1. 1

2. 2

3. 3 1.6.2 Dominance Since Player I is maximizing her security level, she prefers “large” payoffs. If one row is smaller (element- wise) than another, then Player I will not use it with positive probability. Similarly, Player II prefers “small” payoffs, thus will not consider using a column that is larger (element-wise) than another column. This is what is behind the notion of dominance.

4. 4 1.6.1 Definition Row i of the payoff matrix V is dominated by row k if v ij ≤ v kj, for all j and v ij < v kj, for some j. Similarly, column j is dominated by column k if v ij ≥ v ik, for all i and v ij > v ik, for some i.

5. 5 1.6.3 Theorem You can delete dominated rows and columns and solve the reduced game. Then assign 0 weight to the x* and y* elements corresponding to the dominated rows and columns.

6. 6 2.6.8 Example

7. 7

8. 8

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10. 10

11. 11

12. 12 The reduced game is 2x2, since no saddle, we can use the recipe: x’ = (2/3, 1/3) ; y’ = (1/6, 5/6) x*=(2/3, 1/3, 0) ; y*=(0, 1/6, 0, 5/6)

13. 13

14. 14 1.6.1 2x2 Games Theorem 1.6.1 V has no saddle points if and only if either a > b, a > c, d > b, d > c or a < b, a < c, d < b, d < c. Proof: See lecture notes (case analysis)

15. 15 1.6.2 Theorem Assume that V has no saddle points and let r = a + d – (b + c). Then the value of the game is equal to v = (ad – bc)/r and the optimal (mixed strategies) are: x* = (d – c, a – b)/r y* = (d – b, a – c)/r Only for 2x2

16. 16 Proof Since V does not have saddle points, Theorem 1.6.1 entails that that r is not zero, and it is of the same sign as each of d – c, a – b, d – b, and a – c. This ensures that x* is in S and y* is in T. WHY? Next we compute xVy* and x*Vy for an arbitrary pair (x, y) in S 5 T and show that xVy* ≤ x*Vy* ≤ x*Vy thereby establishing that (x*, y*) is indeed in equilibrium.

17. 17 x*Vy = (d – c, a – b)/r (V)(y 1, y 2 ) = (1/r)(ad-bc,ad-bc)(y 1,y 2 ) = (1/r)(ad-bc)y 1 + (1/r)(ad-bc)y 2 = (1/r)(ad-bc)(y 1 +y 2 ) = (1/r)(ad-bc), (recall that y 1 +y 2 =1) Similarly for y*: xVy* = (1/r)(ad-bc) It therefore follows that x*Vy = x*Vy* = xVy* for all (x,y) in S5T and x*Vy*= v = (ad–bc)/r Hence clearly (x*, y*) is an equilibrium point and hence also optimal.

18. 18 1.6.2 Example There is no saddle point. r = 1 + 2 – (5 + 6) = – 8 v = (ad – bc)/r = (2 – 30)/(– 8) = 7/2 x* = (d – c, a – b)/r = (5/8, 3/8) y* = (d – b, a – c)/r = (1/2, 1/2).

19. 19 How might the players decide what to play, on a given play of the game, if playing such mixed strategies? SOME SUGGESTIONS For (5/8, 3/8): Put 8 pieces of paper, 5 red and 3 green, in a hat. Draw one. If its red Player I plays strategy a 1, If its green Player I plays strategy a 2. For (1/2, 1/2): Toss a coin Look at the seconds reading on a digital watch. Player II plays A 1 if it shows odd, Player II plays A 2 if it shows even.