game theory notes

CM32OX: Topics in Mathematics

Game Theory

Andrew Pressley

At first sight. the Theorv of Games’ might appear to be a trivial pursuit. However.game theory aims to model any situation involving conflict where each party (or‘player’) has only partial control over the outcome. This includes not only games inthe usual sense (e.g. chess, bridge, poker), but also many situations in economics.politics, business and management. In fact, game theory was created by John vonNeumann to model problems in economics.Consider the game with two players in which one player chooses a row andsimultaneously the other player chooses a column of the matrix

/0 1 2f—i 0 100\\—2 —100 0

If a is the entry in the chosen row and column. the row player gets La from thecolumn player (if a is negative, the column player receives L(—a) from the rowplayer). Greed might encourage the row player to choose row 2 but then he couldlose Li: similarly, the column player might be tempted to choose column 2 butthen he could lose Li. Avoidance by both players of a worse outcome will ensurethat the first row and column are chosen.We shall see that a wide class of games can be reduced to a ‘matrix game’ ofthis kind (possibly with a different matrix, of course. In general. the analysis willnot be so simple, but eventually we shall obtain an algorithm which allows us to‘solve’ all matrix games.

Contents

1 Games and their Trees 1

1.1 Introduction and Examples 11.2 Game Trees 5Problems

7

2 Pure Strategy Solutions 9

2.1 Pure Strategies 92.2 Examples 92.3 Matrix Games 122.4 Optimal Pure Strategies 122.5 Minimax Theorem (first version) 15Problems 18

3 Mixed Strategy Solutions 20

3.1 Mixed Strategies 203.2 Finding Mixed Strategy Solutions 213.3 Inferior Strategies (again) 243.4 The 2 x n Case 283.5 Convexity 323.6 Proof of Von Neumann’s Minimax Theorem 37Problems 41

4 Solving Matrix Games in general 44

4.1 Simple Solutions 444.2 Convexity (again) 484.3 Shapley-Snow Algorithm 494.4 Proof of the Algorithm 54Problems .56

Chapter 1: Games and their Trees

We begin by describing the kind of games we shall discuss in this course, how toformulate such games in a precise mathematical way, and how to represent thempictorially.

1.1 Introduction and Examples

Before we attempt to formulate a mathematical definition of a game, we give ageneral discussion illustrated by some examples .Any game (or at least any gamethat we shall consider in this course) is characterized by a set of rules governingthe behaviour of certain individuals or groups called players. An n-person gamehas n players (n 2,3... .). Of course, chess is a 2-person game. but so is bridgesince two players in the same team have identical interests and so count as a singleplayer in game theory.When the game is played, a sequence of moves are made, the nature of whichis specified by the rules. Moves are of two kinds: personal moves, which involve achoice by one of the players of one of a finite set of alternatives, and chance moves,in which a choice among a finite set of alternatives is made by a chance mechanism,the probabilities of each alternative being specified by the rules. The rules specifywhich player or chance mechanism is to make each move. For example, all movesin chess are personal moves, while the first move in a game of bridge is the deal,which is a chance move with 52!/(13!) alternatives.Finally, the rules specify which sequences of moves constitute a complete playof the game, and they assign a numerical score, the payoff, to each player for eachplay of the game. In a 2-person win, lose or draw game such as chess one wouldnormally assign a score of + 1 to the winner and —1 to the loser (and 0 to eachplayer in case of a draw).

Example 1.1 In the 2-person game of NIM, one starts with a finite number ofpiles of matches, with at least one match in each pile. The first player selects onepile and removes at least one match from that pile. He can remove any number ofmatches from one up to the total number of matches in the pile. The next playerdoes the same with the remaining piles of matches. This continues until there areno matches left. The last player to remove a match loses.To simplify the discussion, we shall actually consider only the game of (2,2)-NIM, which starts with two piles each with two matches. Let us call the playerwho makes the first move ‘player 1’ and the other player player 2’. We representthis game by a game tree, in which the initial state of the matches is represented bya vertex labelled 1 since player 1 makes the first move. Player 1 can either removeone match from one of the piles or both matches from one of the piles. Hence, atthe initial vertex there are two possible (personal) moves, leading to states whichwe represent by

and

where the horizontal line represents one empty pile of matches. These two newstates are represented by two vertices labelled 2, because it is now the turn ofplayer 2’. Continuing in this way, we get the game tree shown below.C i— I)

.“...“.. —I

The vertices corresponding to two empty piles of matches do not have a label asneither player will then be making a move. They are called terminal vertices andrepresent the end of the game.

It is very important to note that in general there are several different vertices allof which correspond to the same ‘match situation’, Each vertex corresponds notjust to the current match situation’ but to the entire sequence of moves of the playof the game that leads to that vertex.The game tree contains everything there is to know about the game. For example,we can see from the game tree that, if player 2 plays correctly. he will always win

no matter what player 1 does. Indeed, if player 1 removes 2 matches at the firstmove, then player 2 will win by removing 1 match at his first move; if player 1removes 1 match at his first move, player 2 should remove both matches from theother pile. Unfortunately, riot all games can be analyzed so simply.

Example 1.2 The game HI-LO is played with three cards, labelled 1, 2. 3. Player1 picks up the three cards, looks at them, chooses one and lays it face down on thetable. Player 2 guesses ‘high’ or ‘low’. The card on the table is turned face up; ifthe card was 3 and player 2 guessed ‘high’, or if it is 1 and player 2 guessed ‘low’,player 2 gets £3 from player 1; if the card was 3 and player 2 guessed ‘low’, or if itwas 1 and player 2 guessed ‘high’, player 1 gets £2 from player 2. If the card was 2and player 2 guessed ‘low’, player 2 gets £2 from player 1: hut if player 2 guessed‘high’, player 1 must lay one of the two remaining cards face down on the table,

1-li

(—Ii’)

(- i ,i)

[——1

III

H -I

(—I)’)

I

L’ ‘I

H ‘I

and then player 2 guesses ‘high’ or ‘low’ and player 2 gets £1 from player 1 if he isright but loses £3 to player 1 if he is wrong.

The game tree is shown below, At the initial vertex, there are three (personal)moves according to which card player 1 selects; at the next three vertices there aretwo (personal) moves according to whether player 2 guesses ‘high’ (H) or ‘low’ (L),etc. We label each terminal vertex with a pair of numbers representing the payoffsto each player.

(-1,1) (—1,1)

This game differs from NIM in that, after the first move, player 2 does not knowwhich card is on the table. So although player 2 is at one of the vertices labelled2 just above the initial vertex, he does not know which. To indicate this we drawa box surrounding these three vertices. Similarly, the next two moves available toplayer 2 are also enclosed in a box. Since player 1 has complete information aboutthe game at all times, each vertex labelled 1 should be in a separate box. The set of

(2,-2) (2,2)

L

H L L

1

I

I

4

vertices of the game tree in a given box is called an information set. So this gamehas four information sets.

Our next example is of a game that involves chance moves.

Example 1.3 In the game RED-BLACK, player 1 is dealt face down a red orblack card, each card being equally likely. Player 1 looks at the card but does nottell player 2 what it is. If the card is red, player 1 can either call for the toss of afair coin or pass the move to player 2. If the card is black, player 1 must pass themove to player 2. If player 2 gets the move, he has to guess whether the card is redor black. If the coin is tossed, the garrie ends.

The game tree is as follows. We label the vertices at which a chance move is madeO and label the chance moves themselves with the probability of each alternative.Chance moves do nor belong to any inforthation set.

U

B

Player 2 pays player 1 Lp, Lq, Lr, Ls, Lt or Lu, depending on the final stateof the game as indicated on the game tree.

The games in these examples all terminate after a finite number of moves, and inthis course we shall only study games which have this property, although there arecertainly games which do not have it. For example, it is well known that in chessthere are some sequences of moves in the endgame which can be repeated cyclically,arid so could go on for ever. In such cases, however, some rule is normally agreedthat causes the game to terminate; for example, it might be agreed in advance thatif a cyclical sequence of moves has been repeated a certain number of times, thegame is a draw. In this way, games which are potentially infinite’ are convertedinto the ‘finite’ games studied in this course.

toss

1

B

D

1.2 Game Trees

With these examples in mind, we now introduce an abstract mathematical modelof a game tree.

A game tree (sometimes called a game in extensive form) consists of the followingten ingredients.

(i) A finite set V, whose elements are called vertices.

The vertices of the game tree are identified with the states of the game.

(ii) A set M of pairs (vi, v2), where u1, v2 E V and v1 v2; the elements of M arecalled edges.

The interpretation is that in = (vi,v9) E Al if there is a move from the statev1 of the game to state v2. Then v1 is called the initial point of m and v2 its finalpoint. A vertex v E V that is not the initial point of any move is called a terminalvertex. These correspond to the states at the end of the game.

(iii) A distinguished initial vertex v0 E V which is not the final point of any edge.

Of course. v0 corresponds to the state of the game before play has commenced.

(iv) A path in the game tree is a sequence of vertices (v1,v2 vk) such that, ifk> 1, (vi, u÷i) Al for i = 1 k—i. Then u1 is the initial point of the path, Vkis its final point and k — 1 is its length. The rank of the game tree is the maximumof the lengths of all paths in the game tree.

A path whose final point is a terminal vertex corresponds to a (complete) playof the game.

The game tree is required to have the following property (it is this propertywhich makes the game tree a tree).

(v) For any v e V, there exists a nnique path with initial point vo and final pointv.

The players of the game are introduced into our mathematical game tree asfollows:

(vi) The set V of vertices has a partition into n + 1 disjoint subsets ,V1. .. ,

such that V = V U V1 U .. U V and is non-empty if i > 0.

The interpretation is that the game tree is that of an n-player game. 1 beingthe set of states at which a chance move is made (if any), while V, (for 1 < j < n)is the set of personal moves for player i.

(vii) For any v E let rn1,.. . , mk be the edges with initial point v. Then, thereare positive numbers ir1 irk such that in1 + ... ± ir = 1.

Of course. ir is the probability with which the chance move rn.3 is made startingat the state v of the game.

It is also useful to partition V in another way.

(viii) For j = 1, 2, 3,. .., let A be the set of vertices which are the initial points ofprecisely j edges.

Thus. if v e fl A. with j > 1 and 1 < i < ii. then at the state v of the gameplayer i has to make one of j possible moves. If u E .4j. the j moves with initialpoint v are labelled 1,2 j.

6

The set V1 U U V, = V\V0 has yet another partition:

(ix) There is a partition of V\V0 into disjoint non-empty subsets called informationsets such that, if I is any information set, then I C fl A for some i, j withi 0. Moreover, we assume that, if v and w are two different vertices in the sameinformation set, there is no path with initial point u and final point iv.

The final ingredient is the introduction of the payoffs of each player.

(x) There are a real-valued functions pr,. . ., p,., defined on the set of terminal

vertices.

If v is a terminal vertex, pj(V) is the payoff to player i if the game ends in thestate v.

Remark It is important that the indexing of the moves in (viii) is done properly,respecting the structure of the game, and in particular the information sets. Forexample. in the game HI-LO, there are two information sets for player 2. In eachof these, the moves in which player 2 guesses • high’ must all have the same label.

It might appear that games which involve simultaneous choices by two or moreplayers cannot be described by using a game tree, but this is not the case. Forexample, consider the game of MATCHING PENNIES. This is a 2-person gamein which players 1 and 2 simultaneously show a coin. If both coins are ‘heads’ or ifboth are ‘tails’, player 1 receives £1 from player 2; otherwise, player 2 receives £1from player 1. We can interpret this as a game in which player 1 makes the firstmove and then player 2 makes his move without knowing the result of player l’smove. This gives the game tree

(—l) (—1.1) (—1.1) (1,—l)

H T

2

I

This shows the importance of information sets, since the game tree

would represent a very different game. in which player 1 shows his coin and thenplayer 2, having seen player 1 ‘s coin, shows his. Obviously, player 2 will always winthis latter game. whereas it is clear that there is no strategy that either player canadopt in the first game which will always result in a win.

Problems

1.1. In the game of n < n NOUGHTS AND CROSSES, plaYer 1 places an 0 in an n < narray, then player 2 places an X in one of the n2 — 1 unoccupied positions, then player 1 placesan 0 in one of the n2 — 2 unoccupied positions. etc. The winner of the game is the first to forma complete row or column or diagonal of Os or Xs (and the game is a draw if this has not beenachieved by the time the matrix is filled). Draw the game tree of the (completely uninteresting)game of 2 x 2 NOUGHTS AND CROSSES. (If you have a very large piece of paper you couldtry the same for the more usual 3 x 3 case.)

1.2. Modify the game of MATCHING PENNIES by keeping the same rules except that theplayers now toss their coins instead of choosing whether to show ‘heads or ‘tails. Draw the gametree.

1.3. In the game STONE-SCISSORS-PAPER, two players simultaneously call out ‘stone’,scissors’ or paper’, the rule being that ‘stone’ beats ‘scissors’ beats paper’ beats stone’ (and the

game is a draw if the players make the same call). Draw the game tree.

