lecture10Mixed Strategy Nash Equilibrium(博弈论,Carnegie Mellon University)
Static (or Simultaneous-Move) Games of Complete Information Mixed Strategy Nash Equilibrium
73-347 Game Theory--Lecture 10
Outline of Static Games of Complete Information
Introduction to games
Normal-form (or strategic-form) representation
Iterated elimination of strictly dominated strategies Nash equilibrium
Review of concave functions, optimization
Applications of Nash equilibrium
Mixed strategy Nash equilibrium
73-347 Game Theory--Lecture 10
Today’s Agenda
Review of previous class
Example: Rock, paper and scissors (Exercise 1391>.1 of Osborne) How to find mixed strategy Nash equilibria in a 2-player game each
with a finite number of strategies
73-347 Game Theory--Lecture 10
Example: Rock, paper and scissors
Each of two players simultaneously announces either Rock, or Paper,
or Scissors.
Paper beats (wraps) rock
Rock beats (blunts) scissors
Scissors beats (cuts) paper
The player who names the winning object receives $1 from her opponent
If both players name the same choice then no payment is made
73-347 Game Theory--Lecture 10 Example: Rock, paper and scissors Can you guess a mixed strategy Nash equilibrium?
Player 2
0 , 0
-1 , 1
1 , -1
Scissors
Paper
1 , -1
-1 , 1
Scissors
0 , 0
1 , -1
Paper
Player 1
-1 , 1
0 , 0
Rock
Rock
73-347 Game Theory--Lecture 10 Mixed strategy Nash equilibrium: 2-player each with two pure
strategies
Mixed strategy Nash equilibrium:
A pair of mixed strategies ((r*, 1-r*), (q*, 1-q*)) is a Nash equilibrium if (r*,1-r*) is a best response to (q*, 1-q*), and (q*, 1-q*) is a best response to (r*,1-r*). That is, v1((r*, 1-r*), (q*, 1-q*)) ?? v1((r, 1-r), (q*, 1-q*)), for all 0?? r ??1 v2((r*, 1-r*), (q*, 1-q*)) ?? v2((r*, 1-r*), (q, 1-q)), for all 0?? q ??1
Player 2
s22 ( 1- q )
u1(s12, s22), u2(s12, s22)
u1(s12, s21), u2(s12, s21)
s12 (1- r )
Player 1
u1(s11, s22), u2(s11, s22)
u1(s11, s21), u2(s11, s21)
s11 ( r )
s21 ( q )
73-347 Game Theory--Lecture 10
2-player each with two strategies
Theorem 1 (property of mixed Nash equilibrium)
pair of mixed strategies ((r*, 1-r*), (q*, 1-q*)) is a Nash A
equilibrium if and only if v1((r*, 1-r*), (q*, 1-q*)) ?? EU1(s11, (q*, 1-q*)) v1((r*, 1-r*), (q*, 1-q*)) ?? EU1(s12, (q*, 1-q*)) v2((r*, 1-r*), (q*, 1-q*)) ?? EU2(s21, (r*, 1-r*)) v2((r*, 1-r*), (q*, 1-q*)) ?? EU2(s22, (r*, 1-r*))
Player 2
s22 ( 1- q )
u1(s12, s22), u2(s12, s22)
u1(s12, s21), u2(s12, s21)
s12 (1- r )
Player 1
u1(s11, s22), u2(s11, s22)
u1(s11, s21), u2(s11, s21)
s11 ( r )
s21 ( q )
73-347 Game Theory--Lecture 10
Mixed strategy equilibrium: 2-player each with two strategies Theorem 2 Let ((r*, 1-r*), (q*, 1-q*)) be a pair of mixed strategies, where 0 <r*<1, 0<q*<1. Then ((r*, 1-r*), (q*, 1-q*)) is a mixed strategy Nash equilibrium if and only if EU1(s11, (q*, 1-q*)) = EU1(s12, (q*, 1-q*)) EU2(s21, (r*, 1-r*)) =
-r*))EU2(s22, (r*, 1
That is, each player is indifferent between her two pure strategies given the mixed strategy chosen by the other player.
