Maynard

no matter what order you list off rock, paper, and scissors, they will always be "in order" or "in reverse order" in the sense that either (1) each thing beats the next; or (2) each thing will be beaten by the next

this property is due to the fact that rock paper scissors has three possible hands. this property fails with larger numbers of hands. for instance, we can consider the game of rock, paper, scissors, lizard, spock as per the following diagram

now listing off spock, scissors, lizard, rock, paper is in neither order of "beats" or "beaten by"

but hey, you think, is it necessary to have five hands for this property to fail? what happens in the case of four hands? don't ask

ok let's talk about four hands

here's the thing. if we want to extend rock paper scissors to have $\mathrm{N<!--\; -->}$ hands (for our choice of $\mathrm{N<!--\; -->}$), we can't just do it willy-nilly. we want to guarantee that:

1. the only way to tie is if both players choose the same hand. ie, every pair of distinct hands has a chosen winner
2. every hand wins as many pairings as it loses

if $\mathrm{N<!--\; -->}$ is even, such as in the case of $\mathrm{N<!--\; -->}=<!--\; -->\mathrm{4<!--\; -->}$, then condition (2) is impossible to uphold, as each hand will have to win against $\frac{\mathrm{1<!--\; -->}}{\mathrm{2<!--\; -->}}(<!--\; -->\mathrm{N<!--\; -->}-<!--\; -->\mathrm{1<!--\; -->})<!--\; -->$ hands and lose against $\frac{\mathrm{1<!--\; -->}}{\mathrm{2<!--\; -->}}(<!--\; -->\mathrm{N<!--\; -->}-<!--\; -->\mathrm{1<!--\; -->})<!--\; -->$ hands. $\mathrm{N<!--\; -->}-<!--\; -->\mathrm{1<!--\; -->}$ is not divisible by two, so this is a no-go

the implication is that in any game of $\mathrm{N<!--\; -->}$-rock paper scissors (as defined above), it must be that $\mathrm{N<!--\; -->}$ is odd

we can see the failure of even $\mathrm{N<!--\; -->}$ in the variant rock, paper, scissors, glue, a game that is under discussion on, for some reason, a forum for the game Civilization

as is inevitable, rock, paper, scissors, glue is unfair in that rock and glue have the joy of winning $\mathrm{2<!--\; -->}\mathrm{/<!--\; -->}\mathrm{3<!--\; -->}$ possible matchups, where as paper and scissors win only $\mathrm{1<!--\; -->}\mathrm{/<!--\; -->}\mathrm{3<!--\; -->}$.

at first glance this is Bad, and indeed in some mathematical sense actually is Bad. but as noted by the forum post it gives rise to a metagame. a player aiming to win may want to play rock, but they may wonder if their opponent is expecting that (as rock is "objectively" best) and therefore planning on playing paper.