Rock-Paper-Scissors Variants

It is a well-known “fact” that any valid Rock-Paper-Scissors variant with 5 elements must be isomorphic to Rock-Paper-Scissors-Lizard-Spock – that’s the only way for every element to have two things that beat it and two things that it beats. However, there is in fact a second option we invented recently.

It’s called Rock-Paper-Scissors-Bomb-Defuser – Bomb beats all the old things, Defuser beats Bomb, and all the old things beat Defuser. This is still a valid variant; although some elements are more useful than others, every element is useful.

So what really is the rule for a variant being valid? It’s simple: there must be a strategy that ties with any other strategy. For standard Rock-Paper-Scissors, it’s playing every element with a 1 in 3 probability, for Rock-Paper-Scissors-Lizard-Spock it’s playing every element with a 1 in 5 chance, and for Rock-Paper-Scissors-Bomb-Defuser it’s playing Rock, Paper, and Scissors with a 1 in 9 chance and playing Bomb and Defuser with a 1 in 3 chance.

A good rule of thumb for construction is that any pair of elements must be contained within a three-element cycle. This is a necessary condition, although this isn’t a sufficient condition – adding an element to Rock-Paper-Scissors-Bomb-Defuser that beats only Bomb and Scissors satisfies that condition, but if you calculate, it turns out that new element is useless.

Here is a proof that the only 5-element variants are the two already mentioned:
If every element beats two other elements, it must be Rock-Paper-Scissors-Lizard-Spock.
Otherwise, there must be an element that beats one element and/or an element that beats three. If an element beats one thing, then for any element other than it to form a 3-cycle with it, the thing it beats needs to beat three things. There can’t be two things that beat only one element each, because then neither of them could beat the other. So the only remaining combination is that one element beats one element, one element beats three elements, and three elements beat two elements. A quick check shows that to do that it must be Rock-Paper-Scissors-Bomb-Defuser.

I still have plenty of unknown questions:
– Are there any valid variants with an even number of elements?
– How many valid variants are there with 7 elements?
– Is there a valid variant where its tie-with-everything strategy doesn’t use one of the elements?