1. LECTURE 2 MIXED STRATEGY GAME
May 19, 2003
LECTURE 2 MIXED STRATEGY GAME
Notes modified from Yongqin Fudan University

2. Matching pennies Hence, NO Nash equilibrium -1 , 1 1 , -1
Player 1
Player 2
Tail
Head
-1 , 1
1 , -1
Head is Player 1’s best response to Player 2’s strategy Tail
Tail is Player 2’s best response to Player 1’s strategy Tail
Tail is Player 1’s best response to Player 2’s strategy Head
Head is Player 2’s best response to Player 1’s strategy Head
Hence, NO Nash equilibrium

3. Solving matching pennies
Player 2
Head
Tail
Player 1
-1 , 1
1 , -1 q
1-q r
1-r
Randomize your strategies
Player 1 chooses Head and Tail with probabilities r and 1-r, respectively.
Player 2 chooses Head and Tail with probabilities q and 1-q, respectively.
Mixed Strategy:
Specifies that an actual move be chosen randomly from the set of pure strategies with some specific probabilities.

4. Mixed strategy
A mixed strategy of a player is a probability distribution over player’s (pure) strategies.
A mixed strategy for Chris is a probability distribution (p, 1-p), where p is the probability of playing Opera, and 1-p is that probability of playing Prize Fight.
If p=1 then Chris actually plays Opera. If p=0 then Chris actually plays Prize Fight.
Battle of sexes
Pat
Opera
Prize Fight
Chris
Opera (p)
2 , 1
0 , 0
Prize Fight (1-p)
0 , 0
1 , 2

5. Solving matching pennies
Player 2
Head
Tail
Player 1
-1 , 1
1 , -1
Expected payoffs r
1-2q
1-r
2q-1 q
1-q
Player 1’s expected payoffs
If Player 1 chooses Head, -q+(1-q)=1-2q
If Player 1 chooses Tail, q-(1-q)=2q-1

6. Solving matching pennies
Player 2
Head
Tail
Player 1
-1 , 1
1 , -1
Expected payoffs r
1-2q
1-r
2q-1 q
1-q 1 q r
1/2
Player 1’s best response B1(q):
For q<0.5, Head (r=1)
For q>0.5, Tail (r=0)
For q=0.5, indifferent (0r1)

7. Solving matching pennies
Player 2
Head
Tail
Player 1
-1 , 1
1 , -1
Expected payoffs r
1-2q
1-r
2q-1 q
1-q
Expected payoffs
2r-1
1-2r
Player 2’s expected payoffs
If Player 2 chooses Head, r-(1-r)=2r-1
If Player 2 chooses Tail, -r+(1-r)=1-2r

8. Solving matching pennies
Player 2
Head
Tail
Player 1
-1 , 1
1 , -1
1-2q
2q-1
Expected payoffs r
1-r q
1-q
Expected payoffs
2r-1
1-2r 1 q r
1/2
Player 2’s best response B2(r):
For r<0.5, Tail (q=0)
For r>0.5, Head (q=1)
For r=0.5, indifferent (0q1)

9. Solving matching pennies
Player 2
Head
Tail
Player 1
-1 , 1
1 , -1
Player 1’s best response B1(q):
For q<0.5, Head (r=1)
For q>0.5, Tail (r=0)
For q=0.5, indifferent (0r1)
Player 2’s best response B2(r):
For r<0.5, Tail (q=0)
For r>0.5, Head (q=1)
For r=0.5, indifferent (0q1)
Check r = 0.5  B1(0.5) q = 0.5  B2(0.5) r
1-r q
1-q
Mixed strategy Nash equilibrium 1 q r
1/2

10. Mixed strategy: example
Matching pennies
Player 1 has two pure strategies: H and T ( 1(H)=0.5, 1(T)=0.5 ) is a Mixed strategy. That is, player 1 plays H and T with probabilities 0.5 and 0.5, respectively. ( 1(H)=0.3, 1(T)=0.7 ) is another Mixed strategy. That is, player 1 plays H and T with probabilities 0.3 and 0.7, respectively.

11. Mixed strategy: example
Player 2
L (0)
C (1/3)
R (2/3)
Player 1
T (3/4)
0 , 2
3 , 3
1 , 1
M (0)
4 , 0
0 , 4
2 , 3
B (1/4)
3 , 4
5 , 1
0 , 7
Player 1:
(3/4, 0, ¼) is a mixed strategy. That is, 1(T)=3/4, 1(M)=0 and 1(B)=1/4.
Player 2:
(0, 1/3, 2/3) is a mixed strategy. That is, 2(L)=0, 2(C)=1/3 and 2(R)=2/3.

