GAME THEORY: AN INTRODUCTION ERRATA

Transcription

1 GAME THEORY: AN INTRODUCTION ERRATA E. N. BARRON Please notfy me at for any errors. These have been found so far. () p. 2 Lemma..3, second lne of proof should be v + = mn j max a,j max a,j a,j mn j a,j max mn a,j = v. j p. 2 n proof of Lemma..3, Let be such that... j =, 2, m. Should be: Let j be such that v + = max a,j and such that v = mn j a,j. Then a,j v = v + a,j, for any =, 2,, n, j =, 2,, m. (2) p. 6, lne 6, v + = mn x C max y D f(x, y), and v = max y D mn x C f(x, y), should be v + = mn max f(x, y), and y D x C v = max mn x C y D f(x, y). (3) p. 22, The last lne of the thrd paragraph These probablty vectors are called mxed strateges, and wll turn out to be the class correct class of strateges for each of the players. should be These probablty vectors are called mxed strateges, and wll turn out to be the correct class of strateges for each of the players. (4) p. 47, Problem.29, part (a) should have mn j E(X, j) = (5) p. 85, Problem 4.6 : Should be: Suppose that two frms have constant unt costs c = 2, c 2 =, and Γ = 9 n the Stackelberg model. (6) p. 75, Quotaton added (7) p., lne 7 from top E 2 should be E II. (8) p. 25, lne 9 from bottom, E(, Y ) should be E I (, Y ). (9) p. 45, lne 5 from bottom, Y T should be Y T. (0) p. 54, problem 3.23 has the answer fxed on p. 393: should have f(x, y, p, q) = 7x + 7y 6xy 6 p q, and 2 x q should be 2x q. Date: August 30, 2008.

2 2 E. N. BARRON () p. 22, Example 5.(4): but wll take $ mllon..., should be but wll take $00 mllon.... (2) p. 24, Problem 5.0: x 2 should be x 2. (3) p. 246, x+x2+x3=\frac{5}{2} should be x + x2 + x3 = 5/2. (4) p. 400 Problem 5.3 should have 6 x x 2, not 6 x x 2. (5) p. 40 Problem 5.9 soluton n (b) should have x 4 = 3 2, not 32. The followng errors were found by Yan Jn to whom I am grateful. () p. 43, lne 2 from bottom, E(4, Y ) = 5y + 6( y) should be E(4, Y ) = 7y 8( y). (2) p. 44, lne from top, E(, X) should be corrected as E(X, ). Lne 2 from top, E(4, X) should be E(X, 2), and (x = 5/6, /3) should be corrected as (x = 5 6, v = 3 ). (3) p. 68, the second lne of the proof of Theorem 2.3. should read E(X, X) = XAX T = XA T X T = (XA T X T ) T = XAX T = E(X, X). In other words, the thrd A should be A T. (4) p. 69, the thrd lne from the bottom, (aλ, bλ, cλ) should be (cλ, bλ, aλ). I am grateful to Stephen Conwll who found the followng errors. () p.7 In the table at the bottom of the page II3 shout be the strategy: If I, then S; If I2, then S. The strategy II4 should be: If I, then S; If I2, then P. (2) p. 8 lne 5 from the top pass as well should be spn. E-mal address: ebarron@luc.edu

3 2 MATRIX TWO-PERSON GAMES We see that v = largest mn = and v + = smallest max =. Ths says that v + = v =, and so 2 2 Nm has v =. The optmal strateges are located as the (row,column) where the smallest max s and the largest mn s also. Ths occurs at any row for player I, but player II must play column 3, so =, 2, 3, j = 3. The optmal strateges are not at any row column combnaton gvng as the payoff. For nstance, f II plays column, then II wll play row and receve +. Column s not part of an optmal strategy. We have mentoned that the most that I can be guaranteed to wn should be less than (or equal to) the most that II can be guaranteed to lose, (.e., v v + ), Here s a quck verfcaton of ths fact. For any column j we know that for any fxed row, mn j a j a j, and so takng the max of both sdes over rows, we obtan v = max mn j a j max a j. Ths s true for any column j =,..., m. The left sde s just a number (.e., v ) ndependent of as well as j, and t s smaller than the rght sde for any j. But ths means that v mn j max a j = v +, and we are done. Now here s a precse defnton of a (pure) saddle pont nvolvng only the payoffs, whch bascally tells the players what to do n order to obtan the value of the game when v + = v. Defnton..2 We call a partcular row and column j a saddle pont n pure strateges of the game f a j a j a j, for all rows =,..., n and columns j =,..., m. (..) Lemma..3 A game wll have a saddle pont n pure strateges f and only f v = max Proof. If (..) s true, then v + = mn j max mn j a,j max a,j a j = mn j a,j mn j max a j = v +. (..2) a,j max mn a,j = v. j But v v + always, and so we have equalty throughout and v = v + = v = a,j. On the other hand, f v + = v then mn j max a,j = max mn a,j. j Let j be such that v + = max a,j and such that v = mn j a,j. Then a,j v = v + a,j, for any =,..., n, j =,..., m.

4 PROBLEMS 85 frm 2 s producton cost s C 2 (q 2 ) = 2q Fnd the proft functons and the Nash equlbrum quanttes of producton and profts. 4.3 Compare profts n the model wth uncertan costs and the standard Cournot model. Can you fnd a value of 0 < p < that maxmzes frm s profts? 4.4 Suppose that we consder the Cournot model wth uncertan costs but wth three possble costs, P rob(c 2 = c ) = r, =, 2, 3, where r 0, r + r 2 + r 3 =. Solve for the optmal producton quanttes. Fnd the explct producton quanttes when r = 2, r 2 = 8, r 3 = 3 8, Γ = 00, and c = 2, c =, c 2 = 2, c 3 = In the Stackelberg model compare the quantty produced, the proft, and the prces for frm assumng that frm 2 dd not exst so that frm s a monopolst. 4.6 Suppose that two frms have constant unt costs c = 2, c 2 = and Γ = 9 n the Stackelberg model. (a) What are the proft functons? (b) How much should frm 2 produce as a functon of q? (c) How much should frm produce? (d) How much, then, should frm 2 produce? 4.7 Set up and solve a Stackelberg model gven three frms wth constant unt costs c, c 2, c 3 and frm announcng producton quantty q. 4.8 In the Bertrand model show that f c = c 2 = c, then (p, p 2 ) = (c, c) s a Nash equlbrum. 4.9 Determne the entry deterrence level of producton for frm gven Γ = 00, a = 2, b = 0. How much proft s lost by settng the prce to deter a compettor? 4.0 We could make one more adjustment n the Bertrand model and see what effect t has on the model. What f we put a lmt on the total quantty that a frm can produce? Ths lmts the supply and possbly wll put a floor on prces. Let K Γ 2 denote the maxmum quantty of gadgets that each frm can produce and recall that D(p) = Γ p s the quantty of gadgets demanded at prce p. Fnd the proft functons for each frm. 4. Suppose that the demand functons n the Bertrand model are gven by q = D (p, p 2 ) = (a p + bp 2 ) + and q 2 = D 2 (p, p 2 ) = (a p 2 + bp ) +, where b > 0. Ths says that the quantty of gadgets sold by a frm wll ncrease f the prce set by the opposng frm s too hgh. Assume that both frms have a cost of producton c mn{p, p 2 }. (a) Show that the proft functons wll be gven by u (p, p 2 ) = D (p, p 2 )(p c), =, 2.

