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  Nash Equilibrium and Battle of the Sexes

Nash Equilibrium:

Some games do not have a dominant strategy equilibrium. In these cases, we look for a strategically stable solution – one that none of the players would choose to deviate from.

Strategies s * = (s1 *,.... , si* ) are a Nash equilibrium if each strategy

* is a best response to the other strategies. That is, si * solves max πi (s1 *,.... , si-1* s1 *,.... , si* si+1 *,.... , sI*)

To start simply, let’s consider the following 2 × 2 game:

1262_nash equillibrium.jpg

First, there are no dominant strategies. Second, there is a Nash equilibrium. To see this, suppose Jill plays Up, so Jack’s best response is Left and Jill’s best response is Up. Each is a best response to the other, so (Up, Left) is a Nash equilibrium. One can show that (Down, Right) is also a Nash equilibrium. However, the other two pairs are not.

Battle of the Sexes:

In Jack and Jill’s 2 × 2 game, there are multiple Nash equilibria. Likewise, the following Battle of the Sexes game has two Nash equilibria.

2011_battle of sexes.jpg

Fred and Ethel choose strategies simultaneously without coordination. Clearly, they like to be together: going separate ways is not a Nash equilibrium. But there are two Nash equilibria: both go to the boxing match or both go to the opera. Unlike Jack and Jill’s game, one Nash equilibrium dominates the other. So as a refinement of Nash equilibrium, we expect Fred and Ethel to go to the Opera. If multiple Nash equilibria cannot be ranked, then there’s no telling what might happen.

Mixed Strategies:

Must there be at least one Nash equilibrium? Yes, but we might not find it in pure strategies, which is what we’ve analyzed so far. More generally, players can randomize their strategies. These are called mixed strategies. Assume we find no Nash equilibrium. We then need to check best responses that take the form of a probability associated with each strategy. This is best implicit in the context of sports competition.

Consider a simplified version of the game played by a pitcher and a batter in baseball. The pitcher has two strategies concerning the pitch he will throw: Fastball or Curve. The batter’s two strategies are: Rush or Wait. If the batter guesses wrong, he’s sure to make an out; if he correctly guesses curve ball, he’s sure to get a hit; if he correctly guesses fastball, however, his probability of getting a hit is p therefore the game matrix is:

786_mixed stratagies.jpg

For this zero-sum game. One can authenticate that there is no Nash equilibrium in pure strategies. For instance, if the pitcher chose to throw a curve, the batter would wait, which implies that the pitcher wouldn’t want to throw a curve.

To find the Nash equilibrium in mixed strategies, we also assume that each player is an expected utility-maximizer. If the pitcher throws a fastball, his expected payoff would be πR (1− p) + (1− πR) ⋅1, where πR is the probability the batter rushes. Similarly, if the pitcher throws a curve, his expected payoff would be π R . Therefore, the pitcher’s best response function is the solution to:

max πfπf [πr (1− p) +(1− πr)] + (1− πf) πr

given πr. The solution to this problem is

π* r= 1/(1+P) ............(v)

That is, to make the pitcher indifferent between throwing fastballs and curve balls, the batter’s probability of rushing must be π* r

We must also solve the batter’s problem. The batter chooses π r to maximize his expected utility given the pitcher’s mixed strategy.

The solution is:

π*f = 1/(1+P)  ...........(vi)

If p = 1/2, fastballs would be thrown and anticipated two-third of the time. Notice, the better the batter is at hitting an anticipated fastball, the less likely he’ll see one. What’s the batter’s usual batting average? (p/(1+P)) . Do better batters have higher equilibrium batting averages?

(Yes, but the pitcher’s response attenuates the effect of skill on batting average.)

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