Finding Mixed-Strategy Nash Equilibria
Consider the following game.
Player
X
L
N
Player Y
L
N
(-5,-5)
(25,0)
(0,15)
(10,10)
Payoffs: (X,Y)
There are no dominant strategies, but two pure-strategy Nash equilibria: (L,N) and (N,L).
There is also a mixed-strategy Nash equilibrium. Suppose a mixed-strategy Nash exists.
Then it must be the case that the expected gain from playing L is equal to the expected
gain from playing N. (This must be the case for each player.)
Let q denote the probability that Y plays L so that (1-q) is the probability that Y plays N.
Thus (q,1-q) is player Y’s mixed strategy. Against this strategy X expects to earn
q (-5) + (1-q) (25) = 25 – 30q
from playing strategy L and
q (0) + (1-q) (10) = 10 – 10q
from playing strategy N. For player X to be willing to randomize her strategies these
expected outcomes need to be equal; that is,
25 – 30q = 10 – 10q.
Thus, 15 = 10q, or q=15/20 = 3/4.
Now follow the same procedure for player Y. Let p be the probability that player X plays
L and 1-p be the probability that player X plays strategy N. In choosing L, player Y’s
expected outcome is
p (-5) + (1-p)(15) = 15 -20p.
By choosing N, player Y’s expected outcome is
p (0) + (1-p)(10) = 10 – 10p.
In order for player Y to be willing to choose a mixed strategy, these expect outcomes
must be equal; that is,
15 – 20p = 10 – 10p.
Thus, 5 = 10p, or p = ½ .
The mixed-strategy profile ((½, ½), (¾, ¼ ) is a mixed-strategy Nash equilibrium.
Note that, like pure-strategy equilibria, mixed-strategy equilibria never use dominated
strategies.
Nash Equilibrium Theorem: Every game with a finite number of players and a finite
number of strategies has at least one Nash equilibrium (in either pure or mixed
strategies).
Example
Player 2
Player
1
A
A
(3,4)
B
(0,0)
B
(1,6)
(3,7)
Payoffs: (Player 1, Player 2)
Note that there are two pure-strategy Nash equilibria. Find the mixed-strategy Nash.
Assuming 2 plays A and B with probabilities q and 1 – q, player 1’s expected payoff
from A is
q (3) + (1-q) (0) = 3q
and 1’s expected payoff from B is
q (1) + (1-q) (3) = 3 – 2q.
Equating these expected payoffs and solving for q yields
3q = 3 – 2q
or
q = 3/5.
Similarly, assuming 1 plays A with probability p and B with probability 1-p, player 2’s
expected payoff from A is
p (4) + (1-p) (6) = 6 – 2p
And 2’s expected payoff from B is
p (0) + (1-p)(7) = 7 – 7p.
Equating these and solving yields
6 – 2p) = 7 – 7p,
5p = 1,
p = 1/5.
The mixed-strategy profile for this game is: ((1/5,4/5), (3/5,2/5)).
Example (my Game 3)
Firm 2
Increase
Price
(200,1000)
(500,800)
Cut Price
(300,900)
(600,600)
Advertise
Payoffs: (Firm 1, Firm 2)
Cut Price
Firm 1
Advertise
(400,1100)
(200,600)
Let (q1, q2, 1 - q1 - q2 ) be the probabilities that firm 2 plays the three strategies. Firm 1’s
expected payoff from cutting price is
q1 (200) + q2 (500) + (1 - q1 – q2) (400) = 400 – 200 q1 + 100 q2 .
Firm 1’s expected payoff from advertising is
q1 (300) + q2 (600) + (1 - q1 – q2) (200) = 200 + 100 q1 + 400 q2.
Set these equal and solve for q1.
400 – 200q1 + 100 2 = 200 + 100q1 + 400q2
or
200 = 300q1 + 300q2
or
2 = 3 q1 + 3q2 .
Notice that firm 2’s strategy, “increase price,” is dominated in the sense that it would
never be chosen by firm 2. Thus we can assign a probability of zero to this strategy (i.e.,
q2 = 0). Thus, 2 = 3 q1 and q1 = 2/3.
