1

Games in Strategic FormA game in strategic form is defined by a 2n-tuple:G =(X1, X2,…,Xn,u1,u2,…,un),

where the players are 1,2,…, n, the strategy set of i isThe nonempty set Xi, and

ui: X R is the payoff function of i, where

X=X1 x X2 x…x Xn

is the set of strategy profiles.

The game has one stage. In this stage the players are making simultaneous choices, x1,…,xn, and each player i receives the payoff ui(x1,…,xi,…,xn).

A Player, when making her choice does not know the other player’s choices.

2

IISome terminology:

For every player i we denote by X-i the set of all strategyprofiles of all other players. That is:

X-i =X1 x X2 x…Xi-1 x Xi+1 x… x Xn.

E.g., if n=3, X-2= X1 x X3

A generic element of X-i is denoted by x-i.More over if x = (x1,x2,…, xn) in X, thenx-i = (x1,x2,…,xi-1,xi+1,…,xn)

A game in strategic form is finite if all strategy sets Xi, are finite sets.

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IIIExample:

Player 1 chooses a row, Player 2 chooses a column, and Player 3 chooses a matrix.

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-1For example,u1(1,2,B) = 3, u3(1,2,B) = 5

matrix A matrix B

4

Bimatrix GamesTwo-person finite games are sometimes described by a bimatrix (A,B), where A,B are m x n matrices; Player 1 chooses a row i,1≤ i≤ m, Player 2 chooses a column j, 1≤j≤n, and the playersreceive ai,j, bi,j, respectively. It is sometime useful to describeThe pair of matrices as a matrix, each of its cells is a vectorof the form (ai,j,bi,j). That is,

a1,1

b1,1

a1,2

b1,2

ai,j

bi,j

Player 1 chooses a row i , Player 2 chooses a column j.Player 1 receives ai,j, and Player 2 receives bi,j.

i

j

5

Dominated StrategiesLet xi and zi be two strategies of Player i. We say that

xi dominates zi, or that zi is dominated by xi ifxi yields a better payoff than zi for every choice of theother players. That is,

ui(xi,x-i) > ui(zi,x-i), for every x-i in X-i.

6

Implicate the other

4

4

-10

8

8

-10

-2

-2 c

d

c d

Don’t implicate the other

d is dominated by c.That is, each playerimplicates the other,and both will get two years in jail.

Prisoner’s Dilemma

Player 1 and 2 are being held by the police at separate cells.The police knows that the two together robed a bank, but lackevidence to convict. The police offers each of them the followingdeal; each is asked to implicate her partner. If neither does so,then each gets no time in jail. If one implicate the other but isnot implicated, the first gets off, while the one implicatedgo to jail for 10 years. If each implicate the other, then each goes to jail for 2 years.

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Sequential Elimination

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1: 2>3

1 :2>4

2 :2>1

Example:

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II

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1: 2>3

2: 2>4

3: 2>1

4: 1>2

5: 2>3

Player 1 chooses 1, and Player 2 chooses 2

Example:

9

IIILet G=(X1,…,Xn, u) be a game in strategic form, whereu=(u1,…,un). For every player i, let Zi be a nonempty subset of Xi .

We say that the game H=(Z1,…,Zn, u) is obtained by a one-stage elimination from G if at least for one player i, Zi is a strict

subset of Xi, and for every i, every xi in Di=Xi/Zi is dominated in G by some strategy in Xi.

A sequence G0, G1,…,Gk,.. is an elimination sequence of games,if Gs is obtained by a one-stage elimination from Gs-1 for every

s>0 .Note that if G is a finite game, every elimination sequencestarting with G is a finite sequence. in order to deal with infiniteelimination sequences we introduce the following terminology:

We denote by Gmi , the strategy set of i at the game Gm .

Let G∞i

be the intersection of all Gmi, m≥0. Hence, G∞ is a well-defined game

10

IVin strategic form if G∞

i is a nonempty set for every i. An

elimination sequence terminates at the game H, if itis finite and the last game in the sequence is H, or it isinfinite and G∞=H. If an elimination sequence terminates atthe game H, we say that H is obtained from G by a sequential elimination. An elimination sequence

is maximal, if it terminates at a game that does not have dominated strategies.

