1 a1a1 A1A1 a2a2 a3a3 A2A2

2 2.4 Mixed Strategies When there is no saddle point: We’ll think of playing the game repeatedly. We continue to assume that the players use the same basic philosophy and principles as before (that is, to find the “safest” strategy that best protect yourself from loosing). However we now assume that the players can mix up the strategies that they use. Let Player I play strategy a i with probability x i and let Player II play strategy A j with probability y j

3 2.4 Mixed Strategies Player II Player I Relative frequency of application

4 Since x 1, x 2, y 1 and y 2 are probabilities x 1 + x 2 = 1 and y 1 + y 2 = 1 One possibility would be for Player I to use a 1 3/4 of the time and a 2 1/4 of the time whilst Player II to uses A 1 1/2 of the time and A 2 1/2 of the time. BUT, WHAT IS THE BEST COMBO ??? Example

Definition A mixed strategy for Player I is a vector x = (x 1,...,x m ) with x i ≥ 0 for all i and  i x i = 1. Similarly, a mixed strategy for Player II is a vector y = (y 1,...,y n ) with y j ≥ 0 for all j and  j y j = 1. A pure strategy is a vector x, where one component is 1 and all other components are 0. eg. (0, 0, 1, 0, 0). So if a person uses a pure strategy they play the same option all the time. (This is what we do when there is a saddle.)

6 Expected Payoff Let E(x,y) denote the expected value of the payoff to Player I given that she uses strategy x and Player II uses strategy y. By definition then, E(x,y) :=  i,j x i y j v ij = xVy (Convention: in xVy, x is a row vector, y is a column vector, and V is a matrix.) If we play the game repeatedly many times Player 1 expects to get E(x, y) on average.

Example No saddle. E(x,y) = xVy = (x 1,x 2 ) V (y 1,y 2 ) = (x 1, x 2 )(y 1 + 5y 2, 6y 1 + 2y 2 ) = x 1 y 1 + 5x 1 y 2 + 6x 2 y 1 + 2x 2 y 2 For x = (0.5, 0.5) we obtain E(x,y) = 3.5(y 1 + y 2 ) = 3.5 for any y, since y 1 + y 2 = 1. Note that this is better than the optimal security level for Player I (equal to 2) !!!! BUT, CAN WE DO BETTER?

8 Similarly, if y = (0.5, 0.5), we have E(x,y) = 3x 1 + 4x 2 < 4x 1 + 4x 2 = 4 (why?) Note that the optimal security level for Player II is equal to 5 (for pure strategies). Thus, this mixed strategy is (on average) superior to any pure strategy that Player II can use.

9 Notation S := Set of feasible mixed strategies for Player I, ie. S:={(x 1,...,x m ): x i ≥ 0,  i x i = 1} T:= Set of feasible mixed strategies for Player II, ie. T:={(y 1,...,y n ): y j ≥ 0,  j y i = 1} Our aim is to choose “the best” of all the elements in S and T.

Definition The security level of Player I associated with strategy x in S is the minimum feasible expected payoff to Player I given that she uses x (and that Player II is doing sensible things, that is, y in T). We denote the security level for Player I s(x), ie. s(x):= min {E(x,y): y in T}. Similarly, for Player II, let  (y) denote the security level associated with strategy y in T, namely  (y):= max{E(x,y): x in S}.

Theorem There exists the following equalities: s(x) = min{xV.j : j=1,2,...,n} and  (y) = max{V i. y : i=1,2,...,m} In words, if Player I is using a given strategy x, Player II can restrict herself to pure strategies !!!! If Player II is using a given strategy y, Player I can restrict himself to pure strategies!!! An LP-based proof (See Lecture Notes for an alternative direct proof. )

12 LP based Proof Since by definition E(x,y) = xVy, we have s(x) = min {xVy: y in T} Let c: = xV, then s(x) = min {cy : y in T} = min {cy: y y n = 1, y j ≥ 0 } This is a LP problem with one functional constraint. Thus, a basic feasible solution is of the form y=(0, 0,..., 0, 1, 0, 0,... 0), which is in fact a pure strategy! Similarly for  (y). Q: What is the value of j for which y j = 1 ? A: The j that has the least c j value!

Definition The optimal security level for Player I is equal to v 1 := max {s(x): x in S} and the optimal security level for Player II is equal to v 2 := min {  (y): y in T} If v 1 = v 2 we call the common quantity the value of the game.

14 Example (Continued) v 1 = max {s(x): x in S} = max { min{xV.j : j = 1,2,...,n}: x in S} (using Theorem 1.4.1) = max {min{xV.1, xV.2 }: x in S} = max {min{x 1 + 6x 2, 5x 1 + 2x 2 } x in S} = max {min{x – 6x 1, 5x – 2x 1 }: 0 ≤ x 1 ≤ 1} (using x 1 + x 2 = 1 ) = max {min{–5x 1 + 6, 3x 1 + 2}: 0 ≤ x 1 ≤ 1}

x x1x1 –5x 1 + 6

–5 x x x1x1 minimum of the two lines

– 5 x x x1x1 maximum of the minimum function here

18 v 1 = max {min{–5x 1 + 6, 3x 1 + 2}: 0 ≤ x 1 ≤ 1}. x* 1 = 1/2; x* 2 = 1 – x* 1 = 1/2; v 1 = 3x* = 7/ –5x x x1x1

19 For Player II v 2 = min{  (y): y in T} = min{ max {V i. y: i = 1,...,m}: y in T} = min { max {y 1 + 5y 2, 6y 1 + 2y 2 }: y in T} = min { max {–4y 1 + 5, 4y 1 + 2}: 0 ≤ y 1 ≤ 1}

–4y y y1y1

–4y y y1y1 Max of two lines

–4y y y1y1 v 2 = min { max {–4y 1 + 5, 4y 1 + 2}:0 ≤ y 1 ≤ 1} y* 1 = 3/8; y* 2 = 1 – y* 1 = 5/8; v 2 =7/2. Min of max function

23 Is this solution stable? Let us see if Player I is happy with x* given that Player II is using y*. For any feasible strategy x for Player I we thus have: xVy* = (7/2)(x 1 + x 2 ) = 7/2 for all x in S. Thus, given that Player II is using y*, Player I will be happy with x*, in fact she will be as happy with any feasible strategy. Convince yourself that Player II is happy with y* given that Player I is using x*.

Definition A strategy pair (x*,y*) in S  T is said to be in equilibrium if xVy* ≤ x*Vy* ≤ x*Vy for all (x,y) in S  T. Fundamental questions: Do we always have such pairs? How do we construct such pairs if they exist?

25 Example Solve the two person zero sum game whose payoff matrix is See lecture for solution.