# An Answer for the Self-test on Mixed Strategies

GAME 4: HAWK-DOVE
BIRD 2
Hawk         Dove
Hawk           -2, -2        4,0
BIRD 1
Dove              0,4        1,1

This is like the Hawk-Dove Game we looked at in class, except I have changed the numbers so that while the ordinal rankings are the same, Hawks do much better when they meet Doves. (I changed 0,2 to 0,4; and 2,0, to 4,0).

4_1 What is the mixed-strategy Nash equilibrium probability of Hawk for Bird 1?
A. Between 0 and .2, inclusive.
B. Greater than .2 but less than .5.
C. Between .5 and .7, inclusive
D. Greater than .7

C. CORRECT. If a bird chooses a mixing probability of .6 for Hawk, then the payoffs for the other bird are equal for Hawk and for Dove.

Payoff of Bird 2 (Hawk) =.6 (-2) + (1-.6) (4) = .4= Payoff of Bird 2 (Dove) =6 (0) + (1-.6) (1) = .4.

To find the Nash equilibrium, find a mixing probability X that equates the payoffs from the two pure strategies.

Payoff of Bird 2 (Hawk) =X (-2) + (1-X) (4) = Payoff of Bird 2 (Dove) =X (0) + (1-X) (1).

This solves out to -2X + 4 -4X = 1 -X, so 3 = 5X, and X = .6.

This game does have two pure strategy equilibria-- with the payoffs (4,0) and (0,4). Since these are asymmetric, however, the mixed-strategy equilibrium seems more realistic for a meeting between strangers. (If the two birds know each other, though, it seems reasonable that one might always get the advantage just out of tradition-- that is part of what a ``pecking order'' is all about).

Also, in the population interpretation of mixed strategies, asymmetric equilibria don't make sense. The population interpretation says that if 60 percent of birds are pure Hawks, and 40 percent are Doves, then expected payoffs are equal for each type of bird, and neither population will increase or decrease.