Player 2
s22 ( 1- q )
u1(s12, s22), u2(s12, s22)
u1(s12, s21), u2(s12, s21)
s12 (1- r )
Player 1
u1(s11, s22), u2(s11, s22)
u1(s11, s21), u2(s11, s21)
s11 ( r )
s21 ( q )
73-347 Game Theory--Lecture 10 -player each with a finite number of pure strategies 2
Set of players: {Player 1, Player 2} Sets of strategies: player 1: S1= { s11, s12, ..., s1J } player 2:
S2= { s21, s22, ..., s2K }
Payoff functions: player 1: u1(s1j, s2k) player 2: u2(s1j, s2k) for
j = 1, 2, ..., J and k = 1, 2, ..., K 73-347 Game Theory--Lecture 10 2-player each with a finite number of pure strategies
Player 1’s mixed strategy: p1=(p11, p12, ..., p1J )
Player 2’s mixed strategy: p2=(p21, p22, ..., p2K )
Player 2
u2(s1J, s22)
u1(s1J, s22)
.......
u2(s12, s22)
u1(s12, s22)
u2(s11, s22)
u1(s11, s22)
s22 (p22)
u2(s1J, s21)
u1(s1J, s21)
......
u2(s12, s21) u1(s12, s21) u2(s11, s21) u1(s11, s21) s21 (p21)
......
.......
....
u2(s12, s2K) u1(s12, s2K) .......
s12 (p12)
s2K (p2K) u2(s1J, s2K) u1(s1J, s2K) ......
s1J (p1J)
u2(s11, s2K) u1(s11, s2K) .......
s11 (p11)
.......
Player 1
73-347 Game Theory--Lecture 10
Expected payoffs: 2-player each with a finite number of pure strategies
Player 1’s expected payoff of pure strategy s11: EU1(s11, p2)=p21×u1(s11, s21)+p22×u1(s11, s22)+...+p2k×u1(s11, s2k)+...+p2K×u1(s11, s2K)
Player 1’s expected payoff of pure strategy s12: EU1(s12, p2)=p21×u1(s12, s21)+p22×u1(s12, s22)+...+p2k×u1(s12, s2k)+...+p2K×u1(s12, s2K)
.........
Player 1’s expected payoff of pure strategy s1J: EU1(s1J, p2)=p21×u1(s1J, s21)+p22×u1(s1J, s22)+...+p2k×u1(s1J, s2k)+...+p2K×u1(s1J, s2K)
Player 1’s expected payoff from her mixed strategy p1: v1(p1, p2)=p11??EU1(s11, p2)+p12??EU1(s12, p2)+...+p1j??EU1(s1j, p2)+...
+p1J??EU1(s1J, p2)
73-347 Game Theory--Lecture 10
Expected payoffs: 2-player each with a finite number of pure strategies
Player 2’s expected payoff of pure strategy s21: EU2(s21, p1)=p11×u2(s11, s21)+p12×u2(s12, s21)+...+p1j×u2(s1j, s21)+...+p1J×u2(s1J, s21)
Player 2’s expected payoff of pure strategy s22: EU2(s22,
p1)=p11×u2(s11, s22)+p12×u2(s12, s22)+...+p1j×u2(s1j,
s22)+...+p1J×u2(s1J, s22)
...........
Player 2’s expected payoff of pure strategy s2K: EU2(s2K, p1)=p11×u2(s11, s2K)+p12×u2(s12, s2K)+...+p1j×u2(s1j,
s2K)+...+p1J×u2(s1J, s2K)
Player 2’s expected payoff from her mixed strategy p2: v2(p1,
p2)=p21??EU2(s21, p1)+p22??EU2(s22, p1) +...+p2k??EU2(s2k, p1)+....
+p2K??EU2(s2K, p1)
73-347 Game Theory--Lecture 10
Mixed strategy Nash equilibrium: 2-player each with a finite number of pure strategies
A pair of mixed strategies (p1*, p2*), where p1*=(p11*, p12*, ..., p1J* ) p2*=(p21*, p22*, ..., p2K* ) is a mixed strategy Nash equilibrium if player 1’s mixed strategy p1* is a best response to player 2’s mixed strategy p2*, and p2* is a
best response to p1*.
