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Introduction
Game theory was developed for the purpose of analyzing competitive situations involving
conflicting interests. In other words, game theory is used for decision making under
conflicting situations where there are one or more opponents (i.e., players). For example,
chess, poker, etc., are the games which have the characteristics of a competition and are
played according to definite rules. Game theory provides solutions to such games, assuming
that each of the players wants to maximize his profits and minimize his losses.
What are the underlying assumptions of game theory?
1) There are finite numbers of competitors (players).
2) The players act reasonably.
3) Every player strives to maximize gains and minimize losses.
4) Each player has finite number of possible courses of action.
5) The choices are assumed to be made simultaneously, so that no player knows his
opponent's choice until he has decided his own course of action.
6) The pay-off is fixed and predetermined.
7) The pay-offs must represent utilities.
Pay-off Function
Let [aij] be any pay off matrix of order m X n. Then the pay-off function or mathematical
m n
expectation of a game which is denoted by E (p,q) and defined as E(p,q) aij piqj
i 1 j 1
Where p, q are the mixed strategies for A and B respectively.
Value of the Game
It refers to the expected outcome per play, when players follow their optimal strategy. It is
a .a a .a
generally denoted by V and defined byv 11 22 12 21 .
(a a ) (a a )
11 22 12 21
In the subsequent sections of this chapter, we provide several trivial illustrations. It should
be noted that we make no pretense about the realism of these illustrations.
Fair Game:
If the value of the game is zero i.e. there is no loss or gain for any player, the game is
called fair game.
Pure Strategy
The simplest type of game is one where the best strategies for both players are pure
strategies. This is the case if and only if, the pay-off matrix contains a saddle point. To
illustrate, consider the following pay-off matrix concerning zero sum two person game.
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Example-01: For what value of , the game with the following pay-off matrix is strictly
determinable?
Player-A
5 2
-1 -8
-2 3
Solution: We ignoring the value of and determine v and vof the matrix by computing the
minimums and column maximums. For this we have the matrix below.
Player-A
5 2
-1 -8
-2 3
We have maximin for A (v)=2 and minimax for B (v ) =-1.The strategy for A is III and for
B is I. If the game is to be determinable with its value v, then v should be such
that 1 v 2.
Thus for a strictly determinable game, 1 v 2
Example-02: What is the optimal plan for both the players?
Player-B
I II III IV
Player-A I -2 0 0 5
II 4 2 1 3
III
-4 -3 0 -2
5 3 -4 2
IV
Solution: We use the maximin (minimax) principle to analyze the game.
Player-B Minimum
I II III IV
Player-A I -2 0 0 5 -2
II 4 2 1 3 1
III -4 -3 0 -2 -4
5 3 -4 2 -6
IV
Maximum 5 3 1 5
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Select minimum from the maximum of columns. ie. Minimax = 1
Player A will choose II strategy, which yields the maximum payoff of 1.
Select maximum from the minimum of rows. ie. Maximin = 1
similarly, player B will choose III strategy.
Since the maximin = the minimax, therefore, = 1.
The optimal strategies for both players are: Player A must select II strategy and player B
must select III strategy. The value of game is 1, which indicates that player A will gain 1
unit and player B will sacrifice 1 unit.
Mixed Strategy WBUT-08
In situations where a saddle point does not exist, the maximin (minimax) principle for
solving a game problem breaks down. The concept is illustrated with the help of following
example.
A mixed strategy game can be solved by following methods:
1. Algebraic Method
2. Calculus Method
3. Linear Programming Method
Dominance CS-511/2003
The principle of dominance states that if one strategy of a player dominates over the other
strategy in all conditions then the later strategy can be ignored. A strategy dominates over
the other only if it is preferable over other in all conditions. The concept of dominance is
especially useful for the evaluation of two-person zero-sum games where a saddle point
does not exist.
Rule:
1. If all the elements of a column (say ith column) are greater than or equal to the
corresponding elements of any other column (say jth column), then the ith column is
dominated by the jth column and can be deleted from the matrix.
2. If all the elements of a row (say ith row) are less than or equal to the corresponding
elements of any other row (say jth row), then the ith row is dominated by the jth row
and can be deleted from the matrix.
Advantages & Limitations of Game Theory
Advantages
1. Game theory gives insight into several less-known aspects, which arise in situations
of conflicting interests. For example, it describes and explains the phenomena of
bargaining and coalition-formation.
2. Game theory develops a framework for analyzing decision making in such situations
where interdependence of firms is considered.
3. At least in two-person zero-sum games, game theory outlines a scientific
quantitative technique that can be used by players to arrive at an optimal strategy.
Limitations
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1. The assumption that players have the knowledge about their own pay-offs and pay-
offs of others is not practical.
2. The techniques of solving games involving mixed strategies particularly in case of
large pay-off matrix are very complicated.
3. All the competitive problems cannot be analyzed with the help of game theory.
Example-05: Find the value of 2X2 game algebraically by using mixed strategies?
B
A 2 3
4 1
Solution: The problem has no saddle point. It can be solved by mixed strategies. Let
p=(p ,p ) and q=(q ,q ) with p p p ,p > 0 and q1+q2 =1, q ,q >0 be the
1 2 1 2 1+ 2=1nad 1 2 1 2
probabilities for the player A and B respectively. Assuming the existence of the value of the
game, considering from the A’s point of view,
E (p)=2p +3p =2p +3(1-p )
1 1 2 1 1
E (p)=4p - p = 4p -(1-p )
2 1 2 1 1
To determine the optimal values of p ,p , we have 2p +3(1-p )= 4q -(1-q ). Solving
1 2 1 1 1 1
* *
p1 =5/6 and p2 =1/6.
Again for B’s point of view
E (q)=2q +4q =2q +4(1-q )
1 1 2 1 1
E (q)=4q - q = 4q -(1-q )
2 1 2 1 1
To determine the optimal values of p ,p , we have 2q +4(1-q )= 4q -(1-q ). Solving
1 2 1 1 1 1
* * * *
q1 =2/3 and q2 =1/3. And the value of the game (v) =2q1 +3q2 =7/3.
* *
Hence the optimal strategies are p =(5/6,1/6) and q (2/3,1/3) and v=7/3.
* * 5 2 1 2 5 1 1 1 7
Again it can be verified that E(p ,q )=2 6 3 4 6 3 3 6 6 1 6 3 3.
Exercises
01. Solve the following 2X3 game graphically MB/205/07
B
A 1` 3 11
8 5 2
02. Explain the following in terms of game theory: MB/205/08
a) Pay-off matrix b) Pure strategy
c) Mixed strategy d) Saddle point
e) Value of a rectangular game
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