11

Recall:The process of elimination is a virtual process run in theplayers’ minds. The game has only one stage !!!!

Definition: Let G=(X1,…,Xn, u) be a game in strategic form,and let x be a strategy profile in X. We say that x=(x1,…,xn) is a solutionobtained by a sequential elimination of dominated strategies,if there exists an elimination sequence of games beginning withG and terminating with H=({x1},…,{xn},u). If such a solution xexists, we say that G is solvable by a sequential elimination ofdominated strategies.

Note: Every elimination sequence of games that terminatesin H must be a maximal elimination sequence, becauseH does not possess dominated strategies.

12

Order of EliminationTheorem 1:Let G be a finite game. All maximal elimination sequencesThat begin with G terminate at the same game. That is,If G0, …,Gs, and H0,..,Ht are maximal elimination sequenceswith G0=H0=G, then Gs=Ht.

Definition: Let xi, zi, be strategies of i, and let Y-i be a subset of X-i. We say that zi dominates xi w.r.t. Y-i, ifui(zi, y-i)>ui(xi, y-i) for every y-i in Y-i.

Note that by Theorem 1 every finite game has at mostone solution obtained by a sequential elimination ofdominated strategies.

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Example

Process 1:First stage: eliminate Row 2 and Row 3.Second stage: eliminate column 1.

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II

Process 2:First stage: eliminate Row 3.Second stage: eliminate column 1 and Row 2.

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Process 3:First stage: eliminate Row 3.Second stage: eliminate Column 1.Third stage: eliminate Row 2.

16

Infinite GamesWe will be mainly interested in a special type of infinite games.Definition: A game in strategic form is a regular game,if for every player i, Xi is a compact metric space, and the payoff functions are continuous on X.

Non-regular infinite games may possess solutions thatviolate our intuition.For example, consider the two=person game, in which X1=X2=(0,1) – the open intervalof all real numbers r, 0<r<1, and u1(x1,x2)=x1, u2(x1,x2)=x2.In this (non-regular) game every pair (x1,x2)is a solution obtained by a sequential elimination of dominatedstrategies. In particular Theorem 1 is false in this example.

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II

A famous example for an infinite regular game, which issolvable by a sequential elimination of dominated strategies is the linear Cournot games described bellow:

For regular games we have:Theorem 2:Let G be a regular game, and let G0, G1,…, Gm,… be anelimination sequence, then the sequence terminates at a game.If the elimination sequence is maximal, the sequenceterminates at a regular game.

Theorem 3: Let G be a regular game. Any twomaximal elimination sequences terminate at the same game.

18

Linear Cournot GamesTwo similar firms are producing a divisible good, say wine. The cost of producing one unit of wine is c≥0. The market demand function for wine is d(p)=Max(b-ap,0), a>0,b>0, where p is the price per unit, and d(p) is the demand for winewhen the market price is p. Let P(x) be the inverse demand function. That is, P(x) is the minimal price in which it is possible to sell x units of wine. That is,

P(x)=a-1(b-x), if x≤b

0, if x≥b.

19

Linear Cournot Games, an Example In the following example c=10, and:

P(x)=100-x, if x≤100

0, if x≥100.

How much should firm 1 produce? Note that if a firm producesmore than 90 units, the market price will be less than 10, andthe firm cannot possibly make a non-negative profit.Hence, we assume that a firm does not produce more than90 units.

20

IIThe profit of each firm depends also on the other firm’s decision.Hence we have a two-person game G=(X1,X2,u1,u2), where fori=1,2, Xi is the set of all feasible levels of production for Firm i.That is, X1=X2=[0,90], andu1(x1,x2) = x1P(x1+x2)-10x1, u2(x1,x2) = x2P(x1+x2)-10x2.Note that G is an infinite regular game.We show that for player 1, every z1>45 is dominated by 45. That isu1(45,x2) > u1(z1,x2), for every x2 in X2.

21

IIIProof:

u1(x1,x2) =

x1(90 – x2 – x1), if x1 ≤100-x2,

-10x1, if 100-x2 < x1 ≤ 90.

