presentation lecture 2 of game theory.pdf

SHREE M.P. SHAH ART’S AND SCIENCE COLLEGE
SURENDRANAGAR
MATHEMETICS DIPARTMENT
Theory of Game
Project work
Project of Maths 2020-21, Prepared By: Chirag, Brijesh, Hiralal, Bhavik, Dipen,Akshay. [M.P. Shah Art’s And Science College]

INDEX
1. HISTORY................................................................................... 3
2. DEFINITION OF GAME THEORY.................................................... 4
3. BASIC TERM USED IN GAME THOERY........................................... 5
4. FLOW CHART OF GAME THEORY ................................................ 13
5. RULES FOR GAME THEORY........................................................ 14
6. PURE STRATEGIES (MINIMAX AND MAXIMIN PRINCIPLE) ............. 17
7. RULE TO FINDOUT SADDLE POINT............................................. 21
8. DOMINANCE RULES ................................................................. 27
9. MIXED STRATEGY (GAME WITHOUT SADDLE POINT) ................... 34
(1) ALGEBRIC METHOD.............................................................. 35
(2) ARITHMETIC METHOD .......................................................... 41
(3) MATRIX METHOD ................................................................. 46
(4) GRAPHICAL METHOD............................................................ 54

1. HISTORY
 In Game theory came in to existence in 20th
Century.
 In 1944 John Von Neumann and Oscar
Morgenstern published a book Theory of
game and Economic Behavior which they
discussed how businesses of all types may
use this technique to determine the best
strategies given a competitive business
environment
John Von Neumann

2. DEFINITION OF GAME THEORY
In business and economics literature, the term 'game' refers to a
situation of conflict and competition in which two or more
competitor(or participant) are involved in the decision making
process in anticipation of certain outcome over a period of time.

3. BASIC TERM USED IN GAME THOERY
(1) PLAYER:
The competitor is referred to as player A player may be individual, a
group of individual, or an organization.

(1) PLAYER:
(2) TWO-PERSON GAME/ N-PERSON GAME:
 If a game involves only two players (competitors), then it is called a
two-person game.
 If numbers of player are more than two, then game is referred to as
n-person game.

(1) PLAYER:
(2) TWO-PERSON GAME/ N-PERSON GAME:
 If a game involves only two players (competitors), then it is called a
two-person game.
 If numbers of player are more than two, then game is referred to as
n-person game.
(3) ZERO SUM GAME/NON-ZERO SUM GAME:
In a game, if sum of the gain to one player is exactly equal to the
sum of losses to another player, so that the sum of the gains and losses
equals to zero, then the game is said to be a zero-sum game. Otherwise
it is said to be non zero-sum game.

(4) STRATEGY:
The strategy for a player is the list of all possible actions (moves or
courses of action) that he will take for every payoff (outcome) that might
arise.

(4) STRATEGY:
arise.
1. PURE STRATEGY
 A player always chooses the same strategy-same row or column.

(4) STRATEGY:
arise.
1. PURE STRATEGY
2. MIXED STRATEGY
 A player chooses the strategy with some fix probabilities.

(4) STRATEGY:
arise.
1. PURE STRATEGY
2. MIXED STRATEGY
 A player chooses the strategy with some fix probabilities.
3. OPTIMAL STRATEGY
 A strategy that maximizes a player’s expected pay-off.

(5) PAYOFF MATRIX:
The payoff (a quantitative measure of satisfaction that a player gets
at the end of the play) in terms of gains and losses, when player select
their particular strategies, can be represented in the form of a matrix,
called the payoff matrix.
In this pay-off matrix, positive pay-off is the gain to maximizing
player (X) and loss to minimizing player (Y). E.g., if X chooses strategy
X2 and Y chooses strategy Y1, then X’s gain is 32 and Y's loss is 32.
24 36 8
32 20 16
PLAYER
X
PLAYER Y
Y1 Y2 Y3
X1
X2

5. RULES FOR GAME THEORY
RULE 1: Look for pure Strategy (Saddle point)

5.RULES FOR GAME THEORY
RULE 2: Reduce game by Dominance
If no pure strategies exist, the next step is to eliminate certain strategies
(row/column) by law of Dominance.