1-1. Draw a game tree for the following game. The first move is the choice by player 1 of aninteger i from {O, 1}. The second move is a chance move in which an integer j is chosen from(0, 1}, both choices being equally likely. At the third move player 2, knowing j but not i, selectsan integer k from (0, 1} and the game then terminates. If i + j k = 1, player 1 pays player 2£1; otherwise player 2 pays player I Li.

1.5. Draw game trees for the following four variants of the game considered in Problem 1.4, wherethe moves and payoffs are the same hut player 2 has the following information when he makes hismove

T

T

8

(i) he knows neither i nor j;(ii) he knows both i and j;(iii) he knows z + j but not the separate values of i and j;(iv) he knows the value of the product of i and j but not the separate values of i and j.1.6. Draw a game tree for the following 4-person game. The first move is a chance move at whichan integer .r is chosen from (1, 2}, the probabilities that 1 and 2 are chosen being 1/3 and 2/3,respectively. For the second move, if .r = I player 1 chooses a number p from {1. 2, 3}, while ifx = 2 player 2 chooses a number p from {1. 2, 3}. For the third move, if p 1 player 3, notknowing , chooses an integer z from {1,2}. while if p 1 player 4, knowing .r but not knowingwhether p = 2 or 3, chooses an integer z from {f, 2}. The game then terminates.1.7. Draw . game tree for the following 2-person game. Player 2 consists of two agents A and B.Player I and the agents A and B are isolated and play is conducted by an umpire At the startof the game, the umpire requires player 1 to choose an integer z from {i, 2}. If player 1 chooses1. the umpire asks A to choose an integer y from {1.2}, while if player 1 chooses 2 the umpireasks B to choose an integer p from {1,2}. Finally, the umpire asks the agent for player 2 who hasnot yet participated to choose an integer z from {1,2}. Player 1 receives h(xy,z) from player 2,where h is a real-valued function. (In this game, it is important to note that when an agent ofplayer 2 is asked to make a choice he does not know whether he is making the second or thirdmove of the game.)

1.8. Draw a game tree for the 2-person game which starts with each player being dealt a card froma pack of cards labelled 1,2 and 3. Player 1 then calls ‘high’ or ‘low’ to which player 2 respondsby calling ‘high’ or ‘low’. The game then terminates unless player 1 has called low and player 2has called high, in which case player I mtist make one further move, calling either high’ or ‘low’.Denote the play of the game in which player 1 calls ‘high’ and player 2 calls ‘low’ by HL, anddenote the other plays similarly. The payoffs are as follows.HH: the player with the higher card receives £2 from the other player;LL: the player with the higher card receives £1 from the other player:HL: player 1 receives £1 from player 2;LHL: player 2 receives £1 from player 1;LHH: the player with the higher card receives £2 from the other player.

1.9. Show that, in any game tree:(1) there is no edge with final point vu;(ii) if v is a vertex different from v, there is exactly one edge which has v as its final point.

9

Chapter 2: Pure Strategy Solutions

In this chapter we discuss what it means to ‘solve’ a game. i.e. to find ‘optimalstrategies’ for each player.

2.1 Pure Strategies

Roughly speaking, a pure strategy for a player of a game is a plan made in advanceof what he will do in every situation involving him that may arise in the game.(The word ‘pure’ is added for reasons that will appear later.) The formal definitionis as follows. We use the notation introduced in Section 1.2.

Definition 2.1 Let Z be the set of information sets for player i in a game tree (i.e.I is the set of information sets I such that I C I’. A pure strategy for player i isamapu :I —*Nsuch that j(I) E {1,2,... .j} jf I A.

The interpretation of this is as follows. Suppose that the game has reached astate v at which player i is to make a move. Let I be the information set containingv, and suppose that I C A, so that there are j possible moves that player i maymake. Then, the pure strategy oj dictates that player i chooses the move labelledu(I).

The fact that the pure strategy picks a move for each information set, ratherthan for each vertex of the game, reflects the fact that a player knows only whichinformation set he is in at any stage of the game, and not necessarily which vertexhe is at.

2.2 Examples

Example 2.2 In (2,2)-NIM, there are five information sets for player 1, eachconsisting of one vertex, at two of which there are two possible moves and at theother three only one possible move. So there are 2 x 2 x 1 x 1 x 1 = 4 pure strategiesfor player 1. Similarly, there are 2 x 3 x 1 x 1 = 6 pure strategies for player 2.

Example 2.3 In HI-LO, there are two information sets for player 1, one withthree possible moves and one with two, giving 3 x 2 = 6 pure strategies for player1. There are also two information sets for player 2, each with two possible moves,giving 2 x 2 = 4 pure strategies for player 2.

Suppose that each player in a game has selected a pure strategy. If there areno chance moves in the game, then clearly the play is completely determined, andhence so is the terminal vertex of the play and hence the payoff to each player.However, if there are chance moves in the game, there will in general he severalpossible terminal vertices for any given choice of pure strategies for each player.

Suppose that in an fl-person game player i selects pure strategy o, i = 1,... , n.The probability of arriving at a particular terminal vertex v is the product of theprobabilities assigned to each chance move in the unique path in the game treewith initial point v0 and final point v; denote this probability by ,o).

10

Of course, it may he that. for the given choice of pure strategy for each player.o is not a possible terminal vertex; in that case we put ir(ui a,,) = 0. Forexample, if there are no chance moves in the game. we shall have x, (a1 a,,) = 1if v is the unique terminal vertex determined by a1 a,,, and ir (aj a,,) = 0otherwise.

Definition 2.4 Let be the set of pure strategies for player i in an 71-persongame. The pure pay[f funclion for player i is the map P, : x x x- Rgiven by

a,,) = pi(v)xu(ai

where the sum is over the set of terminal vertices v of the game.Remark This definition follows the usual rules for calculating expected values ofrandom variables in probability theory. Recall that, if X is a random variablewith a finite number of possible values r £N, and if pk is the probability thatX = xe, the expected value of X is

E(X) PkXk.

In this course, we shall be particularly interested in 2-person games. In thatcase. let us label the pure strategies for each plover, say =

. .,

= {a1) where ni, n 1. Then, the pure payoff function for player ican conveniently he described by the mYn matrix whose (r. s)-entrv is P, (a’,This is called the payoff niatri.v of the game for player i.Example 2.5 In RED-BLACK, each player has two pure strategies, which canbe described as follows:

player 1: (1) if red then toss; (2) if red then pass move to player 2.player 2: (1) guess red; (2) guess black.

Suppose. for example, that both players adopt their pure strategy (1). Then thereare three possible terminal vertices, with payoffs £p, £q or £t. The probabilitiesof these terminal vertices being reached are 1/4, 1/4 and 1/2. respectively, soP1(1. 1) (p + q + 2t)/4. You should check that the complete payoff matrix forplayer I is((p+q+2f)/4 (p±q±2u)/4

(r+t)/2 (s+u)/2

Of course. the payoff matrix for player 2 is just the negative of this matrix, sincefor an play of the game the payoff to player 2 is minus that to player 1.Example 2.6 A 2-person game starts with the choice by player 1 of an integerx from {l, 2}. Next, an umpire tosses a coin and informs player 2 of x if theoutcome is ‘heads’, but not otherwise. Player 2 then chooses an integer p from{3. 4}. Finally, the umpire selects, by a chance device, one of the integers 1.2 or 3with probabilities 2/5. 1/5 and 2/5, respectively. If z is the choice at the last move,player 1 receives £(i + p + z) from player 2 if ,r + p + z is even, and player 2 receives£(x + p + ) from player 1 if x + p + z is odd.

11

—__-7 --Tç

½

0

‘ -.7 c.

L — i

I

-7 —-?

2

½

G —7 ‘ —7

1/2

The game tree is shown above. We show only the payoffs to player 1 at theterminal vertices, since the payoffs to player 2 are simply the negatives of those toplayer 1.

There is just one information set for player 1 at which there are two possiblemoves, so player 1 has two pure strategies which we denote by 1 or 2 accordingto which integer is chosen. For player 2, there are three information sets, at eachof which there are fwo possible moves, giving S pure strategies for player 2. Wedenote them by triples (i.j,k), this denoting the pure strategy which consists inchoosing i if the coin falls heads’ and player 1 has chosen 1, choosing j if the coinfalls ‘heads’ and and player 2 has chosen 2. and choosing k if the coin fails tails’.We list the pure strategies for each player as follows:

player 1: 1. 2player 2: (3,33). (3,3,4), (3.4.3). (4,3.3), (3.4,4), (43,4). (4.4.3), (4,4.4).

You should check (see Problem 2.8) that the payoff matrix for player 1 is

3/10 —18/5 3/10 3/10 21/5 3/10 21/5—3/10 —3/10 21/5 —24/5 —3/10 —3/10 —24/5

(—18/s\ 21/5

12

2.3 Matrix Games

For the remainder of this course we shall only consider 2-person games that arezero-sum, in the following sense.

Definition 2.7 Let P1 (i = 1,2) be the pure payoff function for player i in a2-person game. Then, the game is said to be zero-sum if Pi + P2 0.

(1) (in) (1) (n)Let {o o ) 2 = {a 2 ) be the sets of pure strategies

for players 1 and 2, respectively, in a 2-person game. Then the zero-sum condition

is

for all i = 1,... , m, j = 1,... , n. This means that the payoffs to both players are

determined by the payoff matrix for player 1; this matrix is called the payoff matrix

of the game.Conversely, any rn x ri matrix (a) is the payoff matrix of some 2-person zero

sum game, namely the game in which two players 1 and 2 independently selectintegers i E {1. 2,. . . , m} and j {1, 2 n}, respectively, and player 2 pays

player 1 £a2.The analysis of 2-person zero-sum games depends only on a knowledge of the

payoff matrix. Accordingly, such games are often called matrix games.

2.4 Optimal Pure Strategies

We now consider the problem of finding the ‘optimal’ strategies for each player in

a given matrix game with rn x n payoff matrix (a). To simplify the notation, we

shall denote the elements of the pure strategy sets and 2 simply by 1, 2,. . . , m

and 1.2,... , n, respectively.If player 1 adopts pure strategy i, his payoff will be at least

mm = b, say.1jn

Let b0 he the maximum of the numbers b for 1 < i <m. Then, by adopting pure

strategy i0, player 1 can ensure that his payoff is not less than i.e. not less than

max mm a3.1<i<m 1<j<n

Similarly, if player 2 employs pure strategy j, his payoff will be at least

min(_aj) mnaxa,

so player 2 has a pure strategy which ensures that his payoff is at least

max ( maxa) = — minmaxa

To summarize, player 1 has a pure strategy which ensures that his payoff is at

leastmax mm

I J

13

and player 2 has a pure strategy which ensures that player l’s payoff is at most

mm max.1 ‘

Suppose that these two numbers are equal, i.e. that

(1) max mm a3 = 1ninrnax0= say.i J j i

Then, player 1 has a pure strategy which gives him a payoff of at least v and player2 has a pure strategy which prevents player 1 from obtaining a payoff of more thanv. These pure strategies would then have to be considered ‘optimal’ for each player,and the game would be ‘solved’.

Unfortunately, condition (1) does not always hold:

Example 2.8 In the game of MATCHING-PENNIES described at the end ofChapter 1, each player has two pure strategies. namely:

1: choose ‘heads’; 2: choose ‘tails’.

The payoff matrix (a) is/1 —1

1

and clearlymax mma = —i, mm max = + 1.

z j 3 ‘

To understand when condition (1) holds, we shall consider a more general situation (which will be required later in any case),

Definition 2.9 Let X and Y be sets of real numbers and let f : X x U — R beany function, An element (x, y*) e X x Y is called a saddle point of f if

(2) f(x,y) <f(x,y) <f(x*,y) for allx e X. yE Y.

Of course, we can regard an rn x n payoff matrix (a) to be a function of thiskind by taking X = {1,2.. ,m}. U = {1,2.... ,n} and f(i,j) =

Proposition 2.10 Let X and Y be sets of real numbers, let f: Xx Y —* R be anyfunction. and assume that maXmiflf(i,y) and miflmaxf() both exist. Then:xX yE zEX

(i) max mm f@ ) <mmnmaxf(1,).zEX YEY yeY sEX

(ii) max mm f(,, ) = mm max f(z, y) if and only if f has a saddle point.xE.V yE yE sEX

Moreover, if (x*, y) is a saddle point of f, then

(3) maxminf(x.y) zmiflmaxf(x,y)f(x*y)sEX yEY sEX

Proof (i) For x E X, letrn(x) = minf(x,y),

YE Y

14

and for y e Y, let

M(y) - rnaxf(x,).E A

Then,rn(x) < f(x,y) < M(y) for all x X, y E Y.

Hence, for all x E X,m(x) <minM(y),

5EY

andsomaxm(x) <miflM(y),rCX yEY

as required.

