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Problem solving methods in game theoryElaine Audrey TerryAtlanta University

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Recommended CitationTerry, Elaine Audrey, "Problem solving methods in game theory" (1988). ETD Collection for AUC Robert W. Woodruff Library. Paper1796.

PROBLEM SOLVING METHODS IN GAME THEORY

A THESIS

SUBMITTED TO THE FACULTY OF ATLANTA UNIVERSITY

IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR

THE DEGREE OF MASTER OF SCIENCE

BY

ELAINE AUDREY TERRY

DEPARTMENT OF MATHEMATICAL AND COMPUTER SCIENCES

ATLANTA, GEORGIA

JULY, 1988

ABSTRACT

MATHEMATICS

TERRY, ELAINE A. B.A. SPELMAN COLLEGE, 1983

PROBLEM SOLVING IN GAME THEORY

Advisor: Professor Negash Medhin

Thesis dated July, 1988

Game theory is the mathematical theory associated

with winning strategic and non-strategic games. In order

to win a game, a player must find an optimal strategy to

play. Strategies may be either pure or mixed. The latter

is used when there are no pure strategies available • •

Games that require mixed strategies may be solved by

various methods.

This study is concerned with the basic theory of

games. Definitions and methods for solving games are

discussed. The methods for solving involve both pure and

mixed strategies. The simplex method for solving linear

programming problems is reviewed. The numerical examples

were solved using the IBM Macintosh with the MacSimplex

package.

ACKNOWLEDGMENTS

I would like to extend sincerest thanks to Dr. Negash

Medhin and Dr. Nazir Warsi for their patience and

sincerity in handling this project.

ii

TABLE OF CONTENTS

Acknowledgement • • •

Table of Contents • •

CHAPTER

I. GAME THEORY:

. . • • • • • • • . . .

. . • • • • • . . . . .

AN OVERVIEW. . . . . .

. . • • • ii

.... . iii

• • • • •

II. METHODS FOR SOLVING SIMPLE GAMES • • • • • • •

1

4

III.

IV.

v.

Payoff Matrix Dominating Strategies Minimax-Maximin

SOLVING COMPLEX GAMES. . . • • • • • • • • • • 13

Mixed Strategies Graphical Solution

LINEAR PROGRAMMING • • • . . . . . . Simplex Method Review

• • • • • 21

Linear Programming Solution for a Game

NUMERICAL EXAMPLES . . . . . . . . . • • • • • 28

BIBLIOGRAPHY •••••••• • • • • • • . . . • • • • • 37

iii

CHAPTER I

GAME THEORY: AN OVERVIEW

A game is a situation involving two or more persons

who are referred to as players: The players are in

conflict with each other as to the other's objectives in

the game. The result of each player's objectives may be a

win, which is a positive payoff, or a loss, a negative

payoff.

Games ~ay be either strategic or games of chance.

Some strategic games include chess or checkers, while

tossing a coin or die may be thought of as a game of

chance. Those games that are strategic in nature require

much skill and some knowledge of the opponent's possible

moves. Games of chance require none or very little skill.

Everyday strategic games can be political or economical by

nature.

Game theory is a mathematical theory of the game as

it is described above. In game theory there are

opponents, the players, whose objectives may be finite or

infinite. The objectives are called strategies. The wins

and losses of each one of the players are defined as

1

2

payoffs. Most research in game theory has been done with

two opponents. This type is defined as two-person game

theory. The opponents may be two persons or even two

warring countries. Labor-management disputes can be

settled using game theory. Some research has been done

for an n-person game, that is, there are at least n>2

players. An example of an n-person game is an inter­

national disarmament conference where the players are the

participating countries. Games may be further described

as constant-sum or non-constant sum games. In the first

case, rewards or payoffs to the players yield a fixed

constant regardless of the strategy chosen. In the latte·r

case, rewards carry various values. As stated before,

most research deals with a two-person zero-sum game. That •

is, there are two players and the winnings or payoffs to

one of the players is a loss for the other. Thus, the

total of the payoffs is zero always. This research will

be concerned with the two~person zero-sum game.

The theory of games was created to approach and solve

some economic problems. Game theory is used for military,

business, finance, and marketing purposes. The theory has

been used in pure mathematics, political science, and even

in psychology. It can be applied to any type of parlor

game, that is, bridge, chess, checkers, and solitaire. It

is a theory that is applied to games as probability is

3

applied to chance. The basic concern of game theory is

with the feasibility of there being a best method to play

a game. The number of ways or methods of playing the game

may be finite or infinite. Methods of playing a game are

termed the strategies of the game. A game usually ends

after a finite number of moves and thus produces a winner

and a loser. Two-person games are characterized by

(1) two players, (2) each player's strategies, and (3) the

payoffs to each player.

