Game Theory SV.docx

Operations Research-
Game Theory
Presentation By: Dr. Santosh Vispute

Games Theory
This theory is helpful when 2 or more individuals
or organizations with conflicting objectives try to
make decisions. In such situations, a decision made
by one decision-maker affects the decision made
by one or more of the remaining decision makers
and the final outcome depends upon the decision
of all the parties. Such situations often arise in the
fields of business, industry, economics, sociology
and military training.

Games Theory
The theory of games is based on the minimax
principle put forward by J. Von Neumann which
implies that each competitor will act so as to
minimize his maximum loss (or maximize his
minimum gain) or achieve best of the worst.
This theory does not describe how a game should
be played; it describes only the procedure and
principles by which plays should be selected.

Characteristics of Games
• There are finite number of participants or competitors. If
the no. of participants is 2, the game is called two-person
game, for no. greater than two, it is called n-person game.
• Each participant has available to him a list of finite number
of possible courses of action. The list may not be same for
each participant.
• Each participant knows all the possible choices available to
others but does not know which of them is going to be
chosen by them.
• A play is said to occur when each of the participants
chooses one of the courses of action available to him. The
choices are assumed to be made simultaneously so that no
participant knows the choices made by others until he has
decided his own.

Characteristics of Games
• Every combination of courses of action determines an
outcome which results in gains to the participants. Thegain
may be positive, negative or zero. Negative gain is called a
loss.
• The gain of a participant depends not only on his own
actions but also those of others.
• The gains (payoffs) for each and every play are fixed and
specified in advance and are known to each player. Thus,
each player knows fully the information contained in the
payoff matrix.
• The players make individual decisions without direct
communication.

Definitions
• Game: It is an activity between two or more persons,
involving actions by each one of them according to a set of
rules, which results in some gain (+ve, -ve or zero) foreach.
If in a game the actions are determined by skills, it is called
a game of strategy, if they are determined by chance, it is
termed as a game of chance. Further a game may be finite
or infinite. A finite game has a finite number of moves and
choices, while an infinite game contains an infinite number
of them.
• Player: Each participant or competitor playing a game is
called a player. Each player is equally intelligent and
rational in approach.
• Play: A play of the game is said to occur when each player
chooses one of his course of action.

Definitions
• Strategy: It is the predetermined rule by which a player
decides his course of action from his list of courses of actions
during the game. To decide a particular strategy, theplayer
need not know the others strategy.
• Pure Strategy: It is the decision rule to always select a
particular course of action. It is usually represented by a
number with which the course of action is associated.
• Mixed Strategy: It is decision, in advance of all plays, to
choose a course of action for each play in accordance with
some probability distribution. Thus, a mixed strategy is a
selection among pure strategies with some fixed
probabilities (proportions). The advantage of a mixed
strategy, after the pattern of the game has become evident,
is that the opponents are kept guessing as to which course
of action will be adopted by a player.

Definitions
• Optimal Strategy: The strategy that puts the player in
the most preferred position irrespective of the
strategy of his opponents is called an optional
strategy. Any deviation from this strategy would
reduce his payoff.
• Zero-sum game: It is a game in which the sum of
payments to all the players, after the play of thegame,
is zero. In such a game, the gain of players thatwin is
exactly equal to the loss of players that lose, e.g., two
candidates fighting elections, wherein the gain of
votes by one is the loss of votes to the other.

Definitions
• Two-person-zero-sum game: It is a game involving only two
players, in which the gain of one player equals the loss to the
other. It is also called a rectangular game or matrixgame
because the payoff matrix is rectangular in form. If there are
n players and the sum of the game is zero, it is called n-
person zero-sum game. The characteristics of a two-person
zero-sum game are:
a. Only two players are involved,
b. Each player has a finite number of strategies to use,
c. Each specific strategy results in a payoff,
d. Total payoff to the two players at the end of each play in
zero.

