A game is any interaction between multiple people in which each person’s payoff is affected

by the decisions made by others. It was pioneered by John Nash. Many practical problems
require decision-making in a competitive situation, where there are two or more opposing
parties with conflicting interests and where the action of one depends upon the one taken by
the opponent. For example, candidates for an election, advertising for an election, advertising
and marketing campaigns by competing business firms, countries involved in military battles
etc. have their conflicting interests. In such a competitive situation the courses of action
(alternatives) for each competitor may be either infinite or finite. A competitive situation will
be called a game; if it has the following characteristics,

I. There are a finite number of competitors

II. Each player has finite number of strategies.
III. A play of the game takes place where each player employs his strategy.
IV. Every game result in an outcome, e.g. loss or gain or a draw, usually called payoff to
some player.

TWO-PERSON ZERO SUM GAME

When there are two competitors playing a game, it is called a two-person game. In case the
number of competitors exceeds to say ‘n’, then game is termed as a ‘n-person’ game.

Games having zero sum character that the algebraic sum of gains and losses of all the players
is zero are called zero – sum games. The play does not add a single paisa to the total wealth
of the players; it merely results in a new distribution of initial money among them. Zero sum
games with two players are called two-person zero sum games. In this case the loss/gain of
one player is exactly equal to the gain/loss of the other. If the sum of the gains or losses is not
equal to zero, then the game is of non-zero-sum character or simply non-zero-sum game.

SOME BASIC CONCEPTS

1. Player: The competitors in the game are known as players. The player may be
individual or group of individuals, or an organisation.
2. Strategy: A strategy of a player is defined as a set of rules or alternative cources of
actions available to him in advance, by which player decides the cource of action that
he should adopt. A strategy may be of two types:
a) Pure Strategy: If the player selects the same strategy each time, then it is
referred to as pure strategy. In this case each player knows exactly what the
 other party is going to do, the objective of the players is to maximise the gains
or to minimise the losses.
b) Mixed Strategy: When the players use a combination of strategies and each
player always kept guessing as to which cource of action is to be selected by
the other player at a particular occasion, then this is known as mixed strategy.
3. Optimum Strategy: A cource of action or play which puts the player in the most
preferred position, irrespective of the strategy of his competitors, is called an optimum
strategy. It is also called Nash Equilibrium.
4. Value of the game: It is the expected payoff of the play when all the players of the
game follow their optimum strategies. The game is called fair if the value of the game
is zero and unfair if it is non-zero.
5. Payoff Matrix: When the players select their particular strategies, the payoff (gain or
losses) can be presented in the form of a matrix called the payoff matrix. Since the
game is zero-sum, therefore the gain of one player is equal to the loss of the other and
vice-versa.
Let Player A have ′𝑚′ strategies 𝐴1 , 𝐴2 , 𝐴3 , … … . 𝐴𝑚 ., and the player 𝐵 have ′𝑛′ strategies
𝐵1 , 𝐵2 , 𝐵3 , … . . 𝐵𝑛 . Here it is assumed that each player has his choices from amongst the pure
strategies. Also, it is assumed that player A is always a gainer and player B is always a loser.
That is the payoffs are assumed in terms of player ′𝐴′. Let ′𝑎𝑖𝑗 ′be the payoff which player ′𝐴′
gains from player ′𝐵′, if player ′𝐴′ chooses strategy ′𝐴𝑖 ′ and player ′𝐵’ chooses strategy ′𝐵𝑗 ’
then the payoff matrix of player ′𝐴′ is,

𝐵1 𝐵2 ……… 𝐵𝑛
𝐴1 𝑎11 𝑎12 ⋯ 𝑎1𝑛
𝐴 = 𝐴2 𝑎21 𝑎22 ⋯ 𝑎2𝑛
( ⋮ ⋮ ⋱ ⋮ )
⋮
𝐴𝑚 𝑎𝑚1 𝑎𝑚2 ⋯ 𝑎𝑚𝑛
The payoff matrix to player ′𝐵′ is −𝑎𝑖𝑗

Example: Consider a two-person coin tossing game. Each player tosses an unbiased coin
simultaneously. Player 𝐵 pays ₹7 to A if {𝐻, 𝐻} occurs and ₹4 if {𝑇, 𝑇} occurs, otherwise
player 𝐴 pays ₹3 to 𝐵. Each player has his choices from amongst two pure strategies 𝐻 and 𝑇.
Write the payoff matrix.

