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Graphical Method for Games

(a) Let us consider a 2 x n game i.e., the payoff matrix will consist of 2 rows and n columns. So player A (or, row-player) will have two strategies. Also assume that there is no saddle point. Then the problem can be solved by using the following procedure :

 

(i)                 Reduce the size of the payoff matrix using the rules of dominance, if it is applicable.

 

(ii)               Let p be the probability of selection of strategy I and 1 - p be the probability of selection of strategy II by player A.

 

Write down the expected gain function of player A with respect to each of the strategies of player B .

 

(iii)             Plot the gain functions on a graph. Keep the gain function on y-axis and p on x-axis. Here p will take the value 0 and 1 .

 

(iv)             Find the highest intersection point in the lower boundary (i.e., lower envelope) of the graph. Since player A is a maximum player, then this point will be a maximum point.

 

(v)               If the number of lines passing through the maximum point is only two, then obtain a 2 x 2 payoff matrix by retaining the columns corresponding to these two lines. Go to step (vii) else go to step (vi).

 

(vi)             If more than two lines passing through the maximum point then identify two lines with opposite slopes and form the 2 x 2 payoff matrix as described in step (v).

 

(vii)           Solve the 2 x 2 game.

Example 6. Consider the following game and solve it using graphical method.

Solution. It is observed that there is no saddle point. Column V is dominated by column I and column II is dominated by column IV. Therefore delete column V and column II and the payoff matrix is reduced as follows :

Let p be the probability of selection of strategy I and (1-p) be the probability of selection of strategy II by player A. Therefore, the expected gain (or payoff) functions to player A with respect to different strategies of player B is given below :

 

B 's strategy                            A's expected                           A's expected gain

gain function                           p = 0                p = 1

I                       3p - 2(1 - p) = 5p – 2                           -2                    3

III                    6p - ( 1 - p) = 7p – 1                            - 1                   6

IV                    -p + 2(1 - p) = -3p + 2                        2                     – 1

 

Now the A's expected gain function is plotted in Fig. 9 . 1 . It is observed that line I and IV passes through the highest point of the lower boundary. Hence we can form 2 x 2 payoff matrix by taking the columns due to I and IV for player A and it is displayed below :

 

Player B

I           IV

I           3          -1

Player A          II         -2         2

 

Let the mixed strategies for A be SA = ( I  II  pp2 )

 

With                                        p2 =1- p1

 

the mixed strategies for  B be SB = ( I II III  IV  V  q1  0  0  q2  0 )

 

q2 = 1- ql

Therefore,

 

P1 = 2 - (-2) / (3 + 2) - (-1 - 2) =1/ 2 , p2 = 1 – p1 = 1/2

 

P2 = 2 - (-1)/ (3 + 2) - (- 1 - 2) = 3/8, q2 = 1 – q1 = 5/8

 

v = 6-2 / (3 + 2) -(- 1 - 2) = 1/ 2

 

The optimal mixed strategies for A is

 

SA  = (  I  II  ½  ½ )

 

the optimal mixed strategies for B is

 

Sb  = ( I  II  III  IV  V  3/8  0  0  5/8  0 )

 

value of game = 1/2

 

(b) Let us consider a m x 2 game i.e., the payoff matrix will consist of m rows and 2 columns. Also assume that there is no saddle point. Then the problem can be solved by using the following procedure :

 

(i)                 Reduce the size of the payoff matrix using the rules of dominance, if it is applicable.

 

(ii)               Let q ·be the probability of selection of strategy I and 1-q be the probability of selection of strategy II by the player B .

 

Write down the expected gain function of player B with respect to each of the strategies of player A.

 

(iii)             Plot the gain functions on a graph. Keep the gain function on y-axis and q on x-axis. Here q will take the value 0 and 1 .

(iv) Find the lowest intersection point in the upper boundary (i.e., upper envelope) of the graph. Since player B is a minimax player, then this point will be a minimax point.

 

(v) If the number of lines passing through the minimax point is only two, then obtain a 2 x 2 payoff matrix by retaining the rows corresponding to these two lines. Go to step (vii) else goto step (vi).

 

(vi) If more than two lines passing through the minimax point then identify two lines with opposite slopes and form a 2 x 2 payoff matrix as described in step (v).

 

(vii) Solve the 2 x 2 game.

 

Example 7. Consider the following game and solve it using graphical method.

 

Player B

I           II

I           2          1

II         1          3

Player A  III        4          -1

IV        5          -2

 

Solution. The given game dose not have saddle p0int. Also it is observed that none of the rows can be deleted using the rules of dominance.

 

Let q be the probability of selection of strategy I and 1 - q be the probability of selection of strategy II by player B. Therefore, the expected gain (or pay respect to different strategies of player A is given below

 

A's strategy                   B 's expected                                     B's expected gain

gain function                           q = 0          q = 1

I                        2q + (1 - q) = q + 1                                 1                  2

II                     q + 3(1 - q) = -2q + 3                             3                  1

III                    4q - ( 1 - q) = 5q – 1                              -1                  4

IV                    5q - 2(1- q) = 7q - 2                                -2                 5

 

Now the B 's expected gain function is plotted in Fig. 9 .2.

 

It is observed that the line II and IV passes through the lowest point of the upper boundary. Hence we can form 2 x 2 payoff matrix by taking the rows due to II and IV for player B and it is displayed below :

Player B

I           II

II         1          3

Player A

IV        5          -2

 

Let the mixed strategies for A be SA =          I    II   III   IV

( 0   P1   0  P2)

 

With                                                                P2 = 1 - P1

 

 

and      the mixed strategies for B be SB =  I   II

q1  q2

with                                                         q2 =1-q1

Therefore,

 

P1 =-2 -5 /(1-2 –(5+3) =7/9 ,p2 =1 –p1 =2/9

 

Q1 = -2 -3 /(1-2) –(5+3) =5/9 ,q2 = 1-q1 =4/9

 

V =-2 -15 /(1-2) –(5+3) =17/9

 

. . The optimal mixed strategies for A is

SA    =           I      II       III      IV

0    7/9       0      2/9

 

the optimal mixed strategies for B is

SB =     I            II

5/9       4/9

 

value of game =  1/2