Rock-paper-scissors, mixed equilibrium

Problem

In the game «rock-paper-scissors» with three actions (Paper, Rock, Scissors), show that there is no equilibrium in pure strategies. Then find the equilibrium in mixed strategies by exploiting the symmetry.

Solution

In rock-paper-scissors each action beats one of the other two and loses to the third (cyclic symmetry: Paper beats Rock, Rock beats Scissors, Scissors beats Paper). There is no stable pair of pure actions: any fixed choice of the opponent can be exploited by changing one’s own, so no pure strategy is dominant and there is no equilibrium in pure strategies. By symmetry, the only equilibrium is the one in mixed strategies in which each player plays the three actions with the same probability: $P(\text{Carta}) = P(\text{Sasso}) = P(\text{Forbici}) = \boxed{\dfrac{1}{3}}$ With this strategy the expected value of the game is $\bar V = 0$: the game is fair.

Links

Topics: Probabilita
Concepts: Equilibrio di nash · Strategia mista
Skills: Calcolo probabilita · Modellizzare
Exercise type: Problema modellizzazione