Historically, mathematical models of a national

economy appeared around the 1870s. This limited theory at

the time was proposed by Leon Walras and Vilfredo Pareto

in Recherches sur les Principes Mathematigues de la

Theorie Des Richesses. However, the models used then were

broadened in the late 1920s and early 1930s by the great

mathematician John von Neumann. The work that he

contributed was made available in later years. In this

work there are a finite number of moves or strategies.

There are also payoffs or rewards to the players dependent

upon the moves made. It should be noted that von Neumann

first published this theory at the age of 25. The theory

of games was available to the public in the treatise

Theory of Games and Economic Behavior in 1944.

CHAPTER II

METHODS FOR SOLVING SIMPLE GAMES

Recall that a two-person game is characterized by

three things, the first being the two players, which has

been discussed somewhat earlier. The second and third

characterizations will be studied in this chapter. The

payoffs to players will be put in tabular form and the

strategies are the essentials to begin solving a game.

The payoff matrix will be discussed first, followed by

strategies used to solve simple games.

Payoff Matrix

The children's game stone-paper-scissors can be set

up as a model game. The game consists of two players with

rules of the game as follows: each player chooses one of

stone, paper, or scissors. If both choose the same, then

the payoff to each is zero, that is, there is a tie. As

for the items, paper will cover stone, stone breaks

scissors, and scissors cut paper. Thus, a player receives

+1 for a win and -1 for a loss. The game may be

represented in matrix form from the position of player

one. The matrix is as follows:

4

----- ------ ---------------

stone

paper

scissors

stone

0

+1

-1

5

paper

-1

0

-1

scissors

+1

-1

0

(2-1)

Matrix 2-1 is an example of a payoff matrix that is used

in game theory. The headings, stone, paper, and scissors,

are the strategies available to each player. The values

of the matrix are the payoffs in the form of winnings,

losses, or ties.

The two players in the game must make moves, actions,

and choices in order to compete with one another. The

players are then said to be strategizing or using a

strategy in order to win the game. Prior to each game,

the two players know the strategies available to one

another. However, upon choosing a strategy, neither

player knows for certain which strategy his opponent will

choose.

The payoffs to player one in matrix 2-1 represent the

wins and losses for player one. In general, the rewards

available to player one is in the form of an m x n matrix

(Aij)' where m rows and n columns represent the strategies

available to players one and two, respectively. The total

number of possible outcomes is the product mn. Thus, the

two-person zero-sum game is characterized by: (1) the

6

strategies of each player, and (2) the payoff matrix

relative to player one. The payoff matrix for the

second player is the negative of the payoff matrix for

player one. This is so since any gain or win for player

one is a loss for player two.

The following 3 x 3 payoff matrix represents player

one's payoffs. Both players have three strategies

available. Since this matrix represents player one's

payoffs, he wins in most cases except where -1 in row

three represents a loss. The zero in row two represents a

tie or no win or loss for either· of the players. The

numbers in the matrix may be used to represent units such

as dollars or customers.

1

2

3

1

1

2

-1

2

3

4

0

3

5

8

2

(2-2)

Because of the possible net gain of zero in strategy

three for player one, it is wise that strategy three not

be used. It would be best that player one play strategy

two in hopes that player two will play strategy three.

This results in a net gain of eight for player one.

However, with the idea that player two is competent, he

7

will notice such a large gain for his opponent and play

the strategy that will minimize his loss.

It may be noticed that each player should develop

some rational criteria for choosing a strategy. Thus,

each player must do the best to adopt the most rewarding

strategy. That is, the players must choose an optimal

strategy. However this is done, player one must do so in

order to maximize winnings. At the same time, player two

must do all possible to minimize his losses. Recall that

the payoff matrix represents the wins of player one.

Dominating ~trategies

In payoff matrix 2-2,. it may be that player one 1s

able to play one strategy only throughout the game to

insure a win. That strategy is the best available

strategy of all those in the matrix. If this is so, then

one strategy is said to dominate all of his other

strategies. Because that is the only one played always,

it is said to_be a pure strategy. In payoff matrix 2-2,

player one's strategy two (2, 4, 8) dominates the other

available strategies since the payoff values are greater

than or equal to the respective entries in the other

strategies.