12. Expected payoffs: 2 players each with two pure strategies
s21 ( q )
s22 ( 1- q )
Player 1
s11 ( r )
u1(s11, s21), u2(s11, s21)
u1(s11, s22), u2(s11, s22)
s12 (1- r )
u1(s12, s21), u2(s12, s21)
u1(s12, s22), u2(s12, s22)
Player 1 plays a mixed strategy (r, 1- r ). Player 2 plays a mixed strategy ( q, 1- q ).
Player 1’s expected payoff of playing s11: EU1(s11, (q, 1-q))=q×u1(s11, s21)+(1-q)×u1(s11, s22)
Player 1’s expected payoff of playing s12: EU1(s12, (q, 1-q))= q×u1(s12, s21)+(1-q)×u1(s12, s22)
Player 1’s expected payoff from her mixed strategy: v1((r, 1-r), (q, 1-q))=rEU1(s11, (q, 1-q))+(1-r)EU1(s12, (q, 1-q))

13. Expected payoffs: 2 players each with two pure strategies
s21 ( q )
s22 ( 1- q )
Player 1
s11 ( r )
u1(s11, s21), u2(s11, s21)
u1(s11, s22), u2(s11, s22)
s12 (1- r )
u1(s12, s21), u2(s12, s21)
u1(s12, s22), u2(s12, s22)
Player 1 plays a mixed strategy (r, 1- r ). Player 2 plays a mixed strategy ( q, 1- q ).
Player 2’s expected payoff of playing s21: EU2(s21, (r, 1-r))=r×u2(s11, s21)+(1-r)×u2(s12, s21)
Player 2’s expected payoff of playing s22: EU2(s22, (r, 1-r))= r×u2(s11, s22)+(1-r)×u2(s12, s22)
Player 2’s expected payoff from her mixed strategy: v2((r, 1-r),(q, 1-q))=qEU2(s21, (r, 1-r))+(1-q)EU2(s22, (r, 1-r))

14. Expected payoffs: example
Player 2
H (0.3)
T (0.7)
Player 1
H (0.4)
-1 , 1
1 , -1
T (0.6)
Player 1:
EU1(H, (0.3, 0.7)) = 0.3×(-1) + 0.7×1=0.4
EU1(T, (0.3, 0.7)) = 0.3× ×(-1)=-0.4
v1((0.4, 0.6), (0.3, 0.7))=0.4 (-0.4)=-0.08
Player 2:
EU2(H, (0.4, 0.6)) = 0.4×1+0.6×(-1) = -0.2
EU2(T, (0.4, 0.6)) = 0.4×(-1)+0.6×1 = 0.2
v2((0.4, 0.6), (0.3, 0.7))=0.3×(-0.2)+0.7×0.2=0.08

15. Expected payoffs: example
Player 2
L (0)
C (1/3)
R (2/3)
Player 1
T (3/4)
0 , 2
3 , 3
1 , 1
M (0)
4 , 0
0 , 4
2 , 3
B (1/4)
3 , 4
5 , 1
0 , 7
Mixed strategies: p1=( 3/4, 0, ¼ ); p2=( 0, 1/3, 2/3 ).
Player 1:
EU1(T, p2)=3(1/3)+1(2/3)=5/3, EU1(M, p2)=0(1/3)+2(2/3)=4/3 EU1(B, p2)=5(1/3)+0(2/3)=5/3. v1(p1, p2) = 5/3
Player 2:
EU2(L, p1)=2(3/4)+4(1/4)=5/2, EU2(C, p1)=3(3/4)+3(1/4)=5/2, EU2(R, p1)=1(3/4)+7(1/4)=5/2. v1(p1, p2) = 5/2

16. Mixed strategy equilibrium
A probability distribution for each player
The distributions are mutual best responses to one another in the sense of expected payoffs
It is a stochastic steady state

17. Mixed strategy vs. Correlated strategy

18. Mixed strategy equilibrium: 2-player each with two pure strategies
s21 ( q )
s22 ( 1- q )
Player 1
s11 ( r )
u1(s11, s21), u2(s11, s21)
u1(s11, s22), u2(s11, s22)
s12 (1- r )
u1(s12, s21), u2(s12, s21)
u1(s12, s22), u2(s12, s22)
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

19. Find mixed strategy equilibrium in 2-player each with two pure strategies
Find the best response correspondence for player 1, given player 2’s mixed strategy
Find the best response correspondence for player 2, given player 1’s mixed strategy
Use the best response correspondences to determine mixed strategy Nash equilibria.

20. Employee Monitoring Employees can work hard or shirk
Salary: $100K unless caught shirking
Cost of effort: $50K
Managers can monitor or not
Value of employee output: $200K
Profit if employee doesn’t work: $0
Cost of monitoring: $10K

21. Employee Monitoring Employee’s best response B1(q): 50 , 90 50 , 100
Shirk (r=0) if q<0.5
Work (r=1) if q>0.5
Any mixed strategy (0r1) if q=0.5
Expected payoffs 50
100(1-q)
Expected payoffs
100r-10
200r-100
Manager
Monitor ( q )
Not Monitor (1-q)
Employee
Work ( r )
50 , 90
50 , 100
Shirk (1-r )
0 , -10
100 , -100

22. Employee Monitoring Manager’s best response B2(r): 50 , 90 50 , 100
Monitor (q=1) if r<0.9
Not Monitor (q=0) if r>0.9
Any mixed strategy (0q1) if r=0.9
Expected payoffs 50
100(1-q)
Expected payoffs
100r-10
200r-100
Manager
Monitor ( q )
Not Monitor (1-q)
Employee
Work ( r )
50 , 90
50 , 100
Shirk (1-r )
0 , -10
100 , -100

