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Game Theory…



-Prof. In charge – Prof. Chitnis
Group No. 1
•   SagarAgarwal                    01
•   Raj Bazaz                       06
•   ZubinChawda                     10
•   ChiragDihrawani(GL)             14
•   AkshayGadkari              17
•   NeenadKadam                24
•   Harshit Patel                   40
•   TusharPatil                     42
•   Tejas Shah                      55
Applications Of Game Theory…
• Example 1 (Principle of Dominance):
In an election campaign, the strategies adopted by the ruling and opposition party along
    with pay-offs (ruling party's % share in votes polled) are given below:


                             Opposition Party's Strategies

          Ruling Party's    Campaign             Campaign     Spend two days
           Strategies       one day in          two days in      in large.
         Campaign one           55                  40             35
         day in each city
          Campaign two          70                  70             55
          days in large
             towns
         Spend two days         75                  55             65
          in large rural
             sectors



Assume a zero sum game. Find optimum strategies for both parties and expected payoff
   to ruling party.
Let A1, A2 and A3 be the strategies of the ruling party and B1, B2 and B3 be those of the
    opposition. Then


                                         Player B
                                             B1             B2             B3
                              A1             55             40             35
           Player A
                              A2             70             70             55
                              A3             75             55             65



   Here, one party knows his strategy as well as other party's strategy and one person's gain is
   another person's loss.
Now, with the given matrix:


                                    Player B
                               B1              B2    B3        Row
                                                              minimax
       Player A      A1        55              40    35         35
                     A2        70              70    55         55
                     A3        75              55    65         65
       Column                  75              70    65
       maxima

As maximin = 55 and minimax= 65 , there is no saddle point.
Row 1 is dominated by row 2 and column 1 is dominated by column 2 giving the
reduced 2 x 2 matrix as :


                                        B2                  B3
                A2                      70                  55
                A3                      55                  65




      P=                   d–c
                     (a + c) – (b + c)
        =                   65 – 55
                     (70 + 65)- (55 + 55)
        =                    2
                             5
  1-P =                    1– 2
                                5
           =                 3
                             5
q   =           d–b
                          (a + d) – (b + c)
                =               65 – 55
                         (70 + 65) – (55 + 55)
                =               2
                                5
    1–q         =               3
                                5

v   =     ad – bc
                      (a + b) – (b + c)
                =        70 * 65 – 55 * 55
                                 25
                =         4550 – 3025
                               25
                =              61
By equating expected losses of opposition party, regardless of what ruling party would
   choose, we get

                   15 q1 + 55 = -l0ql + 65 so that q1 = 2/5 and (1 – q1) = 3/5



Hence opposition party would choose strategy B2 and B3 with a probability of



                                 0.4 and 0.6 respectively.

   The value of the game is determined by substituting the value of p1 and q1 in any of the
   expected values and is determined as 61, i. e.,

                               Expected gain to ruling party:

                                   (i) 15 x 0.4 + 55 = 61
                                   (ii)-10 x 0.4 + 65 = 61

                            Expected loss to opposition party:

                                   (i) 15 x 0.4 + 55 = 61
                                  (ii) -10 x 0.4 + 65 = 61
EXAMPLE 2.

 Even though there are several manufacturers of scooters, two
 firms with branch names Janta and Praja, control their market in Western India. If both
 manufacturers make model changes of the same type for this market segment in the
 same year, their respective market shares remain constant. Likewise, if neither makes
 model changes, then also their market shares remain constant. The pay-off matrix in
 terms of increased/decreased percentage market share under different possible
 conditions is given below:



         JANTA                               PARAJ
                         NO CHANGE        MINOR           MAJOR
                                          CHANGE          CHANGE

         NO CHANGE              0                -4           -10
         MINORCHAN              3                0             5
         GE
         MAJOR                  8                1             0
         CHANGE
Solution.
(i) Clearly, the game has no saddle point. Making use of dominance principle, since the
    first row is dominated by the third row, we delete the first row. Similarly, first
    column is dominated by the second column and hence we delete the first column.
    The reduced pay-off matrix will be as follows:



                                PRAJA
                JANTA           MINOR CHANGE          MAJOR CHANGE
           MINOR CHANGE         0                     5
           MAJOR CHANGE         1                     0




As the reduced pay-off matrix does not possess any saddle point, the players will use
mixed strategies. The optimum mixed strategy for player A is determined by
P         =    d– c
                  (a + d) – (b + c)
=             0-1
                   (0+0)-(5+1)
          =              1
      6

1-p       =              1-1
                           6
=             5
                  6

q         =             d-b
                  (a + d) - (b + c)
          =            0–5
                  (0 – 0) – (5 - 1)
          =              5
                         6

1–q       =       1–5
                     6
          =        1
                   6
V           =            ad-bc
(a + d) - (b + c)

           =        (0 x 0) – (5 x 1)
                     (0 + 0) – (5+1)
           =                5
                            6

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Game theory project...