As for player two, since his earnings must be

negative and his winnings small, he may play strategy two

8

(;~) only since its entries are smaller than the other

payoffs. Thus, player two's dominating strategy should

have payoff values less than or equal to other strategic

entries. To use dominated strategies, each player should

rule out inferior strategies successively until there is

only one remaining choice. Each player should eliminate

those strategies that are dominated.

In matrix 2-2, if player one eliminates strategy

three, then the matrix becomes:

1 2 3

1 1 3 5

2 2 4 8

Player two then eliminates strategy three to obtain:

1 2

1 1 3

2 2 4

Player one in return eliminates strategy one to obtain:

1 2

2 2 4

9

where player two should eliminate strategy two. Player

one will receive a value of two from player one. The

value of the game is said to be two. The method of using

dominated strategies is very useful for reducing the size

of large-sized payoff matrices. In some, the value of the

game can be found.

Minimax-Maximin

In reference to matrix 2-2, if player one were to

choose strategy two and player two chooses his strategy

one, then player one wins two and player two loses two.

The value of the game is said to be two. However, suppose

player one were to choose strategy three, then neither

player wins or loses. The value of the game is zero and

is said to be fair. Because player one's strategy two

dominates the other ones, it would be best to choose

strategy three. Player .two should have guessed that such

would happen, and thus chooses strategy two in order to

minimize loss. This process could continue on and on with

each one trying to outguess the other.

In general, the primary purpose for each player is to

minimize maximum losses to his opponent. This is called

the minimax criterion for choosing strategies. That is,

player one will play that strategy that will give him the

largest payoff; however, he will do so at a minimum.

Player two, on the other hand, wishes to minimize losses,

10

so that he will choose that one strategy with the smallest

payoff to player one but at a maximum. Therefore, player

one has the maximin strategy to play while his opponent

plays a minimax strategy. The maximin and minimax will be

values from the payoff matrix. The maximin is the lower

value of the game while the minimax is the upper value of

the game.

Each player is attempting to minimize losses, but

must do so while gaining a maximum reward. Player one has

to determine the largest minimum reward so as not to lose

much to playe_r two. At the same time, player two should

· choose the maximum values in each of his strategies and

select the smallest of them in order to avoid large

losses.

As stated before, player one is choosing the maximin

of his strategies and player two will select his minimax

strategy. This method allows for optimization of

strategies for each of the players. The following example

illustrates the procedure. Let matrix 2-3 be as follows:

l 2 3

l l 4 2

2 8 9 2 (2-3)

3 8 9 5

11

Thus, the minimax and maximin have the same value.

A game that has a saddle point is said to be stable. Both

players should play their respective strategy three.

1

2

3

maximum in column

1

1

8

7

8

2

4

9

6

9

3

3

2

5

minimum in row

1

2 (2-4)

[5] maximin

[5] minimax

Thus, the minimax and maximin have the same value.

The value of the game is five, and since minimax=maximin,

then the value five is the saddle point of the game. That

is to say that the saddle point is a minimum for player

one at the same time that it is a maximum for player one.

A game that has a saddle point is said to be stable. Both

players in this instance will play strategy three only.

Strategy three is an optimal strategy for both players in

this game. Neither player can improve his winnings

otherwise. In general, minimax ~ value of the game and

maximin ~ value of the game.

Games that had saddle points are easy to solve and

tend to be boring in nature. When a saddle point does not

exist in a game, then pure plays are not available to the

12

players. Thus, other methods for solving games will be

needed.

CHAPTER III

SOLVING COMPLEX GA~lES

The concept of mixed strategies is used with games

that do not have saddle points. Consider the following

payoff matrix with no saddle point:

l 2 3

I 1 I 6 0 7 0

l 2 I 3 1 6 1 (3-1) .

I 3 I 5 4 2 2

6 4 7

Thus, the maximin value < minimax value. There is no

saddle point since equality does not hold. Observing the

matrix, it is noticed that player one has no dominating

strategy available to him. However, if player one plays

strategy three and player two plays strategy three, then

player one wins two. If player one were to instead choose

his first strategy, then player two will be wise to choose

strategy two. He insures himself of no loss and player

one does not gain. The players could continue in this

manner in order to improve their winnings.

