Game Theory OM

7/29/2019 Game Theory OM

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MEHAK WADHERA

ROHIT LUTHRA


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DEFINITION

A type of decision theory based on reasoning, in whichchoice of action is determined after considering the

possible alternatives available to the opponents playing the

same game.

Dates back to 1944 Theory of Games and Economic Behavior, Von Neumann

& Morgenstern

Applications

War strategies

Collective bargaining

Management decisions


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FEATURES

Finite number of competitors

Finite number of courses of action

Knowledge of alternatives available to opponent Choice

Outcome or Gain

Choice of opponent


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BASIC CONCEPTS

TWO PERSONS ZERO-SUM GAME Involves two players

Gains made by one person=Loss incurred by other

Sum of gains & losses = Zero

PAY OFF MATRIX

Gains & losses represented in the form of a matrix

Player on LHS is Maximising/Offensiveplayer

Player on top of matrix is Minimising/Defensive

player

VALUE OF THE GAME

The offensive players gain and the defensive players

loss


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TYPES OF STRATEGY

PURE STRATEGY:

A pure strategy defines a specific move or action that a

player will follow in every possible attainable situation

in a game

MIXED STRATEGY:

A strategy consisting of possible moves and a probability

distribution

when keeping the opponent guessing is desirable - thatis, when the opponent can benefit from knowing the

next move.


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MAXIMIN-MINIMAX PRINCIPLE

MAXIMIN CRITERIA:

Maximising player lists his minimum gains

Selects the maximum out of these

MINIMAX CRITERIA:

Minimising player lists his maximum loss

Selects the minimum out of these


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SADDLE POINT

A value that is simultaneously the minimum of a row

and the maximum of a column

Gives the optimal solution

May or may not exist in a given game

More than one saddle point implies more than one

optimal solution


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A PURE STRATEGY GAME

Each player adopts a single strategy as an optimal

strategy irrespective of the other players strategy

Offensive player selects the strategy with the largest of

the minimum payoffs (maximin)

Defensive player selects the strategy with the smallest of

the maximum payoffs (minimax)

Game is solved when

MAXIMIN VALUE=MINIMAX VALUE


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ILLUSTRATION

There are two lighting fixture stores, Xand Y, who havehad relatively stable market shares. Two new marketing

strategies being considered by store Xmay change this

peaceful coexistence. Thepayoff tablebelow shows the

potential affects on market share if both stores begin to

advertise.

(Offensive player is on left,

Defensive player is on top)

Store X Store Y Strategies

Strategies 1 (radio) 2 (newspaper)

1 (radio) 2 7

2 (newspaper) 6 -4


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Maximin strategy for Store X, the offensiveplayer

Optimal strategy is strategy 1

Store X Store Y Strategies MinimumStrategies 1 (radio) 2 (newspaper) Payoff

1 (radio) 3 5 3

2 (newspaper) 1 -2 -2row

minimums}

maximin: maximum of

the minimum payoffs


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Minimax strategy for StoreY, the defensiveplayer

Optimal strategy is strategy 1

Store X Store YStrategiesStrategies 1 (radio) 2 (newspaper)

1 (radio) 3 5

2 (newspaper) 1 -2

Maximum Payoff 3 5

column

maximums

minimax: minimum of

the maximum payoffs


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MIXED STRATEGY GAMES

If both players are logical and rational, it can be

assumed a minimax criterion will be employed

Existence of a saddle point is indicative of a pure

strategy game

A mixed strategy gameresults if:

Minimax criterion are not employed, or

Each player selects an optimal strategy and they

do not result in a saddle pointwhen the minimax

criterion is used.


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EQUAL GAINS METHOD

A plan of strategies such that the expected gainof the

maximizing player or the expected lossof the minimizing

player will be the same regardless of the opponents

strategy

For 2 X 2 games, an algebraic approach based on the

diagram below can be used to determine the percentage of

the time that each strategy will be played


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EQUAL GAINS METHOD

WhereQand 1 - Q= the fraction of time Xplays strategies X1 and

X2, respectively

Pand 1 - P= the fraction of time Yplays strategies Y1 and

Y2, respectively

Y1 Y2

X1 4 2 Q

X2 1 10 1 - Q

P 1 - P


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STEPS Steps for determining the optimum mixed strategy for a

2 X 2 game algebraically(1) Compute the expected gain for player X

Arbitrarily assume that player Y selects strategy Y1

given this condition, there is a probability qthat

player X selects strategy X1 and a probability 1 - q

that player X selects strategy X2

expected gain = 4q+ 1(1 - q) = 1 + 3q

Arbitrarily assume that player Y selects strategy Y2

given this condition, there is a probability qthat

player X selects strategy X1 and a probability 1 - q

that player X selects strategy X2

expected gain = 2q+ 10(1 - q) = 10 - 8q


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(2) Player X is indifferent to player Ys strategy

Equate the expected gain from each of the strategies

1 + 3q =10 - 8q 11q= 9; q= 9/11

q= thepercentage of timethat strategy X1 is used

Player Xs plan is to use strategy X1 9/11 ofthe time and strategy X2 2/11 of the time


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(3) Compute the expected loss for player Y