23. Employee Monitoring r q Employee’s best response B1(q):
Shirk (r=0) if q<0.5
Work (r=1) if q>0.5
Any mixed strategy (0r1) if q=0.5
Manager’s best response B2(r):
Monitor (q=1) if r<0.9
Not Monitor (q=0) if r>0.9
Any mixed strategy (0q1) if r=0.9
Mixed strategy Nash equilibrium
((0.9,0.1),(0.5,0.5)) 1 q r
0.5
0.9

24. Battle of sexes Chris’ expected payoff of playing Opera: 2q
Chris’ expected payoff of playing Prize Fight: 1-q
Chris’ best response B1(q):
Prize Fight (r=0) if q<1/3
Opera (r=1) if q>1/3
Any mixed strategy (0r1) if q=1/3
Pat
Opera (q)
Prize Fight (1-q)
Chris
Opera ( r )
2 , 1
0 , 0
Prize Fight (1-r)
0 , 0
1 , 2

25. Battle of sexes Pat’s expected payoff of playing Opera: r
Pat’s expected payoff of playing Prize Fight: 2(1-r)
Pat’s best response B2(r):
Prize Fight (q=0) if r<2/3
Opera (q=1) if r>2/3
Any mixed strategy (0q1) if r=2/3,
Pat
Opera (q)
Prize Fight (1-q)
Chris
Opera ( r )
2 , 1
0 , 0
Prize Fight (1-r)
0 , 0
1 , 2

26. Battle of sexes Chris’ best response B1(q): Pat’s best response B2(r):
Prize Fight (r=0) if q<1/3
Opera (r=1) if q>1/3
Any mixed strategy (0r1) if q=1/3
Pat’s best response B2(r):
Prize Fight (q=0) if r<2/3
Opera (q=1) if r>2/3
Any mixed strategy (0q1) if r=2/3
Three Nash equilibria:
((1, 0), (1, 0))
((0, 1), (0, 1))
((2/3, 1/3), (1/3, 2/3)) 1 q r
2/3
1/3

27. Battle of sexes The limitation of best-response approach
Can only deal with simple games
Comments of mixed strategies
Randomization
vNM expected utility
Choice under uncertainty

28. 2-player each with two strategies
Lecture 6
May 27, 2003
2-player each with two strategies
Player 2
s21 ( q )
s22 ( 1- q )
Player 1
s11 ( r )
u1(s11, s21), u2(s11, s21)
u1(s11, s22), u2(s11, s22)
s12 (1- r )
u1(s12, s21), u2(s12, s21)
u1(s12, s22), u2(s12, s22)
Theorem 1 (property of mixed Nash equilibrium)
A pair of mixed strategies ((r*, 1-r*), (q*, 1-q*)) is a Nash 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*))

29. Expected payoffs: 2 players each with two pure strategies
s21 ( q )
s22 ( 1- q )
Player 1
s11 ( r )
u1(s11, s21), u2(s11, s21)
u1(s11, s22), u2(s11, s22)
s12 (1- r )
u1(s12, s21), u2(s12, s21)
u1(s12, s22), u2(s12, s22)
Player 1 plays a mixed strategy (r, 1- r ). Player 2 plays a mixed strategy ( q, 1- q ).
Player 1’s expected payoff of playing s11: EU1(s11, (q, 1-q))=q×u1(s11, s21)+(1-q)×u1(s11, s22)
Player 1’s expected payoff of playing s12: EU1(s12, (q, 1-q))= q×u1(s12, s21)+(1-q)×u1(s12, s22)
Player 1’s expected payoff from her mixed strategy: v1((r, 1-r), (q, 1-q))=rEU1(s11, (q, 1-q))+(1-r)EU1(s12, (q, 1-q))

30. Expected payoffs: 2 players each with two pure strategies
s21 ( q )
s22 ( 1- q )
Player 1
s11 ( r )
u1(s11, s21), u2(s11, s21)
u1(s11, s22), u2(s11, s22)
s12 (1- r )
u1(s12, s21), u2(s12, s21)
u1(s12, s22), u2(s12, s22)
Player 1 plays a mixed strategy (r, 1- r ). Player 2 plays a mixed strategy ( q, 1- q ).
Player 2’s expected payoff of playing s21: EU2(s21, (r, 1-r))=r×u2(s11, s21)+(1-r)×u2(s12, s21)
Player 2’s expected payoff of playing s22: EU2(s22, (r, 1-r))= r×u2(s11, s22)+(1-r)×u2(s12, s22)
Player 2’s expected payoff from her mixed strategy: v2((r, 1-r),(q, 1-q))=qEU2(s21, (r, 1-r))+(1-q)EU2(s22, (r, 1-r))

31. Theorem 1: illustration
Lecture 6
May 27, 2003
Theorem 1: illustration
Matching pennies
Player 2
H (0.5)
T (0.5)
Player 1
-1 , 1
1 , -1
Player 1:
EU1(H, (0.5, 0.5)) = 0.5×(-1) + 0.5×1=0
EU1(T, (0.5, 0.5)) = 0.5× ×(-1)=0
v1((0.5, 0.5), (0.5, 0.5))=0.50+0.50=0
Player 2:
EU2(H, (0.5, 0.5)) = 0.5×1+0.5×(-1) =0
EU2(T, (0.5, 0.5)) = 0.5×(-1)+0.5×1 = 0
v2((0.5, 0.5), (0.5, 0.5))=0.5×0+0.5×0=0

