Math 486
Game
Theory Miderm 3 Dec.3, 2009.
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, 4 |
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 3):
|
0, 1* |
1*, 0 |
0, 0 |
1, 1 |
|
1*, 1 |
1*, 3 |
2*, 4* |
2, 2 |
|
0, 3* |
0, 0 |
1, 3* |
3*, 2 |
|
1*, 0 |
0, 0 |
0, 0 |
1, 4* |
Two Pareto optimal pure payoffs: (3, 2) and (2,4).
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’ |
4* |
3 |
0’ |
|
1*’ |
2 |
2 |
4* |
v(empty)=0, v(Ann,Bob) = 6.
Nash bargaining: (u-1)(v-1) ->
max , u >= 1, v >= 1, (u, v) is
a mixture of (3, 2) and (2,4).,
2u + v =8, 2(u – 1) + (v – 1) =5, 2(u-1)= v - 1=2.5;
u=2.25, v=3.5
optimal solution= Nash solution= the arbitration pair.
This arbitration pair is a mixture of given payoffs: (2.25, 3.5) =(3/4)(2, 4) + (1/4)(3.2).
Ann Bob
Ann Bob
1 5
Bob Ann
5 1
------------------
Shapley values 3 3
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 0
Solution.
Two equilibria, two Pareto optimal triples:
strategies
payoffs
1 1
1
0 1 2
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 4 0 Pareto optimal
v(empty)=0, v(A,B,C)) = 8.
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 |
0 |
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 |
2 |
v(B,C) = 4 = the value of the matrix game
|
B&C vs A |
|
|
c3 |
|
|
3 |
1 |
0 |
|
|
1 |
2 |
0 |
|
r21 |
6 |
6 |
2 |
|
r22 |
5 |
5 |
4*' |
order contribution
A B C
ABC 0 5 3
ACB 0 6 2
BAC 4 1 3
BCA 4 1 3
CAB 2 6 0
CBA 4 4 0
---------------
7/3 23/6 11/6 Shapley
values.
a= 2, b >= 1 = v(B), c >= 0=v(C), (b-1)c -> max, (b,c) a mixture of (4,0) and (3,3).
3b+c = 12, 3(b-1)+c= 9, 3(b-1) = c = 4.5, b=2.5, c = 4.5. This optimal solution on the line 3b+c = 12 is outside the interval of Pareto optimal pairs. The closest point on the interval is (3,3).
So the Nash solution is (2,3,3).