The payoff matrix of player A is given by,

Player B
H T
Player A H 7 −3
[ ]
T −3 4
And the payoff matrix of player B is given by,

Player B
H T
Player A H −7 3
[ ]
T 3 −4
THE MAXIMIN-MINIMAX PRINCIPLE
For player A, minimum value in each row represents, the least gain (payoff) to him, if he
chooses his particular strategy. These are written in the matrix by row minima. He will then
select the strategy that maximises his minimum gains. This choice of Player A is called the
MAXIMIN principle.

For Player B, on the other hand likes to minimise his losses. The maximum value in each
column represents the maximum loss to him if he chooses his particular strategy. These are
written in column maxima. He will then select the strategy that minimises his maximum
losses. The choice of B is known as MINIMAX PRINCIPLE, and the corresponding loss is
the minimax value of the game.

If the MAXIMIN value equals the MINIMAX value, then the game is said to have a saddle
(equilibrium) point and the corresponding strategies are called optimum strategies. The
amount of payoff at an equilibrium point is known as the value of the game.
To illustrate the MAXIMIN-MINIMAX principle, let us consider a two-person zero-sum
game with the following 3 × 2 payoff matrix for player A.

𝑩𝟏 𝑩𝟐
𝑨𝟏
9 2
Player A 𝑨𝟐
[ 8 6]
𝑨𝟑
6 4 Saddle Point or
Solution: equilibrium point

𝑩𝟏 𝑩𝟐 Row Minima
𝑨𝟏
9 2 2
6
Player A 𝑨𝟐 [ 8 6]
𝑨𝟑 6 4 4

Column Maxima 9 6

Minimax Maximin

𝑴𝒊𝒏𝒊𝒎𝒂𝒙 = 𝑴𝒂𝒙𝒊𝒎𝒊𝒏

𝑨𝒏𝒅 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒈𝒂𝒎𝒆 = 𝟔

Example1: Determine which of the two-person zero-sum games are strictly determinable and
fair. Give optimum strategies for each player in case of strictly determinable games.

5 0
(a) [ ]
0 2
0 2
(b) [ ]
−1 4
Example2: Solve the game whose payoff matrix is given by,

1 3 1
[0 −4 −3]
1 5 −1
Example3: Determine the range of value of p and q that will make the payoff element 𝑎22 , a
saddle point for the game whose payoff matrix is given below,

2 4 7
[10 7 𝑞]
4 𝑝 8
Solution (Example1) (a)

Row
𝐵1 𝐵2 Minima
𝐴1 5 0 Maximin
[ ] 0
𝐴2 0 2 0
Column Maxima 5 2

Minimax

Here, 𝑀𝑎𝑥𝑖𝑚𝑖𝑛 ≠ 𝑀𝑖𝑛𝑖𝑚𝑎𝑥

So, the game is not determinable and there is no saddle point.

(b)
Saddle Point
𝐵1 𝐵2
Row
Minima
𝐴1 0 2 0
[ ]
𝐴2 −1 4 -1 Maximin
Column Maxima 0 4

Minimax

Here, 𝑀𝑎𝑥𝑖𝑚𝑖𝑛 = 𝑀𝑖𝑛𝑖𝑚𝑎𝑥

So, the game is determinable. The point of intersection of Minimax and maximin is 0. The
saddle point will be 0. Also, the value of the game is zero. As the value of the game is zero,
the game is fair or cooperative.

