## Applying Nash Equilibrium to Rock, Paper, and Scissors

Nash Equilibrium is a pair of strategies in which each player’s strategy is a best response to the other player’s strategy. In a game like Prisoner’s Dilemma, there is one pure Nash Equilibrium where both players will choose to confess. However, the players only have two choices: to confess or not to confess. What happens if there are more choices? For example, in the classic game of rock, paper, and scissors, there are three choices. How can we find the Nash Equilibrium then? And if we do, is it helpful?

See the following article:

In the above article, the author discusses the application of Nash Equilibrium to games like Rock, Paper, and Scissors. Recall from class that in game theory, games can have:

(1) Only one pure Nash Equilibrium (e.g. in Prisoner’s Dilemma)
(2) Only one mixed Nash Equilibrium and no pure Nash Equilibrium (e.g. Kicker/Goalie Penalty kicks)
(3) Multiple pure Nash Equilibrium (e.g. Hawk-Dove Game)
(4) Pure and mixed (e.g. Hawk-Dove Game)

So which category does the game Rock, Paper, and Scissors fall under?

According to the article, Rock, Paper, and Scissors fall under (2) –  only one mixed Nash Equilibrium. However, you can easily arrive at this conclusion by applying your knowledge of game theory and Nash equilibrium – all topics we learned in INFO 2040.

Let p = player one and q = player two. (For the sake of simplicity, there will only be two players)

First, the reason why there isn’t a pure Nash Equilibrium is that there is no way a player will 100% of the time choose one choice. For example, let’s take player 1. If he consistently plays rock, then player 2 will always choose paper. Player one will never win. Likewise, if player 2 always choose paper, player one will always choose scissors. Player two will always lose. The two players will then fall into a cycle of rock, then paper, then scissors. Thus, there is no equilibrium – it just doesn’t make sense for one player to ALWAYS pick one choice for the whole game – it’s just too predictable.

Now let p(rock) be the probability that player 1 pick rock, p(scissors) be the probability that player 1 pick scissors, and p(paper) be the probability that player 1 chooses paper. Likewise, q(rock), q(scissors), and q(paper) for player 2.

We know that none of these probabilities is fully a 1 (always choose).

(From here on, I will not explain all the math as the article does a wonderful job at doing this, but I will summarize the findings here.)

The expected value for player 2 is:
EV[q(rock)] =  0*p(rock) + (-1)*p(paper) + 1*(p(scissors))
EV[q(paper)] =  1*p(rock) + 0*p(paper) + (-1)*(p(scissors))
EV[q(scissors)] = (- 1)*p(rock) + 1*p(paper) + 0*(p(scissors))

Also, p(rock) + p(paper) + p(scissors) = 1

Using these equations, you will eventually reach that the Nash Equilibrium for the game Rock, Paper, and Scissors is:

For player 1, p(rock) = 1/3, p(paper) = 1/3, and p(scissors) = 1/3

and similarly,

For player 2, q(rock) = 1/3, q(paper) = 1/3, and q(scissors) = 1/3

So that’s the Nash Equilibrium. But how useful is it?

Why Nash Equilibrium may not apply to a game like Rock, Paper, and Scissors

There is another major difference between a game like Prisoner’s Dilemma and Rock, Paper, and Scissors (besides the number of choices) and that is:

The players will play again and again.

In Prisoner’s Dilemma, they play one round and so they must pick the dominant strategy in that game, but in Rock, Paper, and Scissors, the two players repeatedly play. The article states that in such a case, it’s best for the players to stick to about 1/3 for rock, paper, or scissors throughout the game. However, is that really the best?

See the following article:

http://guardianlv.com/2014/05/rock-paper-scissors-and-game-theory/

In this article, a large amount of people repeatedly play rock, paper, and scissors against each other and the results are:

“Upon review of the results, Wang did find numbers that backed up the Nash Equilibrium theory coming into play. He also found the above-mentioned pattern: winners were the players who stayed loyal to their strategy and losers were the players who switched. In game theory, this is called “conditional response.” In fact, the conditional strategy proved to be 10 percent more reliable for winning than did the Nash Equilibrium.”

From this, you should definitely be more cautious in using the Nash Equilibrium. Of course, we did find the Nash Equilibrium for Rock, Paper, and Scissors but we cannot say that will be the best strategy. In fact, often times it’s not (as we have found out in class). Thus,  as shown in class and here, we can find the Nash Equilibrium in cases where there are more than two choices but we also need to be careful when applying it – even if it’s a game as simple as Rock, Paper, and Scissors.

Side Notes:

(1) I recommend readers to look up “conditional response”
(2) For the game Prisoner’s Dilemma, a tournament was held where players repeatedly play Prisoner’s Dilemma and one strategy that did well was called “tit for tat.” See: http://en.wikipedia.org/wiki/Prisoner’s_dilemma