unit ii-chapter 2-game theory

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UNIT II-CHAPTER 2

GAME THEORY

y Introduction

y Characteristics of Games Theory

y Definitions

y Minimax Criteriay Saddle Point

y Rectangular Games without Saddle Point

y Solution of m x n games by LP Method

y Principles of Dominancey Applications of Game Theory

KOPPAR & ASSOCIATES, CHARTERED ACCOUNTANTS5/21/2011



Introduction

y Game is defined as an activity between two or morepersons involving activities by each person according to aset of rules, at the end of which each person receives somebenefit or satisfaction or suffers loss (negative benefit).

y The set of rules defines the game.

y Game theory is a type of decision theory in which oneschoice of action is determined after taking into account allpossible alternatives available to an opponent playing thesame game, rather than just by the possibilities of severaloutcomes.

KOPPAR & ASSOCIATES, CHARTERED ACCOUNTANTS5/21/2011



Introduction

y The game theory finds use in everyday life situations aslife is a series of struggles & competitions.

y The game theory is fundamentally based upon the

minimax or maximin criterion given by J. VanNeumann who is called the father of game theory.

y The criterion implies that each player will act so as tomaximize his minimum gain or minimize hismaximum loss. Here it is assumed that each person

will act rationally.

5/21/2011KOPPAR & ASSOCIATES, CHARTERED

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Characteristics of Game Theory

y

The various types of games can be classified on the basis of thefollowing characteristics.

i. Chance or strategy : If in a game, activities are determined by skill, it is said to be a game of strategy. If they are determined

by chance, it is a game of chance.

ii. Number of Persons: A game is called a n-person game if thenumber of persons playing is n. The person means anindividual or a group aiming at a particularobjective.

iii. Number of Activities: These may be finite or infinite.


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Characteristics of Game Theory

iv. Number of alternatives (choices) available to each person:In a particular activity, choices may be finite or infinite. A finitegame has a finite number of activities, each involving a finitenumberof alternatives. Otherwise the same is said to be infinite.

v. Information to the players about the past activities of other

players: Completely available, partly available or not available atall

vi. Pay off : A quantitative measure of satisfaction a person gets atthe end of each play is called a pay off. If Vi = o , where Vi isthe payoff to the player Pi ( 1 i n ) then the game is said tobe a zero-sumgame.


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Definitions

y Competitive Game: A competitive situation is called a competitive

game if it has the following 4 properties.

There are finite number (n) of competitors called players such thatn 2. In case n=2, it is called a 2 person game and in case n > 2, it isreferred to as n-person game.

Each player has a list of finite number of possible activities.

A play is said to occur when each player chooses one of his activities.The choices are assumed to be made simultaneously i.e. no playerknows the choice of the other until he has decided on his own.

Every combination of activities determines an outcome (which may be points, money or anything else whatsoever) which results in again of payments (+ve, -ve, zero) to each player, provided eachplayer is playing uncompromisingly to get as much as possible.Negative gain implies the loss of same amount.


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Definitions

y Zero-sum & Non-zero-sum Games: Competitive gamesare classified according to the number of players involvedi.e. as a two-person game, three-person game etc. Anotherdistinction is between zero-sum games and non-zero-sumgames. If the players make payments only to each other i.e.the loss of one is the gain of others, and nothing comesfrom outside, the competitive game is said to be zero-sum.Mathematically, it is represented as Vi = o

y A game which is not a zero-sum is called a non zero-sumgame. Most of the competitive games are zero-sum games.


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Definitionsy Strategy : A strategy of a player has been defined as a rule for decision-

making in advance of all the plays he decides the activities he shouldadopt.

y The strategy may be of two kinds.

a) Pure Strategy : If a player knows exactly what the other player is goingto do, a deterministic situation is obtained and the objective function isto maximize the gain. Therefore, a pure strategy is a decision rulealways to select a particularcourse of action.

b) Mixed Strategy : If a player is guessing as to which activity is to beselected by the other on any particular occasion, a probabilisticsituation is obtained and the objective function is to maximize theexpected gain.


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Definitions

y Tw o-persons, zero-sum Games [Rectangular Games]: A game with only 2 players is called a two-person, zero-sum game if the losses of one player are equivalent to thegains of the other, so that the sum of their net gains is zero.

y Two-person, zero-sum games are also called rectangulargames as they are usually represented by a pay off matrix inrectangular form.