5 COALITIONS AND CHARACTERISTIC FUNCTIONS 22 Why? Well, v(23) = d because the car wll be sold for d, v() = M because the car s worth M to player, v(3) = d because player wll sell the car to player 3 for d > M, v(2) = c because the car wll be sold to player 2 for c > M, and so on. The reader can easly check that v s a characterstc functon. 3. A customer wants to buy a bolt and a nut for the bolt. There are three players but player owns the bolt and players 2 and 3 each own a nut. A bolt together wth a nut s worth 5. We could defne a characterstc functon for ths game as v(23) = 5, v(2) = v(3) = 5, v() = v(2) = v(3) = 0, and v( ) = 0. In contrast to the car problem v() = 0 because a bolt wthout a nut s worthless to player. 4. A small research drug company, labeled, has developed a drug. It does not have the resources to get FDA (Food and Drug Admnstraton) approval or to market the drug, so t consders sellng the rghts to the drug to a bg drug company. Drug companes 2 and 3 are nterested n buyng the rghts but only f both companes are nvolved n order to spread the rsks. Suppose that the research drug company wants $ bllon, but wll take $00 mllon f only one of the two bg drug companes are nvolved. The proft to a partcpatng drug company 2 or 3 s $5 bllon, whch they splt. Here s a possble characterstc functon wth unts n bllons: v() = v(2) = v(3) = 0, v(2) = 0., v(3) = 0., v(23) = 0, v(23) = 5, because any coalton whch doesn t nclude player wll be worth nothng. 5. A smple game s one n whch v(s) = or v(s) = 0 for all coaltons S. A coalton wth v(s) = s called a wnnng coalton and one wth v(s) = 0 s a losng coalton. For example, f we take v(s) = f S > n/2 and v(s) = 0 otherwse, we have a smple game that s a model of majorty votng. If a coalton contans more than half of the players, t has the majorty of votes and s a wnnng coalton. 6. In any bmatrx (A, B) nonzero sum game we may obtan a characterstc functon by takng v() = value(a), v(2) = value(b T ), and v(2) =sum of largest payoff par n (A, B). Checkng that ths s a characterstc functon s skpped. The next example works one out. EXAMPLE 5.2 In ths example we wll construct a characterstc functon for a verson of the prsoner s dlemma game n whch we assumed that there was no cooperaton. Now we wll assume that the players may cooperate and negotate. One form

6 246 COOPERATIVE GAMES so that x mnmzes the maxmum excess for any coalton S. When there s only one such allocaton x, t s the far allocaton. The problem s that there may be more than one element n the least core, then we stll have a problem as to how to choose among them. Remark: Maple Calculaton of the Least Core. The pont of calculatng the ε-core s that the core s not a suffcent set to ultmately solve the problem n the case when the core C(0) s () empty or (2) conssts of more than one pont. In case (2) the ssue, of course, s whch pont should be chosen as the far allocaton. The ε-core seeks to address ths ssue by shrnkng the core at the same rate from each sde of the boundary untl we reach a sngle pont. We can use Maple to do ths. The calculaton of the least core s equvalent to the lnear programmng problem Mnmze z subject to v(s) x(s) = v(s) S x z, for all S N. The characterstc functon need not be normalzed. So all we really need to do s to formulate the game usng characterstc functons, wrte down the constrants, and plug them nto Maple. The result wll be the smallest z = ε that makes C(ε ), as well as an mputaton whch provdes the mnmum. For example, let s suppose we start wth the characterstc functon v() = 0, =, 2, 3, v(2) = 2, v(23) =, v(3) = 0, v(23) = 5 2. The constrant set s the ε-core C(ε) = { x = (x, x 2, x 3 ) v(s) x(s) ε, S N} = { x ε, =, 2, 3, 2 x x 2 ε, x 2 x 3 ε, 0 x x 3 ε, x + x 2 + x 3 = 5 2 } The Maple commands used to solve ths are very smple: > wth(smplex): > cnsts:={-x<=z,-x2<=z,-x3<=z,2-x-x2<=z,-x2-x3<=z,-x-x3<=z, x+x2+x3=5/2}; > mnmze(z,cnsts); Maple produces the output x = 5 4, x 2 =, x 3 = 4, z = 4.

7 PROBLEM SOLUTIONS X = C( 0 ) = {x + x 2 = 9 0, 4 0 x, 2 0 x 2}. The next least core s X 2 = C( 4 ) = {( 20, 7 20, 2 20 )}. 5.7 The least core s the set C( ) = {x =, x 2 + x 3 =, x 2, x 3 2}. The nucleolus s the sngle pont {(, 2, 2 )} 5.8 For the least core ε = 2 : Least core = X = C( 2 ) = {x + x 2 = 3 2, x 3 + x 4 = 3 2, x, =, 2, 3, 4, 2 x 2 + x 3 3 2, x + x 4 3 2, x + x 3 5 4, x 2 + x 4 2, x + x 2 + x 3 3 2, x + x 2 + x 4 3 2, x + x 3 + x 4 3 2, x 2 + x 3 + x 4 3 2, x + x 2 + x 3 + x 4 = 3}. Next X 2 has ε 2 =. X 3 has ε 3 = 3, and nucleolus={( 3 4, 3 4, 3 4, 3 4 )}. 5.9 (a) The characterstc functon s the number of hours saved by a coalton. v() = 0, and v(2) = 4, v(3) = 4, v(4) = 3, v(23) = 6, v(24) = 2, v(34) = 2, v(23) = 0, v(24) = 7, v(34) = 7, v(234) = 8, v(234) = 3. (b) Nucleolus={( 3 4, 33 8, 33 8, 3 2 )} wth unts n hours. The least core s X = C( 3 2 ) = {x + x 2 + x 3 = 23 2, x 4 = 3 2, x + x 2 + x 3 + x 4 = 3, x + x 2 + x 4 7 2, x 2 + x 3 + x 4 9 2, x 3 2, x 2 3 2, x + x 2 2, x 3 3 2, x + x 3 2, x 2 + x 3 5 2, x + x 4 9 2, x 2 + x 4 7 2, x 3 + x 4 7 2, x + x 3 + x } The next least core, whch wll be the nucleolus, s X 2 = {( 3 4, 33 8, 33 8, 3 2 )} wth ε 2 = 0. (c) The schedule s set up as follows: () Curly works from 9:00 to :52.5, () Larry works from :52.5 to :45, () Shemp works from :45 to 3:30, and (v) Moe works from 3:30 to 5:00.