32. Theorem 1: illustration
Matching pennies
Player 2
H (0.5)
T (0.5)
Player 1
-1 , 1
1 , -1
Player 1:
v1((0.5, 0.5), (0.5, 0.5))  EU1(H, (0.5, 0.5))
v1((0.5, 0.5), (0.5, 0.5))  EU1(T, (0.5, 0.5))
Player 2:
v2((0.5, 0.5), (0.5, 0.5))  EU2(H, (0.5, 0.5))
v2((0.5, 0.5), (0.5, 0.5))  EU2(T, (0.5, 0.5))
Hence, ((0.5, 0.5), (0.5, 0.5)) is a mixed strategy Nash equilibrium by Theorem 1.

33. Theorem 1: illustration
Employee’s expected payoff of playing “work”
EU1(Work, (0.5, 0.5)) = 0.5× ×50=50
Employee’s expected payoff of playing “shirk”
EU1(Shirk, (0.5, 0.5)) = 0.5× ×100=50
Employee’s expected payoff of her mixed strategy
v1((0.9, 0.1), (0.5, 0.5))=0.950+0.150=50
Employee Monitoring
Manager
Monitor (0.5)
Not Monitor (0.5)
Employee
Work (0.9)
50 , 90
50 , 100
Shirk (0.1)
0 , -10
100 , -100

34. Theorem 1: illustration
Manager’s expected payoff of playing “Monitor”
EU2(Monitor, (0.9, 0.1)) = 0.9×90+0.1×(-10) =80
Manager’s expected payoff of playing “Not”
EU2(Not, (0.9, 0.1)) = 0.9× ×(-100) = 80
Manager’s expected payoff of her mixed strategy
v2((0.9, 0.1), (0.5, 0.5))=0.5×80+0.5×80=80
Employee Monitoring
Manager
Monitor (0.5)
Not Monitor (0.5)
Employee
Work (0.9)
50 , 90
50 , 100
Shirk (0.1)
0 , -10
100 , -100

35. Theorem 1: illustration
Employee
v1((0.9, 0.1), (0.5, 0.5))  EU1(Work, (0.5, 0.5))
v1((0.9, 0.1), (0.5, 0.5))  EU1(Shirk, (0.5, 0.5))
Manager
v2((0.9, 0.1), (0.5, 0.5))  EU2(Monitor, (0.9, 0.1))
v2((0.9, 0.1), (0.5, 0.5))  EU2(Not, (0.9, 0.1))
Hence, ((0.9, 0.1), (0.5, 0.5)) is a mixed strategy Nash equilibrium by Theorem 1.
Employee Monitoring
Manager
Monitor (0.5)
No Monitor (0.5)
Employee
Work (0.9)
50 , 90
50 , 100
Shirk (0.1)
0 , -10
100 , -100

36. Theorem 1: illustration
Use Theorem 1 to check whether ((2/3, 1/3), (1/3, 2/3)) is a mixed strategy Nash equilibrium.
Battle of sexes
Pat
Opera (1/3)
Prize Fight (2/3)
Chris
Opera (2/3 )
2 , 1
0 , 0
Prize Fight (1/3)
0 , 0
1 , 2

37. Mixed strategy equilibrium: 2-player each with two strategies
s21 ( q )
s22 ( 1- q )
Player 1
s11 ( r )
u1(s11, s21), u2(s11, s21)
u1(s11, s22), u2(s11, s22)
s12 (1- r )
u1(s12, s21), u2(s12, s21)
u1(s12, s22), u2(s12, s22)
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*)) = EU2(s22, (r*, 1-r*))
That is, each player is indifferent between her two strategies.
Significance: it gives conditions for a mixed strategy NE in terms of each player’s expected payoffs only to her pure strategies

38. Use indifference to find mixed Nash equilibrium (2-player each with 2 strategies)
Use Theorem 2 to find mixed strategy Nash equilibria
Solve EU1(s11, (q*, 1-q*)) = EU1(s12, (q*, 1-q*))
Solve EU2(s21, (r*, 1-r*)) = EU2(s22, (r*, 1-r*))

39. Use Theorem 2 to find mixed strategy Nash equilibrium: illustration
Matching pennies
Player 2
H ( q )
T ( 1–q )
Player 1
H ( r )
-1 , 1
1 , -1
T ( 1–r )
Player 1 is indifferent between playing Head and Tail.
EU1(H, (q, 1–q)) = q×(-1) + (1–q)×1=1–2q
EU1(T, (q, 1–q)) = q×1 + ×(1–q) (-1)=2q–1
EU1(H, (q, 1–q)) = EU1(T, (q, 1–q)) 1–2q = 2q–1 4q = This give us q = 1/2

40. Use Theorem 2 to find mixed strategy Nash equilibrium: illustration
Matching pennies
Player 2
H ( q )
T ( 1–q )
Player 1
H ( r )
-1 , 1
1 , -1
T ( 1–r )
Player 2 is indifferent between playing Head and Tail.
EU2(H, (r, 1–r)) = r ×1+(1–r)×(-1) =2r – 1
EU2(T, (r, 1–r)) = r×(-1)+(1–r)×1 = 1 – 2r
EU2(H, (r, 1–r)) = EU2(T, (r, 1–r)) 2r – 1= 1 – 2r 4r = 2 This give us r = 1/2
Hence, ((0.5, 0.5), (0.5, 0.5)) is a mixed strategy Nash equilibrium by Theorem 2.

