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.
|Len Guienie||Up||10, 0||0, 10||3, 3|
|Middle||2, 10||10, 2||6, 4|
|Down||3, 3||4, 6||6, 6|
1. What is the payoff to Al when he plays Left and Len plays Up? That is find uAl(UpLen, LeftAl) = 0.
2. What is the payoff to Len for uLen(MiddleLen, RightAl)? 6
Let si (*, *, *) denote the probabilities that the ith player plays her available strategies. For example, sLen(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 uAl(sLen, C) when sLen = (1/3, 2/3 , 0)? uAl(sLen,C) = 10(1/3)+2(2/3)+6(0) = 14/3.
4. What is the payoff for Len for uLen((1/3, 2/3, 0),C)? uLen((1/3, 2/3, 0),C) = 0(1/3)+10(2/3)+4(0) = 20/3.
5. Find the payoff uAl(sLen,sAl), where sLen = (1/2, 1/2, 0) and sAl = (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.
uAl(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 SLen(p1, p2, 1-p1-p2) and Al is playing SAl(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.