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

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