(ii) Suppose that (x*,y*) is a saddle point off. By condition (2),

maxf(x,y*) <f(x*,y*) <minf(x*,y).— yEY

But obviously,

minmaxf(x,y) <rnaxf(x,y), rnaxrninf(z,y) > minf(,),yE x.X x.\ yE}

sominrnaxf(x,y) <f(x*,y*) <maxminf(x,y).yCY xEX xX 5EY

Part (i) now shows that both of these inequalities must be equalities.Conversely, suppose that

maxminf(x,y) minmaxf(x.y).zEX yCY yCY xcX

Let x E X, y e Y be such that, in the notation of part (i) of the proof,

m(x*) = max’m(x), M(y) = minjw(y).xEX

Then, the assumption says that m(x) M(y). But clearly, for all x E X, y E Y,

f(x, ) <f(*)f( ) >

and so

f(x,y) <f(x*,g).

Taking y =y* gives f(x,y*) < f(x*,y*), and taking x = x* gives f(x,y*) <

f(x, y), so (x, y) is a saddle point.

Warning If (x* y*) E X x Y is such that-condition (3) holds, it does not follow that(x, y) is a saddle point (see Problem 2.11).

We now specialize to the case of the payoff matrix (a) of a matrix game. Recallthat the indices i and j label the pure strategies of players 1 and 2, respectively.

15

Definition 2.11 Let (a) be the payoff matrix of a matrix game. A pair (j*,j) iscalled a saddle point of the payoff matrix, or a pure strategy solution of the game.if

(4) < <

for all i,j.A pure strategy i for player 1 is said to be optimal if there is a pure strategy j’

for player 2 such that (j*j1) is a pure strategy solution of the game. Optimal purestrategies fpr player 2 are defined similarly.

Thus, the condition (4) for (j,j*) to be a saddle point is that the entry ofthe payoff matrix is a largest entry in its column and a smallest entry in its row.Proposition 3.4 tells us that, if (i*,j*) is a saddle point, then

= maxmina minmaxa.i j j z

but as we have remarked above the converse of this statement is false.

Example 2.12 The payoff matrix

/2 —3 1 —4

( 6 —4 1 —5j4 3 3 2\2 —3 2 —4

has a saddle point in the (3. 4) position. since the (3. 4)-entry is clearly the smallestin its row and the largest in its column. Inspection shows that there are no othersaddle points. Hence, 3 is the unique optimal pure strategy for player 1, and 4 isthe unique optimal pure strategy for player 2.

2.5 Minimax Theorem (first version)

We have already seen that not all games have pure strategy solutions. There is,however, a simple class of games which do have pure strategy solutions.

Definition 2.13 A game is said to be of perfect information if every informationset for each player contains only a single vertex.

Thus, in a game of perfect information each player knows, at each stage of thegame, the precise state of play and all that has gone before.

Minimax Theorem 2.14 Every matrix game of perfect information has a purestrategy solution.

Proof Let F be a matrix game of perfect information. The proof is by inductionon the rank of F (see Chapter 1). Although we could start the induction when therank off is zero (in which case there is nothing to prove), it is instructive to treatthe rank 1 case in detail first.

Suppose then that the rank of F is 1. If the initial vertex v0 off is a chance move,then each player has a unique pure strategy (do nothing!), so there is nothing toprove. Suppose now that v0 is the initial point of a move for player 1: then player 2

16

has a unique pure strategy (do nothing), which is necessarily optimal. Let v1 vbe the final points of the moves with initial point j; thus. vi r are also theterminal vertices of F in this case. Let pj(v) be the payoff to player i (i = 1, 2) ifthe game ends at the terminal vertex v, and let s be such that

Pi(Vs) max m (v.1<t<r

Then. an optimal pure strategy for player 1 is to choose the move with final pointv. The argument when player 2 is to make a move at v0 is similar.For the inductive step, we observe that any game of perfect information can be‘decomposed’ into the union’ of certain subgames of smaller rank. in the followingsense. (This decomposition is possible for games of perfect information with anynumber of players, although we shall only need it for 2-person games.) Roughlyspeaking, if v1 Cr are the final points of the moves rn1 mr with initialpoint the initial vertex v0 of F, we define games (for s = 1 ) with initialvertex u. consisting of all the branches of F emanating from u5.

The precise definition is as follows. Let 1 be the set of vertices of F, 141 the setof vertices at which a chance move is made. and 14 (i = 1, 2) the set of vertices atwhich player i must make a personal move. For s = 1 r, let p(s) be the gametree defined as follows:

(i) the set V of vertices of F(s) is the set of final points of all paths in F withinitial point v5;(ii) the sets of vertices of at which chance moves (resp. personal moves forplayer i) are to be made are given by

V0

14 fl for i = 0,1,2.

17

(iii) the set Al1 of edges of F(s) is Al n ((5) x note that, except for theedges with initial point u0, every edge of Al is in one of the M, since no edge ofAl can have initial point in some v) and final point in some ‘(‘ with s(iv) the information sets of F> are the sets containing a single element of(v) the initial vertex of F(s) is v5:(vi) the pyffp5) to the ith player (i = 1,2) in F is the restriction to the set ofterminal vertices of of the payoff p to player i in the original game F (theseterminal vertices are simply the terminal vertices of F that are vertices of F(s)).

It is easy to verify that these definitions make n(s) into a game tree in the senseof Chapter 1. But it is important to note that the fact that the information setsof F ard single element sets plays a crucial role here: if this condition were notsatisfied, a given information set of F would not, in general. be contained in the setof vertices of one of the and there would then be no sensible way to define theinformation sets of

Since every information set of F (s = 1 r) is an information set of F, thepure strategies of F1> for player i (i = 1, 2) are simply the restriction of the purestrategies of F to the information sets of We denote the restriction to F1 ofa pure strategy u for a player in F by a(s),

With this construction understood, we can proceed with the inductive step inthe proof of the minimax theorem. The induction hypothesis tells us that eachgame (s = 1,... , r) has a pure strategy solution (u,r). Thus.

(5) P5,r) <P()(u;,;) <P1(u,r5)

for all pure strategies a and T5 for players 1 and 2, respectively, in p(s),

We must now consider three cases:

Case I: the moves with initial point v0 are chance moves.Since Vo does not lie in any information set in this case, there are unique purestrategies u and r for players 1 and 2, respectively, in F such that *(s)

= o; and= r. We shall prove that (o, r*) is a pure strategy solution of F.

Let r5 be the probability that the move m5 is chosen, and let u and r be any purestrategies for players 1 and 2, respectively, in I’. If P is the pure payoff function ofF and p(s) that of it is clear that

(6) P(u,r) =

Hence,

P(,r) =

By (5),

so by (6),

(7) P(u. r*) = p(s) ((5 r) <p(s) (u ;) = P(a*. r).

[8

Similarly,

(8) P(u5r) p(s)(g;T(s)) > p(S);)

Inequalities (7) and (8) show that (*, r*) is a pure strategy solution of F.Case II: the moves with initial point t’0 are personal moves for player 1.In this case. a pure strategy for player 1 in F is determined by its restriction to eachF(s) together with the move for player lit chooses at v0. Let o be the unique purestrategy for player 1 in F such that (s) = o for s = 1 r and which choosesthe move mt at t’, where t is such that

p(t)(u*r) = maxi<s<r

Let r be the unique pure strategy for player 2 in F such that r for alls = 1,... ,r. We shall prove that (o. r5) is a pure strategy solution of F.Note first that

(9) P(u*, T*)

Now let u and T be any pure strategies for players 1 and 2. respectively, in F. andsuppose that u chooses the move m at v0, where 1 < u < r. Then,

(10) P(,r) = p(u)((u)r(u))

so

P(u. *) = p(u)(j(n)r)

But p L) ((u), ) < p) (o, r) since (, r) is a pure strategy solution of p( u)and p( it) (, ) < p() (, T) by the choice of t. Hence, (9) gives

P(u,r5)<P(a,r).

On the other hand,

r) p(q(;, r(t)) > p(t)( r) = P(,r5).

Hence, (o. r*) is a pure strategy solution of F.Case III: the moves with initial point u0 are personal moves for player 2.The argument in this case is similar to that in Case II, arid we leave it as Problem2.15.

Remark It is possible for games to have pure strategy solutions even if they are notof perfect information (see the game in Problem 2.4).

Problems

2.1. Write down the payoff matrix for player 1 in the game of STONE-SCISSORS-PAPER(assume that the winner gets £1 from the loser).

19

2.2 Find the payoff matrix for the game HI-LU in Example 1.2.

2.3. List the pure strategies of each player and find the payoff matrices for player I in each of thefive games described in Problems 1.4 and 1.5.

2.1. List the pure strategies for each player and find the payoff matrix for player 1 in the gamedescribed in Problem 1.7.

2.5. In a 2-person game the first move is a chance move in which an integer x is selected from1.2.3, 4}, all choices being equally likely, Player 1 is informed whether i (I, 2} or (3. 1}; if

x E (1, 2}, player chooses an integer p from {3. 4}, while if .r 5 {3, 4} player I chooses an integerp from {f, 2}. Player 2 is informed only whether p is odd or even and player 2 then chooses aninteger z from {1, 2}. The game then terminates and player 1 receives £(x + p + z) from player 2if x + p + z is even, while player 2 receives £(i ± p ± z) from player 1 otherwise. For this game,draw a game tree, indicate the information sets for each player, list the pure strategies, and findthe payoff matrix for player 1.

2.6. Repeat Problem 2.5 for the following 2-person game. First, an integer x is selected at randomfrom (1,2. 3,4,5}. Then the two players. not knowing x, simultaneously select integers from{1. 2,3,4. 5}. Each player’s objective is to select, an upper bound for a. If only one player issuccessful, he wins. If both are successful, and their choices are not the same, the winner is theplayer who has chosen the smaller upper hound. In all other cases, the game is a draw. Thewinner gets £1 from the loser.

2.1. Repeat Problem 2.5 for the game of MORRA, in which each of two players simultaneouslydisplays either one or two fingers and at the same time guesses how many fingers the opposingplayer will show. If both players guess correctly or if both guess incorrectly, the game is a draw.If only one player guesses correctly, he wins from his opponent an amount (in £) equal to thetotal number of fingers shown by both players.

2.8. Check that the payoff matrix given in Example 2.6 is correct.

2.9. Show that, if (C ,j) and (i’,j’) are pure strategy solutions of a matrix game with payoffmatrix (aj), then (C ,j’) and (,j*) are also pure strategy solutions.

2.10. Show that, if C is an optimal pure strategy for player 1. and j is an optimal pure strategyfor player 2, then (C, j) is a pure strategy solution.

2.11. A matrix game has payoff matrix (aj) given by

(0 01

Show that there exists a pair of indices (C .j) such that

max mm mm max‘ 3 3 ‘

but that the game has no pure strategy solution.

2.12. Find the pure strategy solutions of the games considered in Problems 2.1 to 2.7, if theyexist.

2.13. A matrix game has mm x mm payoff matrix (aq), where a,3 = i—

j. Show that the game has apure strategy solution, and find the optimal pure strategies for each player.

2.14. Let two functions f.g R x —* f be defined by

f(x,p) — —c2 g() —a2 —p2.

Show that f has a saddle point hut p does not. (Thus. the notion of saddle point used in thischapter does not coincide with the notion of saddle point used in calculus, since both f and phave a saddle point in the latter sense.)

2.15. Give the proof of the third case of the Minimax Theorem 2.14.

20

Chapter 3: Mixed Strategy Solutions

In this chapter we consider a new kind of ‘solution’ of a game that is more generalthan the pure strategy solutions we have considered so far.

3.1 Mixed Strategies

We have seen that some matrix games have no pure strategy solution. For example,the payoff matrix of the game of MATCHING PENNIES

(1 —11

has no saddle point. So what should the two players do in this case to maximizetheir payoffs? Nothing can really be done if the game is played only once, but if thegame is played repeatedly each player might try to ‘hedge his bets by sometimescalling ‘heads’ and sometimes ‘tails’. If player 1 calls ‘heads’ in a proportion x ofplays of the game and ‘tails’ in a proportion 1 — x, and if player 2 calls ‘heads’ in aproportiony of plays and ‘tails’ in a proportion l—g of plays (so that 0< x,y < 1),the average payoff to player 1 is

xy—x(1—y)—(l—x)y+(l—x)(l—y)= (2x—1)(2y— 1).

The function f : [0,1] x [0, 1] -4 given by f(x, y) = (2x—

l)(2y — 1) has a saddlepoint at (1/2, 1/2) (in the sense of Definition 29), since

f(x, 1/2) = f(1/2. 1/2) = f(1/2. y)

for all xy. Hence,

max mm f(x,y) mm maxf(x,y)0,E[0,1] y€[0,1 yeo,1] xE0,1

So the optimal strategy for each player is to call ‘heads’ half the time, in which casethe expected payoff for each player is zero.Such combinations of pure strategies are called ‘mixed strategies’:

Definition 3.1 Let a player of a game have rn pure strategies, where in > 1. Then,a mixed strategy for the player is an m-tupie x = (x1 ....x,) of real numbers suchthatIn

O<x<l fori=1,,.,,rn and

The subset of R consisting of the points (xi,,... rn) satisfying these conditionsis denoted by i,,-,.

