How do I find the "perfect strategy" for rock-paper-scissors variants? (see details)
Asked by
PhiNotPi (
12686
)
December 27th, 2011
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).
Observing members:
0
Composing members:
0
9 Answers
I would start by reading up on the mathematics of Game theory. GQ.
The perfect strategy comes from having insight into your opponent’s personality and by understanding these basic principles.
@flutherother Even if that gives me an advantage against imperfect opponents, that does not solve the problem of finding a strategy that ”no strategy used by the opponent can beat” (on average). If I start trying to predict them, theoretically they can detect where my resulting play is not perfect, and can beat me.
@PhiNotPi The only strategy that your opponent cannot predict and beat is a random one. Any ‘system’ is susceptible to being understood and defeated by a more intelligent and insightful opponent.
@flutherother I know that for a normal game, but I am asking how to find it when not all throws are equally powerful or there are many more symbols. Just to elaborate, I am asking how to find the strategy that will at least tie all other strategies. I am not asking how to find a strategy that always wins.
@flutherother just to add to your point, let’s say there’s a strategy A that will defeat the opponent’s strategy on average. What’s to prevent your opponent from also using strategy A? In other words, what happens when two strategy A opponents play each other?
Strategy A might win on average but might also lose on average if your opponent is clever enough to understand strategy A and devise strategy B to counteract it. This can be done in theory for all strategies apart from the strategy of choosing completely at random. There is no strategy that can counter the random strategy but unfortunately it only guarantees that in the long run you will not lose, not that you will win.
I am not asking how to play a better game of RPS, I am asking how to find the strategy that always ties given any version of the game. For normal RPS, the answer is to split each throw 33% (a random game). For RPSLS, the answer is to split every throw 20%.
How would I find this for unbalanced games? How would I find the strategy that effectively cancels out any strategy used by the opponent, even when the answer is not trivial?
Answer this question 
This question is in the General Section. Responses must be helpful and on-topic.
Composing members:
0
