Game Theory

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OPERATIONS RESEARCH

Chapter 2

Game Theory

Prof. Bibhas C. Giri

Department of Mathematics Jadavpur University

Kolkata, India

Email: bcgiri.jumath@gmail.com

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1.0 Introduction

Game theory was developed for decision making under conflicting situations where there are one or more opponents (players). The games like chess, poker, etc. have the characteristics of competition and are played according to some definite rules. Game theory provides optimal solutions to such games, assuming that each of the players wants to maximize his profit or minimize his loss.

Game theory has applications in a variety of areas including business and eco- nomics. In 1994, the Nobel Prize for Economic Sciences was won by John F. Nash, Jr., John C. Harsanyi, and Reinhard Selton for their analysis of equilibria in the theory of noncooperative games. Later, in 2005, Robert J. Aumann and Thomas C. Schelling won the Nobel Prize for Economic Sciences for enhancing our understanding of con- flict and cooperation through game theory analysis.

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MODULE - 1: Basic Concept and Terminologies, Two-person Zero-sum Game, and Game with Pure and Mixed Strategies

In this Module, we will discuss some basic terminologies used in Game Theory, two- person zero-sum game and games with pure and mixed strategies.

1.1 Basic Terminologies

The following terminologies are commonly used in Game theory.

Player : Each participant (interested party) of a game is called a player.

Strategy :The strategy of a player is the predetermined rule by which a player decides his course of action from the list of courses of action during the game. A strategy may be of two types:

• Pure strategy -It is a decision, in advance of all plays, always to choose a particular course of action.

• Mixed strategy - It is a decision, in advance of all plays, to choose a course of action for each play in accordance with some particular probability distribution.

Optimal strategy : The course of action which maximizes the profit of a player or minimizes his loss is called an optimal strategy.

Payoﬀ:The outcome of playing a game is called payoﬀ.

Payoff matrix : When the players select their particular strategies, the payoffs (gains or losses) can be represented in the form of a matrix called the payoffmatrix.

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Saddle point : A saddle point is an element of the payoﬀ matrix, which is both the smallest element in its row and the largest element in its column. Furthermore, the saddle point is also regarded as an equilibrium point in the theory of games.

Value of the game : It refers to the expected outcome per play when players follow their optimal strategy.

1.2 Two-Person Zero-Sum Game

A game with only two players is called a two-person zero-sum game if the losses of one player are equivalent to the gains of the other so that the sum of their net gains is zero. This game also known asrectangular game.

In a two-person game, suppose that the playerAhasmactivities and the playerBhas nactivities. Then a payoﬀmatrix can be formed by adopting the following rules:

(i) Row designations for each matrix are activities available to the playerA.

(ii) Column designations for each matrix are activities available to the playerB.

(iii) Cell entryv_ij is the payment to the playerAinA’s payoﬀmatrix whenAchooses the activityi andBchooses the activityj.

(iv) For a zero-sum game, the cell entry in the playerB’s payoffmatrix will be nega- tive corresponding to the cell entryv_ij in the playerA’s payoffmatrix so that the sum of payoff matrices for the playersAandB is ultimately zero, see Tables 1.1 and 1.2.

PlayerB

1 2 ··· n

PlayerA

1 v11 v12 ··· v1n

2 v₂₁ v₂₂ ··· v_2n ... ... ... ... ... m v_m1 v_m2 ··· v_mn

Table 1.1:PlayerA’s payoﬀmatrix

PlayerB

1 2 ··· n

PlayerA

1 −v11 −v12 ··· −v1n

2 −v21 −v22 ··· −v2n

... ... ... ... ... m −vm1 −vm2 ··· −vmn Table 1.2:PlayerB’s payoﬀmatrix

Consider a two-person coin tossing game. Each player tosses an unbiased coin simul- taneously. Each player selects either a head H or a tail T. If the outcomes match (i.e., (H, H) or (T, T)) then A wins Rs. 4 from B; otherwise, B wins Rs. 3 from A. Player A’s

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payoﬀ matrix is given in Table 1.3. This game is a two-person zero-sum game, since the winning of one player is taken as losses for the other. Each player has his choice from amongst two pure strategies H and T.