Solution (Example2) Saddle Point

Row
𝐵1 𝐵2 𝐵3 Minima
𝐴1 1 3 1 1
𝐴2 [0 −4 −3] -4 Maximin
𝐴3 1 5 −1 -1
Column 1 5 1
Maxima

Minimax
Here, 𝑀𝑖𝑛𝑖𝑚𝑎𝑥 = 𝑀𝑎𝑥𝑖𝑚𝑖𝑛. So, the game is determinable. The value of the game is 1. The
game is non-cooperative. The optimum strategies are (𝐴1 , 𝐵1 ) and (𝐴1 , 𝐵3).

Solution (Example3):
Saddle Point
𝐵1 𝐵2 𝐵3 Row
Minima
𝐴1 2 4 7 2 Maximin
𝐴2 [ 10 7 𝑞] 7
𝐴3 4 𝑝 8 4
Column Maxima 10 7 8

Minimax

Here, we are supposed to find the value of p and q in such a way that 𝑎22 is the saddle point.
In the second row as minimum is 7, 𝑞 ≥ 7. In the third column as the maximum is 8, 𝑞 ≤ 8.

So, 7 ≤ 𝑞 ≤ 8.

Also, in the third row the minimum is 4, 𝑝 ≥ 4. In the second column as maximum is 7, 𝑝 ≤
7. So, 4 ≤ 𝑝 ≤ 7.

Example 4: Assume that two firms are competing for market share for a product. Each firm
is considering what promotional strategy to employ for the coming period. Assume that the
following payoff matrix describes the increase in the market share for firm A and the
decrease in the market share for firm B. Determine the optimum strategies for firm.

FIRM B
No Moderate Much
Promotion Promotion Promotion
No
Promotion 5 0 −10
Moderate
FIRM A Promotion [10 6 2 ]
Much
Promotion
20 15 10
Solution:

FIRM B
No Moderate Much Row Minima
Promotion Promotion Promotion
No
-10
Promotion 5 0 −10
Moderate
FIRM A Promotion [10 6 2 ] 2
Much
Promotion
20 15 10 10
Column Maxima 20 15 10

Minimax Maximin

Saddle Point

𝑀𝑖𝑛𝑖𝑚𝑎𝑥 = 𝑀𝑎𝑥𝑖𝑚𝑖𝑛 . So, the game is determinable. The optimum strategy is found at the
point of intersection of (much promotion, much promotion). The value of the game is found
to be 10.

GAMES WITHOUT SADDLE POINT

For any 2 × 2 two-person zero-sum game without any saddle point having the payoff matrix
for player A.

𝑩𝟏 𝑩𝟐
𝑨𝟏 𝑎11 𝑎12
[𝑎 𝑎22 ]
𝑨𝟐 21

The optimum mixed strategies are,

𝑨 𝑨𝟐 𝑩 𝑩𝟐
𝑺𝑨 = [ 𝟏 ] & 𝑺𝑩 = [ 𝟏 ]
𝑝1 𝑝2 𝑞1 𝑞2
In such a way that,

𝑝1 + 𝑝2 = 1 & 𝑞1 + 𝑞2 = 1
𝑝1 , 𝑝2 , 𝑞1 𝑎𝑛𝑑 𝑞2 are determined by,
 𝑎22 −𝑎21 𝑎22 −𝑎12
𝑝1 = & 𝑞1 =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

Where, 𝑝2 = (1 − 𝑝1 ) & 𝑞2 = 1 − 𝑞1
And the value of the game is,

(𝑎11 𝑎22 −𝑎21 𝑎12 )

𝑉=
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )
Example: For the game with the following payoff matrix, determine the optimum strategies
and the value of the game.

𝐵1 𝐵2
𝐴1 5 1
[ ]
𝐴2 3 4

Solution:

𝐵1 𝐵2 Row
Minima
𝐴1 5 1 1
[ ] Maximin
𝐴2 3 4 3
Column 5 4
Maxima

Minimax

𝑀𝑖𝑛𝑖𝑚𝑎𝑥 ≠ 𝑀𝑎𝑥𝑖𝑚𝑖𝑛

So, the game is not determinable. And we need to solve this problem using mixed strategy.