Fixed z1, 45 < z1 ≤ 90. Fixed x2 , 0 ≤ x2 ≤90,We plot the graph of u1(x1,x2) for 0 ≤ x1 ≤ 90, where

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IV

From the graph, u1(x1,x2) is decreasing when 45 ≤x1 ≤90.As 45<z1, u1(45,x2) >u1(z1,x2) .

QED

0.5(90-x2)90-x2

45

u1(45,x2)

The graph of u1(x1,x2), when x2 is fixed

x190

Note:1. x2 is fixed, but arbitrary.2. 0.5(90-x2)≤45.3. In the graph we plot,45<90-x2. This is not importantfor the conclusion. That is, 45may be located to the right of90-x2.

23

VHence, the interval (45,90] is eliminated for Firm 1 at thefirst stage of our sequential elimination process. Similarly we eliminate (45,90] for the second firm. Hence, after the first stage of elimination X1(1) =X2(1) = [0,45]. In the second stage of elimination we eliminate the interval [0,22.5) for every firm. That is, we show ( for Firm 1) that for everyfixed z1, 0≤z1<22.5,

u1(22.5,x2)>u1(z1,x2), for every x2 in [0,45],

and we show the analogous inequalities for Firm 2. Actually,you show it.

Just draw the graph of u1(x1,x2) for a fixed x2 in [0,45], notethat 0.5(90-x2)≥22.5, and deduce thatu1(x1,x2) is increasing at the interval 0 ≤ x1 ≤ 22.5.

24

VI

We continue inductively. If after the nth round of elimination, n ≥ 0, X1(n)=X2(n) = [an, bn] (a0=0 and b0=90), we define:

Hence, the interval [0,22.5) is eliminated for each firm at thesecond stage of our sequential elimination process. Hence, after the second stage of elimination, X1(2) =X2(2) = [22.5,45].You can guess thatX1(3) =X2(3) = [22.5, 33.75].

bn+1 = (90-an)/2, if n is an even number, and bn+1= bn,if n is an odd number, an+1 = (90-bn)/2, if n is an odd number, and an+1= an,if n is an even number, and X1(n+1)=X2(n+1)=[an+1,bn+1].

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VII

.

It can be easily seen that when n converges to infinity,an 30, and bn 30, and therefore x=(30,30) isthe solution of the Cournot game obtained by a sequentialelimination of dominated strategies.

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Weak Dominationlet xi, zi be strategies of Player i.

xi weakly dominates zi if the following two conditions hold:

• ui(xi, x-i) ≥ ui(zi,x-i), for every x-i in X-i.• There exists x-i in X-i such that ui(xi, x-i) > ui(zi,x-i).

Example2

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Row 1 weakly dominates Row 2

Sequential elimination of weaklydominated strategies yields the solution (Row 1, Column 1)

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IIIn a process of sequential elimination of weakly dominated strategies, we may eliminate the reason for elimination in previous steps.

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Row 1 weakly dominatesRow 2, because of thepossibility that Column 1will be played. However,at the second round of elimination,column 1 is eliminated !!

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IIIEven for a finite game, in the process of sequential elimination of weakly dominated strategies, the order of elimination may make a difference. That is

Two distinct maximal elimination sequences may yield two different solutions.

Please find an example

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Nash EquilibriumDefinition: xi is a best response to x-i if

ui(xi, x-i) ≥ ui(zi, x-i), for all zi in Xi

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Row 3 is a best response ofPlayer 1 to Column 2

Column 1 is a best responseof Player 2 to Row 3

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IIA profile of strategies x = (x1,x2,…,xn) is in equilibriumif xi is a best response versus x-i for every player i.

Definition:

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matrix A matrix B

Example: In the following 3-person game, (Row 2, Column 2, Matrix A) is in equilibrium:

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III

Unlike previous solution concepts, an equilibrium profile is a jointrecommendation. A player should follow the recommendation if he believes that all other players follow their recommendations.

Definition: Let x be in equilibrium in the game G. Theequilibrium payoff of Player i at x is ui(x).