13

14

Mixed Strategies

Matrix 3-1 has no saddle point. When such occurs, it

is best that each player assign a probability distribution

to his set of strategies. That is, let xi = probability

that player one will play strategy i where i = 1,

2, ••• , m and player two has Yj =probability that player

two will play strategy j where j = 1, 2, ••• , n. Herem

and n represent the number of rows and columns of the

matrix, respectively. Since the xi's and Yj's are

probability distributions, then xi, Yj ~ 0 and

I:,x i =I:,Yj = 1.

In vector form the players have (x1 , x2, ••• , xm> and

(yl, Y2' ••• , Yn>' referred to as the mixed strategies.

The given example will be used to illustrate players one

and two using mixed strategies. Matrix 3-1 will be used

in the example. Recall that it has no saddle point and

thus requires mixed strategies to be solved. The

following mixed strategies may be chosen by players one

and two, respectively: (0, ~' ~) and (~, ~' 0). For

player one this means that he is opting to play strategies

two and three only, for he has opted to abandon strategy

one altogether since it has a playing probability of 0%.

As for player two, he has opted to play only strategies

one and two with equal time. He has abandoned the third

strategy. Each player is devoting SO% playing time to the

chosen strategies.

15

To determine the payoff reward to each player, it is

now termed the expected payoff and defined as follows for

each player:

a .. x., lJ 1

Here E1 denotes the expected payoff to player one having

the probability xi and payoff value aij• E2 denotes the

expected payoff to player two with probability Yj and

payoff value aij•

The maximin and minimax criterion are valid here.

Player one should adhere to maximizing the smallest payoff

value in a row. In this case, he will choose the

maximin of the expected payoff values, that is,

max[min( ~ aimxi)], while the other player should

minimize the largest payoff value in the strategic

columns, i.e., min [max ('2: anjYj) 1.

If the optimal solutions or probability values exist,

then the minimax expected payoff=maximin expected payoff.

The value of the game will then be v = 2 ~ aijxiYj. Here

xi and Yj are optimal probabilities for players one and

two, respectively. The value aij is from the payoff

matrix.

Considering the mixed strategies for players one and

two, respectively, (0, j, ~) and (~,~),the expected

value for player one shall be computed. If player two

16

plays strategy one only, then player.one can expect a

value of

1 1 n). En. (O, 2• 2>~ = 4.

If player two chooses strategy two, then player one can

expect

E12 • (0, ~' ~) (!) • 2.5.

For choosing strategy three, player one obtains

Notice that the first and the third expected values are ·

the same. The minimum value is 2.5 and thus is the value

of the maximin expected value.

The same method can be used to compute the expected

values for player two •. However, this time using (~, ~, 0)

and the pure strategies of player one will yield a minimax

of 4.5. It is also possible to improve one's earnings by

changing the values of the probability distribution. It

may be that. player one will play strategy three 90% of the

time while strategy two is played 10% of the time. This

action yields a minimax of 2.4.

------------------------------- - ---

17

The graphical procedure for solving games will be

discussed next. It will be noted that this procedure is

compact and very useful. However, it will prove to be

limiting in the number of strategies available to at least

one of the players. It is used with the concept of mixed

strategies.

Graphical Solution

In order to use the graphical procedure, one of the

players must have only two pure strategies available to

him. Thus, the payoff matrix will be either a 2 x n or

m x 2 matrix. If player one has only two pure strategies

to choose from, then x1 and x2 = l-x1 will be th~ two

probability distributions that will serve as mixed

strategies. Player two's mixed strategies would be the

vector (y1, y2) if he has only two strategies to choose

from. Recall that there is no saddle point in this game.

If player one has two strategies to choose from, then

in general the payoff matrix will appear as:

a12 ••• an

a22 ••• an (3-2)

where the probabilities border the matrix. The expected

value is the weighted average as in probability. In

general, the expected value is given by:

18

• • •

The expected payoffs vary linearly with respect to x1•

Here, x1 is the independent variable while Eln is the

dependent variable. The above linear equations will be

graphed in the "Eln-xl" plane. The highest point on the

lower intersecting lines defines the maximin for player

one and the optimum x1 value. Similarly, if the expected

payoffs of player two were computed and graphed, then the

lowest point on the highest intersecting lines defines the

minimax for player two and the optimum y1 value. The

following example exhibits the maximin expected value.

