Economics 2020b Lecture 3:
Mixed-Strategy Nash Equilibrium
Consider the game
L
R
T
1,0
0,3
B
0,1
2,0
Q. What are the pure-strategy Nash equilibria?
The Mixed Extension of a Strategic Game:
- The set of strategies is expanded to permit not only definite
(“pure”) choices but also probabilistic (“mixed”) choices.
- The payoff functions are extended to indicate each player’s
payoff given a choice of a mixed strategy by all the players.
- Specifically, the player receives the expected payoff given
the joint randomization by all the players, where the
randomizations by different players are assumed to be
independent of each other.
(What is the rationale?)
Note: In order for the expected payoffs to capture the players’
preferences, it is essential that the payoffs be expressed as von
Neumann Morgenstern (Bernoulli) utilities.
.5
.5
T
$10M
0
B
$4M
$4M
.5
T
B
.5
1-r
r
t
1,0
0,3
1-t
0,1
2,0
Strategy sets:
Player 1: { (t,1-t) s.t. 0 ≤ t ≤1 }
Player 2: { (r,1-r) s.t. 0 ≤ r ≤1}
Payoffs:
To P1: t(1-r)*1 + tr*0 + (1-t)(1-r)*0 + (1-t)r*2 = 2+t-3tr
To P2: t(1-r)*0 + tr*3 + (1-t)(1-r)*1 + (1-t)r*0 = 1-t-r+4tr
Note: The payoffs are linear in each strategy separately, but
not in both strategies jointly.
Claim: The pair of (mixed) strategies below constitute a Nash
equilibrium:
2/3
1/3
1/4
1,0
0,3
3/4
0,1
2,0
Check
Assume P2 plays r=1/3
Payoff to T: 2/3
Payoff to B: 2/3
Payoff to any strategy: 2/3 (Why?)
Therefore, every strategy is a best response; in particular, t=1/4.
Assume P1 plays t=1/4
Payoff to L: 3/4
Payoff to R: 3/4
Payoff to any strategy: 3/4
Therefore, every strategy is a best response; in particular, r=1/3.
Graphical Analysis
Consider the best-response graphs:
As r goes from 0 to 1, draw the best response, t.
As to goes from 0 to 1, draw the best response, r.
Intersection?
1-r
r
t
1,
0,
1-t
0,
2,
Best response of Player 1:
(Given r, what is the best t?)
t
1
0
.33
1
r
1-r
r
t
,0
,3
1-t
,1
,0
Best Response of Player 2:
(Given t, what is the best r?)
t
1
.25
0
.33
1
r
1-r
r
t
1, 0
0, 3
1-t
0, 1
2, 0
Nash Equilibrium
t
1
.25
0
.33
1
r
Indifference Principle:
In a Nash equilibrium, all strategies that are assigned positive
probability must have the same expected payoff.
Proof:
In the example:
1-r
r
t
1,0
0,3
1-t
0,1
2,0
t > 0 and 1-t >0
1-r = 2r
r = 1/3
r > 0 and 1-t >0
1-t = 3t
t = 1/4
Consider the one-person maximization
Weight
x1
x2
x3
x4
x5
x6
x7
Strategy
1
2
3
4
5
6
7
Payoff
5.1
7.3
7.3
2.3
7.1
7.3
5.8
Q. What are the x that maximize the weighted average?
Theorem:
A necessary and sufficient condition for Nash Equilibrium:
Given the strategies of the other players
- The payoff is the same for all pure strategies with positive
weight.
- The payoff is
Proof:
Rationales for mixed-strategy equilibrium:
- Self-enforcing beliefs
- Frequency in a population
- Intentional randomization
Battle of the Sexes
Football
Concert
Football
2,1
0,0
Concert
0,0
1,2
Mixed-strategy Nash equilibrium:
M:
2/3F; 1/3 C
W:
1/3F; 2/3 C
(What do they believe?)
Q. What is the MSNE if the diagonal elements were (4,1) and
(1,4)?
M:
W:
The best-response graph:
Chicken
Swerve
Forward
Swerve
6,6
4,7
Forward
7,4
0,0
Mixed-strategy Nash equilibrium:
.8 Swerve; .2 Forward
(both)
Makes sense as prediction? (Pure-strategy equilibria require
breaking the symmetry…)
Probability of collision:
What if off-diagonal elements were (2,12) and (12,2)?
Swerve;
Forward
Probability of collision:
Stag Hunt
Stag
Hare
Stag
5,5
0,3
Hare
3,0
3,3
MSNE:
Stag:
Hare:
Interpretation vs the two states of society:
What if the value of a stag were 10 rather than 5?
Pollution Game
II
II
C
N
C
N
C
2,2,2
2,3,2
C
2,2,3
-1,0,0
N
3,2,2
0,0,-1
N
0,-1,0
0,0,0
I
I
N
C
III
Pure-strategy NE:
Two clean; one doesn’t.
(3 equilibria)
No one cleans (1 equilibrium)
Two pure one mixed:
Assume III mixed
Then III’s payoff has to be the same in a corresponding
entry on the left as on right.
Pollution Game
II
II
C
N
C
N
C
2,2,2
2,3,2
C
2,2,3
-1,0,0
N
3,2,2
0,0,-1
N
0,-1,0
0,0,0
I
I
N
C
III
One pure two mixed:
Assume III pure
(i) III plays C
Then I and II play 2/3C + 1/3N
Check: does III gain by switching to N?
(i) III plays N
Then I and II play 1/3C + 2/3N
Check: does III gain by switching to C?
All mixed:
Assume symmetric. (In fact, can show that it has to
be.)
Let x= prob of clean.
Then
2x^2 + 2x(1-x) + 2(1-x)x – (1-x)^2 = 3x^2
that is
6x^2 – 6x +1 = 0
x= .5 + SQRT(3) /12 = .79 or
x = .5 – SQRT(3) /12 = .21
(2 equilibria)