Definitions
• Nonzero-sum game: Here a third party (e.g. the ‘house or a
‘kitty’) receivesor makes some payment. A payoff matrix for
such a game is shown below. The left-hand entry in eachcell
is the payoff to A,
Table
Player A
Player B
and the right-hand entry is the payoff to B. Note that for play
combinations (1, 1) and (2, 2) the sums of the payoffs are not
equal to zero.
2, 2 - 6, 6
6, -6 -2, -2

Definitions
• Payoff: It is the outcome of the game. Payoff (gain or loss)
matrix is the table showing the amounts received by the
player named at the left-hand side after all possible plays of
the game. The payment is made by player named at the top
of the table.
Let player A have m courses of action and player B have n
courses of action. Then the game can be described by a pair of
matrices which can be constructed as described below:
a. Row designations for each matrix are the courses of
action available to player A.
b. Column designations for each matrix are the courses of
action available to player B.
c. The cell entries are the payments to A for one matrix
and to B for the other matrix. The cell entry aij is the

Definitions
payments to A in A’s payoff matrix when A chooses the
course of action i and B chooses the course of action
j.
d. In a two-person zero-sum game, the cell entries in
B’s payoff matrix will be the negative of the
corresponding cell entries in A’s payoff matrix. A is
called maximizing player as he would try to
maximize the gains, while B is called minimizing
player as he would try to minimize his losses.

1
2
3
.
.
Player A .
i
.
.
.
m
Player B
1 2 3 … j … n
a11 a12 a13 … a1j… a1n a21
a22 a23 … a2j… a2n a31
a23 a33 … a3j... a3n
. . . . . .
. . . . . .
. . . . . .
ai1 ai2 ai3 … aij… ain
. . . . . .
. . . . . .
. . . . . .
am1 am2 am3 … amj… amn
A’s payoff matrix

1
2
3
.
.
Player A .
i
.
.
.
m
B’s payoffmatrix
Thus, the sum of payoff matrices for A and B is a null matrix. Here,
the objective is to determine the optimum strategies of both the
players that result in optimum payoff to each.Irrespective of
the strategy used by the other.
Henceforth, we shall usually omit B’s payoff matrix, keeping in
mind that it is just the negative of A’s payoff matrix.
Player B
1 2 3 … j … n
-a11 -a12 -a13 … -a1j… -a1n
-a21 -a22 -a23 … -a2j… -a2n
-a31 -a23 -a33 … -a3j... -a3n
. . . . . .
. . . . . .
. . . . . .
-ai1 -ai2 -ai3
… -aij… -ain
. . . . . .
. . . . . .
. . . . . .
-am1 -am2 -am3 … -amj… -amn

Saddle Point
The saddle point is the point of intersection of
the two courses of action and the gain at this
point is the value of the game.
The game is said to be fair if
maximin value = minimax value = 0,
and is said to be strictly determinable if
maximin value = minimax value ≠ 0.
Saddle point is the number which is lowest in
its row and highest in its column.

Pure Strategy (Saddle Point)
Example 1: In a certain game, player A has possible choices
L, M and N, while player B has two possible choices P and Q.
Payments are to be made according to the choices made.
Choices Payment
L, P A pays B Rs. 3
L, Q B pays A Rs. 3
M, P A pays B Rs. 2
M, Q B pays A Rs. 4
N, P B pays A Rs. 2
N, Q B pays A Rs. 3
What are the best strategies for player A and B in this game?
What is the value of the game for A and B?

Solution: These payments can be arranged in the matrixform.
Let positive number represent a payment from B to A and
negative number a payment from A to B. We, then, have the
payoff matrix shown in following table:
Player B
P Q Minimum of row
L
Player A M
N
- 3 Minimum gain
guaranteed to A if
- 2 he plays his pure
strategy
2
Maximumof column 2 4
(2) maximin
A selects strategy
that maximizes his
Maximum loss to B
if he plays his pure
strategy
(2) minimax B selects strategy
that minimizes his
maximum loss
minimum gain
Minimax = Maximin = 2 = Value of Game
- 3 3
- 2 4
2 3

Strategies for A = N
Strategies for B = P

Example 2: Consider the game G with the following payoff:
Player B
B1 B2 B3
A1
Player A
A2
Determine the value of game.
4 6 4
2 10 0

Solution:
Player B
B Minimumof row
A1
Player A
A2
1 B2 B3
4
0
Maximumof column 4 10 4 (4) maximin
(4) minimax
Strategies for A = A1
Strategies for B = B1 Or B3
4 6 4
2 10 0

Dominance Rule
If no pure strategies exist, the next step is to eliminate certain
strategies (rows and/or columns) by dominance. Rows
and/or columns of the payoff matrix that are inferior to at
least one of the remaining rows and/or columns are deleted
from further consideration. The resulting game can be solved
for some mixed strategy.
The dominance rule for columns is: Every value in the
dominating column(s) must be less than or equal to the
corresponding value of the dominated column. Delete the
dominated column.
The dominance rule for rows is: Every value in the
dominating row(s) must be greater than or equal to the
corresponding value of the dominated row. Delete the
dominating row.