5.RULES FOR GAME THEORY
RULE 2: Reduce game by Dominance
If no pure strategies exist, the next step is to eliminate certain strategies
(row/column) by law of Dominance.
RULE 3: Solve for mixed Strategy
A mixed strategy game can be solved by different solution method, such
as
1. Algebraic Method
2. Arithmetic Method
3. Matrix Method
4. Graphical method
5. Linear Programming Method

6. PURE STRATEGIES (MINIMAX AND MAXIMIN PRINCIPLE)
(1) Maximin principle:
Maximize the player’s minimum gains. That means select the
strategy that gives the maximum gains among the row minimum value.

6.PURE STRATEGIES (MINIMAX AND MAXIMIN PRINCIPLE)
(2) Minimax Principle:
Minimize the player’s maximum gains. That means, select the
strategy that gives the minimum loss among the column maximum
values.

6.PURE STRATEGIES (MINIMAX AND MAXIMIN PRINCIPLE)
(2) Minimax Principle:
Minimize the player’s maximum gains. That means, select the
strategy that gives the minimum loss among the column maximum
values.
(3) Saddle Point:
If the maximin value equals the minimax value, the game is said to
have a saddle (equilibrium) point and the corresponding strategies are
called "Optimal Strategies".

(4) Value of game:
This is the expected payoff at the end of the game, when each
player uses his optimal strategy.

(4) Value of game:
This is the expected payoff at the end of the game, when each
player uses his optimal strategy.
7. RULE TO FINDOUT SADDLE POINT
 Select the minimum (lowest) element in each row of the payoff
matrix and write them under ‘Row Minimum’ heading. Then, select
the largest element among these element and enclose it in a
rectangle [ ].
 Select the maximum (largest) element in each column of the pay off
matrix and write them under 'Column Maximum' heading. Then,
select the lowest element among these elements and enclose it in a
circle ( ).
 Find out the element(s) that is same in the circle as the well as
rectangle and mark the position of such element(s) in the matrix.
This element represents the value of the game and is called the
Saddle(or equilibrium) point.

Example: Find Solution of game theory problem using saddle
point
Player APlayer B
B1 B2
A1 4 6
A2 3 5
Solution:
 We apply the maximin (minimax) principle to analyze the game,
Player APlayer B
B1 B2
Row
Minimum
A1 [(4)] 6 [4] Maximin
A2 3 5 3
Column
Maximum
(4)
Minimax
6

Select minimum from the maximum of columns
Column MiniMax = (4)

Select maximum from the minimum of rows
Row MaxiMin = [4]

Row MaxiMin = [4]
Here, Column MiniMax = Row MaxiMin = 4
∴ This game has a saddle point and value of the game is 4.

Row MaxiMin = [4]
Here, Column MiniMax = Row MaxiMin = 4
∴ This game has a saddle point and value of the game is 4.
The optimal strategies for both players are,
The player A will always adopt strategy A1.
The player B will always adopt strategy B1.

Why Dominance Rules used?
 To reduce size of payoff matrix in game theory
8. DOMINANCE RULES
Step-1: If all the elements of Column-i are greater than or equal to the
corresponding elements of any other Column-j, then the Column-i is
dominated by the Column-j and it is removed from the matrix.
eg. If all values of Column-2 ≥ Column-4, then remove Column-2
Step-2: If all the elements of a Row-i are less than or equal to the
corresponding elements of any other Row-j, then the Row-i is dominated
by the Row-j and it is removed from the matrix.
eg. If all values of Row-3 ≤Row-4, then remove Row-3
Step-3: If strategy k is dominated by average of any two strategy i and j
than delete column or row strategy k

Example: Given problem Reduced 2×2 matrix using dominance
method
Player APlayer B
B1 B2 B3 B4
A1 3 5 4 2
A2 5 6 2 4
A3 2 1 4 0
A4 3 3 5 2
Solution:
Dominance rule to reduce the size of the payoff matrix
Using dominance property
Player APlayer B
B1 B2 B3 B4
A1 3 5 4 2
A2 5 6 2 4
A3 2 1 4 0
A4 3 3 5 2