Another example: ( 1,1) is in equilibrium.

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Player 1 should play Row 1 if she believes that Player 2 will playColumn 1.

Player 2 should play column 1 if hebelieves that Player 1 will playRow 1.

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IVIf the players can talk before the game starts, and canagree on a joint play, they must agree on an equilibriumprofile, because , these are the only self-enforcing agreements.

If the players cannot communicate, they still play in equilibrium,because they conclude that this is what they would have doneif they could communicate…..

Equilibrium Assumption in Economics:

Economic Agents play in equilibrium.

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Classical Games

1

1

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c

d

c d

(c,c) is the uniqueequilibrium in this

Chicken

1

1

0

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-1

Dove

Hawk

Dove Hawk

There are twoequilibrium profiles.How would you play thisgame??

Prisoner’s Dilemma

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Battle of the Sexes

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Battle of the sexes

Football

Movie

Foot-ball Movie

Woman’s choices

man’s choices

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No Equilibrium1

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1

1

4

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Equilibrium and DominationTheorem 4: Let G be a regular game. Let x in X be a solution

obtained by a sequentialelimination of dominated strategies, or a solution obtained by a sequential elimination of weakly dominated strategies. Then x is in equilibrium.

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The Traveler’s Dilemma

Acrobat Document

38

IIIThe following theorem shows that by eliminating dominatedstrategies we do not lose equilibrium profiles:

Theorem 5: Let G be game in strategic form, and let x be an equilibrium profile in G. Let G0, G1,…,Gm,… be an elimination sequence in G, then x belongs to Gm for every m≥0.

39

IVThe following simple example shows that equilibriumprofiles may be eliminated in the process of eliminationof weakly dominated strategies:

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2

By eliminating Row 2 at the first stage of elimination, weeliminate the equilibrium profile (2,2).

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VMoreover, even if the game has equilibrium profiles, the elimination process that uses the weak relation may eliminate all equilibrium profiles. Moreover, a maximalelimination sequence may terminate in a game that does not haveequilibrium profiles. Example:

1 3

1 3

2 1

0 2

0 2

2 2

7 0

7 1

0 3

Note that (2,3) is the unique equilibriumprofile in this game. However, Row 2is weakly dominated by Row 1.Please check that the elimination processyields a game that does not have equilibrium profiles.

41

Cournot (again!)Let us find a best response of firm 1 versus x2, 0≤x2≤90.

We wish to find a level of production x1 that maximizesu1(x1,x2) = x1P(x1+x2)-10x1, over x1 in X1=[0,90]. The graph reveals that the optimal level of productionis x1=0.5(90-x2).

Similarly, for each x1 in [0,90] the best response of Firm 2is to produce 0.5(90-x1).Thus,

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Cournot Equilibrium if firm 2 produces x2, and firm 1 knows it, it should produce b1(x2)= 0.5(90-x2), 0≤x2≤90.

If firm 1 produces x1, and firm 2 knows it, it should produce b2(x1)= 0.5(90-x1), 0≤x1≤90.

And

Recall that (z1,z2) is in equilibrium if

•z1 is a best response versus z2, that is z1=b1(z2), and•z2 is a best response versus z1, that is z2=b2(z1).

Solving these two linear equations with twounknowns yield z1=30=z2.

Are you surprised?

43

This Graph May Help

90-

45-

9045

x2

x1z1=30

z2=30

The graph of the best responsefunction, b1(x2) of firm 1.

The graph of the best responsefunction, b1(x2) of firm 1.

Equilibriumz1=30=z2

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In General

3 equilibrium points

No equilibrium

We may have:

45

Player i can guarantee the payoff w if it has a strategy xi such that ui(xi,x-i) ≥ w, for all x-i in X-i.

Definition: Let G be a regular game. the safety level of i at the game G is a numberLi=Li(G) that satisfies:1. i can guarantee Li.2. For every e>0, i cannot guarantee Li+e.

Safety Level

Obviously, every player i in a regular game G has a safety level.Moreover, Li = Max{w: player i can guarantee w}, that is, the safety level of Player i is the maximal payoff that i can guarantee.