Example 3: Solve the game by dominance rule and
find the optimum strategies for P and Q and the
value of the game.
Colourchosen by P
Colour chosen by Q
T U V W
X
Y
Z
1 7 1 4
-2 5 2 6
3 6 3 5

Solution: Solving the given problem by dominance rule.
Applying column dominance, dominated column, i.e., U or 2nd
column will be deleted as it is dominating to column ‘T’.
Dominating
T U V W
X
Y
Z
Resultant Matrix T V W
X
Y
Z
1 < 7 2 4
-2 4 6
< 5
3 < 6 3 5
1 2 4
-2 4 6
3 3 5

-2 6
Resultant Matrix
Reduced Matrix
T V W
X
Y
Z
T W
X
Y
Z
Now in this matrix applying
column dominance,
dominated column, i.e., V
or 2nd
column will be
deleted as it is dominating
‘T’ column.
row dominance,
dominating row, i.e., X
or 1st
row will be deleted
as it is dominated by ‘Z’
row.
1
>
4
<
5
3
< <
1 < 2 4
-2 < 4 6
3 = 3 5

Resultant Matrix
Y
Z
Maximumof column
Minimumof row
-2
3
(3) maximin
(3) minimax
Obtained matrix is 2x2 matrix, so to solve this game saddle
point method can be adopted.
Strategies for Player P = Z
Strategiesfor Player Q = T
T W
5
3 6
3
6
-2

-2 4 6
Resultant Matrix
Reduced Matrix
Alternate Solution
T V W
X
Y
Z
T V W
Y
Z
Now in this matrix
applying row
dominance, dominating
row will be deleted.
column dominance,
dominated column, i.e., V
or 2nd
column will be
deleted.
-2 < 4 6
3 = 3 5
1 2 4
> < <
< < <
3 3 5

5
3
6
-2
Alternate Solution
Reduced Matrix
Y
Z
Maximumof column
T W Minimumof row
-2
3
3 6 (3) maximin
(3) minimax
Strategies for Player P = Z
Strategies for Player Q = T

Example 4: Solve the following game by dominance rule:
B1 B2 B3 B4
A1
A2
A3
A4
6 8 3 13
4 1 5 3
8 10 4 12
3 6 7 12

Solution:
A1
Colour chosen by P
A2
A3
A4
Maximumof column
Colour chosen by Q
B1 B2 B3 B4
8 10 7 13
(7) minimax
Minimax ≠ Maximin
Minimumof row
3
1
4
3
(4) maximin
This means saddle point does not exist. Therefore, need to
solve above game by other method.
6 8 3 13
4 1 5 3
8 10 4 12
3 6 7 12

Solution: Now, solving the given problem by dominance rule.
Applying column dominance, dominated column, i.e., B4
column will be deleted as it is dominated to column B2.
A1
A2
A3
A4
B1 B2 B3 B4
6 8 3 13
4 1 5 3
8 10 4 12
3 6 7 12
Dominating

A4
Resultant Matrix B1 B
Now in this matrix
2 B3
Now in this matrix
A2
applying row dominance,
dominating row, i.e., A2
or 1st
row will be deleted
compare columns and as it is dominated by
rows as per dominance A3 average of ‘A3 and A4’
rule to reduce its size rows
A4
Resultant Matrix
A1
A2
A3
B1 B2 B3
Now in this matrix
applying row dominance,
dominating row, i.e., A1
or 1st
row will be deleted
as ‘A3’ row is dominating
it.
Dominating
Dominatin
g
6 8 3
4 1 5
8 10 4
3 6 7

4 1 5
8 10 4
3 6 7
(8+3)/2 (10+6)/2 (4+7)/2
5.5 8 5.5

So now following will be the resultant matrix:
Dominating
B1 B2 B3
A3
A4
Now in this matrix
applying column
dominance, dominated
column, i.e., B2
or 2nd
column will be
deleted as it is
dominating ‘B1’ column.
Reduced matrix:
A3
A4
B1 B3 Minimumof row
4
3
(4) maximin
(7) minimax
8 10 4
3 6 7
8 4
3 7

Mixed Strategy (2 X 2 Games)
Arithmetic and algebraic methods are usedfor
finding optimum strategies as well as game
value for a 2x2 game.

