Rock, Paper, Scissors, and Game Theory

Game theory has long been a fascination of mine, and I'm finally getting a chance to take a class on it here at NYU. We recently had a class where we found the mixed-strategy Nash equilibrium for rock, paper, scissors (RPS), and it made me curious about Nash equilibria for other variants of RPS, which are manifold.

First, of course, I'm going to have to explain some these game theory terms that I'm throwing around. The essence of game theory is exploring decision-making when the outcome of your decision is dependent on someone else's decision. When this is in a two-player game, it's often expressed through a payoff matrix, like so:



This is a payoff matrix for RPS. The left column shows the moves for player one, and the top row shows the moves for player two. The payoffs to each player are listed in the cells, with the format [Player 1's payoff, Player 2's payoff], where a 1 represents a win, a -1 represents a loss, and 0 represents a tie.

Now that we have a payoff matrix, we'll want to find an equilibrium. Specifically, we're going to look for a Nash equilibrium. A Nash equilibrium exists where both players are pursuing a strategy such that neither can improve their payoff if the other play continues to pursue the same strategy. It is obvious that there is no "pure strategy" Nash equilibrium for RPS; that is to say, there's no possible equilibrium where each player plays one move all the time. If player 1 always throws scissors and player 2 always throws rock, player 1 can improve his payoff by always throwing paper instead. However, if he does this, then player 2 can improve his payoff by always throwing scissors, and so on.

When a game has no pure strategy Nash equilibrium, we search for a mixed-strategy Nash equilibrium. A mixed strategy is a grouping of pure strategies, with a proportion assigned to each for often it should be played. In the case of RPS, there is a mixed-strategy Nash equilibrium where each player plays each strategy one-third of the time. This is intuitively unsurprising; each move will win, lose, or tie one-third of the time each.

Another variant of RPS is Rock, Paper, Scissors, Lizard, Spock, which, in an effort to reduce ties, expands upon the original RPS ("RPS Classic", I suppose) by adding in lizard, which eats paper and poisons Spock, but is crushed by rock and decapitated by scissors, and Spock, who vaporizes rock and bends scissors, but is poisoned by lizard and disproved by paper (he is fictional, after all). The payoff matrix for rock, paper, scissors, lizard, Spock is as follows:



It is somewhat more interesting than the classic RPS chart, and certainly does cut down on the number of ties (the probability of a tie is reduced from 1/3 [3/9] to 1/5 [5/25]). However, the Nash equilibrium is essentially the same as for the original RPS: both players play each strategy 1/5 of the time. It's the same game, just slightly expanded.

However, there is another five move variant of RPS that actually adds an additional strategic wrinkle: Rock, paper, scissors, fire, water. In this variant, the original rules hold, except fire beats everything except water, and water, in turn, loses to everything except fire. The payoff matrix looks like this:



The game is said to come with the stipulation that fire can only be used once in one's life time, but this is silly, as the Nash equilibrium can be shown to be, actually, each player playing fire one-third of the time, water one-third of the time, and rock, paper, and scissors each one-ninth of the time. The reason for this is that it's really become a balanced game between three actions, where the three actions are fire, water, and RPS. Fire beats RPS, which beats water, which beats fire. Each of these actions should be played one-third of the time, but "playing" RPS properly means playing rock, paper, and scissors each one-third of the time, and 1/3*1/3=1/9.

The reason I bring this up, and have made such a long blog post about such a ridiculous topic, is because I find this result to be interesting, and worth thinking about, because it is counterintuitive. While water would seem to be arguably the least valuable move, because it only defeats one other action while the other four defeat at least two actions, it actually should be played three times as often as rock, paper, or scissors, even though those would seem to do more. It is solely because the one action that water beats, fire, is the most powerful that water is so valuable move. I think there is a useful lesson to be had here when considering counterintuitive value and pricing. Unfortunately, I don't think I know that lesson yet.