Player B

H T

Player A H 4 -3

T -3 4

Table 1.3

1.3 Pure Strategies (Minimax and Maximin Criterion)

The simplest type of game is one where the best strategies for both players are pure strategies. This is the case if and only if, the payoﬀmatrix contains a saddle point.

Theorem 1.1: Let (v_ij) be the m×npayoﬀ matrix for a two-person zero-sum game. If v denotes the maximin value andv¯the minimax value of the game, thenv¯≥v ¯

¯. That is, minj [max

i {vij}]≥max

i [min

j {vij}].

Proof: We have

maxi {vij} ≥v_ij f or all j = 1,2, ..., n and min

j {vij} ≤vij f or all i= 1,2, ..., m

Let the above maximum and minimum values be attained ati =i₁ and j=j₁, respectively, i.e.,

maxi {vij}=v_i₁_j and min

j {vij}=v_ij₁ Then, we must have

v_i₁_j≥v_ij ≥v_ij₁ f or all j = 1,2, ..., n; f or all i= 1,2, ..., m.

From this, we get

minj v_i₁_j≥v_ij≥max

i v_ij₁ f or all j = 1,2, ..., n; i= 1,2, ..., m.

Therefore, min

j [max

i {vij}]≥max

i [min

j {vij}].

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Note: A game is said to be fair, if v

¯ = 0 = ¯v and it is said to be strictly determinable if

¯v=v= ¯v.

Example 1.1: Consider a two-person zero-sum game matrix which represents payoﬀ to the player A, see Table 1.4. Find the optimal strategy, if any.

Player B

Player A

I II III IV V

I -2 0 0 5 3

II 4 2 1 3 2

III -4 -3 0 -2 6

IV 5 3 -4 2 -6

Table 1.4:Payoﬀmatrix for Example 1.1

Solution: We use the maximin (minimax) principle to determine the optimal strategy.

The playerAwishes to obtain the largest possiblev_ij by choosing one of his activities (I, II, III, IV), while the playerBis determined to makeA’s gain the minimum possible by choice of activities from his list (I, II, III, IV, V). The playerAis called themaximiz- ing playerandB, theminimizing player. If playerAchooses the activity I then it could

Player B

Player A

I II III IV V Row minimum

I -2 0 0 O5 3 -2

II 4 2 O1 3 2 1←Maximin

III -4 -3 0 -2 O6 -4

IV O5 O3 -4 2 -6 -6

Column maximum O5 O3 O1 O5 O6

↑Minimax

Table 1.5:Player A’s payoﬀmatrix

happen that the playerBalso chooses his activity I. In this case, the playerBcan guar- antee a gain of at least−2 to playerA, i.e.,min{−2,0,0,5,3}=−2. Similarly, for other choices of playerA, i.e., activities II, III and IV,B can force the playerAto gain only 1, −4 and−6, respectively, by proper choices from (II, III, IV) i.e.,min{4,2,1,3,2}= 1, min{−4,−3,0,−2,6}=−4 andmin{5,3,−4,2,−6}=−6. For playerA, minimum value in each row represents the least gain to him if he chooses his particular strategy. These

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are written in Table 1.5 by row minimum. Player A will select the strategy that maximizes his minimum gains, i.e.,max{−2,1,−4,−6}= 1 i.e., playerAchooses the strategy II. This choice of playerAis called the maximin principle, and the corresponding gain (here 1) is called the maximin value of the game. In general, the playerAshould try to maximize his least gains or to findmax

i min

j v_ij=v

¯.