41. Use Theorem 2 to find mixed strategy Nash equilibrium: illustration
Employee’s expected payoff of playing “work”
EU1(Work, (q, 1–q)) = q×50 + (1–q)×50=50
Employee’s expected payoff of playing “shirk”
EU1(Shirk, (q, 1–q)) = q×0 + (1–q)×100=100(1–q)
Employee is indifferent between playing Work and Shirk.
50=100(1–q)
q=1/2
Employee Monitoring
Manager
Monitor ( q )
Not Monitor (1–q )
Employee
Work (r)
50 , 90
50 , 100
Shirk (1–r)
0 , -10
100 , -100

42. Use Theorem 2 to find mixed strategy Nash equilibrium: illustration
Manager’s expected payoff of playing “Monitor”
EU2(Monitor, (r, 1–r)) = r×90+(1–r)×(-10) =100r–10
Manager’s expected payoff of playing “Not”
EU2(Not, (r, 1–r)) = r×100+(1–r)×(-100) =200r–100
Manager is indifferent between playing Monitor and Not 100r–10 =200r–100 implies that r=0.9.
Hence, ((0.9, 0.1), (0.5, 0.5)) is a mixed strategy Nash equilibrium by Theorem 2.
Employee Monitoring
Manager
Monitor ( q )
Not Monitor (1–q )
Employee
Work (r)
50 , 90
50 , 100
Shirk (1–r)
0 , -10
100 , -100

43. Use Theorem 2 to find mixed strategy Nash equilibrium: illustration
Use Theorem 2 to find a mixed Nash equilibrium
Battle of sexes
Pat
Opera (q)
Prize Fight (1-q)
Chris
Opera ( r )
2 , 1
0 , 0
Prize Fight (1-r)
0 , 0
1 , 2

44. Use Theorem 2 to find mixed strategy Nash equilibrium: illustration
Use Theorem 2 to find a mixed Nash equilibrium
Example
Player 2
L (q)
R (1-q)
Player 1
T ( r )
6 , 4
2 , 6
B (1-r)
3 , 3
6 , 1

45. Mixed strategy Nash equilibrium
A mixed strategy of a player is a probability distribution over the player’s strategies.
Mixed strategy Nash equilibrium
A probability distribution for each player
The distributions are mutual best responses to one another in the sense of expected payoffs

46. Use Theorem 2 to find mixed strategy Nash equilibrium: illustration
Lecture 6
May 27, 2003
Use Theorem 2 to find mixed strategy Nash equilibrium: illustration
Chris’ expected payoff of playing Opera
EU1(O, (q, 1–q)) = q×2 + (1–q)×0 = 2q
Chris’ expected payoff of playing Prize Fight
EU1(F, (q, 1–q)) = q×0 + (1–q)×1 = 1–q
Chris is indifferent between playing Opera and Prize
EU1(O, (q, 1–q)) = EU1(F, (q, 1–q)) 2q = 1–q 3q = This give us q = 1/3
Battle of sexes
Pat
Opera (q)
Prize Fight (1-q)
Chris
Opera ( r )
2 , 1
0 , 0
Prize Fight (1-r)
0 , 0
1 , 2

47. Use Theorem 2 to find mixed strategy Nash equilibrium: illustration
Lecture 6
May 27, 2003
Use Theorem 2 to find mixed strategy Nash equilibrium: illustration
Pat’s expected payoff of playing Opera
EU2(O, (r, 1–r)) = r ×1+(1–r)×0 = r
Pat’s expected payoff of playing Prize Fight
EU2(F, (r, 1–r)) = r×0+(1–r)×2 = 2 – 2r
Pat is indifferent between playing Opera and Prize
EU2(O, (r, 1–r)) = EU2(F, (r, 1–r)) r = 2 – 2r 3r = 2 This give us r = 2/3
Battle of sexes
Pat
Opera (q)
Prize Fight (1-q)
Chris
Opera ( r )
2 , 1
0 , 0
Prize Fight (1-r)
0 , 0
1 , 2

48. Use Theorem 2 to find mixed strategy Nash equilibrium: illustration
Lecture 6
May 27, 2003
Use Theorem 2 to find mixed strategy Nash equilibrium: illustration
Hence, ( (2/3, 1/3), (1/3, 2/3) ) is a mixed strategy Nash equilibrium. That is,
Chris chooses Opera with probability 2/3 and Prize Fight with probability 1/3.
Pat chooses Opera with probability 1/3 and Prize Fight with probability 2/3.
Battle of sexes
Pat
Opera (q)
Prize Fight (1-q)
Chris
Opera ( r )
2 , 1
0 , 0
Prize Fight (1-r)
0 , 0
1 , 2