8 6 MATRIX TWO-PERSON GAMES Defnton.2. Let C and D be sets. A functon f : C D R has at least one saddle pont (x, y ) wth x C and y D f f(x, y ) f(x, y ) f(x, y) for all x C, y D. Once agan we could defne the upper and lower values for the game defned usng the functon f, called a contnuous game, by v + = mn max f(x, y), and y D x C v = max mn x C y D f(x, y). You can check as before that v v +. If t turns out that v + = v we say, as usual, that the game has a value v = v + = v. The next theorem, the most mportant n game theory and extremely useful n many branches of mathematcs s called the von Neumann mnmax theorem. It gves condtons on f, C, and D so that the assocated game has a value v = v + = v. It wll be used to determne what we need to do n matrx games n order to get a value. In order to state the theorem we need to ntroduce some defntons. Defnton.2.2 A set C R n s convex f for any two ponts a, b C and all scalars λ [0, ], the lne segment connectng a and b s also n C,.e., for all a, b C, λa + ( λ)b C, 0 λ. C s closed f t contans all lmt ponts of sequences n C; C s bounded f t can be jammed nsde a ball for some large enough radus. A closed and bounded subset of Eucldean space s compact. A functon g : C R s convex f g(λa + ( λ)b) λg(a) + ( λ)g(b) for any a, b C, 0 λ. Ths says that the lne connectng g(a) wth g(b), namely {λg(a) + ( λ)g(b) : 0 λ }, must always le above the functon values g(λa + ( λ)b), 0 λ. The functon s concave f g(λa + ( λ)b) λg(a) + ( λ)g(b) for any a, b C, 0 λ. A functon s strctly convex or concave, f the nequaltes are strct. Fgure.4 compares a convex set and a nonconvex set. Also, recall the common calculus test for twce dfferentable functons of one varable. If g = g(x) s a functon of one varable and has at least two dervatves, then g s convex f g 0 and g s concave f g 0. Now the basc von Neumann mnmax theorem. Theorem.2.3 Let f : C D R be a contnuous functon. Let C R n and D R m be convex, closed, and bounded. Suppose that x f(x, y) s concave and y f(x, y) s convex. Then v + = mn max y D x C f(x, y) = max x C mn y D f(x, y) = v.

9 2 MATRIX TWO-PERSON GAMES We see that v = largest mn = and v + = smallest max =. Ths says that v + = v =, and so 2 2 Nm has v =. The optmal strateges are located as the (row,column) where the smallest max s and the largest mn s also. Ths occurs at any row for player I, but player II must play column 3, so =, 2, 3, j = 3. The optmal strateges are not at any row column combnaton gvng as the payoff. For nstance, f II plays column, then II wll play row and receve +. Column s not part of an optmal strategy. We have mentoned that the most that I can be guaranteed to wn should be less than (or equal to) the most that II can be guaranteed to lose, (.e., v v + ), Here s a quck verfcaton of ths fact. For any column j we know that for any fxed row, mn j a j a j, and so takng the max of both sdes over rows, we obtan v = max mn j a j max a j. Ths s true for any column j =,..., m. The left sde s just a number (.e., v ) ndependent of as well as j, and t s smaller than the rght sde for any j. But ths means that v mn j max a j = v +, and we are done. Now here s a precse defnton of a (pure) saddle pont nvolvng only the payoffs, whch bascally tells the players what to do n order to obtan the value of the game when v + = v. Defnton..2 We call a partcular row and column j a saddle pont n pure strateges of the game f a j a j a j, for all rows =,..., n and columns j =,..., m. (..) Lemma..3 A game wll have a saddle pont n pure strateges f and only f v = max Proof. If (..) s true, then v + = mn j max mn j a,j max a,j a j = mn j a,j mn j max a j = v +. (..2) a,j max mn a,j = v. j But v v + always, and so we have equalty throughout and v = v + = v = a,j. On the other hand, f v + = v then mn j max a,j = max mn a,j. j Let j be such that v + = max a,j and such that v = mn j a,j. Then a,j v = v + a,j, for any =,..., n, j =,..., m.

10 PROBLEMS 85 frm 2 s producton cost s C 2 (q 2 ) = 2q Fnd the proft functons and the Nash equlbrum quanttes of producton and profts. 4.3 Compare profts n the model wth uncertan costs and the standard Cournot model. Can you fnd a value of 0 < p < that maxmzes frm s profts? 4.4 Suppose that we consder the Cournot model wth uncertan costs but wth three possble costs, P rob(c 2 = c ) = r, =, 2, 3, where r 0, r + r 2 + r 3 =. Solve for the optmal producton quanttes. Fnd the explct producton quanttes when r = 2, r 2 = 8, r 3 = 3 8, Γ = 00, and c = 2, c =, c 2 = 2, c 3 = In the Stackelberg model compare the quantty produced, the proft, and the prces for frm assumng that frm 2 dd not exst so that frm s a monopolst. 4.6 Suppose that two frms have constant unt costs c = 2, c 2 = and Γ = 9 n the Stackelberg model. (a) What are the proft functons? (b) How much should frm 2 produce as a functon of q? (c) How much should frm produce? (d) How much, then, should frm 2 produce? 4.7 Set up and solve a Stackelberg model gven three frms wth constant unt costs c, c 2, c 3 and frm announcng producton quantty q. 4.8 In the Bertrand model show that f c = c 2 = c, then (p, p 2 ) = (c, c) s a Nash equlbrum. 4.9 Determne the entry deterrence level of producton for frm gven Γ = 00, a = 2, b = 0. How much proft s lost by settng the prce to deter a compettor? 4.0 We could make one more adjustment n the Bertrand model and see what effect t has on the model. What f we put a lmt on the total quantty that a frm can produce? Ths lmts the supply and possbly wll put a floor on prces. Let K Γ 2 denote the maxmum quantty of gadgets that each frm can produce and recall that D(p) = Γ p s the quantty of gadgets demanded at prce p. Fnd the proft functons for each frm. 4. Suppose that the demand functons n the Bertrand model are gven by q = D (p, p 2 ) = (a p + bp 2 ) + and q 2 = D 2 (p, p 2 ) = (a p 2 + bp ) +, where b > 0. Ths says that the quantty of gadgets sold by a frm wll ncrease f the prce set by the opposng frm s too hgh. Assume that both frms have a cost of producton c mn{p, p 2 }. (a) Show that the proft functons wll be gven by u (p, p 2 ) = D (p, p 2 )(p c), =, 2.