Or, v1(p1*, p2*)?? v1(p1, p2*), for all player 1’s mixed strategy p1, and v2(p1*, p2*) ?? v2(p1*, p2), for all player 2’s mixed strategy p2.
That is, given player 2’s mixed strategy p2*, player 1 cannot be
better off if she deviates from p1*. Given player 1’s mixed strategy p1*, player 2 cannot be better off if she deviates from p2*.
73-347 Game Theory--Lecture 10
2-player each with a finite number of pure strategies
Theorem 3 (property of mixed Nash equilibrium)
A pair of mixed strategies (p1*, p2*), where p1*=(p11*, p12*, ..., p1J* ) p2*=(p21*, p22*, ..., p2K* ) is a mixed strategy Nash equilibrium if and only if v1(p1*, p2*) ?? EU1(s1j, p2*), for j = 1, 2, ..., J and v2(p1*, p2*) ?? EU2(s2k, p1*), for k= 1, 2, ..., K
73-347 Game Theory--Lecture 10
2-player each with a finite number of pure strategies
Theorem 4 A pair of mixed strategies (p1*, p2*), where p1*=(p11*, p12*, ..., p1J* ) p2*=(p21*, p22*, ..., p2K* ) is a mixed strategy Nash equilibrium if and only if they satisfies the following conditions:
player 1: for any m and n, if p1m*>0 and p1n*>0 then EU1(s1m, p2*) = EU1(s1n, p2*); if p1m*>0 and p1n*=0 then EU1(s1m, p2*) ?? EU1(s1n, p2*)
player 2: for any i and k, if p2i*>0 and p2k*>0 then EU2(s2i, p1*) = EU2(s2k, p1*); if p2i*>0 and p2k*=0 then EU2(s2i, p1*) ?? EU2(s2k, p1*)
73-347 Game Theory--Lecture 10
2-player each with a finite number of pure strategies
What does Theorem 4 tell us?
A pair of mixed strategies (p1*, p2*), where p1*=(p11*, p12*, ..., p1J* ), p2*=(p21*, p22*, ..., p2K* ) is a mixed strategy Nash
equilibrium if and only if they satisfies the following conditions:
Given player 2’s p2*, player 1’s expected payoff of every pure strategy to which she assigns positive probability is the same, and player 1’s expected payoff of any pure strategy to which she assigns positive probability is not less than the expected payoff of any pure strategy to which she assigns zero probability.
Given player 1’s p1*, player 2’s expected payoff of every pure strategy to which she assigns positive probability is the same, and player 2’s expected payoff of any pure strategy to which she assigns positive
probability is not less than the expected payoff of any pure strategy to which she assigns zero probability.
73-347 Game Theory--Lecture 10
2-player each with a finite number of pure strategies
Theorem 4 implies that we have a mixed strategy Nash equilibrium in the following situation
Given player 2’s mixed strategy, Player 1 is indifferent among the pure strategies to which she assigns positive probabilities. The expected payoff of any pure strategy she assigns positive probability is not less than the expected payoff of any pure strategy she assigns zero probability.
Given player 1’s mixed strategy, Player 2 is indifferent among the pure strategies to which she assigns positive probabilities. The expected payoff of any pure strategy she assigns positive probability is not less than the expected payoff of any pure strategy she assigns zero
probability.
73-347 Game Theory--Lecture 10
Theorem 4: illustration
Check whether ((3/4, 0, 1/4), (0, 1/3, 2/3)) is a mixed strategy Nash
equilibrium
Player 1:
EU1(T, p2) = 0??0+3??(1/3)+1??(2/3)=5/3, EU1(M, p2) =
4??0+0??(1/3)+2??(2/3)=4/3 EU1(B, p2) = 3??0+5??(1/3)+0??(2/3)=5/3.
