Have you ever gotten bored of the classic rock, paper, scissors game? Fear no more! Here is a new twist on the classic that will keep you entertained for hours!

The Rules

Spider

The spider wins against:

• Rock because spiders have been known to withstand the weight of rocks (evolution is amazing)
• Human as humans are terrified of spiders

The spider loses against:

• Paper because spiders can be captured by paper
• Scissors because scissors can cut spider webs (and spiders)

Man

The man wins against:

• Paper because humans can shoot paper with guns
• Scissors because humans can shoot scissors with guns

The man loses against:

• Rock because humans can be crushed by rocks
• Spider because humans are terrified of spiders

Rock

The rock wins against:

• Scissors because rocks can crush scissors
• Human because rocks can crush humans

The rock loses against:

• Paper because rocks can be covered by paper
• Spider because spiders can withstand the weight of rocks (evolution is amazing)

Paper

The paper wins against:

• Rock because paper can cover rocks
• Spider because spiders can be captured by paper

The paper loses against:

• Scissors because scissors can cut paper
• Human because humans can shoot paper with guns

Scissors

The scissors wins against:

• Paper because scissors can cut paper
• Spider because scissors can cut spider webs (and spiders)

The scissors loses against:

• Rock because rocks can crush scissors
• Human because humans can shoot scissors with guns

Cayley Table

	Rock	Paper	Scissors	Spider	Man
Rock	Draw	Paper	Rock	Spider	Rock
Paper	Paper	Draw	Scissors	Paper	Man
Scissors	Rock	Scissors	Draw	Scissors	Man
Spider	Spider	Paper	Scissors	Draw	Spider
Man	Rock	Man	Man	Spider	Draw

On the existence of n-dimensional generalizations

Denote the game as $\mathbf{G}$ with binary operation $\circ$ and identity element $I$.

Given a set $\mathbf{S}$ of $n$ elements ($a_i$), we can define a Cayley table for $\mathbf{G}$ as follows:

Call the game $\mathbf{G}$ balanced if $|\{a_i \ | \ a_i \circ a_j = a_i, \ \forall \ 1 \leq j \leq n, \ \forall \ a_i \in \mathbf{S}\}| = \frac{n-1}{2}$

Indeed for a $\mathbf{G}$ to be nontrivial it must satisfy:

• $\mathbf{G}$ is closed under $\circ$ (by definition)
• $\mathbf{G}$ is balanced

Claim. A balanced game $\mathbf{G}$ exists if and only if $n$ is odd.

Proof.

Disincluding, the identity element, there are $n-1$ elements in $\mathbf{S}$. For $\mathbf{G}$ to be balanced, there must be $\frac{n-1}{2}$ elements that are their own inverse. This is only possible if $n-1$ is even, i.e. $n$ is odd.