Thus, x is the proportion of games in which the player will adopt pure strategyi. Pure strategies then become special cases of mixed strategies: pure strategy i is

21

identified with the mixed strategy 6 = (0,... , 0, 1,0, . . , 0), where the 1 is in theith place.

If (a,) is the m x n payoff matrix of a matrix game, and if x = (Il,.. , x) andY (vi,.. . , y,) arc mixed strategies for players 1 and 2, respectively, the averagepayoff to player 1 is clearly

P(x.y) = cLjxiyj.i=1 .j=1

since the ptoportion of plays of the game in which players 1 and 2 simultaneouslyadopt tle pure strategies i and j. respectively, is Xjj and the payoff in this caseis The function P : Am x A, — R is called the mixed payoff function of thegame. As in the case of pure strategies. player 1 tries to choose x so as to maximizeP(x,y), while player 2 tries to choose y so as to minimize P(x,y). Thus, we havetry to find ‘mixed strategy solutions’ of the game. in the following sense:Definition 3.2 Let P : Am x A be the mixed payoff function of a matrixgame. A pair (x, y”) E Am x A, is called a mixed strategy solution of the game if

(1) P(x.y) <P(x*,y*) <P(x,y) for all XE -sm, yE Am.

In that case, P(x, y) is called the value of the game.A mixed strategy X* E Am is said to be an optimal mixed strategy for player 1 if

there is a mixed strategy y’ E A, for player 2 such that (x*.y’) is a mixed strategy

solution of the game: optimal mixed strategies for player 2 are defined similarly.Thus, the mixed strategy solutions are just the saddle points of the mixed payoff

function P. It follows from Proposition 2.10 (ii) that, if (x*, y) is a mixed strategysolution of the game, then

max mm P(x.y) = mm max P(x.y) = P(x*.y*).yE1 xE.,,,

This implies that the value of the game is well defined.

3.2 Finding Mixed Strategy Solutions

The following theorem, perhaps the most important in game theory, was proved byJohn von Neumann in 1928.

Von Neumann’s Minimax Theorem 3.3 Every matrix game has at least onemixed strategy solution.

Note that this theorem does not assume the game to be of perfect information.We have already seen that a game that is not of perfect information need not havehave any pure strategy solutions.

We shall prove von Neumann’s theorem in Section 3.6. However, although thistheorem is very important, its proof is non-constructive, i.e. it does not tell us howto actually find a mixed strategy solution of a given matrix game. Accordingly,we devote the remainder of this section. and the next two sections, to some simplemethods which can often be used to find mixed strategy solutions.

22

The next two propositions can he used to test whether a candidate for an optimalmixed strategy is actually optimal. To prove them we shall need:

Lemma 3.4 Let P : m x Z —* H be the expected payoff function of a matrixgame. Then,

max P(x,y) = max P(Sj,y) for ally E ,Xrn 1<,<rn

Ifllfl P(x,y) = mm P(x,6) for all x e rnyE lJr

Proof For y E ., let

Q() = max

Then, P(5,y) < Q(y) for i = 1,... ,rn, so for all x (x1,... ,Xrn) E ‘rn, Y(y’,. .y,,) E

P(x. y) = OjjXjj = ( aiii) x = y)xj <Q(y) = Q(y).

Hence,P(x,y) < max P(&,y). for all x E

1< <in

This shows thatmax P(x.y)

XE

exists for all y é , and that

(2) max P(x, y) < max P(&, y) for all Y EXEm 1<m

On the other hand, since every pure strategy is a mixed strategy.

max P(x,y) > max P(6,y) for ally e .xE 1<t<m

Together with (2), this proves the first equation in the lemma. The second equationis proved similarly.

Proposition 3.5 Let P: x . H be the mixed payoff function of a matrixgame with value v. If a mixed strategy x E is optimal for player 1, then

(3) mm P(x. = u for all j = 1.... n.1Jn

Conversely, if a mixed strategy x for player 1 satisfies

(4) P(x*,o) > u for all j = 1,

then x is optimal.

Analogous results hold for player 2.

23

Remark Of course, this shows that if a mixed strategy x for player 1 satisfies (4),then it satisfies the apparently stronger condition (3).

Proof Suppose first that x* is an optimal mixed strategy for player 1. Then, thereexists a mixed strategy 6 z for player 2 such that (x, y*) is a mixed strategysolution of the game. By Definition 32.

P(x,y) P(x,y) = u for ally 6 .

Hence, rrun P(x,y) = v, so by Lemma 4.4,y€

mm P(x*,j)=v.1jn

Conversely, suppose that x 6 satisfies (1). Let (x’.y’) E x Zi,. be amixed strategy solution of the game (this exists by Theorem 3.3). We shall provethat (x*. y’) is also a mixed strategy solution of the game. Of course, this will provethat x* is an optimal mixed strategy for player 1.

By Lemma 3.4,mm p(x*,y) = mmye’ 1jn

and this is > v by (4). Hence.

(5) P(x*,y) > i’ for all yE .

Now, P(x’,y’) = v and since (x’,y’) is a mixed strategy solution,

(6) P(x.y’) <P(x’.y’) <P(x’.y) for all x 6 ‘rn, yE

It follows from (5) and (6) that

P(x, y’) < P(x’, y’) = u < P(x*, I)

and hence that P(x*,y) = P(x’.y’). Thus.

P(x,y’) < P(x,y’) = v < P(x*,y) for all x E Lm, yE

This shows that (x*, y’) is a mixed strategy solution of the game.Proposition 3.6 Let P : m x . —+ be the mixed payoff function of a matrixgame. Let x 6 -sm, Y 6 V ER be such that

P(5.y) <u < P(x5) for all i 1.... ,m, j = 1.. ,n.

Then, (x*, y) is a mixed strategy solution of the game and v is its value.Proof Let y = (y y) E ... and let x*

= (I x). Then, if (a) is thepayoff matrix of the game,

P(x*,y) = aijxyj = (aux:) y = P(x)y > Vj V.

21

Similarly, P(x,y*) < v for all XE Hence,

(7) P(x,y) <v <P(x,y) for all x E ,, yE .

Taking x = x and y = y shows that v = P(x, y*), and then (7) shows that(x, y) is a mixed strategy solution. E

Example 3.7 A game has payoff matrix

/—3 —3 2(—i 3 —2

3 —1 —2\2 2 —3

Let1 1 J 1 1x*=(.O,O,),

Then,

P(x,61)= —3 x +2 x =

Similar calculations show that

(8) P(*,S) — = P(,y*) for all i = 1.2.3,4. j = 1,2,3.

By Proposition 3.6, (x, y*) is a mixed strategy solution of the game and —1/2 isits value. (Note that it is just an accident that equality holds in (8) — to concludethat we have a mixed strategy solution, it is only necessary to know that each termis > the next.)

3.3 Inferior Strategies

In this section we describe a method which can sometimes be used to simplify theanalysis of a matrix game, by reducing the size of the payoff matrix.Definition 3.8 Let (a) be the payoff matrix of a matrix game. A pure strategyi1 for player 1 is said to be inferior to a pure strategy io for player I if

(9) a11 < a0 for all j;

i1 is comparable to io if equality holds for all j; and i1 is strictly inferior to i0 if theinequality in (9) is < for all j. Similarly, a pure strategy j1 for player 2 is said tobe inferior to a pure strategy Jo for player 2 if

ajq > aq0 for all i,

and so on.

Thus, the pure strategy i1 for player 1 is comparable (resp. inferior. resp. strictlyinferior) to the pure strategy i0 if each entry in the i1 th row of the payoff matrix isequal to (resp. less than or equal to, resp. strictly less than) the corresponding entryin the ioth row, while the pure strategy Ji for player 2 is comparable (resp. inferior,

25

resp. strictly inferior) to the pure strategy jo if each entry in the Ji th column of thepayoff matrix is equal to (resp. greater than or equal to. resp. strictly greater than)the corresponding entry in the joth column. In this case, we shall sometimes saythat the i1th row of the payoff matrix is comparable (resp. inferior, resp. strictlyinferior) to the i0th row, etc.

It is intuitively clear that, if the pure strategy i1 for player 1 is inferior to i0,player 1 should adopt i0 in preference to i1 since his payoff when using i0 is atleast as great as his payoff when using i1, whatever pure strategy player 2 adopts.We expect, therefore, that if we delete the i1th row of the payoff matrix, any purestrategy solution of the new matrix game should be a pure strategy solution of theoriginal matrix game. This is indeed the case, as the next proposition shows.Proposition 3.9 Let A be the payoff matrix of a matrix game F, and assume thatsome row (resp. column) of A is inferior to another row (resp. column). Let B bethe submatrix of .4 obtained by deleting the inferior row (resp. column). Then, anypure strategy solution of the game FB with payoff matrix B is also a pure strategysolution of F. Conversely, if the deleted row (resp. column) is strictly inferior toanother row (resp. column) of A. then every pure strategy solution of F arises froma pure strategy solution of FB.

It follows that similar results hold when B is obtained from A by deleting asuccession of inferior rows and columns.

Warning If the deleted rows or columns are only inferior, and not strictly inferior,to other rows or columns, it is possible that not all pure strategy solutions of Farise from pure strategy solutions of FB. For example, for the payoff matrix

/0 1 2A=( 0 1 1

2 0

(1, 1) and (2, 1) are pure strategy solutions. However, pure strategy 2 for player 1is inferior to pure strategy 1, so if we had reduced the payoff matrix by deletingthe second row we would have ‘lost’ the pure strategy solution (2, 1).

The proof of Proposition 3.9 is the subject of Problem 3.16. We shall now provea more general result. To state it we need one more piece of terminology. LetA be the m x n payoff matrix of a matrix game F, let 1 < r < m, 1 < s < ii,1<i1<m2<...<z<m,1<j1<j2<...<j<n,andletBbethesubmatrixof A consisting of the elements in rows i1,.. ,i,. and columns 1, ... ,j, i.e. B isthe r x s matrix whose (k, 1)-entry is akJ. If = (i r) E ,. is a mixedstrategy for player 1 in the game FB with payoff matrix B, one can extend to amixed strategy x E m for player 1 in F in the obvious way, i.e. the ith componentof x is if i = it for some k = 1,... . r and it is 0 otherwise.

Proposition 3.10 Let .4 he the payoff matrix of a matrix game F, and let B be thesubmatrix of A obtained by deleting a row of A corresponding to a pure strategyfor player 1 that is inferior to another pure strategy for player 1 (resp. a columnof 4 corresponding to a pure strategy for player 2 that is inferior to another purestrategy for player 2). Let be a mixed strategy solution of the game FBwith payoff matrix B, and let (x*, y) be its extension to a pair of mixed strategiesfor F. Then, (x, y) is a mixed strategy solution of F. If the row (resp. column)

26

deleted corresponds to a pure strategy for player 1 (resp. 2) that is strictly inferiorto another pure strategy for player 1 (resp. 2). then every mixed strategy solutionof F arises in this way from a mixed strategy solution of FB.

It follows that similar results hold when B is obtained from A by deleting asuccession of inferior rows and columns.

Proof Let A = (a,) be m x n, and suppose that pure strategy i1 for player 1 inthe matrix game F is inferior to pure strategy i0. Let = (ci, , em—i), ?7 =(rh,.. ,r). so that y = i and x (It,... ,Xm) where x = j if 1 < i < i1,

0 and z = if i1 0 and a1 < a0 for all j = 1,... , ri. Let v be the value of PB; since(. i) is a mixed strategy solution of FB.

PB(Si,71*) <u <PB(*,oJ) for all i 11 and all j.By 10), P(x,ö) v for all j, and by (11) P(&, y) < v for all i i1. Finally,by (12),

PQ5j,,y*) < P(0,y) <U.

Hence.

P(&,y) <u <P(x,S) for all i = 1,... rn, j = 1,... a,

so by Proposition 3.6, (x*,y) is a mixed strategy solution of F. and c is its value.Conversely, suppose that pure strategy i1 is strictly inferior to pure strategy i0,

let (x*,y*) be a mixed strategy solution of F, and let x (x ,...,Xm). y =(Yi.... ,y). Since y > 0 for all i and y >0 for at least one value of j. it followsthat

yj(ai0j— auj) > 0

j1

and hence that

= Zyjai1J <Zyjai0j

Since P(ój,y*) < v for all i, we have < v, so by Problem 3.11 x = 0.Define , = y and = (‘ m.-i). where , = x, if 1 < i < i1, j = x, if

27

i1 < i < m —1. The argument in the first part of the proof shows that (*,j*) is amixed strategy solution of the game F8.

Example 3.11 Suppose that the payoff matrix is

f’ 1 —1 2(—1 1 3 *

—2 4

Pure strategy 3 for player 2 is strictly inferior to pure strategy 1, so we delete thethird column:

/1 -1—1 1

—2

In the game with this payoff matrix, pure strategy 3 for player 1 is strictly inferiorto pure strategy 1, so we delete the third row:

(1 —11

We have analyzed this payoff matrix before. and we know that the value of the gameis 0 and the unique optimal mixed strategY for each player is (4, 4). It follows that.for the original game, the value is 0 and the unique optimal mixed strategy for eachplayer is ( 4,0).

