8/28  8/30  10/2  10/4  11/6  11/8  
9/4  9/6  10/9  10/11  11/13  11/15  
9/11  9/13  10/16  10/18  11/20  11/22  
9/18  9/20  10/23  10/25  11/27  11/29  
9/25  9/27  10/30  11/1  12/4  12/6 
Two Results of some importance:
Nash's theorem was announced.


















A counter prudential strategy is the best response to the opponent's prudent strategy. See p 70 for game 11.1.
2  3  1 
0  2  4 
1  2  3 
1  4  0 
4  5  1 
2  4  6 
6  3  2 
4  1  5 
Now we considered the properties a noncooperative equilibrium,
p*..q* must have for this game:
p*=(p1*,p2*) q*=(q1*,q2*,q3*). Let ER*=p*Aq*'
and EC*=p*Bq*, then
from the property that these are Nequilibrium we saw
that
4q1*+5q2*+1q3* <= ER* and when < is true then
p1*=0
2q1*+4q2*+6q3* <= ER* and when < is true then p2*=0
and similarly
6p1*4p2*<= EC* and when < is true then q1*=0
3p1*1p2*<= EC* and when < is true then q2*=0
2p1*5p2*<= EC* and when < is true then q3*=0
Now by dividing these inequalities by ER*>0 and  EC* >0 as appropriate and letting x1=q1*/ER*, etc., we have that
4x1+5x2+1x3 <= 1 and when < is true then y1=0
2x1+4x2+6x3 <= 1 and when < is true then y2=0
and similarly
6y14y2<= 1 and when < is true then x1=0
3y11y2<= 1 and when < is true then x2=0
2y15y2<= 1 and when < is true then x3=0.
Now, if these conditions are satisfied for nonnegative
values of x1,x2,x3,y1,and y2, then by using x$=x1+x2+x3 and y$=y1+y2 and
q1#=x1/x$; etc. p1#=y1/y$,..
then q# and p# are non cooperative equilibrium strategies
and ER(p#,q#)=1/x$ and EC(p#,q#)= 1/y$
In general we have the following theorem due to Lemke and Howson that gives these conditions for the existence of of a noncooperative equilibrium.
These inequalities are translated into equalities by
introducing "slack" variables.
the complementarity conditions also translate into
conditions about the relationship between the slack variables and the corresponding
strategy variables.
The discussion then turned to n  person games in normal form and an example of the visualization of a 3 person 2x2x2 game using a cube.
We then started a discussion of games in characteristic function form,
with superadditivity.
ASSUMING that communication and sidepayments are possible we will
focus on the issue of finding a "fair allocation" of the value of the game
to all the players.
We examined the example in the text where v(A)=v(B)=v(C) = 0 ; v(AB)=2,
v(AC)= 4, and v(BC)=6, while v(ABC)=7.
Here we noted how if the coalition of ABC was formed sequentially,
we could determine what each player added in value to the coalition as
it was being formed. There are 6 ways for the coalition to form, so we
computed the added values
A  B  C  
ABC  0  2  5 
ACB  0  3  4 
BAC  2  0  5 
BCA  1  0  6 
CAB  4  3  0 
CBA  1  6  0 
Totals  8  14  20 
We to saw that these totals gave a proportion of contribution of 8:14:20
or 4:7:10. Allocation the 7 units under this scheme gave an imputation
of (4/3, 7/3, 10/3)... which is also the average of the totals by the 6
permutations used in the considering how the added values could have arisen
sequentially.
This imputation is called the Shapley value of the game, and it is
characterised by the following properties that suggest something of its
fairness.
Axioms for Shapley Value q of a game in characteristic form:
1. q depends only on v.
2. If players have symmetric roles, then their payoffs under
q
should be equal.
3. If a player adds no value to any coalition S (a "dummy"), then that
player has a payoff 0 under q.
4. Adding players who are dummies doesn't effect the payoffs to any
of the other players under q.
5. If we add games A and B that have the same players, then the Shapley
value q of the sum is the sum of the respective Shapley values for
A and B
The major theorem which we will proceed with next class is that there
is always one and only one imputation that will satisfy these axioms.
Def'n.:The core of a game is the set of all imputations y where there is no x that dominates y.
Def'n.: A set V of imputations is called a stable set
for v if
(i) for each x and y in V, it is not the case that x<y, and
(ii) if y is not in V then there is some x in V where x dominates y.