Game theory Questions With Solutions

Game Theory Questions

Game Theory Questions With Solutions

Practice Problems on Game Theory

Game Theory Questions With Solutions

Practice Problems on Game Theory

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Game theory questions with solutions are given here for practice and to understand the concept of game theory as a decision theory. In operations research, game theory is a mathematical theory that deals with some kind of decisions in a competitive situation.

The theory of games started in the 20th century and it was proposed by John Von Neuman and Morgenstern. It uses the minimax principle to decide the strategy within a competitive situation.

Some terms related to game theory:

Terms	Definition
Competitive situations	A situation that has two or more opponent parties/players where each player has a finite list of possible moves with conflicting interests with each other. The action of one depends on the action of the other.
Finite and infinite game	A game which only a finite number of moves, is called a finite game. A game which is not finite is called an infinite game.
Strategy	It is a predetermined plan that a player uses during the course of the game. There are two types of strategy: Pure strategy Mixed strategy
Pure strategy	A particular game plan in a deterministic game situation (where a player what other player is going to do), whose objective is to maximize the gain
Mixed strategy	When a player is guessing the next move of the opponent, a probabilistic situation is created whose objective is to maximise the expected gain. Thus, a mixed strategy is to select among pure strategies with a fixed probability.
Zero-sum game	If the sum of all the payments to all the players after a play of a game is equal to zero.
Two-person zero-sum game (rectangular game)	In a game where there are only two players and the gain of one results in the loss of the other, such that the net gain of both the player is zero.
Pay-off matrix	A matrix that shows the payment of each player for a particular strategy after a play or end of the game
Value of the game	It is the maximum gain to the maximizing player if both the player uses their best strategy. It is denoted by ‘V’ and is unique. If the value of the game for player A (maximizing player) is ε then the value for the opponent player will be –ε (a rectangular game)
Maximin-Minimax principle	Maximin: The maximising player lists all his minimum gains from the respective strategies, and selects a strategy which gives the maximum gain out the selected minimum gains. Minimax: The minimising player lists all his maximum gains from the respective strategies, and selects a strategy which gives the minimum gain out of the selected maximum gains
Saddle point	Saddle point of a pay-off matrix is that position where the maximum of the row minima coincides with the minimum of the column maxima. For a rectangular game, Maximin of A = Minimax of B is the saddle point of the game.

Now let us solve some important questions asked on Game theory.

Question 1:

Solve the following pay-off matrix:

Player A	Player B
	Strategies	I	II	III
	I	6	8	6
	II	4	12	2

Solution:

We shall solve the given pay-off matrix by finding the saddle point,

Player A	Player B
	Strategies	I	II	III	Row Minimum
	I	6	8	6	6 Max.
	II	4	12	2	2
	Column Maximum	6 Min.	12	6 Min.

The matrix has two saddle points at (1, 1)and (1, 3). Thus, the solution of game:

(i) Best strategy for player A is I

(ii) Best strategy for player B is I or III

(iii) Value of the game for A is 6 and for B is –6

Question 2:

Solve the following pay-off matrix:

Player A	Player B
	Strategies	I	II	III	IV
	I	1	7	3	4
	II	5	6	4	5
	III	7	2	0	3

Solution:

We shall solve the given pay-off matrix by finding the saddle point,

Player A	Player B
	Strategies	I	II	III	IV	Row minimum
	I	1	7	3	4	1
	II	5	6	4	5	4 Max.
	III	7	2	0	3	0
	Column maximum	7	7	4 Min.	5

The solution of the game:

(i) Best strategy for player A is II

(ii) Best strategy for player B is III

(iii) Value of the game for A is 4 and for B is –4.

Points to remember:

Saddle points may or may not exist for a game
There may be more than one saddle point, in that case, the game has more than one solution
If the value of the game is zero this means it is a ‘fair game’.

Question 3:

Solve the following pay-off matrix:

Solution:

We shall solve the given pay-off matrix by finding the saddle point,

Player A	Player B
	Strategies	I	II	III	IV	V	Row Minimum
	I	9	3	1	8	0	0
	II	6	5	4	6	7	4 Max.
	III	2	4	3	3	8	2
	IV	5	6	2	2	1	1
	Column Maximum	9	6	4 Min.	8	8

The solution for the game:

(i) Best strategy for player A is II

(ii) Best strategy for player B is III

(iii) Value of the game for A is 4 and for B is –4.

Question 4:

Find a range of values of a and b for which the following pay-off matrix will a saddle point at (2, 2) position.

Player A	Player B
	Strategies	I	II	III
	I	2	4	5
	II	10	7	b
	III	4	a	6

Solution:

Finding the row minimum and column maximum of the given pay-off matrix,

Player A	Player B
	Strategies	I	II	III	Row minimum
	I	2	4	5	2
	II	10	7	b	7
	III	4	a	6	4
	Column Maximum	10	7	6

Now, given (2,2) is a saddle point, for row 2, 7 will be the minimum then b > 7. In the second column, 7 will be the maximum, then a < 7. If b = 7, then also (2,2) is a saddle point. But if b = 7, then (2, 3) will also be a saddle point.