The following result shows that games of perfect information can be solved trivially by successive deletion of inferior strategies.

Theorem 3.12 The payoff matrix of a game of perfect information can be reducedto a 1 x 1 submatrix (i.e. to a single entry) by successive deletion of inferior rowsand columns.

Proof The proof is similar to that of the Minimax Theorem 2.14, so we shall bebrief. We proceed by induction on the rank of the game F, the case in which therank is zero being trivial.

For the inductive step, we use the games s = 1,... , r. defined in the proofof Theorem 2.14, and consider three cases.

Case I: the moves with initial point v0 are chance movesLet = (a), with 1 <i < 1 <j be the payoff matrix of the gamep(s) The payoff matrix .4 of F has rows indexed by r-tuples I = (j(),

1 < < m), columns indexed by r-tuples J = (J(l) j)), 1 < j(S) < a(s)

and its IJ-entry is

ajj=

where it8 is the probability that the sth move is chosen at the initial vertex n0. Itis easy to check that:

(i) if i’ is an inferior row of for some 1 < ? < r, 1 < i’ < then all rowsof .4 for which j(S’)

= i’ are inferior rows in .4:

28

(ii) if row i’ is deleted from A(s), the effect on A is to delete from .4 all rows forwhich (s)

=

Similar facts hold for the columns.Hence, each permissible deletion (i.e. the deletion of an inferior row or column)

in any of the matrices A, 1 < s < r. is associated with a corresponding set ofpermissible deletions in A. By the induction hypothesis, each matrix A( can bereduced to a 1 x 1 matrix by a sequence of permissible deletions. and this has theeffect of reducing .4 to a 1 x 1 matrix. This completes the inductive step in thiscase.

Case II: th moves with initial point u0 are personal moves for player 1In this case, the rows of the matrix A of F are indexed by the set of pairs I = (.s, i),where 1 < s < r and 1 ) as in Case I, and the IJ-entry of .4 is given by

ajj = a(S).

This time, one checks that:(i) if i’ is an inferior row of 4(5’), for some 1 < s’ < r, 1 < i’ < thenI = (s’, i’) is an inferior row of A, and deleting row i’ from A(’ corresponds todeleting row I’ from A:(ii) If j’ is an inferior column of A’>, then all columns J with j(5’> j’ are inferiorcolumns of A, and deleting column j’ from A(’) corresponds to deleting from .4 allcolumns with j(S) — jt

Hence, to each permissible row deletion in some ,4( corresponds a permissible rowdeletion in A, and to each permissible column deletion in some A corresponds aset of permissible column deletions in .4. By the induction hypothesis, there is asequence of permissible deletions which reduces each A> to a 1 x 1 matrix. Thecorresponding sequence of permissible deletions reduces A to a matrix which hasonly one column. Clearly, this latter matrix can be reduced to a 1 x 1 matrixby further deletions of inferior rows until only the maximum entry in the columnremains. This completes the inductive step in Case II.Case III: the moves with initial point v0 are personal moves for player 2The proof in this case is similar to that in Case II.

Theorem 3.12 clearly implies the minimax theorem, since any 1 x 1 payoff matrixobviously has a pure strategy solution (!) which, by Proposition 3.9. will be a purestrategy solution of the original game.

3.4 The 2 x n Case

We specialize now to the case m 2, i.e. player 1 has only two pure strategies. Letthe payoff matrix be

(am a2k\bl b2 •.. b11)

and let the mixed payoff function be P : x —* R. A mixed strategy for player1 is of the form

x = (cm. 1 — cm)

29

for some 0 < c < 1. Then, for j — 1, ,

P(x,ñ2)= aa2 t (1 — = (a2 — b2)a + b.

From Problem 3.7, the value v of the game is given by

= max mm P(x,ö) = max mmxE21jn o1 1jri

and by Proposition 3.5, x* is an optimal mixed strategy for player 1 if and only if

(13) ruin P(x*,) =l<j<a

This means that the value of the game and the optimal mixed strategies forplayer 1 can be found by the following graphical method. We plot the straightline graph of 8 = (a2 — b2)u + b2 in the cr3-plane for j = 1,... , n; the graph of

mm [(a’ — b2)cr + b2] is the polygonal curve which forms the boundary of the1Jn

region which lies below each of the lines. The value L’ is then given by the maximum3-coordinate of the points on this polygonal curve corresponding to 0 <cm < 1. By(13), the value (or values) of cm which give this maximum value of 3 give the optimalmixed strategy (or strategies) for player 1.

The following examples illustrate this method, arid also show how the optimalmixed strategies for player 2 may sometimes be determined.

Example 3.13 We consider the matrix game with payoff matrix

(1 —1 3‘3 5 3

30

We must plot the straight lines

+ 3(1 — = —2n + 3,

— + 5(1 — n) = —6a + 5,3a — 3(1 — = 6n — 3.

The polygonal boundary is clearly formed by segments of the last two lines (see thediagram above). These lines intersect when

Hence, the value of the game is

—6n + 5 6n 3.

i.e. c =

v=—6 - +5=1,

and the (unique) optimal mixed strategy for player 1 is

0

21x* ()

Since P(x*,61) = 1 () + 3 () = > u, Problem 3.11 (applied to player 2)shows that any optimal mixed strategy for player 2 must be of the form

y = (0. n’, 1—

a’).

31

for some 0 a’ < 1. By the analogue of (13) for player 2.

(14) 1 = mill maxO<’<1 1<1<2

Now, P(1 , y) = —a’ + 3(1 — a’) —Ia’ + 3. P(2.y) 5a’ — 3(1 — a’) = 8a’ —3.The graph of the righthand side of (14) against a’ is the polygonal curve whichrepresents the region above the graphs of the two lines

3’ = —4a’ + 3. ‘ = 8a’ — 3

(see the.diagram above). These lines intersect when a’ , and this is clearly thevalue of a’ which satisfies (14). Hence, the (unique) optimal mixed strategy forplayer 2 is

y = (0, , ).Hence, this game has a unique mixed strategy solution.

Example 3.14 We consider the matrix game with payoff matrix

/2 4 11(13 1

4 2

Of course, this is not a 2 x n matrix. However, we observe that pure strategy 2for player 1 is strictly inferior to each of the other pure strategies for player 1. soProposition 3.8 shows that the optimal mixed strategies for the original game canbe obtained from those of the game with payoff matrix

(2 4 114 2

to which the methods of this section can be applied.

0 219 3/5

32

The appropriate diagram for player 1 is shown above, from which it is clear thatany a with < a gives an optimal mixed strategy = (a, 1 — a) for playerI in the reduced game. Hence, the optimal mixed strategies for player 1 in theoriginal game are

9 3x=(a,0,1—a),

The value of the game is v = 4. If P is the mixed payoff function for the originalgame, we observe that

P(x,i)=7—5a>4 ifa<,

P(x*,ñ3)=9a+2>4 ifa>.

It follows from Problem 3.11 that the unique optimal mixed strategy for player 2is (0,1,0), i.e. the pure strategy 2.

Remark The methods of this chapter apply, with obvious modifications, to the caseof in x 2 payoff matrices. This time, one analyzes player 2’s optimal mixed strategiesfirst. If the payoff matrix is

a1 b1a2 b2

am bm

the value of the game is

= mm max [(at — b)a + b].O<a<1 1<i<m

The optimal mixed strategies for player 2 are found by examining the polygonalboundary of the region above each of the straight lines 3 = (a—b)a±b, 1 1). But first we review some concepts from real analysis.

Recall that a subset X of Rtm is said to be bounded if there is a constant D suchthat, if X (xi,... .Xm) is any point of X. we have xj < D for i = 1 in.(You may have been given a different definition in another course; if so, it is anexercise for you to prove that the definitions are equivalent.)

We also recall that a sequence of points x E H. k = 1.2,3,.. ., where(x c), is said to converge to a point x (xi, m) e H

urn x = x for all i = 1,.. . , rn.

A subset X of Rtm is said to be closed if, whenever (x(k)) is a sequence of pointsof X converging to a point x E Htm it follows that x e X. It is obvious thatthe intersection of any family of closed sets is closed. (Again, these definitions are

33

equivalent to aiiy other definitions of convergence and closed sets you might havebeen given.)

The following definition introduces the concept that will be needed to proveTheorem 3.3.

Definition 3.15 The line segment joining two distinct points x and y of Rm(m > 1) is the set of points of the form Ax + (1 — A)y for some 0 < A < 1:

-:7-

-- y

The points x and y are the endpoints of the line segment and (x + y) is itsmidpoint.

A subset C ofm (m > 1) is said to be convex if, whenever x and y are points

of C, the whole of the line segment joining x and y is contained in C.For example, the interior of a triangle is convex

but the interior of the following polygon in the plane is not:

The importance of the idea of convexity for us derives from the following result:Proposition 3.16 If a player in a matrix game has rn pure strategies, the set ofhis optimal mixed strategies is a bounded closed convex subset of R.

x

Proof Let X be the set of optimal mixed strategies for player 1 in a matrix game

31

with mixed payoff function F: m x —* R, and let v be the value of the game.It is clear that Lm is bounded: if x (x1 Xm) E Lm then Xj < 1 for

= 1,... , rn. Since X C X is also bounded.Next, we show that X is closed. Let (x(k)) he a sequence of points of X coti

- (k (k) k)verging to a point x irt . \e have to show that x E A. If x’ (x1,....imand x = (xi x,). then x(k> E m implies that. for all k.

0< < 1 for i = 1,... rn, and Zx = 1.

Letting k —+ , we get

rn

0<x<1fori1....,m.andx=1,

which tells us that x E i.e. that x is a mixed strategy for player 1. ByProposition 3.5, x E V if and only if

(15) P(x,) > v for all j = 1,... ,n.

Since E X for all k.

(16) P(x,) > v for all j = 1. .. ,0.

Now, if (a) is the payoff matrix of the game,

F(x,6) = Zxjajj.

Hence. condition (16) is that

(17) > v

for all j, k. Letting k — in (17) gives (15), so x e X. Hence, X is closed.Finally, we prove that X is convex. Let x = (x1,... , x,,) and y = (yj,... ,y)

be elements of X, and let 0 <) < 1. Then 0< xj,yj < 1 for all i and x =

= 1, so that

0 = \.0 ± (1 — \).0 Ax + (1— A)y <).1 ± (1 — ).1 = 1

for all i, while

m m Tn

(Ax + (1—

)yj) = A x + (1 — A) = A.1 + (1 — A).1 = 1.

It follows that Ax + (1 — A)y E By Proposition 35, the fact that x and y areoptimal is equivalent to

P(x,6) > v and P(y,6) > i; for j = 1, rn.

35

This implies that

P(/\x + (1 — A)y,6j) = (Ax + (1 )yj)ajj

111 Tn

= ± (1— )) yiaj

= )P(x,5)+ (1 —

>Xv+(1—A)v=v.

Hence, .Xx + (1 — ,\)y E X. This proves that X is convex.

Remark In particular, the set Am is bounded, closed and convex for all m 1. Theproof of this is actually contained in the proof just given. Alternatively, consider agame in which player I makes one of rn possible moves at the initial state and thenthe games ends, the payoff to player 1 being the same whichever move he made. Insuch a game, the set of optimal mixed strategies for player 1 is the whole of Am,so our assertion follows from the proposition.

We shall now consider the ‘smallest’ convex set containing a given subset of R.The following (nearly obvious) result shows that this makes sense:

Proposition 3.17 The intersection of any family of convex subsets of R is aconvex subset of R.

Proof Indeed, if x and y are two points in the intersection, they are in each of thesets; then so is the line segment joining x and y since the sets are convex; and thismeans that this line segment is contained in the intersection.

It follows from this proposition that, if X is any subset of Rm, the intersectionof all the convex subsets of R containing X is itself convex. It is called the convexhull of X, is denoted by (X), and is obviously the smallest convex subset of Rmcontaining X (i.e. it is contained in every convex subset of iR that contains X).It can be described explicitly as follows:

Proposition 3.18 The convex hull of a subset X of Rm is the set of points of theform

(18) wherexEX,a>Ofors=1,.. ,randa=1.

A linear combination of points x1, ...,x as in the statement of the propositionis called a convex combination of the points.

Proof Let X’ be the set of points of Rm of the form (18). It is obvious that X C X’(take r = 1). We prove first that X’ is convex; this implies that (X) C X’.

Suppose then that x1 x and 3’i,... , Yq are points of X, and that pr,...,

and va,... , 1q are non-negative real numbers such that IL3 = Vt = 1.The points

XZJLSX8, YVtYt

36

are in X’, and we have to show that, if 0 < A < 1, the point (1 — A)x + Ày is alsoinX’. Definez,. xr forr= 1,...,p. and Zrp=Yr forr= 1, ...,q; defineQrjirforr=1,,par1dcr=Oforrp+1,..,p+q:aflddefiTle3r0forr1....,pand3r±p=rforr1....,q.Then.zrEXforr1 ,p+qand

x=Urzr: YZ3rZr

Now,p+q

(1_A)X+AyZ’yrZr

where 7r (1 A)ar + Ar for r = 1,... ,p + q. Since ar and r are > 0 for all r,we have 7r > 0 for all r, and

p+q p+q p+q

7r = (1 A)n + AZr = 1 A + A = 1.

