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
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20/9  79/18  43/18     Shapley values.

Nash solution = Pareto optimal =  (2,4,3).