𝑎22 −𝑎21
𝑝1 =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )
Putting the values, we get,

4−3 1 1 4
𝑝1 = = ∴ 𝑝2 = 1 − =
5+4−(1+3) 5 5 5
 𝐴1 𝐴2
∴ 𝑆𝐴 = [ 1 4 ]
5 5
Also,

𝑎22 −𝑎12
𝑞1 =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

4−1 3 3 2
𝑞1 = = ∴ 𝑞2 = 1 − =
5+4−(1+3) 5 5 5

𝐵1 𝐵2
∴ 𝑆𝐵 = [ 3 2 ]
5 5
And the value of the game is,

(𝑎11 𝑎22 −𝑎21 𝑎12 )

𝑉=
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

(5×4−3×1) 17
𝑉= =
5+4−(1+3) 5
Example: Consider a modified form of ‘matching unbiased coins’ game problem. The
matching player is paid ₹8 if two coins turn both heads and ₹1, if two coins turn both tails.
The non-matching player is paid ₹3, when two coins do not match. Given the choice of
matching and non-matching player, which one would you choose and what would be your
strategy?

Solution:
Maximin
Non-matching player Row
Minima
H T
H 8 −3 -3
Matching player [ ]
T −3 1 -3
Minimax
Column Maxima 8 1
𝑀𝑖𝑛𝑖𝑚𝑎𝑥 ≠ 𝑀𝑎𝑥𝑖𝑚𝑖𝑛 . So, there is no saddle point and the game need to be solved by
mixed strategy.

Here, 𝑎11 =8

𝑎12 = −3

𝑎21 = −3

And 𝑎22 = 1
𝑎22 −𝑎21
Now, 𝑝1 =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

1−(−3) 4
𝑝1 = =
8+1−(−3−3) 15
4 11
∴ 𝑝2 = 1 − =
15 15
The optimum strategy of the matching player is,

𝑨 𝑨𝟐
𝑺𝑨 = [ 𝟏 ]
𝑝1 𝑝2
𝑨𝟏 𝑨𝟐
i.e., 𝑺𝑨 = [ 4 11 ]
15 15
𝑎22 −𝑎12
Also, 𝑞1 =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

1−(−3) 4
𝑞1 = =
8+1−(−3−3) 15
4 11
∴ 𝑞2 = 1 − =
15 15
The optimum strategy of player B is,

𝑩𝟏 𝑩𝟐
𝑺𝑩 = [ ]
𝑞1 𝑞2
 𝑩𝟏 𝑩𝟐
i.e., 𝑺𝑩 = [ 4 11 ]
15 15

and the value of the game is,

(𝑎11 𝑎22 −𝑎21 𝑎12 )
𝑉=
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

{8×1−(−3)×(−3)}
𝑉=
8+1−(−3−3)
2
∴𝑉=
15
Example: Solve the following game and determine the value of the game,

6 −3
a) [ ]
−3 0
4 1
b) [ ]
2 3
Example: In a game of matching coins with two players, suppose A wins one unit of value,
1
when there are two heads, wins nothing when there are two tails and loses unit of value
2
when there are one tail and one head. Determine the payoff matrix, the best strategies for
each player and the value of the game to A.

Example: Two players A and B match coins. If the coins match, then A wins two units of
value, if the coins do not match then B wins two units of value. Determine the optimum
strategies for the players and the value of the game.

GRAPHIC SOLUTION OF 𝟐 × 𝒏 and 𝒎 × 𝟐 GAMES

With the use of mixed strategy one can solve a game of order 2 × 2. But if the size of the
payoff matrix is supposing that 2 × 𝑛 or 𝑚 × 2. Then one can make use of GRAPHIC
SOLUTION to the given game.
Example: Solve the 2 × 4 game graphically.