Example: Given problem Reduced 2×2 matrix using dominance
method
Player APlayer B
B1 B2 B3 B4
A1 3 5 4 2
A2 5 6 2 4
A3 2 1 4 0
A4 3 3 5 2
Solution:
Dominance rule to reduce the size of the payoff matrix
Using dominance property
Player APlayer B
B1 B2 B3 B4
A1 3 5 4 2
A2 5 6 2 4
A3 2 1 4 0
A4 3 3 5 2
row-3 ≤ row-4, so remove row-3 (A3≤A4: 2≤3,1≤3,4≤5,0≤2)

Player APlayer B
B1 B2 B3 B4
A1 3 5 4 2
A2 5 6 2 4
A4 3 3 5 2
column-2 ≥ column-4, so remove column-2 (B2≥B4:5≥2,6≥4,3≥2)

Player APlayer B
B1 B2 B3 B4
A1 3 5 4 2
A2 5 6 2 4
A4 3 3 5 2
column-2 ≥ column-4, so remove column-2 (B2≥B4:5≥2,6≥4,3≥2)
Player APlayer B
B1 B3 B4
A1 3 4 2
A2 5 2 4
A4 3 5 2
column-1 ≥ column-3, so remove column-1. (B1≥B4:3≥2,5≥4,3≥2)

Player APlayer B
B3 B4
A1 4 2
A2 2 4
A4 5 2
row-1 ≤ row-3, so remove row-1, (A1≤A4: 4≤5,2≤2)

Player APlayer B
B3 B4
A1 4 2
A2 2 4
A4 5 2
row-1 ≤ row-3, so remove row-1, (A1≤A4: 4≤5,2≤2)
Player APlayer B
B3 B4
A2 2 4
A4 5 2

9. MIXED STRATEGY (GAME WITHOUT SADDLE POINT)
In certain case, there is no pure strategy solution for a game. i.e. No
Saddle Point exists.

9. MIXED STRATEGY (GAME WITHOUT SADDLE POINT)
In certain case, there is no pure strategy solution for a game. i.e. No
Saddle Point exists.
(1) ALGEBRIC METHOD
This method is used to determine the probability of using different
strategies by players A and B. This method becomes quite lengthy when a
number of strategies for both the players are more than two.
Example: Find Solution of game theory problem using algebraic
method
Player APlayer B
B1 B2
A1 5 2
A2 3 4

Solution:
 The game does not have a saddle point.
 Solution using algebraic method
Let, p1= probability of selecting the strategy A1,
p2= probability of selecting the strategy A2,
q1= probability of selecting the strategy B1,
q2= probability of selecting the strategy B2
Player APlayer B
B1 B2
A1 5 2
A2 3 4
q1 q2
p1
p2

And V be the value of game,
5 p1+3 p2=v………………………………(1)
2 p1+4 p2=v………………………………(2)
p1+ p2=1……………………………… (3)

5 p1+3 p2=v………………………………(1)
2 p1+4 p2=v………………………………(2)
p1+ p2=1……………………………… (3)
After solving above three equation,
p1=1/4
Therefore optimum strategy for player A is (1/4, 3/4) and value of
game is V=7/2.
p2=3/4
V=7/2

5 p1+3 p2=v………………………………(1)
2 p1+4 p2=v………………………………(2)
p1+ p2=1……………………………… (3)
After solving above three equation,
p1=1/4
Therefore optimum strategy for player A is (1/4, 3/4) and value of
game is V=7/2.
In the same way for player B
5 q1+2 q2=v ……………………………… (1)
3 q1+4 q2=v……………………………… (2)
q1+ q2=1……………………………… (3)
p2=3/4
V=7/2

After solving above equation,
q1=1/2
Therefore optimum strategies for player B is (1/2, 1/2) and value of
game is V=7/2.
q2=1/2
V=7/2

(2) ARITHMETIC METHOD
The arithmetic method (also known as short cut method) provides
an easy method for finding optimal strategies for each player in a payoff
matrix of size 2*2, without saddle point.
The steps of this method are as follows:
Step-1: Find the differences between the two values in the first row
and put it against the second row of the matrix. Neglecting the
negative Sign (if any).
Step-2: Find the differences between the two values in the second
row and put it against the first row of the matrix. Neglecting the
negative sign (if any).
Step-3: Repeat step 1 and step 2 for two columns also.
The values obtained by ‘swapping the difference’ represent the
optimal relative frequency of play for both players’ strategies. These may
be converted to probabilities by dividing each of them by their sum.