Arithmetic method (odds Method or Short cut
Method) for finding optimum strategies and value
of game
It provides an easy method for finding the optimum strategies
for each player in a 2 x 2 game without a saddle point. It
consists of the following steps:
• Subtract the two digits in column 1 and write the
difference under column 2, ignoring the sign.
• Subtract the two digits in column 2 and write the
difference under column 1, ignoring the sign.
• similarly proceed for the two rows.
These values are called oddments. They are the frequencies
with which the players, must use their courses of action in
their optimum strategies.

Player B
B1 B2
A1
O1 = a21 – a22
Player A
A2
O3
= a12 – a22
O2
= a11 – a12
O4
= a11 – a21
A’s Optimal Strategies B’s Optimal Strategies
a11 a12
a21 a22

A1 =
O1
O1 + O2
A2 =
O2
O1 + O2
B1 =
O3
O3 + O4
B2 =
O4
O3 + O4
Value of Game
V =
a11 a22 - a12 a21
(a11 + a22)- (a12 + a21)
a11 O1 + a21 O2
V =
O1 + O2
a12 O1 + a22 O2
V=
O1 + O2
a11 O3 + a12 O4
V=
O3 + O4
a21 O1 + a22 O2
V =
O3 + O4

Two-person zero-sum game without saddle
point
Example: In a game of matching coins, player A
wins Rs. 2 if there are two heads, wins nothing
if there are two tails and loses Rs. 1 when there
are one head and one tail. Determine the payoff
matrix, best strategies for each player and the
value of game to A.

Solution:
H
Player A
T
Player B
H T Minimumof row
-1
-1
(-1) maximin
(0) minimax
Minimax ≠ Maximin
Since there is no saddle point, the optimal strategies will be
mixed strategies. Therefore, need to solve above game by
other method. The order of matrix is 2x2, hence applying
arithmetic method to find the best strategies for both the
players and value of game.
2 -1
-1 0

H
Player A
T
Player B
H T
1
3
1/(1 + 3) = 0.25
3/(1 + 3) = 0.75
1 3
0.25 0.75
Optimal Strategy for player A: (1/4, 3/4),
Optimal Strategy for player B: (1/4, 3/4)
To obtain the value of game any of the following expressions
may be used:
Using A’s Oddments
2 -1
-1 0

-
-
Using A’s Oddments
B plays H; value of the game, V = Rs.
H T
H 1
T 3
1 3
2 * 1 + (– 1)* 3
= Rs.
1 + 3
B plays T; value of the game, V = Rs.
-1 * 1 + (0)* 3
= Rs.
1 + 3
Using B’s Oddments
A plays H; value of the game, V = Rs.
2 * 1 – 1 * 3 1
= Rs.
3 + 1 4
A plays T; value of the game, V = Rs.
-1 * 1 + 0 * 3 1
= Rs.
1
- 4
1
- 4
2 -1
-1 0

Algebraic method for finding optimum
strategies and value of game
While applying this method it is assumed that x
represents the fraction of time (frequency) for
which player A uses strategy 1 and (1 – x)
represents the fraction of time (frequency) for
which he uses strategy 2. Similarly, y and (1 – y)
represent the fraction of time for which player B
uses strategies 1 and 2 respectively.

Two-person zero-sum game without saddle
point
Example: The two armies are at war. Army A has
two airbases, one of which is thrice as valuable
as the other. Army B can destroy an undefended
airbase, but it can destroy only one of them.
Army A can also defend only one of them. Find
the best strategy for A to minimize its losses.

Solution: Since both armies have only two possible
courses of action, the matrix for army A is:
Army B
1 2
Attack the smaller Attack the larger
ArmyA
Defend
smaller 1
airbase
Defend
larger 2
airbus
airbus airbus Minimumof row
-3
-1
(-1) maximin
(0) minimax
Minimax ≠ Maximin
This means saddle point does not exist.
0
Both survive
-3
The larger one
destroyed
-1
The smaller one
destroyed
0
Both survive

Under this method, army A wants to divide itsplays
between the two rows so that the expected
winning by playing the first row are exactly equal
to the expected winnings by playing the second row
irrespective of what army B does. In order to arrive
at the optimum strategies for army A, its necessary
to equate its expected winnings when army B plays
column 1 to its expected winnings when army B
plays column 2.
i.e., when 0.x + (-1).(1 – x) = -3.x + 0.(1 – x)
Or when -1 + x = -3.x i.e., 4x = 1 => x = 1/4
Thus, army A should play first row 1/4th of the
time and second row 3/4th
( = 1 – x) of the time.

