# 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_2 What is the mixed-strategy Nash equilibrium probability of Hawk for Bird 2?
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. The equilibrium probability of Hawk for Bird 2 is .6, just as it was for Bird 1.

The easy way to get to this answer is to realize that this game is completely symmetric. Bird 1 and Bird 2 have the same decisions to make, and face the same payoffs, each from its own point of view. If Bird 1 is the Dove and Bird 2 is the Hawk, Bird 1 gets 0 and Bird 2 gets 4. If Bird 2 is the Dove and Bird 1 is the Hawk, Bird 2 gets 0 and Bird 1 gets 4. Thus, they have the same incentives, unlike in Games 1, 2, and 3.

It's a good idea to confirm guesses though, which we will now do, having guessed that the equilibrium probability of Hawk for Bird 2 is .6. Does this equate the payoffs from the pure strategies?

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

Thus, it seems that .6 is indeed an equilibrium probability for Bird 2. If both birds pick a .6 probability of being a Hawk, then neither one can gain by changing his probability.

In the Hawk-Dove Game I did in class, the payoffs were 2, not 4, from being a Hawk encountering a Dove. That is why the equilibrium probability of playing Hawk has now risen from 1/3 to .6.