two vertices and multiple arcs at each step, while at each step having a vertex-transitive tournament. I'll call these tournaments extended vertex-transitive tournaments.

Counting all the possibilities, we get:

\begin{array}{|l|c|c|c|c|c|c|c|} \hline & 1 & 3 & 5 & 7 & 9 & 11 & 13 \\ \hline \text{Regular tournaments (OEIS A096368)} & 1 & 1 & 1 & 3 & 15 & 1223 & 1495297 \\ \hline \text{Vertex-transitive tournaments (OEIS A346179)} & 1 & 1 & 1 & 2 & 3 & 4 & 6 \\ \hline \text{Extended regular tournaments} & 1 & 1 & 1 & 2 & 8 & 350 & ? \\ \hline \text{Extended vertex-transitive tournaments} & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}

Sequences: OEIS A096368, OEIS A346179

It seems that there is only one way (up to isomorphism) to generalize rock-paper-scissers in a way that it is extended vertex-transitive. I tested the formulas of Vepir and Thomas and it turns out they both generate exactly this graph!