Mixed Strategies: Len Guienie and Al Dente Battle for the Spaghetti Market

Len and Al are locked in a strategic battle for the pasta market in Philadelphia.  The game that confronts them is presented in normal form below. 

1. What is the payoff to Al when he plays Left and Len plays Up?  That is find u_Al(Up_Len, Left_Al) = 0.

2. What is the payoff to Len for u_Len(Middle_Len, Right_Al)? 6

Let s_i (*, *, *) denote the probabilities that the i^th player plays her available strategies.  For example, s_Len(1/6, 4/6, 1/6) means that Len plays U with probability 1/6, he plays M with probability 4/6, and plays D with probability 1/6.

3. What is the payoff for Al for u_Al(s_Len, C) when s_Len = (1/3, 2/3 , 0)? u_Al(s_Len,C) = 10(1/3)+2(2/3)+6(0) = 14/3.

4. What is the payoff for Len for u_Len((1/3, 2/3, 0),C)? u_Len((1/3, 2/3, 0),C) = 0(1/3)+10(2/3)+4(0) = 20/3.

5. Find the payoff u_Al(s_Len,s_Al), where s_Len = (1/2, 1/2, 0) and s_Al = (1/4, 1/4, 1/2).  Let's first figure out the joint probability distribution for possible moves made by Al and Len.

Now we can compute the expected payoff for Al.

u_Al(s₁,s₂) = 1/8(0)+1/8(10)+0(3) + 1/8(10)+1/8(2)+0(6) + 1/4(3)+1/4(4)+0(6) = 36/8.

6. What mixed strategy should Len play? Suppose Len plays Up with probability p₁ and M with probability p₂.  He has to choose these probabilities so that Al is indifferent between any of his own pure strategies.

p₁(0)+p₂(10)+(1-p₁-p₂)(3) = p₁(10)+p₂(2)+(1-p₁-p₂)(6)

p₁(0)+p₂(10)+(1-p₁-p₂)(3) = p₁(3)+p₂(4)+(1-p₁-p₂)(6)

If you solve the two equations for the two unknowns then p₁=2/21 (.095) and p₂=1/3 (.333).

7. Let's see if we can find out where the argument that Len chooses p₁ and p₂ so that Al is indifferent between his pure strategies 'comes from.' First, we'll construct the joint distribution over outcomes when they are both playing mixed strategies. Len is playing S_Len(p₁, p₂, 1-p₁-p₂) and Al is playing S_Al(q₁, q₂, 1-q₁-q₂)

Second, compute Len's expected payoff when they are both playing mixed strategies.

10p₁q₁+2p₂q₁+3(1-p₁-p₂)q₁ + 0p₁q₂+10p₂q₂+4(1-p₁-p₂)q₂+3p₁(1-q₁-q₂)+6p₂(1-q₁-q₂) + 6(1-p₁-p₂)(1-q₁-q₂)

Len's task is to choose p₁ and p₂ as to maximize his expected payoff.  So, the third step is to take the partial derivatives of his expected payoff with respect to p₁ and p₂ and set the results equal to zero.

10q₁-q₂-3 = 0

-q₁+6q₂ = 0

Note that although we set the problem up as though Len were choosing his p's, in the first order conditions we end up solving for Al's q's in a set of equations in which the payoffs belong to Al!

So q₂ = 3/59 = .051 and q₁ = 6(3/59) = .305 are the probabilities that ought to be played by Al when Len is choosing his own mixed strategy.

Just to be sure we understand what happened let's do it again, but for Al this time. First, write down Al's expected payoff when they are both playing a mixed strategy.

0p₁q₁+10p₂q₁+3(1-p₁-p₂)q₁ + 10p₁q₂+2p₂q₂+6(1-p₁-p₂)q₂+3p₁(1-q₁-q₂)+4p₂(1-q₁-q₂) + 6(1-p₁-p₂)(1-q₁-q₂)

Second, find the partial derivatives with respect to q₁ and q₂, and set the results equal to zero.

9p₂-3 = 0

7p₁-2p₂ = 0

Solving the two equations we have p₁ = 2/21 and p₂ = 1/3.

8. These mixed strategies result in a Nash equilibrium since they represent a best response to the mixed strategy being used by the opponent.