This shows that (1 — A)x + Ày E X’.We must now show that, for any r > 1, any convex combination of r points of

X belongs to (X). We proceed by induction on r, the case r = 1 being obvious.Assume next that r > 1 and that the result is true for convex combinations offewer than r points of X. Let x be a convex combination of r points of X, as inthe statement of the proposition. If a,. = 1, then x = x,. and there is nothingto prove. Assume from now on that 0 < a,. < 1. Note that a/(1 — a,.) > 0 fors= 1,... ,r — 1 and that a/(1 — a,.) 1. By the induction hypothesis,

=1

(X).

Since (X) is a convex set,

X (1 —ar)y±arxr E (X).

To prove Von Neumann’s minimax theorem we shall need one more result aboutconvex sets. For this, we shall need the concept of a hyperplane in am.: this is asubset of R” of the form

H = {x E I b.x = d},

where b is a non-zero vector in R and d E R. If rn = 2, a hyperplane is a straightline, and if in = 3 a hyperplane is a plane in the usual sense. Two subsets X andY of “ are said to be on opposite sides of the hyperplane H if either b.x > d forallxfiXandb.y<dforallyfiY,orb.x<dforalixfiXandb.y>dforally E Y. The result we want is the

Separating Hyperplane Theorem 3.19 If X and Y are disjoint convex subsetsof F’’. there is a hyperplane H in 11m such that X and Y are on opposite sides ofH.

37

Of course, this result is false if the sets X and Y are not assumed to be convex:

3.6 Proof of Von Neumann’s Minimax Theorem

To prove Von Neumann’s minimax theorem we shall need, in addition to the resultson convex sets stated in the preceding section, the following simple lemma from realanalysis.

Lemma 3.20 Let X be a subset of R, and let fi,... , fm be continuous real-valuedfunctions on X (m, n 1). Define functions F and f on X by

F(x) max1<1< rn

Then, F and f are continuous.

f(x) mm f(x), for x e X.1 0. Since is continuous at x0, there exists O > 0 suchthat

x — x0 11< jf1(x)— f,(xo)I <E.

Let 5 = mm Ô; then, ó> 0 and1< i < rn

(19) x—xo<rrlfj(x)—fj(xo)I <eforalli=1, rn.

Now (19) implies that, if x - x0 < , then f(x) < f(xo) + < F(xo) + € for alli = 1 7fl and hence that F(x) < F(xo) + e. On the other hand, let io be suchthat

F(xo) =f0(xo) = max f1(xo).1< i< in

Then (19) also implies that, if x — x0 1< . then fj (x) > f0 (xo) — = F(xo) —

and hence that F(x) > F(xo) — e. Hence.

1 x— XO F(x) — F(xo) <.

38

This shows that F is continuous at x0. Since x0 was an arbitrary point of X, thiscompletes the proof.

We are finally ready for the proof of the minimax theorem. Consider a matrixgamer with mx a payoff matrix (a) and mixed payoff function P : m X iiFor each i = 1, ..., in, the function f IR given by

f(y) P(ö.y) gjajj. for y = (yr,.., y) E

is continuous. By Lemma 3.20. the function F R given by

F(y) max f()1< <rn

is continuous. Since L is a closed hounded subset of , the function F is houndedand attains its bounds. In particular,

mm max p(,yE-X, 1<i<rn

exists. By Lemma 3.4, so does

mm max P(x,y).y,.

Similarly,max mm P(x.y)

XEm yE

exists. We have to show that these two quantities are equal.By Lemma 3.4,

mm max P(x,y) = mm max P(&,y) =v (say).yc I<<m

Let y E Z be such thatmax

1<i<m

For j = 1 a, let ,cJ1 = (a1, a2J... . . a) e m (thus, is the jthcolumn of the payoff matrix. but written as a row). Let T be the convex hull of thepoints By Proposition 3.18, T is the set of points of ffTn of the formZ= yJK(3>, where Yj > 0 for j = 1, a and = 1. We can thereforedefine a map —* T by

= if y = (y.... ,n) E

Clearly, (z11) = T. Note that, if y E Ln and i = 1 m, the ith component of(y) is yjajj = P(ö, y), so

(y) = (P(1.y),...

39

By the definition of v,

max P(&.y) v for ally E .

Hence, if t (t1,... , tm) E T, then

max > ,i<i<m —

andso

(20) if t (t1,.... tm) E T. then t > v for some i = 1, . . , m.

On the other hand,max PQ5, y’) =

1< i < rn

so if th(y) t = (t. t), then

(21) t <u foralli=1....,m.

Now let

Z{z=(zi,...,znz)eRmzi<vforalli1,...,m},

and letZ={z=(zi,.,.,z1)ERmIz<vforal1i=1,...,m}.

It is easy to see that Z is a convex subset of and by (20) it does not intersectthe convex set T. By the separating hyperplane theorem, there exists a non-zerovector b = (b1.... , b,) E m and a real number d such that

(22) b.z<difzEZ. b.t>diftET.

On the other hand, if z E Z, it is clear that there is a sequence of points z(k) e Z,for k = 1, 2, 3,. . ., which converges to z as k —+ (for example. one can takez(k) = z — (1/k,... , 1/k)). Since b.z(k) <v for all k, it follows on letting k —* -cthat b.z < d. Hence, (22) actually implies that

(23) b.z<difzE, b.t>diftET.

By (21), t Tn Z, and hence by (23) and the preceding remarks,

b.t = d.

For i = 1. m. let u’ = (u’, . . . u, where

U) It; ifji,

t—1 ifj=i.

By (21), u E Z for all i = 1,... , m, and so b.u <d. But

b.u = b.t* — = d — b.

40

It follows that, for all i = 1,..., rn, we have d — b < d and so b 0. Sinceb = (b1,.. , bm) is a non-zero vector, some b is > 0 and so b > 0. Then

/ b1 b,

=

is well-defined, and clearly x* E Lm. Also, if

ci=

thenby(23)x.t wift e Tandx*.z <wifz E . Now,ifv=(v v) ERtmthen v E Z so x*.v < w. But x.v = u, so v < w. If tj E , then o(y) E T andso x(y) > w > u. Since

x(y) = P(x, y),

it follows that P(x*, y) > v for all y E . Hence,

mm P(x*,y) >v—

and so< max mm P(x.y).— yEs,,

By Proposition 2.10 (i), the reverse inequality holds, and the minirnax theorem isproved.

z

41

Problems

3.1. Verify that, for any A [O,1},

3 A 5 5A 5 5AA 111x =(-+- —-— -), y =(- -)8 8 1.6 16 16 16 2 4 4 2

gives a mixed strategy solution (x* ,y) of the game with payoff matrix

—3 —3 2—1 3 —2

3 —1 —22 2 —3

3.2. The payoff matrix of a matrix game is

0 01 12 —1 1 —2

—1 1 1 31 010

Verify that x (0,0, , ), y = (, ,0,0) gives a mixed strategy solution (x*,y) of the game.

3.3. A matrix game has payoff matrix (a), where a,, 1 if i j and —1 for all i. Find amixed strategy solution of the game and write down its value.

3.4. A matrix game has payoff matrix

/0 0 0(o 1 —1

—1 1

Show that there is a pure strategy solution of this game, and also that each player has an optimalmixed strategy that is not a pure strategy.

3.5. The n x n payoff matrix of a matrix game has the property that the sum of the entries ineach row, and the sum of the entries in each column, are each equal to the same number a. Finda mixed strategy solution of this game and show that its value is s/n.

3.6. Let (a,,) be the payoff matrix of a matrix game and suppose that (i,j) is a pure strategysolution of the game. Show that (ñ ,6-) is a mixed strategy solution of the game. Deduce that

is the value of the game.

3.7. Let P m x . —÷ 1’. be the mixed payoff function of a matrix game. Prove that

max mm P(x,63) mm max P(5,.y.xE,,, 1<j<,, yEz,, 1<,<m

and that both sides are equal to the value of the game.

3.8. A matrix game with square payoff matrix is said to be symmetric if the payoff to plaverlwhen he uses pure strategy i and player 2 uses pure strategy j is the same as the payoff to player2 when he uses pure strategy i and player I uses pure strategy j. For a symmetric game, showthat(i) the payoff matrix (a,,) of the game is skew-symmetric, i.e. that a, = —a3, for all i,j:(ii) (x,y) is a mixed strategy solution of the game if and only if (y ,x) is a mixed strategysolution;(iii) the value of the game is zero.

3.9. A matrix game F has m x ri payoff matrix (a,,), and another matrix game F’ has payoffmatrix (a1). where a, = a,3 + A and .1 is a constant. Show that

42

(i) if the mixed payoff functions of F and F’ are P and F’, respectively,

P’(xy) P(x,y)+A for au XE s,,,, ys

(ii) if ii and u’ are the values of F and F’, respectively, then v’ = u + A;(iii) each player has the same set of optimal mixed strategies in the two games.

3.10. Let (ajj) be the payoff matrix of a matrix game with value v. Show that, if A > 0 is aconstant, the game with payoff matrix (Aa) has value Au and that a player has the same set ofoptimal mixed strategies in the two games. Is the same result true if the condition that A > 0 isomitted I

3.11. Let P x ,, —* IR be the mixed payoff function of a matrix game with value u, andlet x* (x x) S s,-,, and y S be optimal mixed strategies for players 1 and 2,respectively. Show that, if P(5, y) < u for some i = 1 m, then x 0. Write down thecorresponding result when the roles of players 1 and 2 are interchanged.

3.12. A game is played with five ordinary playing cards, the 2.3, and 4 of clubs anti the 2 and 4 ofdiamonds. Player 2 holds 2 and 3 and player 1 holds the remaining cards. Player 1 places twoof his cards face up on the table. Then player 2 plays one of his cards. For each of the two pairsconsisting of a card played from each player’s hand, player 1 receives from player 2 the differencein values of the two cards if they are of the same suit and pays player 2 the difference of the valuesif they are of different suits. Draw a game tree, list the pure strategies, and write down the payoffmatrix for this game. Find the pure strategy solutions, and verify that each one can be obtainedby reducing the payoff matrix to a single entry by successive deletion of inferior rows and columns.

3.13. In a certain game, a coin is tossed and if the outcome is heads’, player I chooses one ofthe integers 1, 6, while if the outcome is ‘tails’ player 1 chooses one of the integers 2, 7. Player2, knowing the choice made by player 1, then chooses one of the integers 3, 9. The choices of thetwo players are then added to yield the sum .s. The final move is the selection of a winner by achance process which selects player 1 with probability 4/5 and player 2 with probability 1/5. Thewinner then receives £s from the loser. Draw a game tree for this game, list the pure strategiesand find the payoff matrix. Find a pure strategy solution of the game and show that the payoffmatrix can be reduced to a single entry by successive deletion of inferior strategies.

3.14. Let ij be an optimal pure strategy for a player of a matrix game, and suppose that i1 isinferior to a pure strategy i0. Show that io is an optimal pure strategy for the player.

3.15. Show that the matrix/2 3 .1(140\i 0 6

is not the payoff matrix of any game of perfect information, although it does have a saddle point.

3.16. Prove Proposition 3.9 starting from the definitions (not by deducing it from Proposition3.10!).

3.17. A matrix game has payoff matrix

(2 3 11(7 5 2

Find the value of the game and the optimal mixed strategies for each player.

3.18. Repeat Problem 3.17 for the following payoff matrices:(i)

(—5 1(1 —22 —3

(ii)

( 4 —3 —1,—2 3 0

43

(iii)

(19 15 17 16\0 20 15 5

(iv)

f-i 11 —2

‘\O —l

(v)

4 —32 1

—1 3o —i

—3 0

(Vi)

(—i 3 —5 7 —92 —4 6 —8 10

3.19. Repeat Problem 3.17 for the following payoff matrices, deleting inferior strategies wherepossible:(i)

fi 7 2(6 2 7

1 6

(ii)

2 4 0 —24 8 2 6

—2 0 4 2—4 —2 —2 0

(iii)

/ 16 14 6 11—14 4 —10 —8

0 —2 12 —6\. 22 —12 5 10

(iv)

0 36515 10 8 910 15 11 75942

3.20. In a certain game, player I holds two Kings and one Ace. He discards one card, lays theother two face down on the table, and announces either ‘two Kings’ or ‘Ace King’. ‘Ace King’ isa better hand than ‘two Kings’. Player 2 must either ‘accept’ or ‘question’ player l’s call. Thehand is then shown and the payoffs are as follows:

if player 1 calls the hand correctly and player 2 accepts, player 1 wins £1 from player 2;if player 1 calls the hand better than it is and player 2 accepts, player 1 wins £2 from player 2:if player 1 calls the hand worse than it is and player 2 accepts, player 2 wins £2 from player 1;if player 2 questions the call, the above payoffs are doubled and reversed.