49. Example 1
Bruce and Sheila determine whether to go to the opera or to a pro wrestling show.
Sheila gets utility of 4 from going to the opera and 1 from pro wrestling.
Bruce gets utility of 1 from going to the opera and 4 from pro wrestling.
They agree to decide what to do in the following way:
Bruce and Sheila each puts a penny below an issue of the TV guide on the coffee table (assume they don’t cheat by looking at the other). They count to 3 and simultaneously reveal which side of their penny is up. If the pennies match (both heads, or both tails), Sheila decides what to watch, while if the pennies don’t match (heads, tails or tails, heads) then Bruce decides.

50. Example 1 Bruce’s expected payoff of playing Head
Sheila
H ( q )
T ( 1–q )
Bruce
H ( r )
1 , 4
4 , 1
T ( 1–r )
Bruce’s expected payoff of playing Head
EU1(H, (q, 1–q)) = q×1 + (1–q)×4 = 4–3q
Bruce’s expected payoff of playing Tail
EU1(T, (q, 1–q)) = q×4 + (1–q)×1 = 1+3q
Bruce is indifferent between playing Head and Tail
EU1(H, (q, 1–q)) = EU1(T, (q, 1–q)) 4–3q = 1+3q 6q = 3 This give us q = 1/2

51. Example 1 1 , 4 4 , 1 Sheila’s expected payoff of playing Head
H ( q )
T ( 1–q )
Bruce
H ( r )
1 , 4
4 , 1
T ( 1–r )
Sheila’s expected payoff of playing Head
EU2(H, (r, 1–r)) = r ×4+(1–r)×1 = 3r + 1
Sheila’s expected payoff of playing Tail
EU2(T, (r, 1–r)) = r×1+(1–r)×4 = 4 – 3r
Sheila is indifferent between playing Head and Tail
EU2(H, (r, 1–r)) = EU2(T, (r, 1–r)) 3r + 1 = 4 – 3r 6r = 3 This give us r = ½
( (1/2, 1/2), (1/2, 1/2) ) is a mixed strategy Nash equilibrium.

52. Example 2 Player 1’s expected payoff of playing T
L ( q )
R ( 1–q )
Player 1
T ( r )
6 , 0
0 , 6
B ( 1–r )
3 , 2
Player 1’s expected payoff of playing T
EU1(T, (q, 1–q)) = q×6 + (1–q)×0 = 6q
Player 1’s expected payoff of playing B
EU1(B, (q, 1–q)) = q×3 + (1–q)×6 = 6-3q
Player 1 is indifferent between playing T and B
EU1(T, (q, 1–q)) = EU1(B, (q, 1–q)) 6q = 6-3q 9q = 6 This give us q = 2/3

53. Example 2 6 , 0 0 , 6 3 , 2 Player 2’s expected payoff of playing L
L ( q )
R ( 1–q )
Player 1
T ( r )
6 , 0
0 , 6
B ( 1–r )
3 , 2
Player 2’s expected payoff of playing L
EU2(L, (r, 1–r)) = r ×0+(1–r)×2 =2- 2r
Player 2’s expected payoff of playing R
EU2(R, (r, 1–r)) = r×6+(1–r)×0 = 6r
Player 2 is indifferent between playing L and R
EU2(L, (r, 1–r)) = EU2(R, (r, 1–r)) r = 6r 8r = 2 This gives us r = ¼
( (1/4, 3/4), (2/3, 1/3) ) is a mixed strategy Nash equilibrium.

54. Example 3:Market entry game
Two firms, Firm 1 and Firm 2, must decide whether to put one of their restaurants in a shopping mall simultaneously.
Each has two strategies: Enter, Not Enter
If either firm plays “Not Enter”, it earns 0 profit
If one plays “Enter” and the other plays “Not Enter” then the firm plays “Enter” earns $500K
If both plays “Enter” then both lose $100K because the demand is limited

55. Example 3:Market entry game
Firm 2
Enter ( q )
Not Enter ( 1–q )
Firm 1
Enter ( r )
-100 , -100
500 , 0
Not Enter ( 1–r )
0 , 500
0 , 0
How many Nash equilibria can you find?
Two pure strategy Nash equilibrium (Not Enter, Enter) and (Enter, Not Enter)
One mixed strategy Nash equilibrium ((5/6, 1/6), (5/6, 1/6)) That is r=5/6 and q=5/6

56. Example 4 How many Nash equilibria can you find?
Player 2
L ( q )
R ( 1–q )
Player 1
T ( r )
1 , 1
1 , 2
B ( 1–r )
2 , 3
0 , 1
How many Nash equilibria can you find?
Two pure strategy Nash equilibrium (B, L) and (T, R)
One mixed strategy Nash equilibrium ((2/3, 1/3), (1/2, 1/2)) That is r=2/3 and q=1/2

57. 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

58. Example: Rock, paper and scissors
Player 2
Rock
Paper
Scissors
Player 1
0 , 0
-1 , 1
1 , -1
Can you guess a mixed strategy Nash equilibrium?