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II

Li(xi) = Min {ui(xi,x-i): x-i in X-i}.

Let G be a regular game.For a strategy xi of i, let Li(xi) be the payoff that i guaranteesby playing xi. That is,

Theorem 6: For regular games, Li=Max{Li(xi): xi in Xi}.

Note that by Theorem 6,

Li=Max{Min{ui(xi,x-i): x-i}: xi}.

Therefore, the saftey level of i is also called the MaxMin valueof i.

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matrix A matrix BThe safety level of Player 3 is L3 = -2,and a maxmin strategyis x3 = B

L1 = 2x1 = 1

L2 = -2x2= 1 or 2

IIIxi is a safety level strategy (or a MaxMin strategy) for Player i if xi guarantees the safety level Li. That is, xi is a safety level strategy for i if Li=Li(xi).

Example:

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Player 1 canguarantee

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-1

L1 = maxmin = 3

The safety level strategyof Player 1

L2 = 0 x2 = 1 or 3

Example:

49

Punishing Player iLet G be a regular game. For x-i in X-i, define Hi(x-i) = max{ui(xi,x-i): xi in Xi.

By jointly using x-i, all other players can guarantee that i does not receive more than Hi(x-i). Define:

Hi=Min{Hi(x-i): x-i in X-i}=Min{Max{ui(xi,x-i): xi in Xi}: x-i in X-i}.

Hi is called the punishment level of i, or the MinMax value of i.Obviously,Hi = min{w: all other players can guarantee that i will not get more than w}.

x-i is a punishing-i profile of strategies if it guarantees that i will not receive more than Hi, that is, if Hi=Hi(x-i).

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23 4 4Value of punishment

y1

H1 = 3 punishing strategy = y1

H2 = 4 punishing strategy = x1 or x=2 or x3

Example:

51

Theorem 7: For every regular game G, and for every player i Li ≤Hi. That is,

Max{Min{ui(xi,x-i): x-i}: xi} ≤ Min{Max{ui(xi,x-i): xi}: x-i}.

Proof: There exists a strategy xi that guarantees Li, and there exists x-i that guarantees that i will not get more than Hi. Hence, Hi ≥ ui(xi,x-i) ≥ Li.

Because xi guaranteesLi versus every choice ofthe other players.

Because x-i guarantees that i will notreceive more than Hi for every choice of player i

QED

MaxMin ≤ MinMax

52

Because Hi ≥ Li , Player i,’s payoff in equilibriumis at least her safety level payoff.

MaxMin, MinMax, and Equilibrium

Theorem 8: For regular games, For every equilibrium profile x, ui(x)≥Hi.

Proof: Let z-i be a punishing strategy of all otherplayers. Therefore, Max{ui(zi,z-i): zi}=Hi, and forevery other joint strategy, and in particular for x-i,Max{ui(zi,x-i): zi}≥Hi. As xi is a best response to x-i,ui(x)=ui(xi,x-i)= Max{ui(zi,x-i): zi}. Hence, ui(x)≥Hi. QED

53

Mixed StrategiesExample:

Player 1 can draw a fair coin:

Row 1 Row 2

If the coin falls on the green side she will choose Row 1, and otherwise she will choose Row 2.

Whatever player 2 chooses, the outcome of the game is not deterministic for Player 1 (who makes her calculation beforeshe draws the coin).

6

*

0

*

2

*

6

* The safety level of Player 1is L1=2.

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IILet us denote the mixed strategy of drawing the coin by p1

=(0.5,0.5). From the point of view of player 1, her payoff matrix is:

Hence, using mixed strategies may be a good idea

62

06

34

1

2

p1

where the numbers in the third row represent expected payoffs.However, as a risk-neutral player, Player 1 relates to expectedpayoffs as payoffs. Hence, without using mixed strategies player 1 can guarantee a payoff of 2. By using the coin p1, she canguarantee 3.

55

V

Assume p1 = (z1,1-z1), p2 = (w1,1-w1), p=(p1,p2). Hence,

p(1,1)=z1w1, p(1,2)=z1(1-w1), p(2,1)=(1-z1)w1, p(2,2)=z2w2.