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Definitionsy Pay off Matrix: Suppose the player A has m activities and the player B

has n activities then a payoff matrix can be formed by adopting thefollowing rules.

y Row designation of each matrix are activitiesavailable to player A

y Column designations for each matrix are activities available to player B.

y Cell entry V ij is the payment to player A in As pay off matrix when A chooses the activity i and B chooses the activity j.

y With a zero-sum, two-person game, the cell entry in the player Bs pay off matrix will be negative of the corresponding cell entry V ij in theplayer As pay off matrix so that the sum of pay off matrix for player A and player B is ultimately zero.


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Minimax (Maximin) Criterion and

Optimal Strategy

y The minimax criterion of optimality states that if a playerlists the worst possible outcomes of all his potentialstrategies, he will choose that strategy to be most suitablefor him which corresponds to the best of these worst

outcomes. Such a strategy is called an optimal strategy.

y This can be illustrated with the help of the followingexample.


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Concept Problem

y

Table below illustrates a game, where competitors A & Bare assumed to be equal in ability & intelligence. A has achoice of strategy 1 & strategy 2 while B can select strategy 3 or strategy 4.

Competitor B

Competitor A


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Strategy Strategy

Strategy +4 +6

Strategy +3 +5



Concept Problem

y

Both competitors know the pay offs for every possible strategy.The game favours A since all values are positive. Values thatfavour B would be negative. Based upon these conditions, gameis biased against B. However, since B must play the game he willplay to minimize his losses.

y

The various possible strategies for the two competitors are

A wins the highest game value if he plays strategy 1 all the time sinceit has higher values than strategy 2

B realizes this situation and plays strategy 3 in order to minimize his

losses since the value of 4 in strategy 3 is lower than the value of 6 instrategy 4.


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Concept Problem

y The game value is 4 since A wins 4 points while B losses 4 points

each time the game is played, the game value is the average winnings per play over a long number of plays.

y The above is a zero-sum game since A wins as much as B loses.

y From the above we can conclude that

The strategy that A should use is one that ensures that hisaverage gain per play is at least equal to the value of the game(maximizing his minimum gains)

The strategy that B should use is one that his average loss perplay is no more than the value of the game (minimizing hismaximum losses).


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Saddle Point (Equilibrium Point)

y The concept of saddle point is illustrated with the help of the following example.


ACCOUNTANTS

Player B

Player A

P Q Minimumof row

L -3 3 -3

M -2 4 -2

N 2 3 2

Maximumof column

2 4 2 [Maximin &Minimax]



Saddle Point (Equilibrium Point)

y Player A would gain (-3) if uses strategy L, (- 2) if he uses strategy M &

(+2) if he uses strategy N. So he will play strategy N as it will maximizehis minimum gain. This strategy is maximin strategy and hiscorresponding gain is called the maximin value of the game.

y Player B, on the other hand, wants to minimize his losses. If he playsstrategy P, he can lose no more than max (-3, -2, 2) = 2 regardless of Asselection. If he plays his second strategy Q, he stands to lose max (3, 4,

3)=4 .y He will, therefore, select that strategy that minimizes his maximum

loss. So he chooses strategy P and his max loss will be (2). This iscalled minimax value of the game.

y In the instant case, minimax value = maximin value. When the twoare equal, the corresponding pure strategies are called optimal

strategies and the game is said to have a saddle point orequilibrium point.

y Saddle point is the number w hich is lo w est in its ro w and highestin its column.

5/21/2011 KOPPAR & ASSOCIATES, CHARTERED ACCOUNTANTS



Concept Problem 1y Find the saddle point


Player B

Player A

X Y Z Minimum

of row

P 1 13 11

Q -9 5 -11

R 0 -3 13

Maximum of column



Concept Problem 2

Player B

Player A

X Y Z Minimumof row

P 3 -4 8

Q -8 5 -6

R 6 -7 6

Maximumof column


Find the saddle point



Rectangular Games without Saddle Point

y In cases where there is no saddle point, players will resort tomixed strategies.

y In a 2 x 2 game, the following method provides an easy way of finding the optimum strategies for each player. It consists of thefollowing steps.

y Step 1: Subtract the two digits in column 1 and write them undercolumn 2, ignoring sign.

y Step 2: Subtract the two digits in column 2 and write them undercolumn 1, ignoring sign.

y Step 3: Similarly proceed for 2 rows. These values are calledoddments. They are the frequencies with which the players mustuse their courses of action in their optimum strategies.




