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Dominance Rules

 

(a) For rows : (i) In the payoff matrix if all the entries ·in a row i1 are greater than or equal to the corresponding entries of another row i2, then row i2 is said to be dominated by row i 1. In this situation row i2 of the payoff matrix can be deleted.

 

e.g., i2 = ( 1 , 2, -1) is dominated by i1 = (2, 2, 1), hence (1, 2, - 1 ) can be deleted.

 

(ii) If sum of the entries of any two rows is greater than or equal to the corresponding entry of a third row, then that third row is said to be dominated by the above two rows and hence third row can be deleted.

 

(b) For columns : (i) In the payoff matrix if all the entries in a column j1 are less than or equal to the corresponding entries of another column h· then column h is said to be dominated by column j 1. In this situation column h of the payoff matrix can be deleted.

 

e.g., j2 = (2  4) is dominated by j1 =(1   2 ). Hence (2    4) can be deleted.

 

(ii) If sum of the entries of any two columns is less than or equal to the corresponding entry of a third column, then that third column is said to be dominated by the above two columns and hence third column can be deleted.

 

Example 4. Using the rules for dominance solve the following game :

Solution . The given game has no saddle point. Let us apply the rules for dominance. It is observed that column 1 is dominated by column 3 . Hence delete column 1 and the payoff matrix is reduced as follows :

Again, row 1 is dominated by row 2. Hence delete row 1 and the payoff matrix is reduced to a 2 x 2 matrix.

Let the mixed strategy for player A be SA = ( I II III   0 p1  p2) with p2 = 1 - p 1 and the mixed strategy for player B be

 

P1 = 3-( -2) / (3 + 3) - (- 1 - 2)  =  5/9 = p2 =  4/9

 

Q3 =  3-(- 1) / (3 + 3) - (-1 - 2 ) =  4/9,  q2  =  5/9

 

V = 9-2 / (3 + 3) - (-1 - 2) = 7/ 9

 

Hence the optimal mixed strategies are

 

SA = (I II III  0   5/9  4/9 )

SB = (I II III  0  4/9  5 /9 )

 

v = 7/9.

 

Example 5. Solve the following game:

PlayerB

I           II        III        IV

I           4         3          0         3

II         3         4         3         0

Player A                      III        4         3         4         3

IV         0         5         4         4

 

Solution. The given game has no saddle point. Let us apply the rules for dominance to reduce the size of the payoff matrix. It is observed that row II is dominated by row III, hence row I can be deleted and the payoff matrix reduces as follows

It is observed that column III is dominated by column I. Hence column III can be deleted.

Also it is observed that column II is dominated by column IV. Hence column IV can also be deleted. Hence the payoff matrix reduces as follows :

Here row II is dominated by row III. Hence row II can be deleted and the payoff matrix reduces to 2 x 2 matrix.

Let the mixed strategies for A be SA = ( I  II  III IV  0  0  p1  p2 )

 

With                                        P2 = 1 – p1

 

and the mixed strategies for Sb = (I II III IV  q1  0  0  q)

with                                                     q2 = 1-q1

               

p1 = 4-0 / (4+4) – (0-3) = 4/5 , p2 = 1-p1 = 1/5

p2 = 4-3 / (4+4) – ( 0 +3) = 1/5 , q2 = 1-q1 = 4/5

 

v = 16-0/ (4+4) – (0+3)

 

Hence the optimal mixed strategies are

 

SA = ( I  II  III  IV  0  0  4/5  1/5 )

 

SB = ( I  II  III  IV  1/5  0  0  4/5 )

 

And

v = 16/5

 

Note. If we add a fixed number x to each element of the payoff matrix, then the strategies remain unchanged while the value of the game is increased by x.