∴ the range of values for a and b are a ≤ 7 and b > 7.

Question 5:

Solve the following pay-off matrix:

Player A	Player B
	Strategies	I	II	III
	I	1	3	1
	II	0	–4	–3
	III	1	5	–1

Solution:

We shall solve the given pay-off matrix by finding the saddle point,

Player A	Player B
	Strategies	I	II	III	Row minimum
	I	1	3	1	1 Max.
	II	0	–4	–3	–3
	III	1	5	–1	–1
	Column maximum	1 Min.	5	1 Min.

The given pay-off matrix has two saddle points at (1, 1) and (1, 3).

The solution for the game:

(i) Best strategy for player A is I

(ii) Best strategy for player B is I or III

(iii) Value of the game for A is 1 and for B is –1

Mixed Strategies Games:

A pay-off matrix without any saddle point comes under a game of mixed strategy. Methods of solving such a problem:

Odds method ( 2 × 2 game without saddle point)
Dominance method
Sub-games method – (for m × 2 or 2 × n matrices)
Graphical method

Odds Method – 2 × 2 Game

Steps:

Find the difference between the element in (1, 1) and (1, 2), and place it in front of the second row.
Find the difference between the element in (2, 1) and (2, 2), and place it in front of the first row.
Find the difference between the element in (1, 1) and (2, 1), and place it in front of the second column.
Find the difference between the element in (1, 2) and (2, 2), and place it in front of the first column.

\(\begin{array}{l}\begin{matrix} & & Player B & & \\ & Strategy & I & II & Odds \\Player A & I & a_{1} & a_{2} & |b_{1}-b_{2}| \\ &II &b_{1} &b_{2} & |a_{1}-a_{2}| \\ & Odds & |a_{1}-b_{2}| & |a_{2}-b_{2}| & \\\end{matrix}\end{array} \)

Ignore the negative sign of the odds, then the solution of game:
- Value of the game

\(\begin{array}{l}V=\frac{a_{1}|b_{1}-b_{2}|+b_{1}|a_{1}-a_{2}|}{|b_{1}-b_{2}|+|a_{1}-a_{2}|}\end{array} \)

- Probability for the strategies of A:

\(\begin{array}{l}P(Strategy\:I)=\frac{|b_{1}-b_{2}|}{|b_{1}-b_{2}|+|a_{1}-a_{2}|}\end{array} \)

\(\begin{array}{l}P(Strategy\:II)=\frac{|a_{1}-a_{2}|}{|b_{1}-b_{2}|+|a_{1}-a_{2}|}\end{array} \)

- Probability for the strategies of B:

\(\begin{array}{l}P(Strategy\:I)=\frac{|a_{2}-b_{2}|}{|a_{2}-b_{2}|+|a_{1}-b_{1}|}\end{array} \)

\(\begin{array}{l}P(Strategy\:II)=\frac{|a_{1}-b_{1}|}{|a_{2}-b_{2}|+|a_{1}-b_{1}|}\end{array} \)

Question 5:

Solve the following pay-off matrix:

Player A	Player B
	Strategy	I	II
	I	1	5
	II	4	2

Solution:

The matrix have no saddle points, thus solving by the method of odds

Player A	Player B
	Strategy	I	II	Odds
	I	1	5	\|4 – 2\| = 2
	II	4	2	\|1 – 5\| = 4
	Odds	\|5 – 2\| = 3	\|1 – 4\| = 3

Value of game, V = [1 × 2 + 4 × 4]/[2 + 4] = 18/6 = 3

Probabilities of selecting strategies:

Probability of strategies →	I	II
Players ↓	I	II
A	1/3	2/3
B	1/2	1/2

Question 6:

Solve the following pay-off matrix:

Player A	Player B
	Strategy	I	II
	I	–2	8
	II	5	–1

Solution:

The matrix have no saddle points, thus solving by the method of odds

Player A	Player B
	Strategy	I	II	Odds
	I	2	8	\|5 – 1\| = 4
	II	5	1	\|2 – 8\| = 6
	Odds	\|8 – 1\| = 7	\|2 – 5\| = 3

Value of game, V = [2 × 4 + 5 × 6]/[4 + 6] = (6 + 30)/10 = 3.6

Probabilities of selecting strategies:

Probability of strategies →	I	II
Players ↓	I	II
A	2/5	3/5
B	7/10	3/10

Dominance Method:

The main principle behind the dominance method is that if the strategy of a player dominates over the other in all conditions, then the later strategy is ignored.

Rules of dominance method:

If all the elements of the ith row of a payoff matrix are less than or equal to the corresponding jth row, then the jth row strategy dominates over the ith row strategy. Hence, ith row is deleted.
If all the elements of the r-th column are greater than or equal to the corresponding s-th column, then the s-th column dominates over the r-th column. Hence the r-th column is deleted.
A pure strategy may be dominated if it is inferior to an average of two or more other pure strategies.