Concept Problem 1

y

In a game of matching coins, player A wins Rs 2 if there are2 heads, wins nothing if there are two tails & loses Rs 1 when there is one head and one tail. Determine the payoff matrix, best strategies for each player and the value of gameto A.




Solution

y

The pay off matrix for A will be


Player B

Player A

H T Oddments

H 2 -1 1 [¼ = 0.25]

T -1 0 3 [¾ = 0.75]

Oddments 1

[¼= 0.25]

3

[¾ = 0.75]



Solutiony Thus for optimum gains, player A should use strategy H for 25% of the

time and strategy T for 75% of the time while player B should use

strategy H 25% of the time and strategyT 75% of the time.y To obtain the value of the game, we can use either As oddments or Bs

oddments.

y Using Bs oddments:B plays H, value of game = V = {1 x 2 3 x 1} / {3 + 1} = -1 / 4

B plays T, V = [1 x -1 + 3 x 0] / [3 + 1] = -1 / 4y Using As oddments:

A plays H, V = [1 x 2 1 x 3] / [3 + 1] = -1 / 4 A plays T, V = [-1 x 1 + 0 x 3] / [3 + 1] = -1 / 4

y

Thus the final solutionof the game is A (1, 3) B (1, 3) V = - 1 / 4

y This is the value of the game to A, i.e. A gains Rs (-1 / 4) i.e. he loses Rs1/4 which B, in turn gets.




Concept Problemsy Find the optimal strategies for the games for which the pay off matrices

are given below also find the value of the game.


Player 2

Player 1

I II

I 1 3

II 4 2

Player Y

Player X

I II

I -4 6

II 2 -3



Solutions

Player 2

Player 1

I II Oddments

I 1 3 2 [2/4]

II 4 2 2[2/4]

Oddments 1[1/4]

3[3/4]


Using 2s oddments:Player 2 plays I, value of game = V = {1 x 2/4}+{4 x 2/4} = 5/2Player 2 plays II, V = [3x 2/4]+ [2 x 2/4] = 5/2



Principles of Dominance to reduce the size of the game

y Rule 1: If each element in one row, say rth of the pay off matrix[ v ij ] is less than or equal to the correspondingelement in the other row, say sth, then the player A willnever choose the rth strategy.

y

The value of the game and the non-zero choice of probabilities remain unaltered even if rth row is deletedfrom the pay off matrix.

y Such rth row is said to be dominated by sth row.




Principles of Dominance- Rule 1

y Consider the following matrix

y In this case every element of 2nd row is less than or equal tothe corresponding element of 3rd row hence 2nd row is said

to be dominated by 3rd row.y The value of the game will remain unaltered even if the 2nd

row is deleted from the pay off matrix.


I II III

I -4 6 3

II -3 -3 4III 2 -3 4




y Rule 2: If each element in one column Cr is greater than orequal to the corresponding element in the other column,say Cs, then player B will never use the strategy corresponding to column Cr.

y In this case column Cs dominates column Cr.





y Consider the same example given above.

y Every element in 3rd column is grater than or equal to thecorresponding element in column 1

y Hence column 1 dominates column 3.y Hence deleting column 3 will not affect the nature of the

game.


I II III

I -4 6 3

II -3 -3 4

III 2 -3 4




y Rule 3: Dominance need not be based on the superiority of pure strategies only. A given strategy can be dominated if itis inferior to an average of two or more other purestrategies.

y

In general, if some convex linear combination of some rowsdominates the ith row, then ith row will be deleted.

y Conversely if the ith row dominates the convex linearcombinations of some other rows, then one of the rowsinvolved in the combination may be deleted.

y Similar arguments for columns also.





y Consider the following game

y None of the pure strategies of player A is inferior to any other of his pure strategies. However, the average of the players As firstand second pure strategies gives

y { 5-1/2, 0+8/2, 2+6/2} (2, 4, 4)y This is superior to player As third pure strategy; so the third row

may be deleted from the matrix.


