Mixed Strategy Solutions to a Game

To avoid complicating matters with cute names and "realistic" scenarios we will consider a very simple game between Fred and Clara.  They each have two strategies: Left and Right.

Preliminaries:

In trying to find the solution to any game we begin with several straightforward questions:  We pose these questions keeping in mind the Venn diagram for the relationship between dominant strategies solutions, IEDS solutions and Nash equilibria.

Is there a dominant strategy solution for the game between Fred and Clara?

Are there dominated strategies that can be used to find an IDES solution?

Are there Nash equilibria?  To answer this one we'll check for their best responses.

Fred's best response to a play of L by Clara is to play L, but when she sees him play L she will play Right.  There is no coincidence of conjectured and actual play.

Fred's best response to a play of R by Clara is a play of R, but when she sees him play R she wants to play L, not the R that he conjectured.  Again, there is no coincidence of conjectured and actual play.

The answers to all three questions are NO.

Method 1

Suppose Fred plays the mixed strategy (L,p) and Clara plays the mixed strategy (L,q).

The joint probability distribution over game outcomes is

Fred's expected payoff from the game is

V_F = 3pq+(1-p)q+2p(1-q)+3(1-p)(1-q) 

He wants to choose his p to give him a best response to the mixed strategy played by Clara. That is, he wants to maximize his payoff from the strategy played by Clara.  Differentiate with respect to p, set the result equal to zero, and then solve for q.  The derivative is 

3q+(-1)q+2(1-q)+3(1-q)(-1)

Solving for q 

q = 1/3

Do the same thing for Clara.  Her expected payoff is

V_C = 2pq + 4p(1-q) + 3(1-p)q + 2(1-p)(1-q)

Clara wants to maximize her payoff with respect to the mixed strategy being played by Fred.  Differentiate with respect to q, set the result equal to zero and then solve for p.

2p+4p(-1)+3(1-p)+2(1-p)(-1)

p=1/3

So Fred plays the mixed strategy (L, 1/3) and Clara plays the mixed strategy (L, 1/3).

Method 2

2.1 When Clara is playing (L,q) then Fred earns 3q+2(1-q) from playing left.  He earns q+3(1-q) from playing right.  If he is indifferent between these strategies then 3q+2(1-q)=q+3(1-q).  So Clara should choose q=1/3 in order to leave Fred indifferent between the two pure strategies he could play.

2.2 When Fred is playing (L,p) then Clara earns 2p+3(1-p) by playing left and 4p+2(1-p) by playing right. If she is indifferent between these strategies then 2p+3(1-p) = 4p+2(1-p). So Fred should choose p=1/3.