2 1 0 −2
[ ]
1 0 3 2
Solution: 𝑩𝟑
3 3
𝑩𝟏
2 2

1 1
𝑩𝟐 Maximin

0 0

-1 -1
𝑩𝟒
-2 Lower -2
Envelope

The maximin is the point of intersection of strategies 𝐵2&𝐵4. So, the reduced matrix is,

1 −2
[ ] . it can be solved using mixed strategy.
0 2
Now,

𝑎22 −𝑎21
𝑝1 =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )
2−0 2 2 3
𝑝1 = = , ∴ 𝑝2 = 1 − 𝑝1 = 1 − =
1+2−(−2+0) 5 5 5
So, the strategy of A is

𝑨𝟏 𝑨𝟐
𝑺𝑨 = [ 2 3 ]
5 5
Again,

𝑎22 −𝑎12
𝑞2 =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )
 2−(−2) 4 1
𝑞2 = = , ∴ 𝑞4 = 1 − 𝑞2 =
1+2−(−2+0) 5 5
So, the strategy of B is,

𝑩𝟏 𝑩𝟐 𝑩𝟑 𝑩𝟒
𝑺𝑩 = [ 4 1 ]
0 0
5 5
And the value of the game is,

(𝑎11 𝑎22 −𝑎21 𝑎12 )

𝑉=
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

{1×2−(−2)×0} 2
𝑉= =
1+2−(−2+0) 5

Example: Solve the following game,

B’s Strategy
𝐵1 𝐵2
𝐴1 3 −4
A’s Strategy 𝐴2 [2 5]
𝐴3 −2 8
Solution:
The MINIMAX is found to be at (𝐴1 , 𝐴2 ). So, 𝐴3 will be eliminated. The reduced matrix is,

3 −4
[ ]
2 5
Now,

𝑎22 −𝑎21
𝑝1 =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

5−2 3 3 7
𝑝1 = = , ∴ 𝑝2 = 1 − =
3+5−(−4+2) 10 10 10
Hence the strategy of A is,

𝐴1 𝐴2 𝐴3
𝑆𝐴 = [ 3 7 ]
0
10 10
Again,

𝑎22 −𝑎12
𝑞1 =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

5−(−4) 9 9 9
𝑞1 = = = , ∴ 𝑞2 = 1 − 𝑞1 = 1 − =
3+5−(−4+2) 10 10 10
1
10
Hence, the strategy of B is,

𝐵1 𝐵2
𝑆𝐵 = [ 9 1 ]
10 10
And the value of the game is,

(𝑎11 𝑎22 −𝑎21 𝑎12 )

𝑉=
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

{3×5−2×(−4)} 23
𝑉= =
3+5−(−4+2) 10
Example: Obtain the optimal strategies for both-persons and the value of the game for zero-
sum two-person game whose payoff matrix is as follows,

1 −3
3 5
−1 6
4 1
2 2
[−5 0 ]

Solution:

The reduced matrix is,

3 5
[ ]
4 1
Now,

𝑎22 −𝑎21 1−4 3

𝑝2 = = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 3+1−(5+4) 4
3 1
∴ 𝑝4 = 1 − =
4 4
So, the strategy of A is,
 𝑨𝟏 𝑨𝟐 𝑨𝟑 𝑨𝟒 𝑨𝟓 𝑨𝟔
𝑺𝑨 = [ 3 1 ]
0 0 0 0
4 4

Again,

𝑎22 −𝑎12
𝑞1 =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

1−(5) 4 4 1
𝑞1 = = , ∴ 𝑞2 = 1 − 𝑞1 = 1 − =
3+1−(5+4) 5 5 5
Hence, the strategy of B is,

𝐵1 𝐵2
𝑆𝐵 = [ 4 1 ]
5 5
And the value of the game is,

(𝑎11 𝑎22 −𝑎21 𝑎12 )

𝑉=
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

{3×1−4×5} 17
𝑉= =
3+1−(5+4) 5
Example: Use graphical method to solve the following game:

Player B
2 2 3 −2
Player A [ ]
4 3 2 6

Solution: The reduced matrix is,

3 −2
[ ]
2 6
Here, 𝑎11 = 3 , 𝑎12 = −2, 𝑎21 = 2 & 𝑎22 = 6
Now,

𝑎22 −𝑎21 6−2 4

𝑝1 = = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 3+6−(−2+2) 9
4 5
∴ 𝑝2 = 1 − =
9 9
Hence,