  • 1. Game Theory… -Prof. In charge – Prof. Chitnis
  • 2. Group No. 1 • SagarAgarwal 01 • Raj Bazaz 06 • ZubinChawda 10 • ChiragDihrawani(GL) 14 • AkshayGadkari 17 • NeenadKadam 24 • Harshit Patel 40 • TusharPatil 42 • Tejas Shah 55
  • 3. Applications Of Game Theory… • Example 1 (Principle of Dominance): In an election campaign, the strategies adopted by the ruling and opposition party along with pay-offs (ruling party's % share in votes polled) are given below: Opposition Party's Strategies Ruling Party's Campaign Campaign Spend two days Strategies one day in two days in in large. Campaign one 55 40 35 day in each city Campaign two 70 70 55 days in large towns Spend two days 75 55 65 in large rural sectors Assume a zero sum game. Find optimum strategies for both parties and expected payoff to ruling party.
  • 4. Let A1, A2 and A3 be the strategies of the ruling party and B1, B2 and B3 be those of the opposition. Then Player B B1 B2 B3 A1 55 40 35 Player A A2 70 70 55 A3 75 55 65 Here, one party knows his strategy as well as other party's strategy and one person's gain is another person's loss.
  • 5. Now, with the given matrix: Player B B1 B2 B3 Row minimax Player A A1 55 40 35 35 A2 70 70 55 55 A3 75 55 65 65 Column 75 70 65 maxima As maximin = 55 and minimax= 65 , there is no saddle point.
  • 6. Row 1 is dominated by row 2 and column 1 is dominated by column 2 giving the reduced 2 x 2 matrix as : B2 B3 A2 70 55 A3 55 65 P= d–c (a + c) – (b + c) = 65 – 55 (70 + 65)- (55 + 55) = 2 5 1-P = 1– 2 5 = 3 5
  • 7. q = d–b (a + d) – (b + c) = 65 – 55 (70 + 65) – (55 + 55) = 2 5 1–q = 3 5 v = ad – bc (a + b) – (b + c) = 70 * 65 – 55 * 55 25 = 4550 – 3025 25 = 61
  • 8. By equating expected losses of opposition party, regardless of what ruling party would choose, we get 15 q1 + 55 = -l0ql + 65 so that q1 = 2/5 and (1 – q1) = 3/5 Hence opposition party would choose strategy B2 and B3 with a probability of 0.4 and 0.6 respectively. The value of the game is determined by substituting the value of p1 and q1 in any of the expected values and is determined as 61, i. e., Expected gain to ruling party: (i) 15 x 0.4 + 55 = 61 (ii)-10 x 0.4 + 65 = 61 Expected loss to opposition party: (i) 15 x 0.4 + 55 = 61 (ii) -10 x 0.4 + 65 = 61
  • 9. EXAMPLE 2. Even though there are several manufacturers of scooters, two firms with branch names Janta and Praja, control their market in Western India. If both manufacturers make model changes of the same type for this market segment in the same year, their respective market shares remain constant. Likewise, if neither makes model changes, then also their market shares remain constant. The pay-off matrix in terms of increased/decreased percentage market share under different possible conditions is given below: JANTA PARAJ NO CHANGE MINOR MAJOR CHANGE CHANGE NO CHANGE 0 -4 -10 MINORCHAN 3 0 5 GE MAJOR 8 1 0 CHANGE
  • 10. Solution. (i) Clearly, the game has no saddle point. Making use of dominance principle, since the first row is dominated by the third row, we delete the first row. Similarly, first column is dominated by the second column and hence we delete the first column. The reduced pay-off matrix will be as follows: PRAJA JANTA MINOR CHANGE MAJOR CHANGE MINOR CHANGE 0 5 MAJOR CHANGE 1 0 As the reduced pay-off matrix does not possess any saddle point, the players will use mixed strategies. The optimum mixed strategy for player A is determined by
  • 11. P = d– c (a + d) – (b + c) = 0-1 (0+0)-(5+1) = 1 6 1-p = 1-1 6 = 5 6 q = d-b (a + d) - (b + c) = 0–5 (0 – 0) – (5 - 1) = 5 6 1–q = 1–5 6 = 1 6
  • 12. V = ad-bc (a + d) - (b + c) = (0 x 0) – (5 x 1) (0 + 0) – (5+1) = 5 6