Consider the given payoff matrix and its expected

values:

1

2

3

5

-3

4

7

-6

{3-3)

19

E11 = -x1 + 2, E12 = -2x1 + 5, E13 = -7x1 + 4, and

E14 = 13x1 - 6,

The graph of the four lines yield:

E

Figure 1

In the graph, the lower portion of the graph where

lines 3 and 4 intersect give the x1 needed. Equating the

two equations gives a value of ~ for x1 • This implies

that x2 = ~ also, since the sum of the two is one. The

mixed strategy probability distribution is (~, ~), and the

expected payoff is also ~ upon substituting x1 back into

equation 3 or 4. The probabilities obtained suggests that

player one should play strategies three and four for the

best payoffs.

20

It should be noted that the graphical procedure is

limiting. It only allows for two strategies for one of

the players. Although it is compact and interesting, one

must begin to realize that not all games involve only two

strategies. The next chapter will use the simplex method

used in linear programming to solve games that have more

than two strategies available to both players.

CHAPTER IV

LINEAR PROGRAMMING

The graphical procedure is limiting as to the number

of strategies available to a player. There are games in

which each player has three or more pure strategies. The

graphical procedure cannot be used in these cases, and

thus the simplex method used to solve linear programming

problems is utilized. Most games can be converted to

linear programming problems and thus solved. Linear

programming and the simplex method are discussed, followed

by a game converted to a linear.programming problem.

Simplex Method Review

The theory of linear programming is concerned with

determining maximum and minimum values for linear

functions. The function that is maximized or minimized is

the objective function. The solution of a linear pro­

gramming problem is the vector x = (x1 , x2, x3 , ••• , xn>

for the given:

21

22

Maximize F = c1x1 + c2x2 + ••• + cnxn

Subject to: a11x1 + a12~2 + ••• + a1nxn ~ bl

a21x1 + a22x2 + ••• + a2nxn ~ b2

• • •

where x1 , x2 , ••• , xn ~ 0.

The inequalities are constraints and the xj's are

decision variables. In the linear programming problem,

one is seeking a feasible solution or an optimal solution.

The optimal_ solution·· is the most favored one. The simplex

method is the general procedure used to solve a linear

programming problem. The following example illustrates a

linear programming problem solved by the simplex method.

Maximize:

Subject to:

where x1 , x2 ~ 0.

F = x1 + x2

x1 + 2x2 ~ 1

3x3 + x2 ~ 2

Step one is to change the constraints to equations by

adding slack variables x3 and x4 to each to obtain:

23

F = x1 + x2 + ox3 + ox 4

x1 + 2x2 + x3 = 1

3x1 + x2 + x4 = 2

(0}

(1)

(2}

for x1 , x2, x3, x4 ~ 0. The coefficients of the above are

written in tabular form as follows:

Basic Variables Equation x2 x2 x3 x4 Solution

F 0 -1 -1 0 0 0

x3 1 1 2 1 0 1

x4 2 3 1 0 1 1

Steps 1 through 3 outline the simplex method as follows:

(1} Choose the column with the largest negative

number. In case of a tie, choose either. Put a

box around the column and call it the pivot

column.

(2} Using the boxed column, divide the solutions at

the right by each positive numQer in the boxed

column. Put a box around the row with the

smallest ratio and call this row the pivot row.

The number in the box is the pivot number.

(3} Construct a new tableau below the current one.

The first three columns remain the same except

that the variable in the pivot column replaces

24

the basic variable in the pivot row. The new

rows are computed as follows:

New pivot row = Old :Q;i.vot ~ow Pivot number

New row = old row - (pivot column number) x (new pivot row)

The example continues to give:

Basic Variables Equation F xl x2 I x3 x4 Solution I I

F 0 1 -1 -1 I 0 0 0 I

x3 l 0 1 2 I 1 0 1 I

x4 2 0 3 1 ·. 0 1 2

.... n 1 0 =1 0 , + 3

x3 1 0 0 ..5.. 1 + + 3

xl 2 0 1 + 0 + + F 0 1 0 0 -t _g_ + 15

x2 1 0 0 1 -t + -t xl 2 0 1 0 -t % +

Thus, x1 =-!- and x2 =+·

25

Linear Programming Solution for a Game

Games that require mixed strategies may be solved by

linear programming. From player two's standpoint, the

optimum mixed strategy is found by the minimax criterion.

The following example is a maximization problem of a game.