𝑨𝟏 𝑨𝟐
𝑺𝑨 = [ 4 5 ]
9 9
Again,

𝑎22 −𝑎12
𝑞3 =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

6−(−2) 8
𝑞3 = =
3+6−(−2+2) 9
8 1
∴ 𝑞4 = 1 − =
9 9
Hence,

𝑩𝟏 𝑩𝟐 𝑩𝟑 𝑩𝟒
𝑺𝑩 = [ 8 1 ]
0 0
9 9
And the value of the game is,

(𝑎11 𝑎22 −𝑎21 𝑎12 )

𝑉=
𝑎11 +𝑎22 −(𝑎12 +𝑎21 )

{3×6−2×(−2)} 22
𝑉= =
3+6−{−2+2} 9
 Dominance Principle
Sometimes, it is observed that one of the pure strategies of either player is always inferior to
at least one of the remaining ones. The superior strategies are said to dominate the inferior
ones. Clearly, a player would have no incentive to use inferior strategies which are dominated
by the superior ones. In such cases of dominance, we can reduce the size of the payoff matrix
by deleting those strategies which are dominated by the others. Thus, if each element in one
row say kth of the payoff matrix (𝑎𝑖𝑗 ) is less than or equal to the corresponding elements in
some other row, say rth, then Player A will never choose kth strategy. In other words,
probability, 𝑝𝑘 = 𝑃(𝑐ℎ𝑜𝑜𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑘𝑡ℎ 𝑠𝑡𝑟𝑎𝑡𝑒𝑔𝑦) is zero, if 𝑎𝑘𝑗 ≤ 𝑎𝑟𝑗 for all 𝑗 =
1,2, … . . , 𝑛.

The value of the game and the non-zero choice of probabilities remain unchanged even after
the deletion of kth row from the payoff matrix. In such a case kth strategy is said to be
dominated by rth one.

General rules for dominance are,

(a) If all the elements of a row, say kth, are less than or equal to the corresponding
elements of any other row, say rth, then kth row is dominated by rth row.
(b) If all the elements of a column, say kth are greater than or equal to the corresponding
elements of any other column, say rth then kth column is dominated by rth column.
(c) Dominated rows or columns may be deleted to reduce the size of the payoff matrix, as
optimal strategies will remain unaffected.

Example: Two firms are competing for business under the condition so that the one firm’s
gain is another firm’s loss. Firm A’s payoff matrix is given below:

Firm B
No Ad. Medium Heavy
Ad. Ad.
No Advertisement 10 5 −2
Firm A
Medium
Advertisement
[13 12 15 ]
Heavy 16 14 10
Advertisement
Suggest optimum strategies for the two firms and the net outcome thereof.
Solution: Step 1

10 5 −2
[13 12 15 ]
16 14 10
Step 2

13 12 15
[ ]
16 14 10
Finally, we have the reduced matrix of order 2 × 2 which can be solved using mixed strategy.

12 15
[ ]
14 10
Now,

𝑎22 −𝑎21 10−14 4

𝑝2 = = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 12+10−(15+14) 7
4 3
∴ 𝑝3 = 1 − =
7 7
Hence the strategy of player A is,

𝑨𝟏 𝑨𝟐 𝑨𝟑
𝑺𝑨 = [ 4 3 ]
0
7 7

Also,
𝑎22 −𝑎12 10−15 5
𝑞2 = = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 12+10−(15+14) 7

5 2
∴ 𝑞3 = 1 − =
7 7
The strategy of player B is

𝐵1 𝐵2 𝐵3
𝑆𝐵 = [ 5 2 ]
0
7 7
And the value of the game is,

(𝑎11 𝑎22 −𝑎21 𝑎12 ) {12×10−14×15} 90

𝑉= = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 12+10−(15+14) 7
Example: Solve the game whose payoff matrix is given below,

𝐵1 𝐵2 𝐵3 𝐵4
𝐴1 4 −2 3 −1
𝐴2 [−1 2 0 1]
𝐴3 −2 1 −2 0
𝟒 −𝟐 𝟑 −𝟏
Solution: [−𝟏 𝟐 𝟎 𝟏]
−𝟐 𝟏 −𝟐 𝟎
The reduced matrix is,

