Math 486  Game Theory Miderm 3  April 20, 2004.  2 problems, 32 pts each.
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.   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, 0
(payoffs for  Ann and Bob)

Solution.   Two equilibria:
 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, 0*
--- (row 2, column 2) and (row 4, column 1).

Two Pareto optimal payoffs: (3, 2) and (1,3).

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) = 0 = the value of the matrix game
 1 1 3 0*' 0 3 0 0*' 0 0 3* 0*' 1 2 2 0*'

v(empty)=0, v(Ann,Bob) = 5.

Nash bargaining:   (x-1)y -> max , x >= 1, y >= 0,    (x,y) is a mixture of  (3, 2) and (1,3).,
x+2y=7,   (x-1)+2y=6,  x-1=2y=3;  x=4, y=1.5  optimal solution on line,; x=3, y=2 the arbitration pair.

Ann  Bob
Ann Bob           1        4
Bob Ann           5        0
------------------
Shapley values   3       2

2.   Players: A, B, C
strategies                                                payoffs
1  1  1                      0  1  2
1  1  2                      1  0  1
1  2  1                      2  3  4
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, two Pareto optimal:
strategies                                                payoffs
1  1  1                      0  1  2
1  1  2                      1  0  1
1  2  1                      2*  3*  4*  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
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 4 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*' 6 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 7 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 bargaining:
x(y-1)z -> max, x >=0 , y>= 1,  z > = 0,  (x,y,z) is a mixture of  (2,3,4) and (2,4,3), so x = 2.
y+z=7, (y-1)+z=6,  y-1=z=3, (x,y,z)= (2,4, 3) the arbitration triple.

one student came late