In general, the standard form will look like.so:

Maximize:

Subject to: a11Y1 + a12Y2 + ••• + a1nYn ~ 1

a21Y1 + a22Y2 + ••• + a2nYn ~ 1

• • •

Where Y1, Y2, Y3, ••• , Yn ~ 0 andY= 1/v and Yj = Yj/v.

Here v is the value of the game. Recall that in this case

one is looking for the vector (yl, Y2' ••• Yn>•

The following game will be solved by the simplex

method for player two. The payoff matrix is as follows:

1

2

3

1

5

1

10

2

50

1

1

3

50

0.1

10

(4-1}

26

Thus, player one must:

Maximize: y = yl + y2 + y3

Subject to: SY1 + S~Y2 + SOY3 ~ 1 (0)

yl + y2 + O.lY3 ~ 1 (l)

lOY1 + y2 + lOY3 ~ 1 (2)

The simplex tableau is as follows:

Basic Variable I Eq. No. I Y0 I yl y2 y3 y4 Ys y6 I Solution I I_ I I I I I

I I I I I Yo I 0 I 1 -l -1 -1 I 0 0 0 I 0

f I I I y4 I 1 I 0 : 5 so so I l 0 0 1

I I I 1 I

Ys I . 2 0 : i 1 "1:0 0 l 0 1 I I I

y6 I 3 I 0 10 1 10 0 0 1 1 I_ I

I -9 _J._ _J._ Yo 0 1 I 0 0 I 0 0 -ro I 10 10 I

99 -1 1 y4 1 0 : 0 2 45 1 0 T 2 I

_.9- -9 -1 9 Ys 2 o I 0 10 J:O 0 1 -ro TO I 1 -1 9 y1 3 o I 1 1]"" 1 0 0 J:O TO

_I I

9 _l_ J_ 6 Yo 0 1 I 0 0 ll 55 0 11 ss-I

1* _.2_ .:1. _J..._

y2 1 o I 0 99 0 99 99 I

o ua -sg -1 _ft_ Ys 2 o I 0 1 J:r 55 I

"* I~ .JJL 49 y1 3 I o I 1 0 0 99 ~ I_ I I

27

Thus, v =+ = ~ is the value of the game, and the vector

is thus ~ ~' 0). This is player one's best mixed

strategy.

CHAPTER V

NUMERICAL EXAMPLES

The following three examples were solved by computer.

The first example is a simple linear programming problem.

Problems two and three are games. All three problems were

solved by the simplex method. The IBM Macintosh computer

with the MacSimplex package was used.

28

29

Solution:

I

Cl C2 C3 C4 cs C6 r:7 C8 C9 ClO

Rl -5.00 -3.00 -1.00 -4.00 -2.00 0 0 0 0 0

R2 3.000 2.000 4.000 1.000 1.000 0 0 0 0 6.000

R3 1.000 3.000 1.000 2.000 3.000 0 1.000 0 0 8.000

R4 2.000 4.000 5.000 2.000 1.000 0 0 1.000 0 12.000

R5 1.000 3.000 2.000 4.000 1.000 0 0 0 1.000 8.000

II

Cl C2 C3 C4 cs C6 r:7 C8 C9 ClO

Rl 0 0.33 5.666 -2.33 -0.33 0 0 0 0 10.00

R2 1.000 0.67 1.333 0.33 0.33 0 0 0 0 2.000

R3 0 2.333 -0.33 1.666 2.666 0 1.000 0 0 6.000

R4 0 2.666 2.333 1.333 0.33 0 0 1.000 0 8.000

R5 0 2.333 0.67 3.666 0.67 0 0 0 1.000 6.000

III

Cl C2 C3 C4 cs C6 r:7 C8 C9 ClO

Rl 0 1.818 6.090 0 0.09 0 0 .o 0.64 13.81

R2 1.000 0.45 1.272 0 0.27 0 0 0 -0.09 1.454

R3 0 1.272 -0.64 0 2.363 0 1.000 0 -0.45 3.272

R4 0 1.818 2.090 0 0.09 0 0 1.000 -0.36 5.818

R5 0 0.64 0.18 1.000 0.18 0 0 0 0.27 1.636

The solution vector is (1.454, o, o, 1.636, 0) •

Example 2:

Find

game.

