Question

Discuss the Nash equilibrium and Mixed Strategy Nash equilibrium.

a) Provide the definitions.

b) Provide two games, and illustrate the Nash equilibrium / equilibria

and Mixed Strategy Nash Equilibrium / equilibria for those games.

Answer 1

A mixed strategy Nash equilibrium involves at least one player playing a randomized strategy and no player being able to increase his or her expected payoff by playing an alternate strategy

In many games players choose unique actions from the set of available actions. These are called pure strategies .In some situations though a player may want to randomise over several actions. If a player is choosing which action to play randomly, we say that the player is using a "mixed strategy" as opposed to a pure strategy. In a pure strategy a player chooses an action for sure, whereas in a mixed strategy, he chooses a probability distribution over the set of actions available to him.

Consider the matching pennies game:

Player 2

Heads Tails

Player 1 Heads 1,-1 -1,1

Tails -1,1 1,-1

There is no (pure strategy) Nash equilibrium in this game. If we play this game, we should be “unpredictable.” That is, we should randomize (or mix) between strategies so that we do not get exploited. • But not any randomness will do: Suppose Player 1 plays .75 Heads and .25 Tails (that is, Heads with 75% chance and Tails with 25% chance). Then Player 2 by choosing Tails (with 100% chance) can get an expected payoff of 0.75 ×1 + 0.25 ×(-1) = 0.5. But that cannot happen at equilibrium since Player 1 then wants to play Tails (with 100% chance) deviating from the original mixed stategy. • Since this game is completely symmetric it is easy to see that at mixed strategy Nash equilibrium both players will choose Heads with 50% chance and Tails with 50% chance. • In this case the expected payoff to both players is 0.5 ×1 + 0.5 ×(-1) = 0 and neither can do better by deviating to another strategy (regardless it is a mixed strategy or not). • In general there is no guarantee that mixing will be 50-50 at equilibrium.

There can be mixed strategy Nash equilibrium even if there are pure strategy Nash equilibria.

  Player 2

q (1-q)

L R

Player 1 p U 3,1 0,0

(1-p) D 0,0 1,3

At the mixed Nash equilibrium Both players should be indifferent between their two strategies: • Player 1: E(U) = E(D) ⇒ 3q = 1 − q ⇒ 4q = 1 ⇒ q = 1/4, • Player 2: E(L) = E(R) ⇒ p = 3 × (1 − p) ⇒ 4p = 3 ⇒ p = 3/4. Therefore Player 1 plays (3/4U+1/4D) and Player 2 plays (1/4L+3/4R) at mixed strategy Nash equilibrium.