Example:

56

Example:matrix A matrix B

If Player1 uses p1=(0.6,0.4), Player 2 uses p2=(0.5,0.5),and Player 3 uses p3=(0.8,0.2), the probabilities of the cellsare shown above.

0.4*0.5*0.2

0.24 0.24

0.16 0.16

0.06 0.06

0.04 0.04

57

Still MixingLet Ui(p) be the expected payoff of i when all players, use the profile of mixed strategies p. That is,

Ui(p)=x in X

ui(x)p(x)

w 1-w

1

4

3

2

3

-1

4

2 z

1-z

p1 = (z,1-z)p2 = (w,1-w)

U1(p1,p2) = 4zw +3z(1-w) +3(1-z)w +1(1-z)(1-w)

U2(p1,p2) = 2zw -1z(1-w) +2(1-z)w +4(1-z)(1-w)

Example:

58

Pure StrategiesAssume player i has 3 strategies a,b, and d. What is the meaning of the mixed strategy pi(a)=0, pi(b)=1, pi(d)= 0 (pi=(0,1,0))?

The player conducts a lottery whose outcome is b with probability 1. This is obviously equivalent to the usage of the pure strategy b.

For xi in Xi we denote by exi in Mi, the mixed strategy that assignsprobability 1 to xi. exi is called a pure strategy of i.

From the practical point of view, using the pure strategy exi and using the strategy xi are equivalent.

Note that: Ui(ex)=ui(x) for every x in X,where ex=(ex1,…,exn).

59

Let G =(X1, X2,…,Xn,u1,u2,…,un) be a finite game in strategic form.

We define the mixed extension of G as the game Gm, in whichthe strategy set of player i is Mi =M(Xi), and the payoff functionof i is Ui. That is,

The mixed extension

Gm =(M1,M2,…,Mn,U1,U2,…,Un).

Note: Gm is an infinite game.

60

Safety Level and Punishment LevelDefinitions:

The safety level of Player i in the game Gm, Li(Gm), is called

the safety level of i in mixed strategies in G,

and it is denoted by Lim=Li

m(G). That is,

Lim=MaxpiMinp-iUi(pi,p-i).

Him= Minp-i Maxpi Ui(pi,p-i), is the

punishment level of i in mixed strategies

Similarly,

Theorem 12: Hi≥Him≥Li

m≥Li.

61

Example:

6

*

0

*

0

*

6

* Let p1=(0.5,0.5). U1(p1,e1)=3≥3, and U1(p1,e2)=3≥3.Hence L1

m≥3.Let p2=(0.5,0.5). U1 (e1 ,p2)=3≤3, and U1(e2 ,p2)=3≤3.Hence H1

m≤3. Therefore:3≤ L1

m ≤ H1m ≤3, that is

L1m =3= H1

m.

In this game L1= 0, H1=6. Let us find L1m and H1

m.

Note that p1 is a safety level strategy in mixed strategies,and p2 is a punishing-of-1 strategy (to be used by 2) in mixed strategies.Is the equality between the safety level and punishment levelin mixed strategies is a coincidence? No and Yes

62

The MinMax Theorem:Theorem 13 (The MinMax Theorem):For finite two-person games, Li

m=Him, for every player i=1,2.

It does not work with more players

A Useful Corollary

Theorem 14: Consider a two-person finite game. If for everymixed strategy p2, of Player 2, Player 1 has a strategy x1,such that U1(ex1,p2)≥ c, then Player 1 has a mixed strategy p1,such thatU1(p1,p2)≥ c, for every mixed strategy p2 of Player 2.

63

Mixed Strategy EquilibriumDefinition: Let p=(p1,p2,…,pn) be a mixed strategy profilein M. p is a mixed strategy equilibrium in the game G,if p is in equilibrium in the game Gm.Thus, p is a mixed strategy equilibrium if for every i,pi is a best response to p-i in the game Gm. We proceed to explore properties of best-response strategies in Gm:

Theorem 15: Let pi be a best response to q-i. Let zi be a strategy of i such that the associated pure strategy ezi is not a best response to q-i. Then pi(zi)=0.Theorem 16: Let q-i belongs to M-i. Let pi belongs to Mi.pi is a best response to q-i if the following condition holds:For every zi in Xi, pi(zi)>0 implies zee is a best response to q-i.