Solution:

Rl

R2

R3

R4

Cl

-1.00

6.000

4.000

3.000

Cl

0

5.000

30

the optimum strategy for Y and the value of the

X

C2

-1.00

1.000

4.000

-1.00

C2

0

0

6 1

4 4

3 -1

C3 C4

-1.00 -1.00

6.000 1.000

5.000 -2.00

3.000 2.000

C3 C4

0.25 -1.50

y

6

5

3

I

C5

1.00

4.000

4.000

-2.00

II

C5

0

1

-2

2

C6

0

1.000

0

0

C6

0

4

4

-2

C7

0

0

1.000

0

C7

0.25

4.750 1.500 3.000 1.000 -0.25

(5-1)

cs C9

0 0

0 1.000

0 1.000

1.000 1.000

Rl

R2

R3

R4

1.ooo 1.ooo 1.250 -o.5o 1.000 0

0

0.25

0.25

C8

0

0

0

C9

0.25

0.75

0.25

4.000 0 4.250 1.500 -1.00 1.000 1.250

31

III

Cl C2 C3 C4 cs C6 c:7 C8 C9

Ri s.ooo 0 s.ooo 0 3.000 1.000 0 0 1.000

R2 3.333 0 3.166 1.000 2.000 0.67 -0.17 0 o.so

R3 2.666 1.000 2.833 0 2.000 0.33 0.17 0 o.so

R4 -1.00 0 -o.so 0 -4.00 -1.00 o.so 1.000 o.so

The sol uti on vector is (1,.., 0, +, 0) and the game value

is f.

Example 3:

In an experiment, two-year old girls and boys have

learned to recognize the ten digits. The following is·

proposed: The girls' payoff values are represented by

columns, and the boys' are represented by rows. The

children are shown cards labeled with one, two, three,

five, and eight dots. If the two groups choose the same

card, then the boys receive from the girls ·a number of

points equal to twice the number of dots on the card. If

the cards are different, then the girls receive a number

of points equal to the difference of the dots on the

cards. Find: (a) the payoff matrix, (b) the optimal

strategy for the boys, and (c) the value of the game.

32

Solution:

The payoff matrix is the 5x5 matrix.

1 2 3 5 8

1 2 -1 -2 -4 -7

2 -1 4 -1 -3 -6

3 -2 -1 6 -2 -5 (5-2)

5 -4 3 -2 10 -3

8 -7 -6 -5 -3 16

The value of the game may be zero or negative since the .

matrix has a maximin of •4. A constant K=8 is ~dded to

the entries of the matrix to obtain the following matrix

with positive values.

1 2 3 5 8

1 10 7 6 4 1

2 7 12 7 5 2

3 6 7 14 6 3 (5-3)

5 4 5 6 18 5

8 1 2 3 5 24

The computations for the game are done in six iterations.

The six computer charts are given on the following three

pages.

I

C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 en

R1 -1.000 -1.000 -1.000 -1.000 -1.00 0 0 0 0 0 0

R2 10.000 7.000 6.000 4.000 1.000 1.000 0 0 0 0 1.000

R3 7.000 12.000 7.000 5.000 2.000 0 1.000 0 0 0 1.000

R4 6.000 7.000 14.000 6.000 3.000 0 0 1.000 0 0 1.000

R5 4.000 5.000 6.000 18.000 5.000 0 0 0 1.000 0 1.000

R6 1.000 2.000 3.000 5.000 24.000 0 0 0 0 1.000 1.000

LU LU

II

C1 C2 C3 C4 C5 .

C6 C7 C8 C9 C10 Cll

R1 0 -0.30 -0.40 -0.60 -0.90 0.10 0 0 0 0 0.10

R2 1.000 0.70 0.60 0.40 0.10 0.10 0 0 0 0 0.10

R3 0 7.10 2.80 2.20 1.30 -0.70 1.000 0 0 0 0.30

R4 0 2.80 10.40 3.60 2.40 -0.60 0 1.000 0 0 0.40

R5 0 2.20 3.60 16.40 4.60 -0.40 0 0 1.000 0 0.60

R6 0 '1.30 2.40 4.60 23.90 -0.10 0 0 0 1.000 0.90

~-

III

C1 C2 C3 C4 C5 C6 C7 c8 C9 C10 Cll

R1 0 -0.2510 -0.3096 -0.4268 0 0.0962 0 0 0 0.0376 0.1339

R2 1.000 0.6946 0.5899 0.3808 0 0.10042 0 0 0 -0.0042 0.09623

R3 0 7.0292 2.6694 1.9497 0 . -0.6945 1.000 0 0 -0.05434 0.25099

R4 0 2.6694 10.1590 3.138 0 -0.5899 0 1.000 0 -0.10032 0.3095

R5 0 1.9498 3.1382 15.5145 0 -0.38068 0 0 1.000 -0.19228 0.42658

R6 0 0.0544 0.1004 0.1925 1.000 -0.0042 0 0 0 0.0418 0.0377

w 1:.