Let p be the probability that firm 1 chooses “decrease price,” so that 1 – p is the
probability that firm 1 chooses “advertise.” Firm 2’s expected payoff from “cut price” is
p (1000) + (1-p)(900) = 900 + 100p,
and firm 2’s expected payoff from “advertise” is
p (1100) + (1-p)(600) = 600 + 500p .
Equate these and solve.
900 + 100p = 600 + 500p
300 = 400p
p=3/4.
The mixed-strategy profile is ((¾, ¼), (2/3, 0, 1/3)).
Player
1
Example
U
C
D
L
(8,3)
(3,3)
(5,2)
Player 2
M
R
(3,5) (6,3)
(5,5) (4,8)
(3,7) (4,9)
Payoffs: (Player 1, Player 2)
Note:
 There is no solution in pure strategies.
 Strategy “D” is dominated for player 1.
 Strategy “L” is dominated for player 2.
Let q be the probability that 2 chooses “M” and 1 – q be the probability that 2 chooses
“R.” In this case the probability that 2 chooses “L” is zero (because “L” can be
eliminated by domination). Player 1’s expected payoff from choosing U is
q (3) + (1 – q)(6) = 6- 3q,
and player 1’s expected payoff from choosing “C” is
q (5) + (1-q)(4) = 4 + q.
Solving,
6- 3q = 4 + q
2 = 4q
q=1/2.
Player 2’s expected payoff from “M” is
p (5) + (1-p)(5) = 5.
Player 2’s expected payoff from “R” is
p (3) + (1-p)(8) = -5p + 8.
Equating these and solving,
5 = 8 – 5p
5p = 3
p = 3/5.
Solution: ((3/5, 2/5, 0), (0, ½, ½))
Example (my game 4)
Firm 2
Cut Price
Increase
Price
(300,1100)
(600,800)
Cut Price
(200,900)
(500,600)
Advertise
Payoffs: (Firm 1, Firm 2)
Firm 1
Advertise
(400,1200)
(500,600)
Note that “increase price” is dominated for firm 2. Let q denote the probability that firm 2
chooses “decrease price” and 1 – q the probability that firm 2 chooses “advertise.” Firm
1’s expected profit from “decrease price”:
q (300) + (1 – q)(400) = 400 – 100q
Firm 1’s expected profit from “advertise”:
q (200) + (1 – q)(500) = 500 – 300q
Solve as before:
400 – 100q = 500 – 300q
100 = 200q
q=½.
Similarly, firm 2’s expected payoff for “decrease price” is
p (1100) + (1 – p) (900) = 900 + 200p
and the expected payoff for “advertise” is
p (1200) + (1 – p) (600) = 600 + 600p.
Thus,
900 + 200p = 600 + 600p
300 = 400p
p = ¾.
Solution:
(( ¾ , ¼ ), (½ , 0, ½))
General example
Player
1
A
Player 2
A
B
(a,b)
(c,d)
B
(e,f)
(g,h)
Payoffs: (Player 1, Player 2)
Assuming 2 plays A and B with probabilities q and 1 – q, player 1’s expected payoff
from A is
q (a) + (1-q) (c) = (a - c) q + c
and 1’s expected payoff from B is
q (e) + (1-q) (g) = (e – g) q + g.
Equating these expected payoffs and solving for q yields
(a - c) q + c = (e – g) q + g
or
q = (g –c)/(a – c – e +g).
Similarly, assuming 1 plays A with probability p and B with probability 1-p, player 2’s
expected payoff from A is
p (b) + (1-p)(f) = (b - f) p + f
And 2’s expected payoff from B is
p (d) + (1 - p)(h) = (d – h) p + h.
Equating these and solving yields
(b - f) p + f = (d – h) p + h,
p = (h – f)/(b – f – d + h).
The mixed-strategy profile for this game is:
([(g–c)/(a – c – e +g), (a–e)/(a – c – e +g)], [(h–f)/(b – f – d + h), (b– d)/(b – f – d + h)])).
Note that player one’s probabilities depend only on player two’s payoffs and player 2’s
probabilities depend only on player one’s payoffs.