59. Mixed strategy Nash equilibrium: 2-player each with two pure strategies
s21 ( q )
s22 ( 1- q )
Player 1
s11 ( r )
u1(s11, s21), u2(s11, s21)
u1(s11, s22), u2(s11, s22)
s12 (1- r )
u1(s12, s21), u2(s12, s21)
u1(s12, s22), u2(s12, s22)
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

60. 2-player each with a finite number of pure strategies
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

61. 2-player each with a finite number of pure strategies
s21 (p21)
s22 (p22)
s2K (p2K)
s11 (p11)
u2(s11, s21)
u1(s11, s21)
u2(s11, s22)
u1(s11, s22)
u2(s11, s2K)
u1(s11, s2K)
s12 (p12)
u2(s12, s21)
u1(s12, s21)
u2(s12, s22)
u1(s12, s22)
u2(s12, s2K)
u1(s12, s2K)
....
......
s1J (p1J)
u2(s1J, s21)
u1(s1J, s21)
u2(s1J, s22)
u1(s1J, s22)
u2(s1J, s2K)
u1(s1J, s2K)
Player 1
Player 1’s mixed strategy: p1=(p11, p12, ..., p1J )
Player 2’s mixed strategy: p2=(p21, p22, ..., p2K )

62. 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)

63. 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)

64. 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*.

65. 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

66. 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*)

67. 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.

68. 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.

69. Theorem 4: illustration
Player 2
L (0)
C (1/3)
R (2/3)
Player 1
T (3/4)
0 , 2
3 , 3
1 , 1
M (0)
4 , 0
0 , 4
2 , 3
B (1/4)
3 , 4
5 , 1
0 , 7
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)

70. Theorem 4: illustration
Player 2
L (0)
C (1/3)
R (2/3)
Player 1
T (3/4)
0 , 2
3 , 3
1 , 1
M (0)
4 , 0
0 , 4
2 , 3
B (1/4)
3 , 4
5 , 1
0 , 7
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.

71. Example: Rock, paper and scissors
Player 2
Rock (p21)
Paper (p22)
Scissors (p23)
Player 1
Rock (p11)
0 , 0
-1 , 1
1 , -1
Paper (p12)
Scissors (p13)
Check whether there is a mixed strategy Nash equilibrium in which p11>0, p12>0, p13>0, p21>0, p22>0, p23>0.

72. Example: Rock, paper and scissors
Player 2
Rock (p21)
Paper (p22)
Scissors (p23)
Player 1
Rock (p11)
0 , 0
-1 , 1
1 , -1
Paper (p12)
Scissors (p13)
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.

73. Example: Rock, paper and scissors
Player 2
Rock (p21)
Paper (p22)
Scissors (p23)
Player 1
Rock (p11)
0 , 0
-1 , 1
1 , -1
Paper (p12)
Scissors (p13)
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.

74. Example: Rock, paper and scissors
Player 2
Rock (p21)
Paper (p22)
Scissors (p23)
Player 1
Rock (p11)
0 , 0
-1 , 1
1 , -1
Paper (p12)
Scissors (p13)
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

75. Example: Rock, paper and scissors
Player 2
Rock (p21)
Paper (p22)
Scissors (p23)
Player 1
Rock (p11)
0 , 0
-1 , 1
1 , -1
Paper (p12)
Scissors (p13)
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.

76. Example: Rock, paper and scissors
Player 2
Rock (p21)
Paper (p22)
Scissors (p23)
Player 1
Rock (p11)
0 , 0
-1 , 1
1 , -1
Paper (p12)
Scissors (p13)
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

77. Example: Rock, paper and scissors
Player 2
Rock (1/3)
Paper (1/3)
Scissors (1/3)
Player 1
0 , 0
-1 , 1
1 , -1
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.

78. Example: Rock, paper and scissors
Player 2
Rock (p21)
Paper (p22)
Scissors (p23)
Player 1
Rock (p11)
0 , 0
-1 , 1
1 , -1
Paper (p12)
Scissors (p13)
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.

79. Example: Rock, paper and scissors
Player 2
Rock (p21)
Paper (p22)
Scissors (p23)
Player 1
Rock (p11)
0 , 0
-1 , 1
1 , -1
Paper (p12)
Scissors (p13)
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.

80. Example: Rock, paper and scissors
Player 2
Rock (p21)
Paper (p22)
Scissors (p23)
Player 1
Rock (p11)
0 , 0
-1 , 1
1 , -1
Paper (p12)
Scissors (p13)
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.

81. Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
We first consider pure-strategy Nash equilibria. How many can you find?
In order to find all Nash equilibria, we need to consider 15 more cases by Theorem 4.
We first consider complicated cases. Some cases are very easy.

82. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 1: check whether there is a mixed strategy in which p11>0, p12>0, p21>0, p22>0, p23>0 By theorem 4, we should have 2p11+2 p12= 3p11+1 p12 = 3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these equations. If we can get a solution that satisfies p11>0, p12>0, p21>0, p22>0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21>0, p22>0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

83. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 2: check whether there is a mixed strategy in which p11>0, p12>0, p21>0, p22>0, p23=0 By theorem 4, we should have 2p11+2 p12= 3p11+1 p12  3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11>0, p12>0, p21>0, p22>0, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21>0, p22>0, p23=0, then we do not have such a mixed strategy Nash equilibrium.

84. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 3: check whether there is a mixed strategy in which p11>0, p12>0, p21>0, p22=0, p23>0 By theorem 4, we should have 2p11+2 p12 = 3p11+2 p12  3p11+1 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11>0, p12>0, p21>0, p22=0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21>0, p22=0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

85. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 4: check whether there is a mixed strategy in which p11>0, p12>0, p21=0, p22>0, p23>0 By theorem 4, we should have 2p11+2 p12 3p11+1 p12 = 3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11>0, p12>0, p21=0, p22>0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21=0, p22>0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

86. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 5: check whether there is a mixed strategy in which p11>0, p12>0, p21>0, p22=0, p23=0 (Note this implies p21=1) By theorem 4, we should have 2p11+2 p12 3p11+1 p12 and 2p11+2 p12 3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11>0, p12>0, p21>0, p22=0, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21>0, p22=0, p23=0, then we do not have such a mixed strategy Nash equilibrium.

87. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 6: check whether there is a mixed strategy in which p11>0, p12>0, p21=0, p22>0, p23=0 (Note this implies p22=1) By theorem 4, we should have 3p11+1 p12  2p11+2 p12 and 3p11+1 p12  3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11>0, p12>0, p21=0, p22>0, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21=0, p22>0, p23=0, then we do not have such a mixed strategy Nash equilibrium.

88. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 7: check whether there is a mixed strategy in which p11>0, p12>0, p21=0, p22=0, p23>0 (Note this implies p23=1) By theorem 4, we should have 3p11+2 p12  2p11+2 p12 and 3p11+2 p12  3p11+1 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11>0, p12>0, p21=0, p22=0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21=0, p22=0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

89. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 8: check whether there is a mixed strategy in which p11>0, p12=0, p21>0, p22>0, p23>0 (Note this implies p11=1) By theorem 4, we should have 2p11+2 p12= 3p11+1 p12 = 3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11>0, p12=0, p21>0, p22>0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12=0, p21>0, p22>0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

90. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 9: check whether there is a mixed strategy in which p11>0, p12=0, p21>0, p22>0, p23=0 (Note this implies p11=1) By theorem 4, we should have 2p11+2 p12= 3p11+1 p12  3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11>0, p12=0, p21>0, p22>0, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12=0, p21>0, p22>0, p23=0, then we do not have such a mixed strategy Nash equilibrium.

91. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 10: check whether there is a mixed strategy in which p11>0, p12=0, p21>0, p22=0, p23>0 (Note this implies p11=1) By theorem 4, we should have 2p11+2 p12 3p11+1 p12  3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11>0, p12=0, p21>0, p22=0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12=0, p21>0, p22=0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

92. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 11: check whether there is a mixed strategy in which p11>0, p12=0, p21=0, p22>0, p23>0 (Note this implies p11=1) By theorem 4, we should have 2p11+2 p12 3p11+1 p12 = 3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11>0, p12=0, p21=0, p22>0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12=0, p21=0, p22>0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

93. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 12: check whether there is a mixed strategy in which p11=0, p12>0, p21>0, p22>0, p23>0 (Note this implies p12=1) By theorem 4, we should have 2p11+2 p12= 3p11+1 p12 = 3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11=0, p12>0, p21>0, p22>0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11=0, p12>0, p21>0, p22>0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

94. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 13: check whether there is a mixed strategy in which p11=0, p12>0, p21>0, p22>0, p23=0 (Note this implies p12=1) By theorem 4, we should have 2p11+2 p12= 3p11+1 p12  3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23 3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11=0, p12>0, p21>0, p22>0, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11=0, p12>0, p21>0, p22>0, p23=0, then we do not have such a mixed strategy Nash equilibrium.

95. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 14: check whether there is a mixed strategy in which p11=0, p12>0, p21>0, p22=0, p23>0 (Note this implies p12=1) By theorem 4, we should have 2p11+2 p12 3p11+1 p12  3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23  3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11=0, p12>0, p21>0, p22=0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11=0, p12>0, p21>0, p22=0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

96. Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1
Lecture 6
May 27, 2003
Exercise of Osborne
Player 2
L (p21)
M (p22)
R (p23)
Player 1
T (p11)
2 , 2
0 , 3
1 , 3
B (p12)
3 , 2
1 , 1
0 , 2
Case 15: check whether there is a mixed strategy in which p11=0, p12>0, p21=0, p22>0, p23>0 (Note this implies p12=1) By theorem 4, we should have 2p11+2 p12 3p11+1 p12 = 3p11+2 p12 and p11+p12=1.
We should have 2p21+0 p22+1 p23  3p21+1 p22+0 p23 and p21+ p22+ p23 = 1
Solve these. If we can get a solution that satisfies p11=0, p12>0, p21=0, p22>0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11=0, p12>0, p21=0, p22>0, p23>0, then we do not have such a mixed strategy Nash equilibrium.