11 COALITIONS AND CHARACTERISTIC FUNCTIONS 22 Why? Well, v(23) = d because the car wll be sold for d, v() = M because the car s worth M to player, v(3) = d because player wll sell the car to player 3 for d > M, v(2) = c because the car wll be sold to player 2 for c > M, and so on. The reader can easly check that v s a characterstc functon. 3. A customer wants to buy a bolt and a nut for the bolt. There are three players but player owns the bolt and players 2 and 3 each own a nut. A bolt together wth a nut s worth 5. We could defne a characterstc functon for ths game as v(23) = 5, v(2) = v(3) = 5, v() = v(2) = v(3) = 0, and v( ) = 0. In contrast to the car problem v() = 0 because a bolt wthout a nut s worthless to player. 4. A small research drug company, labeled, has developed a drug. It does not have the resources to get FDA (Food and Drug Admnstraton) approval or to market the drug, so t consders sellng the rghts to the drug to a bg drug company. Drug companes 2 and 3 are nterested n buyng the rghts but only f both companes are nvolved n order to spread the rsks. Suppose that the research drug company wants $ bllon, but wll take $00 mllon f only one of the two bg drug companes are nvolved. The proft to a partcpatng drug company 2 or 3 s $5 bllon, whch they splt. Here s a possble characterstc functon wth unts n bllons: v() = v(2) = v(3) = 0, v(2) = 0., v(3) = 0., v(23) = 0, v(23) = 5, because any coalton whch doesn t nclude player wll be worth nothng. 5. A smple game s one n whch v(s) = or v(s) = 0 for all coaltons S. A coalton wth v(s) = s called a wnnng coalton and one wth v(s) = 0 s a losng coalton. For example, f we take v(s) = f S > n/2 and v(s) = 0 otherwse, we have a smple game that s a model of majorty votng. If a coalton contans more than half of the players, t has the majorty of votes and s a wnnng coalton. 6. In any bmatrx (A, B) nonzero sum game we may obtan a characterstc functon by takng v() = value(a), v(2) = value(b T ), and v(2) =sum of largest payoff par n (A, B). Checkng that ths s a characterstc functon s skpped. The next example works one out. EXAMPLE 5.2 In ths example we wll construct a characterstc functon for a verson of the prsoner s dlemma game n whch we assumed that there was no cooperaton. Now we wll assume that the players may cooperate and negotate. One form

12 246 COOPERATIVE GAMES so that x mnmzes the maxmum excess for any coalton S. When there s only one such allocaton x, t s the far allocaton. The problem s that there may be more than one element n the least core, then we stll have a problem as to how to choose among them. Remark: Maple Calculaton of the Least Core. The pont of calculatng the ε-core s that the core s not a suffcent set to ultmately solve the problem n the case when the core C(0) s () empty or (2) conssts of more than one pont. In case (2) the ssue, of course, s whch pont should be chosen as the far allocaton. The ε-core seeks to address ths ssue by shrnkng the core at the same rate from each sde of the boundary untl we reach a sngle pont. We can use Maple to do ths. The calculaton of the least core s equvalent to the lnear programmng problem Mnmze z subject to v(s) x(s) = v(s) S x z, for all S N. The characterstc functon need not be normalzed. So all we really need to do s to formulate the game usng characterstc functons, wrte down the constrants, and plug them nto Maple. The result wll be the smallest z = ε that makes C(ε ), as well as an mputaton whch provdes the mnmum. For example, let s suppose we start wth the characterstc functon v() = 0, =, 2, 3, v(2) = 2, v(23) =, v(3) = 0, v(23) = 5 2. The constrant set s the ε-core C(ε) = { x = (x, x 2, x 3 ) v(s) x(s) ε, S N} = { x ε, =, 2, 3, 2 x x 2 ε, x 2 x 3 ε, 0 x x 3 ε, x + x 2 + x 3 = 5 2 } The Maple commands used to solve ths are very smple: > wth(smplex): > cnsts:={-x<=z,-x2<=z,-x3<=z,2-x-x2<=z,-x2-x3<=z,-x-x3<=z, x+x2+x3=5/2}; > mnmze(z,cnsts); Maple produces the output x = 5 4, x 2 =, x 3 = 4, z = 4.

13 PROBLEM SOLUTIONS X = C( 0 ) = {x + x 2 = 9 0, 4 0 x, 2 0 x 2}. The next least core s X 2 = C( 4 ) = {( 20, 7 20, 2 20 )}. 5.7 The least core s the set C( ) = {x =, x 2 + x 3 =, x 2, x 3 2}. The nucleolus s the sngle pont {(, 2, 2 )} 5.8 For the least core ε = 2 : Least core = X = C( 2 ) = {x + x 2 = 3 2, x 3 + x 4 = 3 2, x, =, 2, 3, 4, 2 x 2 + x 3 3 2, x + x 4 3 2, x + x 3 5 4, x 2 + x 4 2, x + x 2 + x 3 3 2, x + x 2 + x 4 3 2, x + x 3 + x 4 3 2, x 2 + x 3 + x 4 3 2, x + x 2 + x 3 + x 4 = 3}. Next X 2 has ε 2 =. X 3 has ε 3 = 3, and nucleolus={( 3 4, 3 4, 3 4, 3 4 )}. 5.9 (a) The characterstc functon s the number of hours saved by a coalton. v() = 0, and v(2) = 4, v(3) = 4, v(4) = 3, v(23) = 6, v(24) = 2, v(34) = 2, v(23) = 0, v(24) = 7, v(34) = 7, v(234) = 8, v(234) = 3. (b) Nucleolus={( 3 4, 33 8, 33 8, 3 2 )} wth unts n hours. The least core s X = C( 3 2 ) = {x + x 2 + x 3 = 23 2, x 4 = 3 2, x + x 2 + x 3 + x 4 = 3, x + x 2 + x 4 7 2, x 2 + x 3 + x 4 9 2, x 3 2, x 2 3 2, x + x 2 2, x 3 3 2, x + x 3 2, x 2 + x 3 5 2, x + x 4 9 2, x 2 + x 4 7 2, x 3 + x 4 7 2, x + x 3 + x } The next least core, whch wll be the nucleolus, s X 2 = {( 3 4, 33 8, 33 8, 3 2 )} wth ε 2 = 0. (c) The schedule s set up as follows: () Curly works from 9:00 to :52.5, () Larry works from :52.5 to :45, () Shemp works from :45 to 3:30, and (v) Moe works from 3:30 to 5:00.