Arbitrarily assume that player X selects strategy X1 given this condition, there is a probabilitypthat

player Y selects strategy Y1 and a probability 1 - p

that player Y selects strategy Y2

expected loss = 4p+ 2(1 -p) = 2 + 2p

Arbitrarily assume that player X selects strategy X2

given this condition, there is a probabilitypthat

player Y selects strategy Y1 and a probability 1 - p

that player Y selects strategy Y2

expected loss = 1p+ 10(1 -p) = 10 - 9p


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(4) Player Y is indifferent to player Xs strategy

Equate the expected gain from each of the strategies

2 + 2p =10 - 9p 11p= 8; p= 8/11

p= thepercentage of timethat strategy Y1 is used

Player Ys plan is to use strategy Y1 8/11 ofthe time and strategy Y2 3/11 of the time


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VALUE OF A MIXED STRATEGY

GAME

Once optimum strategies are determined, the value of the

game can be calculated by multiplying each game outcome

times the fraction of time that each strategy is employed

Value of the game is the average or expected game

outcome after a large number of plays

Game Outcome Q P

4 x 9/11 x 8/11 = 2.38

2 x 9/11 x 3/11 = 0.45

1 x 2/11 x 8/11 = 0.1310 x 2/11 x 3/11 = 0.50

Value of the Game 3.46

Y1 Y2

X1 4 2 9/11X2 1 10 2/11

8/11 3/11


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ODDS METHOD

Possible only in case of games with 2x2 matrix.

Sum of column odds and row odds is equal.

METHOD OF FINDING OUT ODDS Find out the difference in value of cell (1,1) and (1,2)

of 1st row and place it in front of 2nd row.

Find out the difference in value of cell (2,1) and (2,2)

of 2nd row and place it in front of 1st row.


of 1st column and place it in front of 2nd column.


of 2nd column and place it in front of 1st column.


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Mathematically

Y1 Y2 ODDS

X1 a1 a2 (b1 - b2)

X2 b1 b2 (a1 - a2)

ODDS (a2 - b2) (a1 - b1)

Value of game = a1(b1

b2) + b1(a1

a2)

(b1b2) + (a1a2)

Probabilities for X1 = b1 - b2 , X2 = a1a2 .

(b1-b2) + (a1-a2) (b1-b2) + (a1-a2)

Probabilities forY1 = a2 - b2 ,Y2 = a1b1 .

(a2-b2) + (a1-b1) (a2-b2) + (a1-b1)


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DOMINANCE

The principle ofdominancecan be used to reduce the size

of games by eliminating strategies that are dominated by

other strategies in all conditions

A strategy is dominated, and can therefore be eliminated,

if all of its payoffs are worse or no better than thecorresponding payoffs for another strategy

By eliminating some of the rows & columns, if the game is

reduced to 2x2 form it can be solved by Odds Method.


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DOMINANCE

CASE 1

Y1 Y2

X1 4 3

X2 2 20

X3 1 1

Y1 Y2 Y3 Y4

X1 -5 4 6 -3

X2 -2 6 2 -20 CASE 2


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DOMINANCE

X3 will never be played because player X can always

do better by playing X1

or X2

Y1 Y2

X1 4 3X2 2 20

X3 1 1

Y2 and Y3 will never be played because player Y can

always do better by playing Y1 or Y4

Y1 Y2 Y3 Y4

X1 -5 4 6 -3

X2 -2 6 2 -20


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SUB-GAMES METHOD

One of the players has 2 alternatives while other more

than 2.

Useful when no saddle point and cannot be reduced by

dominance method.

PROCEDURE

Divide the m*2 or 2*n game matrix into as manygames as possible.

Taking each game one by one, value of each sub game

is found.

Select the best sub game from point of view of the

player who has more than two alternatives. The strategies for this selected sub game will hold good

for both the players for the whole game and the value

of the so selected sub game will be the value of

complete game.


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APPROXIMATION METHOD OR

ITERATIVE METHOD

Can be applied to solve 3*3 or higher games.

Gives an approximate solution for the value of game.

PROCEDURE

Player A chooses the superior strategy over the other

and places that row below the matrix. Player B examines this row and chooses a column

corresponding to the smallest number in that

row.This column is placed to the right of matrix.

Player A examines this column and selects a row

corresponding to the largest number in thiscolumn.This row is then added to the row last chosen

and then sum of the two rows is placed below the

previous row selected.


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Player B chooses a column corresponding to the

smallest number in the row and adds this column to

the last chosen and place it below the previous

column selected.

In case of tie, the player should choose the row or

column different from his last choice.The procedure is

repeated for a number of iterations.

The upper limit of the game is calculated by dividing

the highest number in the last column by the total

number of iterations.Similarly the lower limit of the

game is determined by dividing the lowest number in

the last row by the total no of iterations.


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SUMMARY

Develop Strategiesand Payoff Matrix

Solve Problem for

Saddle Point

Solution and Value

Solve for Mixed

Strategy Probabilities

and Value of Game

Use algebraic,sub games,

Or approximate method

Is

There a Pure Strategy/

Solution?

Is

Game

2 x 2?

Can

Dominance be Used to

Reduce Matrix?

Yes

Yes

No

No

No

Yes


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Game Theory OM