Math 486 Game Theory Miderm 3 April 26, 2007.
Name______________________________________
Find the equilibria in pure strategies, the Pareto optimal solutions in pure strategies,
the characteristic function, the Nash bargaining solution, and the Shapley values.
1 (30 pts). Players: the row player Ann; the column player Bob.
|
0, 1 |
1, 0 |
0, 0 |
1, 1 |
|
1, 1 |
1, 3 |
2, 0 |
2, 2 |
|
0, 3 |
0, 0 |
1, 3 |
3, 2 |
|
1, 0 |
0, 0 |
0, 0 |
1, 4 |
(payoffs for Ann and Bob)
Solution. One equilibrium in pure joint strategies ((row 2, column 2):
|
0, 1* |
1*, 0 |
0, 0 |
1, 1* |
|
1*, 1 |
1*, 3* |
2*, 0 |
2, 2 |
|
0, 3* |
0, 0 |
1, 3* |
3*, 2 |
|
1*, 0 |
0, 0 |
0, 0 |
1, 4* |
Two Pareto optimal payoffs: (3, 2) and (1,4).
v(Ann) = 1 = the value of the matrix game
|
0 |
1* |
0 |
1 |
|
1 |
1*' |
2* |
2 |
|
0 |
0 |
1 |
3* |
|
1 |
0 |
0 |
1 |
v(Bob) = 1 = the value of the matrix game (where Bob is the row player)
|
1* |
1 |
3* |
0' |
|
0’ |
3* |
0’ |
0' |
|
0’ |
0’ |
3* |
0’ |
|
1*’ |
2 |
2 |
4* |
v(empty)=0, v(Ann,Bob) = 5.
Nash bargaining: (u-1)(v-1) -> max , u >= 1, v >= 1, (u, v) is a mixture of (3, 2) and (1,4).,
u + v =5, (u – 1) + (v – 1) =3, u-1= v - 1=1.5; u=2.5, v=2.5 optimal solution = Nash solution= the arbitration pair.
Ann Bob
Ann Bob 1 4
Bob Ann 4 1
------------------
Shapley values 2.5 2.5
2 (45 pts). Players: A, B, C
strategies payoffs
1 1 1 0 1 2
1 1 2 1 0 1
1 2 1 2 3 3
1 2 2 1 2 3
2 1 1 0 0 1
2 1 2 1 1 1
2 2 1 2 3 3
2 2 2 2 3 2
3 1 1 0 0 0
3 1 2 0 0 0
3 2 1 1 1 1
3 2 2 2 4 3
Solution.
Three equilibria, one Pareto optimal:
strategies payoffs
1 1 1 0 1 2
1 1 2 1 0 1
1 2 1 2* 3* 3* equilibrium
1 2 2 1 2 3
2 1 1 0 0 1
2 1 2 1 1 1
2 2 1 2* 3* 3*equilibrium
2 2 2 2 3 2
3 1 1 0 0 0
3 1 2 0 0 0
3 2 1 1 1 1
3 2 2 2* 4* 3* equilibrium Pareto optimal
v(empty)=0, v(A,B,C)) = 9.
v(A) = 0 = the value of the matrix game
|
0*' |
1 |
2 |
1 |
|
0 |
1 |
2 |
2 |
|
0 |
0 |
1 |
2 |
v(B) = 1 = the value of the matrix game
|
1 |
0 |
0 |
1 |
0 |
0 |
|
3 |
2 |
3 |
3 |
1*' |
4 |
v(C) = 0 = the value of the matrix game
|
2 |
3 |
1 |
3 |
0*' |
1 |
|
1 |
3 |
1 |
2 |
0 |
3 |
v(A,B) = 5= the value of the matrix game
|
1 |
1 |
|
5 |
3 |
|
0 |
2 |
|
5*' |
5 |
|
0 |
0 |
|
2 |
6 |
v(A,C) = 2= the value of the matrix game
|
2*' |
5 |
|
2 |
4 |
|
1 |
5 |
|
2 |
4 |
|
0 |
2 |
|
0 |
5 |
v(B,C) = 16/3 = the value of the matrix game
|
B&C vs A |
c1 |
c2 |
c3 |
(5c2+c3)/6 |
|
r11 |
3 |
1 |
0 |
|
|
r12 |
1 |
2 |
0 |
|
|
r21 |
6 |
6 |
2 |
16/3 |
|
r22 |
5 |
5 |
7 |
16/3 |
|
(r21+2r22)/3 |
|
16/3 |
16/3 |
16/3*' |
order contribution
A B C
ABC 0 5 4
ACB 0 7 2
BAC 4 1 4
BCA 11/3 1 13/3
CAB 2 7 0
CBA 11/3 16/3 0
---------------
20/9 79/18 43/18 Shapley values.
Nash solution = Pareto optimal = (2,4,3).