14 6 MATRIX TWO-PERSON GAMES Defnton.2. Let C and D be sets. A functon f : C D R has at least one saddle pont (x, y ) wth x C and y D f f(x, y ) f(x, y ) f(x, y) for all x C, y D. Once agan we could defne the upper and lower values for the game defned usng the functon f, called a contnuous game, by v + = mn max f(x, y), and y D x C v = max mn x C y D f(x, y). You can check as before that v v +. If t turns out that v + = v we say, as usual, that the game has a value v = v + = v. The next theorem, the most mportant n game theory and extremely useful n many branches of mathematcs s called the von Neumann mnmax theorem. It gves condtons on f, C, and D so that the assocated game has a value v = v + = v. It wll be used to determne what we need to do n matrx games n order to get a value. In order to state the theorem we need to ntroduce some defntons. Defnton.2.2 A set C R n s convex f for any two ponts a, b C and all scalars λ [0, ], the lne segment connectng a and b s also n C,.e., for all a, b C, λa + ( λ)b C, 0 λ. C s closed f t contans all lmt ponts of sequences n C; C s bounded f t can be jammed nsde a ball for some large enough radus. A closed and bounded subset of Eucldean space s compact. A functon g : C R s convex f g(λa + ( λ)b) λg(a) + ( λ)g(b) for any a, b C, 0 λ. Ths says that the lne connectng g(a) wth g(b), namely {λg(a) + ( λ)g(b) : 0 λ }, must always le above the functon values g(λa + ( λ)b), 0 λ. The functon s concave f g(λa + ( λ)b) λg(a) + ( λ)g(b) for any a, b C, 0 λ. A functon s strctly convex or concave, f the nequaltes are strct. Fgure.4 compares a convex set and a nonconvex set. Also, recall the common calculus test for twce dfferentable functons of one varable. If g = g(x) s a functon of one varable and has at least two dervatves, then g s convex f g 0 and g s concave f g 0. Now the basc von Neumann mnmax theorem. Theorem.2.3 Let f : C D R be a contnuous functon. Let C R n and D R m be convex, closed, and bounded. Suppose that x f(x, y) s concave and y f(x, y) s convex. Then v + = mn max y D x C f(x, y) = max x C mn y D f(x, y) = v.

15 22 MATRIX TWO-PERSON GAMES In fact, defne y = ϕ(x) as the functon so that f(x, ϕ(x)) = mn y f(x, y). Ths functon s well defned and contnuous by the assumptons. Also defne the functon x = ψ(y) by f(ψ(y), y) = max x f(x, y). The new functon g(x) = ψ(ϕ(x)) s then a contnuous functon takng ponts n [0, ] and resultng n ponts n [0, ]. There s a theorem, called the Brouwer fxed-pont theorem, whch now guarantees that there s a pont x [0, ] so that g(x ) = x. Set y = ϕ(x ). Verfy that (x, y ) satsfes the requrements of a saddle pont for f..3 MIXED STRATEGIES Von Neumann s theorem suggests that f we expect to formulate a game model whch wll gve us a saddle pont, n some sense, we need convexty of the sets of strateges, whatever they may be, and convexty-concavty of the payoff functon, whatever t may be. Now let s revew a bt. In most two-person zero sum games a saddle pont n pure strateges wll not exst because that would say that the players should always do the same thng. Such games, whch nclude 2 2 Nm, tc-tac-toe, and many others, are not nterestng when played over and over. It seems that f a player should not always play the same strategy, then there should be some randomness nvolved, because otherwse the opposng player wll be able to fgure out what the frst player s dong and take advantage of t. A player who chooses a pure strategy randomly chooses a row or column accordng to some probablty process that specfes the chance that each pure strategy wll be played. These probablty vectors are called mxed strateges, and wll turn out to be the correct class of strateges for each of the players. Defnton.3. A mxed strategy s a vector X = (x,..., x n ) for player I and Y = (y,..., y m ) for player II, where n m x 0, x = and y j 0, y j =. = The components x represent the probablty that row wll be used by player I, so x = P rob(i uses row ), and y j the probablty column j wll be used by player II, that s, y j = P rob(ii uses row j). Denote the set of mxed strateges wth k components by j= S k {(z, z 2,..., z k ) z 0, =, 2,..., k, k z = }. In ths termnology, a mxed strategy for player I s any element X S n and for player II any element Y S m. A pure strategy X S n s an element of =

16 PROBLEM SOLUTIONS 385 Once column 2 s gone, row may be dropped. Then X = (0, 4 Y = ( 2 7, 0, 5 7 ). 7, 3 7 ) and.23 v(a) = = E(X, Y ), but E(X, Y ) = 2x, where X = (x, x), 0 x, and t s not true that 2x < v(a) for all x n that range..24 (a) X = ( 5 22, 7 22 );(b) Y = ( 7 9, 2 9 );(c) Y = ( 6 0, 4 0 )..25 Any 3 8 λ 7 6 wll work for a convex combnaton of columns 2 and..26 Let max b = b k. Then x b b k = x (b b k ) = z snce x =. Now b b k for each, so z 0. Its maxmum value s acheved by takng x k = and x = 0, k. Hence max X x b b k = 0, whch says max X x b = b k = max b..27 Ths uses v = mn Y max E(, Y ) = max X mn j E(X, j)..28 By defnton of saddle and Now put them together. E(X 0, Y ) E(X, Y ) E(X, Y 0 ) E(X, Y 0 ) E(X 0, Y 0 ) E(X 0, Y )..29 (a) The gven strateges n the frst part are not optmal because max E(, Y ) = 3 9 and mn j E(X, j) = (b) The optmal Y s Y = ( 52 99, , 0, 99 )..30 Y = ( 5 7, 2 7, 0)..3 X = ( 8, 3 ), Y = ( 6, 5 48 ), v(a) =..32 X = ( 2 3, 3 ), Y = ( 2 3, 3, 0, 0), v(a) = 4 3.The best response for player I to Y = ( 4, 2, 8, 8 ) s X = (0, )..33 Snce XA = (0.28, , 0.27), the smallest of these s 0.27, so the best response s Y = (0, 0, )..34 Best responses are x(y) = (C y)/2, y(x) = (D x)/2, whch can be solved to gve x = (2C D)/3, y = (2D C)/3..35 max X E(X, Y n ) = E(X n, Y n ), mn Y E(X n, Y ) = E(X n, Y n+ ), n = 0,, Then E(X, Y n ) E(X n, Y n ) and E(X n, Y n+ ) E(X n, Y ),, X, Y.