For player B, on the other hand, likes to minimize his losses. The maximum value in each column represents the maximum loss to him if he chooses his particular strategy. These are written in Table 1.5 by column maximum. Player B will then select the strategy that minimizes his maximum losses. This choice of playerBis called the minimax principle, and the corresponding loss is the minimax value of the game. In this case, the value is also 1 and playerBchooses the strategy III. In general, the player Bshould try to minimize his maximum loss or to findmin

j max

i v_ij= ¯v.

If the maximin value equals the minimax value then the game is said to have a saddle point (here (II, III) cell) and the corresponding strategies are called optimum strategies. The amount at the saddle point is known as the value of the game.

Example 1.2: Solve the game whose payoﬀmatrix is given below:

Player B

Player A

I II III I -2 15 -2 II -5 -6 -4 III -5 20 -8

Table 1.6:A’s payoﬀmatrix

Solution: We use the maximin (minimax) principle to determine the optimal strategy.

The game has two saddle points at positions (1, 1) and (1, 3).

Player B

Player A

I II III Row minimum

I -2O 15 -2O -2←Maximin

II -5 -6O -4 -6

III -5 20 -8O -8

Column maximum -2O 20O -2O

↑Minimax ↑Minimax

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(i) The best strategy for playerAis I.

(ii) The best strategy for playerBis either I or III.

(iii) The value of the game is−2 for playerAand +2 for playerB.

1.4 Mixed Strategy: Game without A Saddle Point

Ifmaxmin valueis not equal tominimax valuethen the game is said to have no saddle point. In such a case, both the players must determine an optimal mixture of strategies to find an equilibrium point. The optimal strategy mixture for each player may be determined by assigning to each strategy its probability of being chosen. The strategies so determined are calledmixed strategies.

The value of game obtained by the use of mixed strategies represents the least payoffwhich playerAcan expect to win and the least payoffwhich playerBcan expect to lose. The expected payoff to a player in a game with payoffmatrix [v_ij]_m×ncan be defined as

E(p,q) =

∑m

i=1

∑n

j=1

p_iv_ijq_j =pvq^T

wherep= (p₁, p₂, ..., p_m) andq= (q₁, q₂, ..., q_n) denote probabilities or relative frequency with which a strategy is chosen from the list of strategies associated with m strategies of player A and n strategies of player B, respectively. Obviously, p_i ≥ 0 (i = 1,2, ...m), q_j≥0(j= 1,2, ..., n) andp₁+p₂+...+p_m= 1;q₁+q₂+...+q_n= 1.

Theorem 1.2: For any2×2two-person zero-sum game without any saddle point having

the payoﬀ matrix for player A given by





B₁ B₂ A₁ v₁₁ v₁₂ A₂ v₂₁ v₂₂



, the optimal mixed strategies S_A=



 A₁ A₂

p₁ p₂



 andS_B =



 B₁ B₂

q₁ q₂



 are determined by ^p_p¹

2 = ^v_v²²^−v²¹

11−v12, ^q_q¹

2 = ^v_v²²^−v¹²

11−v21 where p₁+p₂= 1andq₁+q₂= 1. The valuevof the game toAis given byv=_v ^v¹¹^v²²^−v²¹^v¹²

11+v₂₂−(v12+v₂₁). Proof: Let a mixed strategy for playerAbe given byS_A=



 A₁ A₂

p₁ p₂



, wherep₁+p₂= 1.

Thus, if playerBmovesB₁then the net expected gain ofAwill beE₁(p) =v₁₁p₁+v₂₁p₂ and ifBmovesB₂, the net expected gain ofAwill beE₂(p) =v₁₂p₁+v₂₂p₂.

Similarly, ifBplays his mixed strategyS_B =



 B₁ B₂ q₁ q₂



, whereq₁+q₂ = 1, thenB’s net expected loss will beE₁(q) =v₁₁q₁+v₁₂q₂ifAplaysA₁, andE₂(q) =v₂₁q₁+v₂₂q₂ifA

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playsA₂. The expected gain of playerA, whenBchooses his moves with probabilities q₁ and q₂, is given by E(p,q) =q₁[v₁₁p₁+v₂₁p₂] +q₂[v₁₂p₁+v₂₂p₂]. Player A would always try to mix his moves with such probabilities so as to maximize his expected gain.