4 −2 3 −1
[ ]
−1 2 0 1
The reduced matrix is as follows,

4 −1
[ ]
−1 1
Now,

𝑎22 −𝑎21 1−(−1) 2

𝑝1 = = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 4+1−(−1−1) 7

2 5
∴ 𝑝2 = 1 − =
7 7
 𝐴1 𝐴2 𝐴3
𝑆𝐴 = [ 2 5 ]
0
7 7
Again,

𝑎22 −𝑎12 1−(−1) 1

𝑞1 = = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 4+1−(−1−1) 4
1 3
∴ 𝑞4 = 1 − 𝑞1 = 1 − =
4 4
Hence the strategy of B is,

𝑞1 𝑞2 𝑞3 𝑞4
[𝑆𝐵 = 1
0 0
3 ]
4 4
And the value of the game is,

(𝑎11 𝑎22 −𝑎21 𝑎12 ) {4×1−(−1×−1} 5

𝑉= = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 4+1−(−1−1) 7
Example: Find the optimal strategies for two companies competing based on the following
payoff matrix showing gain or loss of customers:

Company B
Company A Quantity CSR Events
Discounts Initiatives Sponsorship
Quantity Discounts 20 60 30
CSR Initiatives
[−10 30 25 ]
Events Sponsorship
40 50 −30
I. State the required assumptions for finding optimal strategies of the two companies.
II. Find out the value of the game. Use dominance rule.

Solution: Player A always tries to maximise his minimum gain and player B always tries to
minimise his maximum loss. All the elements of the second column are of ≥ type as
compared to the first column, so the first column dominates the second column and the
second column is eliminated from the payoff matrix, as per the dominance principle.

20 60 30
[−10 30 25 ]
40 50 −30
Now the reduced payoff matrix is,

20 30
[−10 25 ]
40 −30
Now, if we analyse row wise, we find that all the elements of the second row are of ≤ type as
compared to the first row. So, we eliminate the second row as per the dominance principle.

20 30
[−10 25 ]
40 −30
The reduced matrix is given as follows,

20 30
[ ]
40 −30
Now we can solve the above matrix using mixed strategy. Here we are supposed to find out
𝑝1 , 𝑝3 , 𝑞1 𝑎𝑛𝑑 𝑞3 and the value of the game.
Now,

𝑎22 −𝑎21 −30−40

𝑝1 = = =∝
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 20+(−30)−(30−40)
The probability comes out to be infinite. So, what we can infer is that the game is not
determinable. We can not find the value of the game.

Example: Determine the optimum strategies and the value of the game from the following
2 × 5 payoff matrix for X.

Y
6 3 −1 0 −3
X [ ]
3 2 −4 2 −1
Solution: If we analyse the columns, we find that column-2 dominates column-1 as all
elements of column-1 are ≥ all the elements of column-2. So, column-1 will be eliminated
from the payoff matrix.

6 3 −1 0 −3
[ ]
3 2 −4 2 −1
The reduced matrix is,

3 −1 0 −3
[ ]
2 −4 2 −1
In the reduced matrix, all the elements of column-1 are ≥ column-2. So we will eliminate
column-1.

3 −1 0 −3
[ ]
2 −4 2 −1
The reduced matrix is,

−1 0 −3
[ ]
−4 2 −1
In the reduced matrix all the elements of column-2 are ≥ all the elements of column-1. So,
column-2 is eliminated from the matrix.

