1

a 1A1

a2

a3A2

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 ai with probability xi and

• let Player II play strategy Aj with probability yj

3

2.4 Mixed Strategies

Player II

Player I

Relative frequencyof application

4

a1 x1

a2 x2

A1 A2

y1 y21 56 2

⎡ ⎣ ⎢

⎤ ⎦ ⎥

• Since x1, x2, y1 and y2 are probabilities• x1 + x2 = 1 and y1 + y2 = 1 • One possibility would be for Player I to use a1 3/4 of

the time and a2 1/4 of the time whilst • Player II to uses A1 1/2 of the time and A2 1/2 of the

time.• BUT, WHAT IS THE BEST COMBO ???

Example

5

2.4.1 Definition• A mixed strategy for Player I is a vector

x = (x1,... ,xm) with xi ≥ 0 for all i and i xi = 1.

• Similarly, a mixed strategy for Player II is a vector y = (y1,... ,yn) with yj ≥ 0 for all j and j yj = 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 xi yj vij = 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.

7

2.4.2 Example

• No saddle.

• E(x,y) = xVy = (x1,x2) V (y1,y2) = (x1, x2)(y1 + 5y2, 6y1 + 2y2) = x1y1 + 5x1y2 + 6x2y1 + 2x2y2

• For x = (0.5, 0.5) we obtain E(x,y) = 3.5(y1 + y2) = 3.5 for any y, since y1 + y2 = 1.• Note that this is better than the optimal security level

for Player I (equal to 2) !!!!• BUT, CAN WE DO BETTER?

V =

1 5

6 2

È

Î Í

˘

˚ ˙

8

• Similarly, if y = (0.5, 0.5), we have E(x,y) = 3x1 + 4x2 < 4x1 + 4x2 = 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.

V =

1 5

6 2

È

Î Í

˘

˚ ˙

9

Notation

• S := Set of feasible mixed strategies for Player I, ie.

S:={(x1,... ,xm): xi ≥ 0, i xi = 1}• T:= Set of feasible mixed strategies for Player

II, ie. T:={(y1,... ,yn): yj ≥ 0, j yi = 1}

• Our aim is to choose “the best” of all the elements in S and T.

10

2.4.3 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}.

11

1.4.1 Theorem

• There exists the following equalities:s(x) = min{xV.j : j=1,2,...,n} and

(y) = max{Vi . 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: y1 + ... + yn = 1, yj ≥ 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 yj = 1 ?• A: The j that has the least cj value!

13

2.4.5 Definition

• The optimal security level for Player I is equal to

• v1 := max {s(x): x in S}• and the optimal security level for Player

II is equal to• v2 := min {(y): y in T}• If v1 = v2 we call the common quantity the

value of the game.

14

Example 2.4.2 (Continued)

• v1 = 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{x1 + 6x2, 5x1 + 2x2} x in S} = max {min{x1 + 6 – 6x1, 5x1 + 2 – 2x1}: 0 ≤ x1 ≤ 1}

(using x1 + x2 = 1 ) = max {min{–5x1 + 6, 3x1 + 2}: 0 ≤ x1 ≤ 1}

V =

1 5

6 2

È

Î Í

˘

˚ ˙

15

01

23456

0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1

3 x1 + 2

x1

–5x1 + 6

16

01

23456

0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1

–5 x1 + 6 3 x1 + 2

x1

minimum of the two lines

17

01

23456

0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1

– 5 x1 + 6 3x1 + 2

x1

• maximum of the minimum function here

18

•v1 = max {min{–5x1 + 6, 3x1 + 2}: 0 ≤ x1 ≤ 1}.x*1 = 1/2; x*2 = 1 – x*1 = 1/2; v1 = 3x*1 + 2 = 7/2.

01

23456

0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1

–5x1 + 6 3x1 + 2

x1

19

For Player II

• v2 = min{(y): y in T} • = min{ max {Vi . y: i = 1,...,m}: y in T}• = min { max {y1 + 5y2 , 6y1 + 2y2}: y in T}• = min { max {–4y1 + 5, 4y1 + 2}: 0 ≤ y1 ≤ 1}

20

01

23456

0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1

–4y1 + 5 4y1 + 2

y1

21

01

23456

0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1

–4y1 + 54y1 + 2

y1

Max of two lines

22

01

23456

0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1

–4y1 + 5 4y1 + 2

y1

• v2= min { max {–4y1 + 5, 4y1 + 2}:0 ≤ y1 ≤ 1}

y*1 = 3/8; y*2 = 1 – y*1 = 5/8; v2=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)(x1 + x2) = 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*.

24

1.4.4 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.

−1 32 1 ⎛ ⎝ ⎜ ⎞

⎠