Saturday 14 December 2013

Game Theory Example

Q1.Solve the game whose Pay off Matrix is given by

                           B1          B2          B3           Player B
Player A  A1   [  1              3            1
               A2       0             -4          -3
               A3       1              5           -1]


Soln:
STEP I:
Find out the Row minima and column maxima.

              [1       3       1        1
               0       -4      -3      -4        maximin=1
               1        5       -1]     -1

               1         5        1
  minimax=1

STEP II:
maximin=minimax=1
There exists a saddle point value of game = 1
The optimal strategy is (A1,B1)    -->  this  value is the intersection of maxima and minima.
The game is strictly determined since maximin=minima   is not equal to  0

Q2. Solve:

[ 0      2
 -1      4]

Soln:
STEP I:
     [ 0      2   0
      -1      4]  -1     maximin=0
       0       4
minimax=0

STEP II:
maximin=minimax=0
there exists a saddle point value of game=0
The optimum strategy is (A1,B1)
The game is fair since maximin=minimax=0

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