−1 0 −3
[ ]
−4 2 −1
The reduced matrix is,

−1 −3
[ ]
−4 −1
Finally, we have a matrix of order 2 × 2, which we can solve with use of mixed strategy. The
columns of this reduced matrix are column-3 and column-5 of the original matrix.
Now,

𝑎22 −𝑎21 −1−(−4) 3

𝑝1 = = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) −1+(−1)−(−3−4) 5
2
𝑝2 = 1 − 𝑝1 =
5
So, the strategy of A is,

𝐴1 𝐴2
𝑆𝐴 = [ 3 2 ]
5 5
Again,

𝑎22 −𝑎12 −1−(−3) 2

𝑞3 = = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) −1+(−1)−(−3−4) 5
3
∴ 𝑞5 = 1 − 𝑞3 =
5
So, the strategy of B is,

𝐵1 𝐵2 𝐵3 𝐵4 𝐵5
𝑆𝐵 = [ 2 3 ]
0 0 0
5 5
And the value of the game is,

(𝑎11 𝑎22 −𝑎21 𝑎12 ) {(−1)×(−1)−(−4)×(−3)} 13

𝑉= = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) −1−1−(−3−4) 5
Example: The following matrix represents the payoff to P1 in a rectangular game between
two persons P1 and P2:

𝑃2
8 15 −4 −2
𝑃1 [19 15 17 16 ]
0 20 15 5
By the notion of dominance, reduce the game to 2 × 4 game and solve it graphically.
Solution: The given payoff matrix is,

8 15 −4 −2
[19 15 17 16 ]
0 20 15 5
In the above game, all the elements of row-1 are ≤ all the elements of row-2. So, row-2
dominates row-1 and row-1 will be eliminated from the matrix.

8 15 −4 −2
[19 15 17 16 ]
0 20 15 5
The reduced matrix is,

19 15 17 16
[ ]
0 20 15 5
As per the requirement of the question, the matrix has been reduced to 2 × 4 order. Solving
the game graphically,

Now the matrix reduces into 2 × 2 order, making it comfortable to solve it using mixed
strategy. The reduced matrix is,

𝐵2 𝐵4
𝐴2 [15 16]
𝐴3 20 5
The probability of A2 can be calculated as,

𝑎22 −𝑎21 5−20 15

𝑝2 = = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 15+5−(16+20) 16
 1
∴ 𝑝3 = 1 − 𝑝2 =
16
Hence the strategy of A is,

𝐴1 𝐴2 𝐴3
𝑆𝐴 = [ 15 1 ]
0
16 16
In the same way,

𝑎22 −𝑎12 5−16 11

𝑞2 = = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 15+5−(16+20) 16

11 5
∴ 𝑞4 = 1 − 𝑞2 = 1 − =
16 16
Hence the strategy of B is,

𝐵1 𝐵2 𝐵3 𝐵4
𝑆𝐵 = [ 11 5 ]
0 0
16 16
And the value of the game is,

(𝑎11 𝑎22 −𝑎21 𝑎12 ) {15×5−20×16} 245

𝑉= = =
𝑎11 +𝑎22 −(𝑎12 +𝑎21 ) 15+5−(16+20) 16
Example: In a small town, there are two discount stores ABC and XYZ. They are the only
stores that handle sundry goods. The total number of customers is equally divided between
two, because the price and quality of goods sold are equal. Both stores have good reputation
in community, and they render equally good customer services. Assume that a gain of
customers by ABC is a loss to XYZ, and vice-versa.

Both stores plan annual pre-Diwali sales during the first week of October. Sales are
advertised through the local newspaper, radio and television media. With the aid of an
advertising firm ABC store constructed the game matrix given below (Figures in the matrix
represent a gain or loss of customer.) Find the optimum strategies for both stores and the
value of the game.

Store XYZ
Newspaper Radio Television
Newspaper 30 40 -80
Store ABC Radio 0 15 -20
Television 90 20 50
Example: Solve the following ‘2-person, zero-sum game’:

Player B
Player A
B1 B2 B3
A1 15 10 4
A2 10 -3 10
A3 12 23 0

Example: Two-breakfast food manufacturers, ABC and XYZ are competing for an increased
market share. The pay-off matrix, shown in the following table, describes the increase in the
market share of ABC and decrease in market share of XYZ.

XYZ
ABC Give Decrease Maintain present Increase
Coupons Price strategy advertising
Give Coupons 2 -2 4 1

Decrease Price 6 1 12 3

Maintain present -3 3 0 6
strategy
Increase advertising 2 -3 7 1