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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.
Find out the difference in value of cell (1,1) and (2,1)
of 1st column and place it in front of 2nd column.
Find out the difference in value of cell (1,2) and (2,2)
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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