The objective is to reduce the given pay-off matrix into a 2 × 2 matrix which can be solved by the odds method.

Also Read:

Question 7:

Solve the following pay-off matrix:

\(\begin{array}{l}\begin{bmatrix} -1& -2 & 8 \\7 & 5 & -1 \\6 & 0 & 12 \\\end{bmatrix}\end{array} \)

Solution:

Clearly, the given pay-off matrix has no saddle points. Let apply the dominance rules to reduce the given pay-off matrix to a 2 × 2 matrix

Elements of row 1 < Elements of row 3 ⇒ Row 1 is deleted

The resultant pay-off matrix

Player A	Player B
	Strategies	I	II	III
	II	7	5	–1
	III	6	0	12

Elements of column 1 > Elements of column 2 ⇒ Column 1 is deleted

The resultant pay-off matrix

Player A	Player B
	Strategies	II	III
	II	5	–1
	III	0	12

Now let us apply the odds method, to find the value and probability of strategies.

Player A	Player B
	Strategies	II	III	Odds
	II	5	–1	12
	III	0	12	6
	Odds	13	5

Value of the game = [5 × 12 + 0 × 6]/[12 + 6] = 60/18 = 10/3

Probability selecting strategies:

Player A:

I = 0

II = ⅔

III = ⅓

Player B:

I = 0

II = 13/18

III = 5/18

Question 8:

Solve the following pay-off matrix:

\(\begin{array}{l}\begin{bmatrix}3 & 2 & 4 & 0 \\2 & 4 & 2 & 4 \\4 & 2 & 4 & 0 \\0 & 4 & 0 & 8 \\\end{bmatrix}\end{array} \)

Solution:

Clearly the given pay-off matrix has no saddle point, let us reduce the given matrix using dominance rules:

Elements of row 1 < Elements of row 3 ⇒ Row 1 is deleted

The resultant pay-off matrix

Player A	Player B
	Strategies	I	II	III	IV
	II	2	4	2	4
	III	4	2	4	0
	IV	0	4	0	8

Elements of column 1 = Elements of column 3 ⇒ Column 1 is deleted

The resultant pay-off matrix

Player A	Player B
	Strategies	II	III	IV
	II	4	2	4
	III	2	4	0
	IV	4	0	8

Now, average of column III and IV (3, 2, 4) is dominated by column II, the matrix is reduced to

Player A	Player B
	Strategies	III	IV
	II	2	4
	III	4	0
	IV	0	8

The average of row III and IV (2, 4) is equal to the elements of row II, the matrix reduced to

Player A	Player B
	Strategies	III	IV
	III	4	0
	IV	0	8

Now let us apply the odds method, to find the value and probability of strategies.

Player A	Player B
	Strategies	II	III	Odds
	III	4	0	8
	IV	0	8	4
	Odds	8	4

Value of the game = [4 × 8 + 0 × 4]/[8 + 4] = 32/12 = 8/3

Probability selecting strategies:

Player A:

I = 0

II = 0

III = ⅔

IV = ⅓

Player B:

I = 0

II = 0

III = ⅔

IV = ⅓

Also Read:

Question 9:

Solve the following pay-off matrix:

\(\begin{array}{l}\begin{bmatrix} 30& 40 & -80 \\0 & 15 & -20 \\90 & 20 & 50 \\\end{bmatrix}\end{array} \)

Solution:

Clearly, the given pay-off matrix has no saddle point, let us reduce the given matrix using dominance rules:

All elements of row 2 < all elements of row 3 ⇒ deleting the row 2, we get the resultant pay-off matrix

Player A	Player B
	Strategy	I	II	III
	I	30	40	–80
	III	90	20	50

All elements of column I > all elements of column III ⇒ deleting column I, we get the reduced pay-off matrix

Player A	Player B
	Strategy	II	III
	I	40	–80
	III	20	50

Now let us apply the odds method, to find the value and probability of strategies.

Player A	Player B
	Strategies	II	III	Odds
	I	40	–80	30
	III	20	50	120
	Odds	130	60

Value of the game = [40 × 30 + 20 × 120]/[30 + 120] = 3600/150 = 24

Probability selecting strategies:

Player A:

I = 1/5

II = 0

III = 4/5

Player B:

I = 0

II = 13/15

III = 2/15.

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1. Solve the following pay-off matrices

(i)

Player A	Player B
	Strategies	I	II	III
	I	1	2	3
	II	–3	1	2
	III	1	3	2

(ii)

Player A	Player B
	Strategies	I	II	III
	I	–2	2	–1
	II	1	1	1
	III	3	0	1

(iii)

Player A	Player B
	Strategies	I	II	III	IV
	I	10	0	7	4
	II	2	6	4	7
	III	5	2	3	8

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Player A

Player B

Strategies