17 CHAPTER 2 SOLUTION METHODS FOR MATRIX GAMES I returned, and saw under the sun, that the race s not to the swft, nor the battle to the strong,...; but tme and chance happeneth to them all. Ecclesates 9: 2. SOLUTION OF SOME SPECIAL GAMES Graphcal methods reveal a lot about exactly how a player reasons her way to a soluton, but t s not a very practcal method. Now we wll consder some specal types of games for whch we actually have a formula gvng the value and the mxed strategy saddle ponts. Let s start wth the easest possble class of games that can always be solved explctly and wthout usng a graphcal method Games Revsted We have seen that any 2 2 matrx game can be solved graphcally, and many tmes that s the fastest and best way to do t. But there are also explct formulas gvng the Game Theory: An Introducton, Frst Edton. By E.N. Barron Copyrght c 2008 John Wley & Sons, Inc. 55

18 56 SOLUTION METHODS FOR MATRIX GAMES value and optmal strateges wth the advantage that they can be run on a calculator or computer. Also the method we use to get the formulas s nstructve because t uses calculus. Each player has exactly two strateges, so the matrx and strateges look lke A = [ ] a a 2 a 2 a 22 player I :X = (x, x); player II :Y = (y, y). For any mxed strateges we have E(X, Y ) = XAY T, whch, wrtten out, s E(X, Y ) = xy(a a 2 a 2 + a 22 ) + x(a 2 a 22 ) + y(a 2 a 22 ) + a 22. Now here s the theorem gvng the soluton of ths game. Theorem 2.. In the 2 2 game wth matrx A, assume that there are no pure optmal strateges. If we set x a 22 a 2 =, y a 22 a 2 =, a a 2 a 2 + a 22 a a 2 a 2 + a 22 then X = (x, x ), Y = (y, y ) are optmal mxed strateges for players I and II, respectvely. The value of the game s v(a) = E(X, Y a a 22 a 2 a 2 ) =. a a 2 a 2 + a 22 Remarks.. The man assumpton you need before you can use the formulas s that the game does not have a pure saddle pont. If t does, you fnd t by checkng v + = v, and then fndng t drectly. You don t need to use any formulas. Also, when we wrte down these formulas, t had better be true that a a 2 a 2 + a 22 0, but f we assume that there s no pure optmal strategy, then ths must be true. In other words, t sn t dffcult to check that when a a 2 a 2 + a 22 = 0, then v + = v and that volates the assumpton of the theorem. 2. A more compact way to wrte the formulas and easer to remember s ] X = ( )A ] and Y = A [ ] ( )A [ ( )A [ value(a) = det(a) ( )A [ ] where [ A = ] a 22 a 2 a 2 a and det(a) = a a 22 a 2 a 2.

19 THE BASICS We need to defne a concept of optmal play that should reduce to a saddle pont n mxed strateges n the case B = A. It s a fundamental and far-reachng defnton due to another genus of mathematcs who turned hs attenton to game theory n the mddle twenteth century, John Nash. Defnton 3.. A par of mxed strateges (X S n, Y S m ) s a Nash equlbrum f E I (X, Y ) E I (X, Y ) for every mxed X S n and E II (X, Y ) E II (X, Y ) for every mxed Y S m. If (X, Y ) s a Nash equlbrum we denote by v A = E I (X, Y ) and v B = E II (X, Y ) as the optmal payoff to each player. Wrtten out wth the matrces, (X, Y ) s a Nash equlbrum f E I (X, Y ) = X AY T XAY T = E I (X, Y ), for every X S n, E II (X, Y ) = X BY T X BY T = E II (X, Y ), for every Y S m. Ths says that nether player can gan any expected payoff f ether one chooses to devate from playng the Nash equlbrum, assumng that the other player s mplementng hs or her pece of the Nash equlbrum. On the other hand, f t s known that one player wll not be usng hs pece of the Nash equlbrum, then the other player may be able to ncrease her payoff by usng some strategy other than that n the Nash equlbrum. The player then uses a best response strategy. In fact, the defnton of a Nash equlbrum says that each strategy n a Nash equlbrum s a best response strategy aganst the opponent s Nash strategy. Here s a precse defnton for two players. Defnton 3..2 A strategy X 0 S n s a best response strategy to a gven strategy Y 0 S m for player II, f E I (X 0, Y 0 ) = max X S n E I (X, Y 0 ). Smlarly, a strategy Y 0 S m s a best response strategy to a gven strategy X 0 S n for player I, f E II (X 0, Y 0 ) = max Y S m E II (X 0, Y ). In partcular, another way to defne a Nash equlbrum (X, Y ) s that X maxmzes E I (X, Y ) over all X S n and Y maxmzes E II (X, Y ) over all Y S m. X s a best response to Y and Y s a best response to X. If B = A, a bmatrx game s a zero sum two-person game and a Nash equlbrum s the same as a saddle pont n mxed strateges. It s easy to check that from the defntons because E I (X, Y ) = XAY T = E II (X, Y ). Note that a Nash equlbrum n pure strateges wll be a row and column j satsfyng a j a j and b j b j, =,..., n, j =,..., m.

20 2 2 BIMATRIX GAMES 25 y Nash pont m/m=2/5 Nash pont Nash pont r/r=3/5 x Nash Equlbra Fgure 3.3 Ratonal reacton sets for both players Ths s curous because the expected payoffs to each player are much less than they could get at the other Nash ponts. We wll see pctures lke Fgure 3.3 agan n the next secton when we consder an easer way to get Nash equlbra. Remark: A drect way to calculate the ratonal reacton sets for 2 2 games. Ths s a straghtforward dervaton of the ratonal reacton sets for the bmatrx game wth matrces (A, B). Let X = (x, x), Y = (y, y) be any strateges and defne f(x, y) = E I (X, Y ) and g(x, y) = E II (X, Y ). The dea s to fnd for a fxed 0 y, the best response to y. Accordngly, max f(x, y) = max xe I(, Y ) + ( x)e I (2, Y ) 0 x 0 x = x[e I (, Y ) E I (2, Y )] + E I (2, Y ) E I (2, Y ) at x = 0 f E I (, Y ) < E I (2, Y ); = E I (, Y ) at x = f E I (, Y ) > E I (2, Y ); E I (2, Y ) at any 0 < x < f E I (, Y ) = E I (2, Y ). Now we have to consder the nequaltes n the condtons. For example, E I (, Y ) < E I (2, Y ) My < m, M = a a 2 a 2 + a 22, m = a 22 a 2. If M > 0 ths s equvalent to the condton 0 y < m/m. Consequently, n the case M > 0, the best response to any 0 y < M/m s x = 0. All remanng cases