Example: Find Solution of game theory problem using arithmetic
method
Player APlayer B
B1 B2
A1 2 -1
A2 -1 0
Solution:
 using arithmetic method to get optimal mixed strategies for both the
players,
Player APlayer B
B1 B2
A1 2 -1
A2
-1 0
│-1-0│=1 │2-(-1) │=3
∴ q1=
1
3+1
=
1
4
∴ q2=
3
3+1
=
3
4
│-1-0│=1 ∴ p1=
1
3+1
=
1
4
│2-(-1) │=3 ∴ p2=
3
3+1
=
3
4

1.Find absolute difference between the two values in the first row
and put it against second row of the matrix
│2-(-1) │=3
2.Find absolute difference between the two values in the second
row and put it against first row of the matrix
│-1-0│=1
∴ p1=
1
3+1
=
1
4
∴ p2=
3
3+1
=
3
4
Hence, Player A should adopt A1 and A2 with 25% of time and 75% of
time respectively.

3.Find absolute difference between the two values in the first
column and put it against second column of the matrix
│2-(-1) │=3
4.Find absolute difference between the two values in the second
column and put it against first column of the matrix
│-1-0│=1
∴ q1=
1
3+1
=
1
4
∴ q2=
3
3+1
=
3
4
Hence, Player B should adopt B1 and B2 with 25% of time and 75% of
time respectively.

Expected gain of Player A
(1) 2×
1
4
+ (-1)×
3
4
= -
1
4
, Player B adopt B1
(2) (-1)×
1
4
+ 0×
3
4
= -
1
4
, Player B adopt B2
Expected loss of Player B
(1) 2×
1
4
+ (-1)×
3
4
= -
1
4
, Player A adopt A1
(2) (-1)×
1
4
+ 0×
3
4
= -
1
4
, Player A adopt A2

(3) MATRIX METHOD
If the game matrix is in the form of a square matrix, then the
optimal strategy mix as well as value of the game may be obtained by
the matrix method. The solution of a two-person zero-sum game with
mixed strategies with a square payoff matrix may be obtained by
using the following formula:
Player A's optimal strategy =
[1 1]𝑃𝑎𝑑𝑗
[1 1] 𝑃𝑎𝑑𝑗[
1
1
]
=[𝑝1 𝑝2]
Player B's optimal strategy =
[1 1]𝑃𝑐𝑜𝑓
[1 1] 𝑃𝑎𝑑𝑗[
1
1
]
=[𝑞1 𝑞2]
Value of the game = (Player A's optimal strategies) x (Payoff matrix
pij ) × (Player B's optimal strategies)
V=[𝑝1 𝑝2] × 𝑝 × [
𝑞1
𝑞2
]

Where padj = adjoint matrix, Pcof= cofactor matrix. Player A's optimal
strategies are in the form of a row vector and B's optimal strategies
are in the form of a column vector.
This method can be used to find a solution of a game with size of
more than 2x2. However, in rare cases, the solution violates the non-
negative condition of probabilities, i.e. pi ≥ 0, qi ≥ 0, although the
requirement p1 + p2 +…+ pm = 1 or q1 + q2 +…+ qn= 1 is met.
Example: Find Solution of game theory problem using matrix
method
Player APlayer B
B1 B2
A1 1 7
A2 6 2
Solution:

 For reduced matrix, calculated PAdj and PCof
PAdj =[
2 −7
−6 1
]
PCof =[
2 −6
−7 1
]

PAdj =[
2 −7
−6 1
]
PCof =[
2 −6
−7 1
]
[1 1]× 𝑃𝑎𝑑𝑗×[
1
1
]

PAdj =[
2 −7
−6 1
]
PCof =[
2 −6
−7 1
]
[1 1]× 𝑃𝑎𝑑𝑗×[
1
1
]
=
[1 1][
2 −7
−6 1
]
[1 1][
2 −7
−6 1
][
1
1
]
=
[−4 −6]
−10
=[
2
5
3
5
]
P1=
2
5
and P2=
3
5
,