Similarly, army B wants to divide its time between
columns 1 and 2 so that the expected winnings are
same by playing each column, no matter what army
A does. Optimum strategies for army B will be
found by equating its expected winnings when
army A plays row 1 to its expected winnings when
army A plays row 2.
i.e., when 0.y + (-3) (1 – y) = -1.y + 0.(1 – y)
Or when -3 + 3.y = -1.y i.e., 4.y = 3 => y = ¾
Thus, army B should play first column 3/4th
of the
time and second column 1/4th ( = 1 – y) of the time.
The optimum strategies can be shown on thegain-
matrix, which becomes -

1 3
0 * - 1 *
4 4
1
Army A
2
Army B
1 2
1/4
3/4
3/4 1/4
The game value can be found either for army A or for army B.
Game value for army A: While army B plays column 1, 3/4 of
time, army A wins zero for 1/4 time and –1 for 3/4 time, also
while army B plays column 2 for 1/4 of time, army A wins –3
for 1/4 time and zero for 3/4 time.
i.e. Total expected winnings for army A are –
Game value =
3
+
1
4 4
= -
9
-
3
16 16
= -
3
4
1 3
- 3 * + 0 *
4 4
0 -3
-1 0

Game value for army B: While army B plays row 1, 1/4 of time,
army B wins zero for 3/4 time and –3 for 1/4 time, also while
army A plays row 2 for 3/4 of time, army B wins –1 for 3/4
time and zero for 1/4 time.
i.e. Total expected winnings for army A are –
Game value =
1
4
3
0 *
4
- 3 *
1 3
+
4 4
3
- 1 *
4
+ 0 *
1
4
1 -3
=
4 4
3
+
4
-3
4
= -
3
-
16
9
16
12 3
= - = -
16 4
Thus, the full solution of the game is:
Army A:
1
4
,
3
4
, Army B:
3
4
,
1
4
3
, Game value = -
4

Mixed Strategy
(2 x n Games or M x 2 games)
These are the games in which one player has only two courses
of action open to him while his opponent may have any
number. To solve such games following steps are performed:
• To look for a saddle point, if there is one, the game is
readily solved.
• If there is no saddle point, then need to reduce the given
matrix to 2x2 size by the rules of dominance.
• If 2x2 matrix obtained then apply arithmetic method or
other method to solve the game.
• If however, the given matrix cannot be reduced to 2x2 size,
it can be still solved by algebraic method, method of sub-
games and graphical method.

Graphical Method for 2xn or mx2
Games
Graphical method is applicable to only thosegames
in which one of the players has twostrategies only.
It reduces the 2xn or mx2 game to 2x2 size by
indentifying and eliminating the dominated
strategies and then solves it by the analytical
(algebraic or arithmetic) methods. The resulting
solution is also the solution to the originalproblem.

Example: Solve the game given in table by the
graphical method:
B
y1 y2 y3 y4
x1
x2
A
x3
x4
19 6 7 5
7 3 14 6
12 8 18 4
8 7 13 -1

Solution: Firstly, check the saddle point.
B
y1 y2 y3 y4
x1
x2
A
x3
x4
Minimumof row
5
3
4
-1
(5) maximin
Maximumof column 19 8 18 6
(8) minimax
Minimax ≠ Maximin
There is no saddle point, so now will apply dominance rule to
reduce the size of matrix.
19 6 7 5
7 3 14 6
12 8 18 4
8 7 13 -1

All the cell values in the column 2 as well as 4 are less than the
corresponding values in columns 1 and 3. hence columns 1
and 3 are dominated by columns 2 and 4.
B
x1
x2
A x3
x4
Dominating
y1 y2
Dominating
y3 y4
19 > 6 7 > 5
7 > 3 14 > 6
12 > 8 18 > 4
8 > 7 13 > -1

Resultant Matrix
A
B
y2 y4
x1
x2
x3
x4
Both the cell values for row 3 are higher than those for row 4.
hence, row 3 dominates row 4.
Reduced Matrix
B
y2 y4
x1
x2
A
x3
6 5
3 6
8
>
4
>
7 -1
6 5
3 6
8 4