Draw a game tree for this game, list the pure strategies. write down the payoff matrix, and findan optimal mixed strategy solution. What is the value of the game?

4.1

Chapter 4: Solving Matrix Games in General

In this final chapter we describe an algorithm which allows one to find all the mixedstrategy solutions of any matrix game.

4.1 Simple Solutions

In this section we discuss certain special mixed strategy solutions of matrix games,called ‘simple’ solutions. Although not every game has a simple solution, it turnsout that all the mixed strategy solutions of any matrix game can be constructedstarting from the special solutions of other games ‘contained’ in the original game.Definition 4.1 A simple solution of a matrix game with mixed payoff functionP -m X L —* R is a pair of mixed strategies (x, y) E X Zn such that, forsome v E III.

(1) P(x*,6j) = u = P(6,y) for all i = 1,... , m., j = 1,... ,n.

It follows from Proposition 3.6 that a simple solution of a matrix game is a mixedstrategy solution of the game and that the value of the game is v.Proposition 4.2 Assume that the payoff matrix of a game is invertible (and hencesquare). Then, if a simple solution of the game exists, it is unique.Proof Let the payoff matrix B of the game be r x r, and suppose that (x*, y*)and (x*. y) are two simple solutions of the game. If we think of x*,x (resp.y*y*) as row (resp. column) matrices, the equations in (1) can be written inmatrix form as

11x*B = v (1 1 ... 1) = x**B, By* = v By*.

1

Hence,(x* — x**)B 0. B(y* — y) = 0.

Since B is invertible, these equations imply that x = x and y= y.

We shall now establish the conditions under which simple solutions exist. Weshall need to introduce some notation.

Namely, if B = (b) is an r x r matrix, let p(B) (resp. K,(B)) be the sum ofthe entries in the ith row (resp. column) of B (i = 1, . . . , r), and let (B) be thesum of all the entries of B. Note that

/1(1(B) ... kr(B)) = (1 ... 1)B, = B

pr(B))

15

Now recall that the adjoint adj(B) of B is the transpose of the matrix of cofactorsof B, and that Badj(B) = adj(B)B = det(B)I, where clet(B) is the determinant ofB and I denotes the r x r identity matrix. We have

1 1 ... 1b21 b29 ba,.

1(adj(B)) = det :

bri br2

since on expanding the determinant along its first row we get the sum of the cofactors of the entries in the first row of B, i.e. the sum of the entries on the firstcolumn of adj(B). Similar formulas hold for the other column sums and the rowsums of adj(B).

Lemma 4.3 Let r > 1, let B = (b) be an r x r matrix, and let C = (b + A),where A E R. Then.

i(adj(C)) = k(adj(B)), p(adj(C)) = p(adj(B)). for i 1,... ,r.

anddet(C) = det(B) + A(adj(B)).

Proof From the above remarks,

1 1 ... 1b21±A b9+A ... b2r±A

1(adj(C)) = det :

bri+A brr+ASubtract A times the first row of this matrix from each of the other rows; this doesnot change the determinant of the matrix, so

1 1 1b21 b22 . .

ti(adj(C)) = det ... kj(adj(B)).

bri br

Similar arguments apply for the other column sums and for the row sums.Next,

b11 birb91+A b22+A ... b2r+A

det(C) det :

b.ri±A br2±A brr±A1 1 1b21+A b22+A b2r+A

+ . .

±A br2+A brr±Ab11 b12 . . birb91+A b)9+A ...

=det : -

+At1(adj(B)).

bri + A br2 ± A ... brr ± A

46

Iterating this calculation, we get

det(C) det(B) + \(tci(adj(B) ± ± r(idj(B))) det(B) ± (adj(B)).

Note that Lemma 4.3 is not true when r = 1! (Where does the proof break downwhen r = 1?) This trivial remark will have a non-trivial consequence in Section4.3.

We can now state the main result about simple solutions:

Proposition 4.4 Suppose that a matrix game F has r x r payoff matrix B, wherer 1.

(2) (adj(B)) o, (a> 0, and >0, for i = 1

the game has a unique simple solution (x*,y*), which is given by

* - (i(adj(B)) r(adj(B))X

- E(adj(B))’’U(pi(adj(B)) pr(adj(B))

(adj(B))’’

and the value of the game is— det(B)

VE(adj(B))

Conversely, suppose that B is invertible and that the game with payoff matrixB has a simple solution (which is necessarily unique by Proposition 4.2). Then, Bsatisfies the conditions in (2).

We shall say that an r x r matrix B satisfying the conditions in (2) is a simplematrix. Thus, the first part of Proposition 4.4 says that a matrix game with asimple payoff matrix has a simple solution.

Proof Assume first that B is simple. Let x* = (x1,... ,xr), where

k(adj(B))X=

E(adj(B))fori=1 r.

By(2),x>0foralliand

= (adj(B))i(adj(B)) = 1,

50 X* ELet B = (b) and let P : . x R be the mixed payoff function ofF. Let

= for j = 1,... ,r.

47

Then,

((1 (r)=X

(adj(B))(in (adj(B)) inr(adj(B))) B

= F(adj(B))(1 •. 1)adj(B)B.

Since adj(B)B = det(B)I, we get

det(B)f 1j)

— F(adj(B))or j —

r.

Similarly, y E r and

det(B)P(&,y

= E(adj(B))’fori 1,... ,r.

We have now proved that the game has the stated simple solution and value,and it remains to prove that the solution is unique. We have already noted inProposition 4.2 that this is true if B is invertible, so assume now that det(B) = 0.Then we must have r > 1, for if r = 1 we would have B = 0 and then B would notbe simple. From (2) the value of the game is 0. Let C = (ej) be the r x r matrixsuch that = + 1 for all i, j = 1. . . . , r. By Lemma 4.3. det(C) = E(adj(B)),so by (2), C is invertible. It follows from Problem 3.9 that, if (x’, y’) is a simplesolution of F, it is also a simple solution of the game F0 with payoff matrix C, thevalue of the latter game being 1. By the first part of the proof, Pc’ has a uniquesimple solution, hence so does F.

The converse is proved by essentially reversing the above argument. Let B beinvertible and let (x*, y) be a simple solution of F; let v be the value of F. Letx* = (x1,... , xv), and let P be the mixed payoff function of F. For j = 1,... , r,we have P(x,6) = v, i.e.

xB=v(1 ... 1),

sox*Badj(B) =v(inj(adj(B)) ... inr(adj(B))).

Since Badj(B) det(B)I, we get

det(B)x = vin(adj(B)) for i = 1, . . . , r.

Since = 1, this gives

det(B) = vF(adj(B)),

so since B is invertible it follows that F (adj(B)) 0. Also,

in(adj(B))S(adj(B))

0 for z = 1,... r.

A similar argument shows that

p(adj(B))F(adj(B))

0 for z = 1,... ,.

Hence, B is simple.

48

4.2 Convexity (again)

If X is a finite set of points of 1:m, the convex hull of X is called a concex pol’tope.These are the generalizations of convex polygons in the plane. For example, a cubeis the convex hull of the set of its eight vertices:

In fact, we can define the analogue of vertices’ for any convex set:

Definition 4.5 Let C be a convex subset of Rm (m > 1). A point x E C is saidto be an extreme point of C if x is not the midpoint of any line segment whoseendpoints are in C.

Example 4.6 The extreme points of Am are precisely the pure strategies. To seethis, we show first that = (1.0.... .0) is an extreme point of Am: a similarargument works for each of the m. pure strategies. Suppose for a contradictionthat ö is the midpoint of a line segment with endpoints x (x1,... , x), y(yr,... ,Ym) EArn. Then, Xi+i = 0 for i >1, so since xj,yj >0 for all i, we have

= = 0 for all i > 1. Hence, x. = = 1, contradicting the fact that x y.Conversely, let a (a1, am) be a proper mixed strategy, i.e. a mixed strategy

that is not a pure strategy. Then, at least two of the components of a are non-zero,say a and a (i <j). Let

= miii 1— a, 1— a}

then, e> 0. Let

x=(ai,....a—e ,a+e. arn).y=(ai a,+e....a—e. am).

It is easy to see that x, y E Am and that a = 4(x + y). Hence, a is not an extremepoint of A1.

We shall accept the following result without proof.

49

Theorem 4.7 Let C be a bounded closed convex subset of R’ (m > 1), andassume that the set E of extreme points of C is finite. Then. C (B) (arid henceC is a convex polytope).

For example, since the set rn of mixed strategies is bounded, closed and convex(see the remark following the proof of Proposition 3.16), the set of mixed strategiesis the convex hull of the set of pure strategies (it is easy to see this directly).

4.3 Shapley-Snow Algorithm

Proposition 3.16 shows that the sets of optimal mixed strategies for the two playersin a matrix game are bounded closed convex sets. By Theorem 4.7, such a setis the convex hull of the set of its extreme points. Let us say that an optimalmixed strategy for a player is an extreme optimal strategy for the player if it isan extreme point of the set of his optimal mixed strategies. If x’ and y areextreme optimal strategies for players 1 and 2: respectively, then (x,y) is calledan extreme solution of the game. Solving the game therefore reduces to determiningthe extreme solutions.

Proposition 4.8 Let A be the payoff matrix of a matrix game LA, let B be asimple submatrix of A, and let LB be a game with payoff matrix B. Let (,i)be a simple solution of LB, and let (x*, y*) be its extension to a pair of mixedstrategies for LA. Then, if (x, y*) is a mixed strategy solution of F.4, it is anextreme solution of LA. Moreover, F4 and LB have the same value.

Conversely, any extreme solution of LA is either a pure strategy solution of LAor is the extension to FA of the simple solution of a game with payoff matrix asimple submatrix B of A.

Remark Note that any pure strategy solution is obviously an extreme solution,since any pure strategy is an extreme point of the set of all mixed strategies byExample 4.6, and a fortiori any optimal pure strategy is an extreme point of theset of optimal mixed strategies.

Before we spell out the algorithm which this proposition gives us, we point outsome corollaries. Let us say that a mixed strategy x = (x1,... , Xm) for a playeruses the pure strategy i if x > 0.

Corollary 4.9 Each of the players of a matrix game with m x mm payoff matrix hasan optimal mixed strategy which uses at most min(m, mm) pure strategies.

Proof Let .4 be the payoff matrix of the game and let x* be an extreme optimalstrategy for player 1 (a similar argument works for player 2). If x is an optimal purestrategy, the conclusion obviously holds. Otherwise, it is obtained by extending asimple solution of a game with payoff matrix a simple submatrix B of .4, say. ThenB is r x i for some r < min(rn, mm), and x uses at most r pure strategies.

Corollary 4.10 The set of optimal mixed strategies for a player of a matrix gameis a convex polytope.

Proof The payoff matrix of the game has only a finite number of simple submnatrices,and by Proposition 4.4 each such submatrix has a unique simple solution. Sincethe original game obviously has only a finite number of pure strategy solutions,

50

by Proposition 4.8 it has only finitely many extreme solutions. Hence, the set ofoptimal strategies for each player is the convex hull of a finite set of points. fl

Proposition 4.8 translates into a method which can be used to solve any matrixgame, called the Shapiey-Snow algorithm. Consider a matrix game F with m x npayoff matrix A. Let be the set of all simple submatrices of A. (Note that, if Aitself is a square matrix (i.e. if m = n) it may be a simple submatrix of itself.)

Step I Determine the pure strategy solutions of F, if any.

Step II If B E , let (,11*) be the simple solution, and v the value, of the gamewith payoff matrix B given by Proposition 4.4. Let (x*,y*) be the extension of

,11*) to a pair of mixed strategies for players 1 and 2, respectively, of the gameF.

Step III Determine which of the pairs of mixed strategies (x*, y*) found in Step Hare mixed strategy solutions of F.

Step IV For i = 1,2, let E be the set consisting of the optimal pure strategies forplayer i found in Step I together with the set of optimal mixed strategies found inStep III. Then, the convex hull (E) of E is the set of all optimal mixed strategiesfor player i.

All of this is an immediate consequence of Proposition 4.8. However, let us spellout in more detail what must be done in Step III. Let B B, and assume thatB consists of the elements of rows i1,. .. , i and columns ,Jr of A. Then,the mixed strategies X = (x1,... ,Xm) and ( ,y) for players land 2,respectively, associated to B as in Step II are given by

— K3(adj(B)) — p(adj(B))f — 1

— (adj(B))’ — (adj(B))’or s — ,...

and Xi = 0 if j {i1,.. ,ir}, Yj 0 if j {ji,... ,jr}. By Proposition 3.5,(x*, y*) is a mixed strategy solution of F if and only if

(4) P(x*.)

forj=l,... ,nand

det(B)

(adj(B))

for i = 1,. . . , m. However, we already know that (4) holds for j = Ji, . . , j sincefor such values of j we have Pn(*,) = P(x*,) as is a mixed strategysolution of the game with payoff matrix B; similarly, and we know that (5) holdsfor i = ii Hence, we need only verify () for j {ji,... ,jr} and (5) for

{ii,... ,ir}.It is important to note that one often does not need to consider all the simple

subrnatrices of A. For, once the value v of F has been determined, for example byfinding a pure strategy solution as in Step I or an extreme solution as in Step III,one need only consider the simple submatrices B E B such that

det(B)

E(adj(B))

51

Note finally that we do not have to consider the 1 x 1 simple submatrices of A,since the corresponding extreme solutions (if any) are pure strategy solutions. (SeeProblem 1.8.)

