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 c1 c2 c3 r11 3 1 0 r12 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
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7/3  23/6   11/6     Shapley values. The mixed Pareto optimal payoffs = mixtures of  (2,4,0) and (2,3,3). >Nash solution: the optimal solution of

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).