Let A1, A2, A3 be the strategies which A mixes with probabilities
x1, x2 and x3 and B2, B4 be the strategies which B mixes with
probabilities y2 and y4 = 1 – y2. When B adopts strategy B2, y2 =
1 and the probability with which he willadopt strategy B4 i.e.,
y4 = 0.
This matrix can be solved by now graphical method. B’s
expected payoffs corresponding to A’s pure strategies are
given below:
A’s pure strategies B’s Expected payoffs y2 = 0 y2 = 1
A1
6y2 + 5(1 – y2) = y2 + 5 5 6
A2
3y2 + 6(1 – y2) = -3y2 + 6 6 3
A3 8y2 + 4(1 – y2) = 4y2 + 4 4 8
These three straight lines can be plotted as function of y2.

Draw 2 lines B4 and B2 parallel to each other one unit
apart and mark a scale on each of them. Torepresent
A’s strategy, join mark 6 on B2 with mark 5 on B4; to
represent A’s second strategy, join mark 3 on B2 with
mark 6 on B4; and so on and bound the figure from
above since B is a minimizing player.
Since player B wishes to minimize his maximum
expected losses, the two lines which intersect at the
lowest point of the upper bound (envelope) show the
two courses of action A should choose in his best
strategy i.e., A1 and A2. We can, thus, immediately
reduce the 3x2 game to 2x2 game which can be easily
solved by arithmetic method.

B4
Minimax
8
7
(6)
(5)
(4)
3
2
0
y2 = 0
B2
(8)
7
(6)
5
4
(3)
2
0
y2 = 1
6 5
3 6

6 * 1 + 5* 3
1 + 3
B
Y2 y4
x1
A
2
Minimumof row
5
3
(5) maximin
(6) minimax
Minimax ≠ Maximin
Arithmetic Method
B
Y2 y4
Optimal Strategies are
A = (3/4, 1/4, 0, 0)
B = (0, 1/4, 0, 3/4)
x1 3
A
x2 1
1 3
3/(1+3) = 3/4
1/(1+3) = 1/4
Value of the game, V
1/(1+3) = 1/4 3/(1+3) = 3/4 = =
21
4
x
6 5
3 6
6 5
3 6

Example: Solve the following 2x5 game by
graphical method:
PlayerA
x1 1
x2 = 1 – x1 2
Player B
1 2 3 4 5
-5 5 0 -1 8
8 -4 -1 6 -5

Solution: Solve the following 2x5 game by
graphical method:
PlayerA
x1 1
x2 = 1 – x1 2
Player B
Minimum of row
-5
-5
Maximumof column
(0) minimax
Minimax ≠ Maximin
(-5) maximin
Saddle point does not exist. Also matrix cannot be reduced
by dominance rule.
1 2 3 4 5
-5 5 0 -1 8
8 -4 -1 6 -5
8 5 0 6 8

A’s expected payoffs corresponding to B’s pure strategies
are:
B’s pure strategies A’s Expected payoffs x1 = 0 x1 = 1
1 -5x1 + 8(1 – x1) = -13x1 + 8 8 -5
2 5x1 - 4(1 – x1) = 9x1 - 4 -4 5
3 0x1 - 1(1 – x1) = x1 - 1 -1 0
4 -1x1 + 6(1 – x1) = -7x1 + 6 6 -1
5 8x1 - 5(1 – x1) = 13x1 – 5 -5 8

A2 8
7
6
5
4
3
2
1
0
8 A1
7
6
5
4
3
2
1
0
-1
-2
-3
-4
-5
x1 = 0 -6
Maximin
-5 0
8 -1
-1
-2
-3
-4
-5
-6 x1 = 1

-5 * 1 + 0* 13
(9 + 5)
B
1 3 Minimumof row
1 -5
A
2 -1
(-1) maximin
(0) minimax
Minimax ≠ Maximin
Arithmetic Method
B
1 3
Optimal Strategies are
A = (9/14, 5/14)
B = (1/14, 0, 13/14, 0, 0)
1
A
2
1 13
9 9/(9+5) = 9/14
5 5/(9+5) = 5/14
Value of the game, V
1/(1+13) = 1/14 13/(1+13) = 13/14 = =
5
- 14
-5 0
8 -1
-5 0
8 -1

Game Theory SV.docx

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