64

Characterizations of Best Response Strategies:

By theorems 15,16, and by the remark preceding the proofof Theorem 15,

Theorem 17: pi is a best response to q-i if and only if pi

assigns positive probabilities only to pure strategies that are best response to q-i.

Corollary 18: Let q-i be in M-i. Let r=Maxxi Ui(exi,q-i). Then,pi is a best response to q-i if and only if pi(xi)=0for every xi, for which Ui(exi,q-i)<r. Consequently, exi isa best response to q-i if and only if Ui(exi,q-i)=r.

65

Exercises:

Consider a two-person 3x19 game, in which p=(0.7,0,0.3) is a best response to q, and U1(p,q)=6. Then:

1. U1(e1,q)≥6. 2. U1(e2,q)<6. 3. U1(e3,q)=6.

4. U1((0.1,0.4,0.5),q) <6.

T F T

F

Exercise 1: Reply with a “True” or a “False”

66

II

Exercise 2:

Consider a game G with 250 players in which Player 1 has four strategies, 1,2,3.4 Let q be a profile of mixed strategies of all other players.

1. U1(e1,q)=6,2. U1(e2,q)=63. U1(e3,q)=64. U1(e4,q)=5

Suppose the following holds:

AND U1(p,q)≤6 for everymixed strategy p of player 1

p=(0.25,0,0.75,0) is a best response to q

True or False:

p=(0.25,0.25,0.25,0.25) is a best response to q. FT

67

Exercise 3:

Consider a game G with 19 players in which Player 1 has four strategies, 1,2,3. 4Let q be a profile of mixed strategies of all other players.

1. U1(1,q)=6,2. U1(2,q)=63. U1(3,q)=64. U1(4,q)=5

Suppose the following holds:

p=(0.25,0,0.75,0) is a best response to q

True or False:

p=(0.25,0.25,0.25,0.25) is a best response to q. FT

III

68

A Characterization of Mixed Strategy Equilibrium

Theorem 19: (p1,p2,…,pn) is a mixed-strategy equilibrium if and only if the following holds for every player i:For every pure strategy xi, either xi is a best response to p-i

orpi(xi)=0.Proof: The proof is a simple consequence of Theorem 17. QED

A simple corollary of Theorem 19 is that by extending the gameG to the game Gm, we do not loose equilibrium profiles:

Corollary 20: Let G be a finite game.ex is a mixed strategy equilibrium if and only if xis an equilibrium strategy.

69

Example: In this example you learn how to prove that a certainprofile of strategies is a mixed-strategy equilibrium.

0

0

0

0

0

1

0

1

0

1

0

0

1

0

0

0

1

0

0

0

1

0

0

0

Matrix 1 Matrix 2

In the following three-person game, player i wishes to be withplayer i+1 and without player i+2.

70

II

We show that (p1,p2,p3) is in equilibrium, where pi=(0.5,0.5) for every i=1,23.We have to show that pi is a best response to p-i for i=1,2,3.However, because the game is symmetric it suffices to showthat p3 is a best response to p-3.

We show that: U3(p1,p2, ematrix 1)= U3(p1,p2, ematrix 2).

Therefore, by Corollary 17, both ematrix 1 and ematrix 2 arebest response to (p1,p2). Hence, p3 assigns a positiveprobability only to pure strategies, which are best responsesto (p1,p2), and thus, p3 is a best response to (p1,p2).

71

An ExerciseExercise 4: True or False?

Is ((1,0),(0,1)) is a mixed-strategy equilibrium in G?

3

2

1

4

9

-8

0

7

G =

Solution:

True;As (Row 1, Column 2) is in equilibrium in G, then by Corollary 17,(eRow1,eColumn2) is in equilibrium in Gm.

72

Existence of Mixed Strategy Equilibrium

Methods for computing mixed-strategy equilibriumwill be given at the TA class.

Theorem 21 (Nash, 1951): Every finite game has a mixed-strategy equilibrium.Proof: Can be found in the Appendix.

