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Game Theory in Rock, Paper, Scissors

In the popular game called Rock, Paper, Scissors, each of the two players’ objectives is to choose the option, out of rock, paper, and scissors, that beats his opponent’s option, rock beating scissors, paper beating rock, and scissors beating paper. Let us examine the payoff matrix for such a game:

Player1\Player2

Rock

Paper

Scissors

Rock

0/0

-1/1

1/-1

Paper

1/-1

0/0

-1/1

Scissors

-1/1

1/-1

0/0

In this version of the game, we assume that losing a round of the game results in a payoff of -1, winning a round results in a payoff of 1, and tying in a round results in a payoff of 0. Let us examine this in terms of pure strategy Nash equilibrium. In a pure strategy Nash equilibrium, each player’s option must be the dominant strategy to the other player’s dominant strategy. If we examine the payoff table for the game of rock, paper, scissors, it becomes evident that there is no such equilibrium. There is no option in which both players’ options are the best response to the other player’s option. Thus, there are no pure strategy Nash equilibria.

Now let us examine this game in terms of mixed strategy Nash equilibria. Since each option is nearly symmetrical with each of the other options, it should be obvious that the mixed strategy Nash equilibrium for this game turns out to be that each player should choose each option with equal probability. Thus, both players should choose rock with 1/3 probability, paper with 1/3 probability, and scissors with 1/3 probability. This result can be obtained through mathematical calculation, however, the symmetry of the game makes it incredibly obvious that each option should have an equal expected payoff if the opponent’s option is unknown.

However, if the game is altered slightly by simply altering the payoffs for a single situation, for example making the payoff for winning with paper against rock 2 and losing with rock against paper -2, the strategy for the game will change. The resulting payoff matrix would look like:

Player1\Player2

Rock

Paper

Scissors

Rock

0/0

-2/2

1/-1

Paper

2/-2

0/0

-1/1

Scissors

-1/1

1/-1

0/0

This game has a different mixed strategy Nash equilibrium because of the fact that winning with paper against rock has a higher payoff than any other situation, and losing with rock against paper has a lower payoff than any other situation. Thus, if you assumed that your opponent was playing each option with equal probability, it would be beneficial to play paper with higher probability and rock with lower probability.

Source: http://www.gametheory101.com/Rock_Paper_Scissors.html

 

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