How do I find the "perfect strategy" for rock-paper-scissors variants? (see details)

Before I ask my question, I must define a “variant.” For each variant, there is a list of possible throws. For each possible pairing of throws, either one throw beats the other or there is a tie. There are no restrictions as to what pairings have what results. The game can be unbalanced. You can introduce a dynamite throw that beats almost anything. You can make it so that lizard and Spock tie each other. The game must be commutative (rock-scissors has the exact opposite result as scissors-rock, but the rock always wins).

Next I must define a “strategy.” A strategy is an assignment of a probability to each different throw. When you make a move, you randomly pick a throw based on the preassigned probabilities. A “perfect strategy” is one in which, on average, no strategy used by the opponent can beat. This doesn’t mean that you will beat the opponent, just that the opponent can’t beat you (on average). For the normal version, the perfect strategy is a ⅓ probability for each. If we were to make rock and paper tie, then the perfect strategy would be 100% rock.

MAIN QUESTION:

Given any variant, how can I find out the strategy that will never lose on average?

This question was inspired by the version of RPS played here, which adds in two throws with unequal power (along with some other restrictions not relevant here).