73

Domination

We will not deal with a complete theory of domination withmixed strategies. Just note the following example:

4

*

1

*

4

*

1

*

3

*

0

*

0

*

6

*

1

*

In G, player 1 does nothave dominated strategies.However, with mixed strategies:(0.5,0,0.5) dominates Row 2.

Hence, the elimination process ofstrategies in the game Gmay yield a smaller game, whenmixed strategies are taken intoaccount!

74

Two-Person Zero-Sum Games

A two-person game G=(X1,X2,u1,u2)) is zero-sum, ifu2(x1,x2) = -u1(x1,x2), for every x=(x1,x2) in X.

A finite zero-sum game can be described by a simple matrix,in which we write the payoffs of Player 1.

Thus, the following zero sum game has the two equivalent forms:

1

-1

-2

2

0

0

-3

3

OR

1

-2

0

-3

75

IIPlease note:A zero-sum game is first of all a (two-person) game,and all previous theorems hold.!!!!

Theorem 22: Let G=(X1,X2,u1,u2) be a regular 2-person zero-sum game. Then, L2=-H1, and H2= -L1.

Proof: Let x2 be a punishing-1 strategy. That is, u1(x1,x2) ≤ H1, for every x1 in X1. Hence, -u1(x1,x2) ≥ -H1 for every x1. That is, u2(x1,x2) ≥ -H1 for every x1. Therefore, L2≥-H1.Let z2 be a safety level strategy of Player 2. Therefore, u2(x1,z2) ≥ L2 for all x1, yielding u1(x1,z2) ≤ -L2. Therefore H1≤-L2,or L2≤-H1. Hence, L2=-H1. The proof of the other equality is similar, and therefore it is omitted. QED

76

III

Corollary 23: Let G be a regular two-person zero-sum game. xi is a safety level strategy for i iff xi is a punishment-of-[-i] strategy, i=1,2.

Example:-1

1

3

-3

2

-2

1.5

-1.5

L1=1.5, H2=-1.5 and Row 2 is both, a safety levelstrategy for 1, and a punishing-player 2 strategy.

Similarly,L2=-2, H1=2, and Column 1 is a safety level strategy for 2,which is also a punishing-Player 1 strategy.

77

IVDefinition:A 2-person regular zero-sum game has a value, if Li=Hi, i=1,2.The value of G denoted by v(G) is defined to be L1.

Note that by Theorem 22, it suffices to check only one of the equalities, that is the game has a value iff L1=H1 iff L2=H2.

Definition: A finite 2-person zero-sum game G has a value in mixed strategies, if Gm has a value, that is, if L1

m=H1m.

Theorem 24: Every finite 2-person zero sum game has a valuein mixed strategies.Proof: By the Minmax theorem (Theorem 13), L1

m=H1m. QED

Definition: in a 2-person zero sum game with a value, a safetylevel strategy for i is called an optimal strategy for i.

78

Examples

-1

1

3

-3

2

-2

1.5

-1.5

4

-4

3

-3

2

-2

1.5

-1.5

L1=1.5<2=H1.Hence, the game does nothave a value.

L1=3=H1.Hence, the game has a value, v=3. Row 1is optimal for 1, and Column 2is optimal for 2.

79

Value and Equilibrium

Theorem 25:

Let G be a regular2-person zero-sum game in strategicform. Then,1. G has a value iff it has an equilibrium.2. If G has a value, then (x1,x2) is in equilibriumiff xi is optimal for i, for every i=1,2.3. If G has a value, and (x1,x2) is in equilibrium, then ui(x1,x2)=Li for every i=1,2,and in particular u1(x1,x2)= v(G).

80

A Few Exercises

Question: Let G be a finite 2-person zero-sum game, and let(p,q) and (p*,q*) be mixed strategy equilibrium profiles.Then, (p,q*) is a mixed strategy equilibrium profile. True or False?

True4

-4

3

-3

2

-2

1.5

-1.5

Question: Consider the following game, G=

If (p,q) is a mixed strategy equilibrium in G, then U2(p,q)=-3.True or False?

True