# 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

D. Try again. If Bird 1 chooses too High a mixing probability for Hawk, then the payoffs for Bird 2 would be greater for Dove than for Hawk. Thus, Bird 2 would not be willing to play a mixed strategy-- he would play pure Dove-- and we would not end up with the mixed-strategy equilibrium we were trying to calculate. (Bird 1 would not want to mix either, if Bird 2 did not.)

If, for example, Bird 1 chose to play Hawk with probability .9, then Bird 2 would always play Dove:

Payoff of Bird 2 (Hawk) =.9 (-2) + (1-.9) (4) = -1.4 < Payoff of Bird 2 (Dove) =.9(0) + (1-.9) (1) = .1.

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).

If this led to a pure-strategy equilibrium with Bird 1 always choosing Hawk and Bird 2 choosing Dove, that would be fine with Bird 1. An asymmetric equilibrium in pure strategies like that, however, is not what we are looking for. See if you can find the symmetric equilibrium, in which both birds choose to mix.