Where p1 and p2 represent the probabilities of player A’s, using his
strategies A1 and A2 respectively.
[1 1] 𝑃𝑎𝑑𝑗[
1
1
]

Where p1 and p2 represent the probabilities of player A’s, using his
strategies A1 and A2 respectively.
[1 1] 𝑃𝑎𝑑𝑗[
1
1
]
=
[1 1][
2 −6
−7 1
]
[1 1][
2 −7
−6 1
][
1
1
]
=
[−5 −5]
−10
=[
1
2
1
2
]
q1=
1
2
and q2=
1
2
,
Where q1 and q2 represent the probabilities of player B’s, using his
strategies B1 and B2 respectively.

Value of the game = (Player A's optimal strategies) x (Payoff matrix
pij ) × (Player B's optimal strategies)
V=[𝑝1 𝑝2] × 𝑝 × [
𝑞1
𝑞2
]
V=[
2
5
3
5
] [
1 7
6 2
] [
1
2
1
2
]=4

(4) GRAPHICAL METHOD
The graphical method is useful for the game where the payoff
matrix is of the size 2 x n or m x 2.
Example: Find Solution of game theory problem using graphical
method
Player APlayer B
B1 B2 B3 B4
A1 2 2 3 -2
A2 4 3 2 6
Solution:

 If the probability of player A’s playing A1 and A2 in the strategy
mixture is denoted by p1 and p2, respectively, where p2 =1-p1,
then the expected payoff (gain) to player A will be:

 If the probability of player A’s playing A1 and A2 in the strategy
mixture is denoted by p1 and p2, respectively, where p2 =1-p1,
then the expected payoff (gain) to player A will be:
B’s Pure Strategies A’s Expected Payoff
B1 2 p1 + 4 p2
B2 2 p1 + 3 p2
B3 3 P1 + 2 P2
B4 -2 p1 + 6 p2
 These four expected payoff lines can be plotted on a graph to solve
the game.

presentation lecture 2 of game theory.pdf

Now the original (2×4) game reduces to that of the game with payoff
matrix of size (2×2) as given below:
Player APlayer B
B3 B4
A1 3 -2
A2 2 6
q3 q4
P1
P2=1-p1

Now the original (2×4) game reduces to that of the game with payoff
matrix of size (2×2) as given below:
The optimal payoff to player A,
3 p1+2(1-p1) =v……………………………… (1)
-2 p1+6(1-p1) =v……………………………… (2)
p1+ p2=1……………………………… (3)
Player APlayer B
B3 B4
A1 3 -2
A2 2 6
q3 q4
P1
P2=1-p1

Now solving above three equation,
p1=4/9 p2=1-p1=5/9
V=3×4/9 + 2×5/9=22/9
Therefore optimum strategy for player A is (4/9, 5/9) and value of game
is V=22/9.

p1=4/9 p2=1-p1=5/9
V=3×4/9 + 2×5/9=22/9
is V=22/9.
The optimal payoff to player B can be found in the same way,
3 q3-2(1-q3) =v……………………………… (1)
2 q3+6(1-q3) =v……………………………… (2)
q3+ q4=1……………………………… (3)

p1=4/9 p2=1-p1=5/9
V=3×4/9 + 2×5/9=22/9
is V=22/9.
The optimal payoff to player B can be found in the same way,
3 q3-2(1-q3) =v……………………………… (4)
2 q3+6(1-q3) =v……………………………… (5)
q3+ q4=1……………………………… (6)
After solving above equation,
q3=8/9 q4=1-q3=1/9
V=3×8/9+(-2)×1/9=22/9

Therefore optimum strategies for player B is (0, 0, 8/9, 1/9) and value of
game is V=22/9.

Reference
 Book
6TH
EDITION OPERATIONS RESEARCH THEORY AND APPLICATION
BY JK SHARMA
 Website
1)http://www.slideshare.net/neelamkushwaha904/game-theory-
47179069?from_m_app=android
2)http://www.slideshare.net/RushabhShah350/game-theory-
operation-research-241185431?from_m_app=android
3)https://youtu.be/m8QYjaF4WPA

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