IV

C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11

R1 0 -0.1974 -0.2233 0 0 0.08574 0 0 0.02751 0.03231 0.1456

R2 1.000 0.6467 0.5128 0 0 0.10975 0 0 -0.02455 0.0005 0.08576

R3 0 6.7841 2.2749 0 0 -0.6467 1.000 0 -0.1256 -0.03018 0.1974

R4 0 2.2749 9.5242 0 0 -0.5130 0 1.000 -0.2023 -0.0605 0.2232

R5 0 0.1257 0.2023 1.000 0 -0.0245 0 0 0.06446 -0.01239 0.02749

R6 0 0.03020 0.0611 0 1.000 0.0005 0 0 -0.01241 0.0439 0.03241

v

C1 C2 C3 C4 C5 C6 C7 co C9 C10 Cll

R1 0 -0.14405 0 0 0 0.0737 0 0.0235 0.02276 0.03088 0.1508

R2 1.000 0.5242 0 0 0 0.1374 0 -0.0538 -0.01366 0.0038 0.07375

R3 0 6.2406 0 0 0 -0.5241 1~000 -0.2389 -0.0773 -0.01562 0.1441

R4 0 0.2389 1.000 0 0 -0.0539 0 0.1050 -0.02124 -0.0064 . 0.02343

R5 0 0.0774 0 1.000 0 -0.0136 0 -0.0213 0.0688 -0.0137 0.02275

R6 0 0.0156 0 0 1.000 0.0033 0 -0.0064 0.0137 0.002 0.03097

LU U1

VI

C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 Cll

R1 0 0 0 0 0 0.0616 0.02307 0.0179 0.02097 0.03052 0.15087

R2 1.000 0 0 0 0 0.1814 -0.0839 -0.0337 -0.00717 0.00511 0.06165

R3 0 1.000 0 0 0 -0.0839 0.1602 -0.0383 -0.01238 -0.0025 0.02308

R4 0 0 1.000 0 0 -0.0338 -0.0382 0.1141 -0.0183 -0.0058 0. 01791

R5 0 0 0 1.000 0 -0.0071 -0.0124 -0.0183 0.0698 -0.0135 0.02096

R6 0 0 0 0 1.000 0.0046 -0.0025 -0.0058 0.0139 0.002 0.03061

L

36

The value of the game matrix 5-3 is v = o.r!087 • The game

value for the original game is v* = 0•15087-8 = -1:372.

The solution vector x* = <x 1*, x2*, ••• , x5*> is 1.1!os1 (0.06165, 0.02308, 0.01791, 0.02096, 0.03061), which gives

X*= (0.40863, 0.15298, 0.11871, 0.1389, 0.20289). This

yields that the boys will ensure a loss most of the time.

They should play strategy one at least 40% of the time

while playing strategy three only 11% of the time.

Game theory deals with competitive situations between

two or more persons. Most research concentrates on the

two-person zero-sum game. ·However, there are n-person

games and even infinite games. Most of the research for

such theory is limited.

Two-person zero-sum games consist of very arbitrary

theory. It requires complete knowledge of the payoff

matrix and, in some cases, much skill •. Game theory has

its limitations in such a fast-growing world. Research is

continuing to adapt the theory to complex situations.

BIBLIOGRAPHY

Chiang, Alpha c. Fundamental Methods of Mathematical Economics. 2nd ed. New York: McGraw-Hill Book Company, 1974.

Grossman, Stanley I. Applications for Elementary Linear Algebra. California: Wadsworth Publishing Company, 1980.

Hillier, Frederick s. and Gerald J. Lieberman. Introduction to Operations Research. 3rd ed. California: Holden-Day, Inc, 1980.

Klekamp, Robert c. and Robert J. Thierauf. Decision Making Theory Through Operations Research. 2nd ed. New York: John Wiley and Sons, 1975.

Owen, Guillermo. Finite Mathematics. Philadelphia: W.B. Saunders Company, 1970.

Steen, Lynn Arthur, ed. Mathematics Today. New York: Vintage Books, 1978.

Taha, Hamdy A. Operations Research: An Introduction. 2nd ed. New York: Macmillan Publishing Co., Inc., 1976.

37