Now, E(p,q) =q₁[v₁₁p₁+v₂₁(1−p₁)] + (1−q₁)[v₁₂p₁+v₂₂(1−p₁)]

= [v₁₁+v₂₂−(v₁₂+v₂₁)]p₁q₁+ (v₁₂−v₂₂)p₁+ (v₂₁−v₂₂)q₁+v₂₂

=λ (

p₁−v₂₂−v₂₁ λ

)(

q₁−v₂₂−v₁₂ λ

)

+v₁₁v₂₂−v₁₂v₂₁ λ

whereλ=v₁₁+v₂₂−(v₁₂+v₂₁).

We see that ifAchoosesp₁= ^v²²^−v_λ ²¹, he ensures an expected gain of at least (v₁₁v₂₂− v₁₂v₂₁)/λ. Similarly, ifBchoosesq₁ = ^v²²^−v_λ ¹², then he can limit his expected loss to at most (v₁₁v₂₂−v₁₂v₂₁)/λ. These choices ofp₁ and q₁ will thus be optimal to the two players. Thus, we get

p₁=v₂₂−v₂₁

λ = v₂₂−v₂₁

v₁₁+v₂₂−(v₁₂+v₂₁) andp₂= 1−p₁= v₁₁−v₁₂ v₁₁+v₂₂−(v₁₂+v₂₁) q₁=v₂₂−v₁₂

λ = v₂₂−v₁₂

v₁₁+v₂₂−(v₁₂+v₂₁) andq₂= 1−q₁= v₁₁−v₂₁ v₁₁+v₂₂−(v₁₂+v₂₁) andv= v₁₁v₂₂−v₁₂v₂₁

v₁₁+v₂₂−(v₁₂+v₂₁) Hence, we have

p₁

p₂ = v₂₂−v₂₁ v₁₁−v₁₂, q₁

q₂ =v₂₂−v₁₂

v₁₁−v₂₁; and v= v₁₁v₂₂−v₂₁v₁₂

v₁₁+v₂₂−(v₁₂+v₂₁). (1.1) Note:The above formula forp₁,p₂,q₁,q₂andvare valid only for 2×2 games without saddle point.

Example 1.3: Suppose that in a game of matching coins with two players, one player wins Rs. 2 when there are 2 heads, and gets nothing when there are 2 tails and looses Re 1 when there are one head and one tail. Determine the best strategies for each player and the value of the game.

Solution: The payoﬀmatrix for playerAis given in Table 1.7. The game has no saddle point. Let the playerAplaysH with probabilityxandT with probability 1−x. Then, if the playerBplaysH, thenA’s expected gain is

E(A, H) =x(2) + (1−x)(−1) = 3x−1.

If the playerBplaysT,A’s expected gain is

E(A, T) =x(−1) + (1−x).0 =−x.

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Player B

H T

Player A H 2 -1

T -1 0

Table 1.7:Player A’s payoﬀmatrix

If the player Achoosesx such that E(A, H) =E(A, T) =E(A) say, then this will deter- mine best strategy for him. Thus we have 3x−1 =−x orx= 1/4. Therefore, the best strategy for the playerAis to playH andT with probability 1/4 and 3/4, respectively.

Therefore, the expected gain for playerAis E(A) =1

4(2) +3

4(−1) =−1 4.

The same procedure can be applied for playerB. If the probability ofB’s choice ofH isy and that ofT is 1−ythen for the best strategy of the playerB,

E(B, H) =E(B, T) which givesy= 1/4. Therefore, 1−y= 3/4.

Thus A’s optimal strategy is (1/4,3/4) and B’s optimal strategy is (1/4,3/4). The expected value of the game is−1/4 to the playerA.

This result can also be obtained directly by using the formulae (1.1).