Hence, EU1(T, p2) = EU1(B, p2) > EU1(M, p2)
Player 2
0 , 7
2 , 3
1 , 1
R (2/3)
C (1/3)
5 , 1
3 , 4
B (1/4)
0 , 4
4 , 0
M (0)
Player 1
3 , 3
0 , 2
T (3/4)
L (0)
73-347 Game Theory--Lecture 10 Theorem 4: illustration
Player 2:
EU2(L, p1)=2??(3/4) + 0??0 + 4??(1/4)=5/2, EU2(C, p1)=3??(3/4) +
4??0 + 1??(1/4)=5/2, EU2(R, p1)=1??(3/4) + 3??0 + 7??(1/4)=5/2.
Hence, EU2(C, p1)=EU2(R, p1)??EU2(L, p1) Therefore, ((3/4, 0, 1/4), (0, 1/3, 2/3)) is a mixed strategy Nash
equilibrium by Theorem 4.
Player 2
0 , 7
2 , 3
1 , 1
R (2/3)
C (1/3)
5 , 1
3 , 4
B (1/4)
0 , 4
4 , 0
M (0)
Player 1
3 , 3
0 , 2
T (3/4)
L (0)
73-347 Game Theory--Lecture 10 Example: Rock, paper and scissors Check whether there is a mixed strategy Nash equilibrium in which
p11>0, p12>0, p13>0, p21>0, p22>0, p23>0.
Player 2
0 , 0
-1 , 1
1 , -1
Scissors (p23)
Paper (p22)
1 , -1
-1 , 1
Scissors (p13)
0 , 0
1 , -1
Paper (p12)
Player 1
-1 , 1
0 , 0
Rock (p11)
Rock (p21)
73-347 Game Theory--Lecture 10 Example: Rock, paper and scissors If each player assigns positive probability to every of her pure
strategy, then by Theorem 4, each player is indifferent among her three
pure strategies.
Player 2
0 , 0
1 , 1 -
1 , -1
Scissors (p23)
Paper (p22)
1 , -1
-1 , 1
Scissors (p13)
0 , 0
1 , -1
Paper (p12)
Player 1
-1 , 1
0 , 0
Rock (p11)
Rock (p21)
73-347 Game Theory--Lecture 10
Example: Rock, paper and scissors
Player 1 is indifferent among her three pure strategies: EU1(Rock,
p2) = 0??p21+(-1)?? p22+1?? p23 EU1(Paper, p2) = 1?? p21+0?? p22+(-1)??
p23 EU1(Scissors, p2) = (-1)?? p21+1?? p22+0?? p23 EU1(Rock, p2)= EU1(Paper, p2)= EU1(Scissors, p2)
Together with p21+ p22+ p23=1, we have three equations and three
unknowns.
Player 2
0 , 0
1 , 1 -
1 , -1
Scissors (p23)
Paper (p22)
1 , -1
-1 , 1
Scissors (p13)
0 , 0
1 , -1
Paper (p12)
Player 1
-1 , 1
0 , 0
Rock (p11)
Rock (p21)
73-347 Game Theory--Lecture 10 Example: Rock, paper and scissors 0??p21+(-1)?? p22+1?? p23= 1?? p21+0?? p22+(-1)?? p23 0??p21+(-1)??
p22+1?? p23 = (-1)?? p21+1?? p22+0?? p23 p21+ p22+ p23=1
The solution is p21= p22= p23=1/3
Player 2
0 , 0
-1 , 1
1 , -1
Scissors (p23)
Paper (p22)
1 , -1
-1 , 1
Scissors (p13)
0 , 0
1 , -1
Paper (p12)
Player 1
-1 , 1
0 , 0
Rock (p11)
Rock (p21)
73-347 Game Theory--Lecture 10
Example: Rock, paper and scissors
Player 2 is indifferent among her three pure strategies: EU2(Rock,
p1)=0??p11+(-1)?? p12+1?? p13 EU2(Paper, p1)=1?? p11+0?? p12+(-1)?? p13
EU2(Scissors, p1)=(-1)?? p11+1?? p12+0?? p13 EU2(Rock, p1)= EU2(Paper, p1)=EU2(Scissors, p1) Together with p11+ p12+ p13=1, we have three equations and three
unknowns.