21 NONLINEAR PROGRAMMING METHOD FORNONZERO SUM TWO-PERSON GAMES 45 because XJn T = J m Y T =. But ths s exactly what t means to be a Nash pont. Ths means that (X, Y ) s a Nash pont f and only f X AY T J T n AY T, (X BY T )J m X B. We have already seen ths n Proposton Now suppose that (X, Y ) s a Nash pont. We wll see that f we choose the scalars p = E I (X, Y ) = X AY T and q = E II (X, Y ) = X BY T, then (X, Y, p, q ) s a soluton of the nonlnear program. To see ths, we frst show that all the constrants are satsfed. In fact, by the equvalent characterzaton of a Nash pont we just derved, we get X AY T J T n = p J T n AY T and (X BY T )J m = q J m X B. The rest of the constrants are satsfed because X S n and Y S m. In the language of nonlnear programmng, we have shown that (X, Y, p, q ) s a feasble pont. The feasble set s the set of all ponts that satsfy the constrants n the nonlnear programmng problem. We have left to show that (X, Y, p, q ) maxmzes the objectve functon f(x, Y, p, q) = XAY T + XBY T p q over the set of the possble feasble ponts. Snce every feasble soluton (meanng t maxmzes the objectve over the feasble set) to the nonlnear programmng problem must satsfy the constrants AY T pj T n and XB qj m, multply the frst on the left by X and the second on the rght by Y T to get XAY T pxj T n = p, XBY T qj m Y T = q. Hence, any possble soluton gves the objectve f(x, Y, p, q) = XAY T + XBY T p q 0. So f(x, Y, p, q) 0 for any feasble pont. But wth p = X AY T, q = X BY T, we have seen that (X, Y, p, q ) s a feasble soluton of the nonlnear programmng problem and f(x, Y, p, q ) = X AY T + X BY T p q = 0 by defnton of p and q. Hence ths pont (X, Y, p, q ) both s feasble and gves the maxmum objectve (whch we know s zero) over any possble feasble soluton and so s a soluton of the nonlnear programmng problem. Ths shows that f we have a Nash pont, t must solve the nonlnear programmng problem.

22 PROBLEM SOLUTIONS X = (, 0) Y = (, 0, 0) E I = 2, E II = X 2 = (, 0) X 3 = (0, ) Y 2 = ( 2, 0, 2 ) Y 3 = (0, 0, ) E I = 2, E II = E I =, E II = Take B = A. The Nash equlbrum s X = ( 5 8, 3 8, 0), Y = (0, 5 8, 3 8 ), and the value of the game s v(a) = The objectve functon s f(x, y, p, q) = 7x + 7y 6xy 6 p q wth constrants 2y p, 5y 3 p, 2x q, 5x 3 q, and 0 x, y. X = (, 0) Y = (0, ) E I =, E II = 2 X 2 = (0, ) Y 2 = (, 0) E I = 2, E II = X 3 = ( 2 3, 3 ) Y 3 = ( 2 3, 3 ) E I = 3, E II = X = ( 2, 3, ) 6, Y = ( 5 3, 5 3, ) 2 3,EI = 0 3, E II =. X 2 = ( 3 4, 0, 4) = Y2 wth payoffs E I = 5 4, E II = 3 2. X 3 = Y 3 = (0,, 0) The matrces are A = , B = One Nash equlbrum s X = (0.7, 0, 0, 0.29), Y = (0, 0, 0.74, 0.26). So Perre fres at 0 paces about 75% of the tme and wats untl 2 paces about 25% of the tme. Bll, on the other hand, wats untl 4 paces before he takes a shot but out of 4 tmes wats untl 2 paces (a) The Nash equlbra are X = (, 0) Y = (0, ) E I =, E II = 2 X 2 = (0, ) Y 2 = (, 0) E I = 2, E II = X 3 = ( 2 3, ) 3 Y 3 = ( 2 3, ) 3 E I = 3, E II = 3 They are all Pareto-optmal because t s mpossble for ether player to mprove ther payoff wthout smultaneously decreasng the other player s payoff, as you can see from the fgure:

23 400 PROBLEM SOLUTIONS Snce x s n the core, we have v(n) > n v() = = n v(n) v(n ) = nv(n) = = n v(n)(n ) < v(n ) = = nv(n) n x j = = j n v(n) x = n x = (n )v(n). = n v(n ), 5.8 Snce the game s nessental, v(n) = n = v(). It s obvous that x = (v(),..., v(n)) C(0). If there s another y C(0), y x, there must be one component y < v() or y > v(). Snce y C(0), the frst possblty cannot hold and so y > v(). Ths s true at any j component of y not equal to v(j). But then, addng them up gves n = y > n = v() = v(n), whch contradcts the fact that y C(0). 5.9 Suppose =. Then x + j x j = v(n) = v(n ) j x j, = and so x 0. But snce x = v() x 0, we have x = Let x C(0). Snce v(n ) x x n = v(n) x, we have x v(n) v(n ). In general, x v(n) v(n ), n. Now add these up to get v(n) = x δ < v(n), whch says C(0) =. 5.3 The core s C(0) = {(x, x 2, 6 x x 2 ) : 3 5 x 3, 2 x 2 2, 5 x + x 2 5}. The least core: ε = 62 5, C(ε ) = {( 7 5, 92 5, 5)} 77. ( 5.4 q = ) 3 0 = 2. The characterstc functon s v() = 0, v(2) = v(3) = v(23) =, v(23) =. 5.5 (b) To see why the core s empty, show frst that t must be true x + x 2 = 2, and x 3 + x 4 = 2. Then, snce x + x 2 + x 3 = 2 + x 3, we have x 3. Smlarly x 4. But then x 3 + x 4 2 and that s a contradcton. (c) A coalton that works s S = {2}.

