1. 1

2. 2 APPENDIX B A PROCEDURE FOR GENERATING AN EQUILIBRIUM POINT FOR 2-PERSON GAMES (That sometimes works!)

3. 3 Basic Idea Player II speaking: "I'll design my strategy, call it y* such that the expected payoff to Player I will be constant regardless of what she will do! This will produce a stable situation for me!!!!!! Thus, to compute y* I have to solve the following equation: xAy* = Constant Any such y* will certainly satisfy xAy* ≤ x*Ay*, for all x in S, regardless of what Player I is selecting for x* (her best strategy)."

4. 4 Needless to say, Player I will attempt to do the same thing and construct a strategy x* such that x*By is independent of y. How then do we solve xAy* = Constant ??? Let z = Ay* Then xAy* = xz = (x 1, x 2,..., x m ) (z 1, z 2,..., z m ) t = (x 1 z 1 + x 2 z 2 +... x m z m ) Now if z 1 = z 2 =... = z m, this becomes xAy* = (x 1 + x 2 +... x m ) z m = z m = constant REGARDLESS OF WHAT x IS!! How do we use this? We can try putting all components of Ay* equal to each other.

5. 5 Particular case: n = m = 2 y* = (y* 1, y* 2 )= (y* 1, 1 – y* 1 ) = (ay* 1 + b(1 – y* 1 ), cy* 1 + d(1 – y* 1 ))

6. 6 ( z 1, z 2 ) = (ay* 1 + b(1 – y* 1 ), cy* 1 + d(1 – y* 1 ) For z 1 = z 2 we obtain ay* 1 + b(1 – y* 1 ) = cy* 1 + d(1 – y* 1 ) So, provided a + d – (b + c) = 0 NOTE: Beware! This y* formula is obtained from matrix A. The x* will use matrix B. y* 1  d  b a  d  (b  c) y* 2  a  c a  d  (b  c) since y* 2 = 1 – y* 1

7. 7 Example 1.8.2 (Continued)

8. 8 checking z=Ay* is constant

9. 9 a b c d Exercise Derive the recipe for x* in the case n = m = 2. Answer: If B = ( ) x* = ((d–c)/(a+d–b–c), (a–b)/(a+d–b–c)) Example: Find an equilibrium pair (X*, Y*) of mixed strategies for the 2-person non-zero sum game with payoff matrix See lecture for solution.

10. 10 Appendix A See lecture for discussion and examples.

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12. 12 Solution concepts of NON-zero-sum games We have two concepts –Security level pairs, Equilibrium pairs The security level idea is not really very good here, because it assumes a player is simultaneously trying to maximize their own payoff, whilst minimizing their opponents payoff. These two objectives are sometimes diametrically opposed. It is no longer true that a player can get rich only by keeping their opponent poor. Also we know that the payoff for secuity level is ≤ that for equilibrium pair.

13. 13 Equilibrium pairs - more acceptable concept, but difficulties with these too. E.g. Only one equilibrium pair, payoff (1,1). But clearly the payoff (5,5) better for both. No satisfactory simple notion of ‘optimal strategy’ and ‘value’ as there is with zero sum games.