Player 2
0 , 0
-1 , 1
1 , -1
Scissors (p23)
Paper (p22)
1 , -1
-1 , 1
Scissors (p13)
0 , 0
1 , -1
Paper (p12)
Player 1
-1 , 1
0 , 0
Rock (p11)
Rock (p21)
73-347 Game Theory--Lecture 10 Example: Rock, paper and scissors 0??p11+(-1)?? p12+1?? p13=1?? p11+0?? p12+(-1)?? p13 0??p11+(-1)??
p12+1?? p13=(-1)?? p11+1?? p12+0?? p13 p11+ p12+ p13=1
The solution is p11= p12= p13=1/3 Player 2
0 , 0
-1 , 1
1 , -1
Scissors (p23)
Paper (p22)
1 , -1
-1 , 1
Scissors (p13)
0 , 0
1 , -1
Paper (p12)
Player 1
-1 , 1
0 , 0
Rock (p11)
Rock (p21)
73-347 Game Theory--Lecture 10
Example: Rock, paper and scissors
Player 1: EU1(Rock, p2) = 0??(1/3)+(-1)??(1/3)+1??(1/3)=0 EU1(Paper, p2) = 1??(1/3)+0??(1/3)+(-1)??(1/3)=0 EU1(Scissors, p2) = (-1)??(1/3)+1??(1/3)+0??(1/3)=0
Player 2: EU2(Rock, p1)=0??(1/3)+(-1)??(1/3)+1??(1/3)=0 EU2(Paper,
p1)=1??(1/3)+0??(1/3)+(-1)??(1/3)=0 EU2(Scissors, p1)=(-1)??(1/3)+1??(1/3)+0??(1/3)=0
Therefore, (p1=(1/3, 1/3, 1/3), p2=(1/3, 1/3, 1/3)) is a mixed
strategy Nash equilibrium by Theorem 4.
Player 2
0 , 0
-1 , 1
1 , -1
Scissors (1/3)
Paper (1/3)
1 , -1
-1 , 1
Scissors (1/3)
0 , 0
1 , -1
Paper (1/3)
Player 1
-1 , 1
0 , 0
Rock (1/3)
Rock (1/3)
73-347 Game Theory--Lecture 10
Example: Rock, paper and scissors
Check whether there is a mixed strategy Nash equilibrium in which one
of p11, p12, p13 is positive, and at least two of p21, p22, p23 are
positive.
The answer is No.
Player 2
0 , 0
-1 , 1
1 , -1
Scissors (p23)
Paper (p22)
1 , -1
-1 , 1
Scissors (p13)
0 , 0
1 , -1
Paper (p12)
Player 1
-1 , 1
0 , 0
Rock (p11)
Rock (p21)
73-347 Game Theory--Lecture 10
Example: Rock, paper and scissors
Check whether there is a mixed strategy Nash equilibrium in which two
of p11, p12, p13 is positive, and at least two of p21, p22, p23 are
positive.
The answer is No.
Player 2
0 , 0
-1 , 1
1 , -1
Scissors (p23)
Paper (p22)
1 , -1
-1 , 1
Scissors (p13)
0 , 0
1 , -1
Paper (p12)
Player 1
-1 , 1
0 , 0
Rock (p11)
Rock (p21)
73-347 Game Theory--Lecture 10
Example: Rock, paper and scissors
Therefore, (p1=(1/3, 1/3, 1/3), p2=(1/3, 1/3, 1/3)) is the unique
mixed strategy Nash equilibrium by Theorem 4.
Player 2
0 , 0
-1 , 1
1 , -1
Scissors (p23)
Paper (p22)
1 , -1
-1 , 1
Scissors (p13)
0 , 0
1 , -1
Paper (p12)
Player 1
1 , 1 -
0 , 0
Rock (p11)
Rock (p21)
73-347 Game Theory--Lecture 10 Summary
Find mixed strategy Nash equilibrium in a 2-player game each with a
finite number of pure strategies
Next time
Review HW1
Reading lists
Cha 4.10 of Osborne
Solutions to HW1
73-347 Game Theory--Lecture 10
Lecture 10
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