We now give some examples of the algorithm in action.

Example 4.11 Consider the game F with mixed payoff function P and payoffmatrix

f’ 1 —1 2 1

.=1 1 0 —1 —12 —1 0

For the submatrix

B=2 2)’

consisting of the entries in rows 1 or 3 and columns 1 or 2 of A, we find that

adj(B)=( ),so E(adj(B)) = 6 is positive and so are the row sunis (3 and 3) and column sums(4 and 2) of adj(B). Hence, B is a simple subrnatrix of A. By Proposition 4.4, thesimple solution of the game PB with payoff matrix B is where = (a., ).

(. ), and since det(B) = 0. the value of the game isO. Extending to solutionsof the original game gives the mixed strategies x = (,0, ), y = (, ,O,0) forplayers 1,2. respectively, in F. Then (x, y) will be a mixed strategy solution ofF if and only if

P(x*,3)> 0, P(x4)Z 0, P(S2,y) <0.

We find that P(x,3)= 4/3 — 4/3 = 0, P(x,ó4)= 2/3, but P(,y*)= 1/2, so

(x, y) is not a mixed strategy solution of P.Next, consider the submatrix

B=(1 )consisting of the entries in rows 1 or 3 and columns 2 or 3 of A. We find that

adi(B’)=(j 1)so E(adj(B’)) = —9 is negative and so are the row sums (—6 and —3) and columnsums (—6 and —3) of adj(B). Hence, B is a simple submatrix of A. The simplesolution of the game PB with payoff matrix B is where = (, ), , =

(. ). and since det(B’) = 0, the value of the game is 0. Extending to solutionsof the original game gives the mixed strategies x = (, 0. i.), y = (0, ., , 0) forplayers 1.2. respectively, in F. Then (x*, y) will be a mixed strategy solution of Fif and only if

P4(x*,5i) >0. P4(x,64)>0. Pt(.y) <0.

52

We find that PA(x*,oi) = 0, PA(x,Sl) = 2/3,P1(62,y) = —1/3, so (x,y) isa mixed strategy solution off. Hence, ((i, O ), (0, , , 0)) is an extreme solutionof P.

If you have enough energy, you can consider the remaining sixteen 2 x 2 submatrices and the four 3 x 3 submatrices of A There are no pure strategy solutions.

Example 4.12 A more manageable example is the game P with payoff matrix

/1 —2—2.4=(—3 0 —1

\\3 —3 0

There are no pure strategy solutions.We find that

/—3 6 2adj(A)= (—3 6 7

\9 —3 —6

Thus, S (adj(A)) = 15 and

ic1(adj(A)) = 3.i2(adj(A)) = 9,i3(aclj(A)) = 3.

p1(adj(=4)) = 5, p.2(adj(.4)) = 10, p3(ad(A)) = 0.

Thus, A is a simple submatrix of itself. Since det(A) = — 15, the value of the gameis —15/1.5 = —1 and extreme optimal strategies for players 1 and 2 are (i-, , )and (i., , 0), respectively.

We must now consider the nine 2 x 2 submatrices:

B (‘2): adj(B)

= ( ), E(adj(B)) = 6. det(B)/dj(B)) = -1;

B= ( 1): adj(B)

= (1 (adj(B)) = -3. det(B)/(adj(B)) -1;

B (3 ) adj(B)=(j °), (adj(B)) = -9, det(B)/(adj(B)) = -1;

B= (‘ ): adj(B)

= ), E(adj(B)) = 5, det(B)/(adj(B)) = -7/5;

B=( _2): adj(B)= ), E(adj(B))=O;

B= (3

‘) : adj(B)= ), (adj(B)) -5. det(B)/(adj(B)) = —3/5;

B= ( ) adj(B)

= (‘2

(adj(B)) = —1. det(B)/(adj(B)) = -2:

B= ; ) : adj(B)

(3 2)‘ (adj(B)) = 3, det(B)/(adj(B)) = —2;

B= (° ‘) adj(B) ( ), E(adj(B)) = 4, det(B)/E(adj(B)) -3/4.

The second, fifth, sixth and seventh matrices are not simple, and the fourth, eighthand ninth, although simple, give the wrong value for the game. So only the firstand third niatrices could give rise to extreme solutions.

53

The submatrix(1 —2—3 0

gives rise t.o pair of mixed strategies x = (4, 4,0) and y = (k , 0) for players 1

and 2 in F, respectively, but since

= (—2) x + (—1) x = <—1,

this is not a mixed strategy solution of F.The siibrnatrix

(—3 03 —3

gives the possible extreme solution (x y), where

21 1x=(0,), y=(,0).

The fact that A is a simple submatrix of itself already showed that y* is an optimal

mixed strategy for player 2, and since

4(x,S3)= (—1) x— > —1,

x* is an optimal mixed strategy for player 1. Hence, we have found the secondextreme solution.

The most general optimal mixed strategy for player 1 is the convex hull of the

subset of 1R3 consisting of the two points

131 21and

namely the straight line segment

= for 0<A< 1.

y

z

(0, Z/s ‘13)

x

Player 2 has a unique optimal mixed strategy (, , 0)

54

4.4 Proof of the Algorithm

In this section we give the probf of Proposition 4.8. We use the notation in thestatement of the proposition. Let A be rn x n and let B he the r x r submatrix of4 consisting of the entries in rows i1,.. ,i,. and columns j, Let P4 aridPB be the mixed payoff functions of L’,4 and TB, respectively. Since P4(x. y*) =

PB(,rJ*). it follows that F..1 and T8 have the same value. savu.We now show that x* is an extreme optimal strategy for piaer 1 in F.4 (a similar

argument will, of course, show that y* is extreme). Let x = 4(x’ ± x”), wherex’ (xi,: . ,x) and x” = (xv,. .. ,x,) are optimal mixed strategies for player1 in TA. Fo i {Ii,. .. ,ir}, (x + x’) = 0, so since x and x7 are > 0, we havex=x’Ofori{iiwhere = x arid ‘ x’ for s = 1,... ,r. Then, = 4(’ +“), so that

= PB(’,) +

for s = 1,... ,r. But PB(*,6s) = v, and since x’ and x” are optimal mixedstrategies for player 1 in TA,

PA(x’,5Jj > V, PB(C’,5s) = E4(x”,j) v.

Hence,= V

for s = 1, . This shows that (‘,i) and (“.i) are simple solutions of TB.But, since B is simple, Proposition 4.4 shows that TB has a unique simple solution.Hence, e = “ = , and so x’ = x” x*. This proves that x is extreme.

For the converse, let (x*. y) be an extreme solution of TA that is not a purestrategy solution. say X = (x1 Xm), y = (yi,... ,y,), and let u be the valueof F4. We consider two cases:

Case I: v 0.By Definition 3.2, y*) < v for alli, and by Problem 3.11, P(3, y8) = v if z >0. If the rows of A are permuted to give a matrix A’ and the same permutation isapplied to the components of x to give an element x’ it is clear that (x’, y)is an extreme solution of the game with payoff matrix A’. Hence, we might as wellassume that there exist positive integers p and s such that p < s < m, x > 0 if andonly if i <p, PA(6j,y*) = v if i <s. and PA(j,y*) <v if s 0if and only if j <q, F.i(x8,öi) = v if j <t, and F(x,ô) > v if t <j <a.

Consider the submatrix

a11 a

C = a1 am

a81 a a8t

of A. We shall prove that the first q columns of C are linearly independent. Similarly, the first p rows of C are linearly independent. Assuming this for the moment,the rank r of C satisfies

mnax{p.q} < r < min{s,t},

30

so we can choose an invertible r x r submatrix B of C which contains

a11 a

a1 apq

as a submatrix. Let ‘B and JB be the sets of integers which index the rows andcolumns of B, respectively. Since {1, . .

. , p} C ‘B, there is a mixed strategy forplayer 1 in the game LB with payoff matrix B which extends to the mixed strategyx for player 1 in F. Similarly, there is a mixed strategy , for player 2 in LB whichextends to y* If j E JB, then 1 < j < t so Pt(x*,6j) = v. Similarly, if i E ‘B thenF4(6,y’) v. Thus, is a simple solution of LB. Since B is an invertiblematrix such that LB has a simple solution, it follows from Proposition 4.4 that Bis a simple submatrix of A.

To complete the proof in the case u 0, we must now show that the first qcolumns of C are linearly independent. For j = 1, , q, let i be the jth columnof C, so that

a13

i= apj

a3

and suppose for a contradiction that there exist real numbers cii, . . . , c, not allzero, such that

i.e.,

na=O fori=1,...,s.

Then,

(aui)

Since u 0, we must have cij = 0. Put a = 0 if q < j < n and consider thepoint

y = (y +An1,y2+A2 y0+Aa,) E

for A E R. Note that

(y +An)= yj +Aa = Yj = 1.

Since y > 0 if j < q, there exists € > 0 such that

(6) y3±Ac3>O forl<j<q

56

if .! < e; arid (6) also holds for q < j < n since in that case y 0 and cj = 0.Thus, y E ifJ <e1.

We now show that there exists 0 < e e such that y, is an optimal mixedstrategy for player 2 if Aj < e. For this we compute

= aj(yj + t)

=

+ A

= PA(6j,y*) +

If 1 < < s, then PA(oi,y*) = v and1cia = 0, so P(&,y>) = v. Onthe other hand, PA(Sj,

y*) < v if s < i < m, so there exists 0 < e < e. such thatPA(ó,yA) <v for all s <i < m provided that I) <e. We have thus shown thaty,, is an optimal mixed strategy for player 2 if A < e. In particular, y and yare distinct optimal mixed strategies for player 2. But this is impossible, since

y = (y, ± y),

and y is extreme. This contradiction shows that the first q columns of the matrixC are indeed linearly independent.

Case II: v 0Let A1 be the matrix obtained by adding 1 to each entry of A. The game FA1 withpayoff matrix -4 has value 1 and (x, y) is an extreme solution of it. It followsfrom Case I that there exists a simple submatrix D of .4 (which is r x r, say) anda simple solution (,i) of the game 1’D with payoff matrix D such that x* andy* are the extensions of and i to mixed strategies for players 1 and 2 of FA1,respectively. Let B be the submatrix of A obtained by subtracting 1 from eachentry of D. By Lemma 4.3, (adj(B)) = (adj(D)) and9(adj(B))p8(adj(B)) =p5(adj(D)) for s = 1,... ,r. It follows that B is a simple subrnatrixof .4. Moreover, (,j*) is a simple solution of the game with payoff matrix B.

Problems

dl. Show that the game with payoff matrix

ft —1 —1—1 —1 3

2 —1

has a unique simple solution, and find it.

4.2. Show that the game with payoff matrix

/i —4 5— 2 0

0 1

has a unique mixed strategy solution.

4.3. Repeat Problem 4.2 for the pa’off matrix

f—i —6 3—5 0 —2

4 —2 2

4.4. Find the value of the game with payoff matrix

/1 —1 21 0 —1

\,—2 2 —4

Show that player 1 has a unique optimal mixed strategy and find all optimal mixed strategies forplayer 2. Sketch the set of optimal mixed strategies for player 2 as a subset of.

1.5. Find the value of the game with payoff matrix

o 0 —1—1 0 0

1 —2 0o —3 2

given that each extreme strategy for player 2 uses three pure strategies. Show that player 2 hasa unique optimal mixed strategy and find all optimal mixed strategies for player 1.

4.6. Use the methods of Section 3.4 to find the value, and the optimal mixed strategies for player1, of the game with payoff matrix

(i 2 4 0k,.o —2 —3 2

Find all the optimal mixed strategies for player 2.

4,7. A matrix game is said to be completely mixed if every optimal mixed strategy for each of theplayers uses every one of his pure strategies. Show that(i) a completely mixed game has a square payoff matrix;(ii) a completely mixed game has a unique mixed strategy solution;(iii) the value of a completely mixed game is non-zero if and only if its payoff matrix is invertible.4.8. Show that every pure strategy solution of a game with payoff matrix .4 arises from a 1 x 1simple submatrix of .4, provided the value of the game is non-zero. Why does the proof ofProposition 4.8 not show that this result is true whatever the value of the game ? Give anexample to show that the result is not, in fact, true if the value is zero.

4.9. Show that the open interval (0, 1) is a convex subset of that has no extreme points. Givean example of a non-empty closed convex subset of with no extreme points. (In fact, there isonly one such subset — can you prove that?)