Math 486 Game Theory. Miderm 3. April 21, 2011.
Name______________________________________
Find the equilibria in pure strategies, the pure Pareto optimal payoffs,
the characteristic function, the Shapley values, and the Nash bargaining solution.
1 (30 pts). Players: the row player Ann; the column player Bob.
|
0, 1 |
1, 0 |
0, 0 |
1, 3 |
|
1, 1 |
2, 3 |
2, 2 |
2, 3 |
|
0, 3 |
0, 0 |
1, 3 |
3, 3 |
|
1, 0 |
0, 0 |
0, 0 |
1, 4 |
Solution. Two equilibriua in pure joint strategies, (row 2, column 2; row 3, column 4):
|
0, 1 |
1, 0 |
0, 0 |
1, 3* |
|
1*, 1 |
2*, 3* |
2*, 2 |
2, 3* |
|
0, 3* |
0, 0 |
1, 3* |
3*, 3* |
|
1*, 0 |
0, 0 |
0, 0 |
1, 4* |
Two Pareto optimal pure payoffs: (3, 3) and (1,4).
v(Ann) is the value the value of the following matrix game
|
0 |
1* |
0 |
1 |
|
1*' |
2* |
2* |
2 |
|
0 |
0 |
1 |
3* |
|
1 |
0 |
0 |
1 |
There is a saddle point, row 2 column 1, so v(Ann) = 1.
v(Bob) = 3 = the value of the matrix game (where Bob is the row player)
|
1* |
1 |
3 |
0' |
|
0’ |
3 |
0' |
0' |
|
0’ |
2* |
3 |
0’ |
|
3* |
3 |
3*' |
4* |
v(empty)=0, v(Ann,Bob) = 6.
Nash bargaining: (u-1)(v-3) -> max , u
>= 1, v >= 3, (u, v) is a mixture of (3, 3) and (1,4).,
u + 2v =9, (u – 1) + 2(v – 3) =2, (u-1)=
2( v -3)=1; u=2, v=3.5 optimal solution= Nash solution=
the arbitration pair.
This arbitration pair is a mixture of given payoffs: (2, 3.5) =(1/2)(1, 4) + (1/2)(3.3).
Ann Bob
Ann Bob 1 5
Bob Ann 3 3
------------------
Shapley values 2 4
2 (45 pts). Players: A, B, C
strategies payoffs
1 1
1 2
3 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 3 0
Solution.
Three equilibria and one Pareto optimal triple (2,3,3):
strategies payoffs
1 1
1
2* 3* 2* equilibrium strategy
1 1
2
1 0 1
1 2
1
2* 3* 3* equilibrium & Pareto optimal
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 & Pareto optimal
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 3 0
v(empty)=0, v(A,B,C)) = 8.
v(A) = 1 = the value of the matrix game
|
2 |
1*' |
2 |
1 |
|
0 |
1 |
2 |
2 |
|
0 |
0 |
1 |
2 |
v(B) = 1 = the value of the matrix game
|
3 |
0 |
0 |
1 |
0 |
0 |
|
3 |
2 |
3 |
3 |
1*' |
3 |
v(C) = 0 = the value of the matrix game
|
2 |
3 |
1 |
3 |
0*' |
1 |
|
1 |
3 |
1 |
2 |
0 |
0 |
v(A,B) = 5= the value of the matrix game
|
5 |
1 |
|
5 |
3 |
|
0 |
2 |
|
5 |
5*' |
|
0 |
0 |
|
2 |
5 |
v(A,C) = 4= the value of the matrix game
|
4*' |
5* |
|
2 |
4 |
|
1 |
5 |
|
2 |
4 |
|
0 |
2 |
|
0 |
2 |
v(B,C) = 3 = the value of the matrix game
|
B&C vs A |
c1 |
c2 |
c3 |
|
r11 |
5 |
1 |
0 |
|
r12 |
1 |
2 |
0 |
|
r21 |
6 |
6 |
2 |
|
r22 |
5 |
5 |
3*' |
order contribution
A B C
ABC 1 4 3
ACB 1 4 3
BAC 4 1 3
BCA 5 1 2
CAB 4 4 0
CBA 5 3 0
---------------
10/3 17/6 11/6 Shapley values.
The Nash solution is (2, 3, 3)., the only Pareto optimal payoff.