I II III

I 5 0 2

II -1 8 6

III 1 2 3



Principles of Dominance-Concept Problems

y Solve the following game by using the rules of dominance


Player B

Player A

I II IIII 1 7 2

II 6 2 7

III 5 2 6



Principles of Dominance-Concept Problem

Solution

y Consider the rows first, row 3 is inferior to the row 2, so row 3 can bedeleted from the pay off matrix. Now consider the columns, column 3is dominated by column 1, therefore, column three will be deleted.

y The resultant pay off matrix will be


Player B

Player A I II

I 1 7

II 6 2



Principles of Dominance-Concept Problem

Solution

y This is a 2x2 game without saddle paint; this can be solvedeither by LPP(simplex method) or by system of simultaneous equations.

y x1+6x2 = v ; x1+x2 =1 ; 7 x1+2x2 = v [for player A]

y y 1+y 2 = v; 6y 1 + 2y 2 = v; y 1+y 2=1 [for player B]y x1 = 2/5, x2 = 3/5

y so strategy for A = (2/5, 3/5, 0)

y y 1 = 1/2, y 2 = 1/2

y So strategy for B = (1/2, 1/2, 0)




Solution of m x n games by

LP (Simplex method)y The procedure to solve the game by simplex method isdescribed below by means of an example.

y Solve the following 3 x 3 games by simplex method


Player B

Player A

I II III

I 3 -1 -3

II -3 3 -1

III -4 -3 3



Solution

y S

tep 1 : First apply Minimax (Maximin) criteria to find the value of thegame


Ro w Min

3 -1 -3 -3

-3 3 -1 -3 Maximin= -3

-4 -3 3 -4

ColumnMax

3 3 3

Minimax=3



Solution

y

Since maximin value is -3, the value of the game may be negative orZero as -3< v <3

y A constant C is added to all the elements of the matrix which is at leastequal to the negative of the maximin value.

y i.e., C 3; let c = 5, then the matrix can be rewritten as


Player B

Player A

I II III

I 8 4 2

II 2 8 4

III 1 2 8



Solution

y Players Bs LPP is

y Maximize , y 0 = Y 1 + Y 2 + Y 3 Subject to

8Y 1 + 4Y 2 + 2Y 3 1

2Y 1 + 8Y 2 + 4Y 3 1

Y 1 + 2Y 2 + 8Y 3 1

Y 1, Y 2, Y 3 0

y Step 2 : Introduce slack variables and construct the simplextable

8Y 1 + 4Y 2 + 2Y 3 + S1 = 1

2Y 1 + 8Y 2 +4Y 3 + S2 = 1

Y 1 + 2Y 2 + 8Y 3 +S3 = 1


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Solution

BV CB

YB

Y1

Y2

Y3

S1

S2

S3

Min

Ratio

S1 0 1 8 4 2 1 0 0 1/8

S20

1 2 8 40

10 ½

S3 0 1 1 2 8 0 0 1 1

y0

= 0 -1 -1 -1 0 0 0


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CJ = 1 1 1 0 0 0



Solution

BV CB

YB

Y1

Y2

Y3

S1

S2

S3

Min

ratio

Y1 1 1/8 1 1/2 1/4 1/8 0 0 1/2

S2 0 3/4 0 7 7/2 -1/4 1 0 3/14

S3 0 7/8 0 3/2 31/4 1/8 0 1 7/62

y0 = 1/8 0 -1/2 -3/4 1/8 0 0


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CJ = 1 1 1 0 0 0



Solution

BV CB

YB

Y1

Y2

Y3

S1

S2

S3

Min

ratio

Y11 3/31 1 14/31 0 4/31 0 -1/31 3/14

S2 0 11/31 0 196/31 0 -6/31 1 14/31 11/196

Y31 7/62 0 6/31 1 -1/62 0 4/31 7/12

y0 = 13/62 0 -11/31 0 7/62 0 3/31


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CJ = 1 1 1 0 0 0



Solution

BV CB

YB

Y1

Y2

Y3

S1

S2

S3

Min

ratio

Y1 1 1/14 1 0 0 1/7 -1/14 0

Y2 1 11/196 0 1 0 -3/98 31/196 -1/14

Y3 1 5/49 0 0 1 -1/98 -3/98 1/7

y0

= 45/196 0 0 0 5/49 11/196 1/14


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Solutiony Solution for Bs Problem isy

y 1 = Y 1/y 0 = 1/14 / 45/196 = 14/45y y 2 = Y 2/y 0 = 11/196 / 45/196 = 11/45y y 3 = Y 3/y 0 = 5/49 / 45/196 = 20/45

y The optimal strategies for player A are given by duality rulesy x0 = y 0 = 45/196y X

1=

4= 5/49

y X 2 = 5 = 11/196y X 3 = 6 = 1/14

y x1 = X 1/x0 = 20/45y x2 = X 2/x0 = 11/45y x3 = X 3/x0 = 14/45

y Therefore V = [1/y 0] C = [196/45] -5 = -29/45y Value of the game = -29/45 [for B]y Value of the game for A = 29/45.


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unit ii-chapter 2-game theory

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