24 PROBLEM SOLUTIONS X = C( 0 ) = {x + x 2 = 9 0, 4 0 x, 2 0 x 2}. The next least core s X 2 = C( 4 ) = {( 20, 7 20, 2 20 )}. 5.7 The least core s the set C( ) = {x =, x 2 + x 3 =, x 2, x 3 2}. The nucleolus s the sngle pont {(, 2, 2 )} 5.8 For the least core ε = 2 : Least core = X = C( 2 ) = {x + x 2 = 3 2, x 3 + x 4 = 3 2, x, =, 2, 3, 4, 2 x 2 + x 3 3 2, x + x 4 3 2, x + x 3 5 4, x 2 + x 4 2, x + x 2 + x 3 3 2, x + x 2 + x 4 3 2, x + x 3 + x 4 3 2, x 2 + x 3 + x 4 3 2, x + x 2 + x 3 + x 4 = 3}. Next X 2 has ε 2 =. X 3 has ε 3 = 3, and nucleolus={( 3 4, 3 4, 3 4, 3 4 )}. 5.9 (a) The characterstc functon s the number of hours saved by a coalton. v() = 0, and v(2) = 4, v(3) = 4, v(4) = 3, v(23) = 6, v(24) = 2, v(34) = 2, v(23) = 0, v(24) = 7, v(34) = 7, v(234) = 8, v(234) = 3. (b) Nucleolus={( 3 4, 33 8, 33 8, 3 2 )} wth unts n hours. The least core s X = C( 3 2 ) = {x + x 2 + x 3 = 23 2, x 4 = 3 2, x + x 2 + x 3 + x 4 = 3, x + x 2 + x 4 7 2, x 2 + x 3 + x 4 9 2, x 3 2, x 2 3 2, x + x 2 2, x 3 3 2, x + x 3 2, x 2 + x 3 5 2, x + x 4 9 2, x 2 + x 4 7 2, x 3 + x 4 7 2, x + x 3 + x } The next least core, whch wll be the nucleolus, s X 2 = {( 3 4, 33 8, 33 8, 3 2 )} wth ε 2 = 0. (c) The schedule s set up as follows: () Curly works from 9:00 to :52.5, () Larry works from :52.5 to :45, () Shemp works from :45 to 3:30, and (v) Moe works from 3:30 to 5:00.

25 GRAPHICAL SOLUTION OF 2 m AND n 2 GAMES 43 EXAMPLE.5 Let s consder A = Ths s a 4 2 game wthout a saddle pont n pure strateges snce v =, v + = 6. There s also no obvous domnance, so we try to solve the game graphcally. Suppose that player II uses the strategy Y = (y, y), then we graph the payoffs E(, Y ), =, 2, 3, 4, as shown n Fgure.0. Payoff to player II optmal pont E(4,Y) E(2,Y) y E(,Y) E(3,Y) Fgure.0 Mxed for player II versus 4 rows for player I. You can see the dffculty wth solvng games graphcally; you have to be very accurate wth your graphs. Carefully readng the nformaton, t appears that the optmal strategy for Y wll be determned at the ntersecton pont of E(4, Y ) = 7y 8( y) and E(, Y ) = y + 2( y). Ths occurs at the pont y = 5 9 and the correspondng value of the game wll be v(a) = 3. The optmal strategy for player II s Y = ( 5 9, 4 9 ). Snce ths uses only rows and 4, we may now drop rows 2 and 3 to fnd the optmal strategy for player I. In general, we may drop the rows (or columns) not used to get the optmal ntersecton pont. Often that s true because the unused rows are domnated, but not always. To see that here, snce and , we see that row 2 s domnated by a convex combnaton of rows and 4; so row 2 may be dropped. On the other hand, there s no λ [0, ] so that 5 7λ ( λ) and 6 8λ + 2( λ). Row 3 s not domnated by a convex combnaton of rows and 4, but t s dropped because ts payoff lne E(3, Y ) does not pass through the optmal pont.

26 44 MATRIX TWO-PERSON GAMES Consderng the matrx usng only rows and 4,we now calculate E(X, ) = x+7( x) and E(X, 2) = 2x 8( x) whch ntersect at (x = 5 6, v = 3 ). We obtan that row should be used wth probablty 5 6 and row 4 should be used wth probablty 6, so X = ( 5 6, 0, 0, 6 ). Agan, v(a) = 3. A verfcaton that these are ndeed optmal uses Theorem.3.7(c). We check that E(, Y ) v(a) E(X, j) for all rows and columns. Ths gves [ ] 2 [ ] 2 [ ] = and = Everythng checks. We end ths secton wth a smple analyss of a verson of poker, at least a small part of t. EXAMPLE.6 Ths s a modfed verson of the endgame n poker. Here are the rules. Player I s dealt a card that may be an ace or a kng. Player I sees the result but II does not. Player I may then choose to fold or bet. If I folds, he has to pay player II $. If I bets, player II may choose to fold or call. If II folds, she pays player I $. If player II calls and the card s a kng, then player I pays player II $2, but f the card comes up ace, then player II pays player I $2. Why wouldn t player I mmedately fold when he gets dealt a kng? It s the rule that I must pay II $ when I gets a kng and he folds. Player I s hopng that player II wll fold f I bets whle holdng a kng. Ths s the element of bluffng, because f II calls whle I s holdng a kng, then I must pay II $2. Fgure. s a graphcal representaton of the game. Now player I has four strateges: F F =fold on ace and fold on kng, F B=fold on ace and bet on Kng, BF =bet on ace and fold on kng, and BB =bet on ace and bet on kng. Player II has only two strateges, namely, F =fold or C =call. Assumng that the probablty of beng dealt a kng or an ace s 2 we may calculate the expected reward to player I and get the matrx as follows: I/II C F FF FB BF 2 0 BB

27 68 SOLUTION METHODS FOR MATRIX GAMES ball, whle II antcpates where the ball wll be ht. Suppose that II can return a ball ht rght 90% of the tme, a ball ht left 60% of the tme, and a ball ht center 70% of the tme. If II antcpates ncorrectly, she can return the ball only 20% of the tme. Score a return as + and not return as. Fnd the game matrx and the optmal strateges. 2.3 SYMMETRIC GAMES Symmetrc games are mportant classes of two-person games n whch the players can use the exact same set of strateges and any payoff that player I can obtan usng strategy X can be obtaned by player II usng the same strategy Y = X. The two players can swtch roles. Such games can be quckly dentfed by the rule that A = A T. Any matrx satsfyng ths s sad to be skew symmetrc. If we want the roles of the players to be symmetrc, then we need the matrx to be skew symmetrc. Why s skew symmetry the correct thng? Well, f A s the payoff matrx to player I, then the entres represent the payoffs to player I and the negatve of the entres, or A represent the payoffs to player II. So player II wants to maxmze the column entres n A. Ths means that from player II s perspectve, the game matrx must be ( A) T because t s always the row player by conventon who s the maxmzer; that s, A s the payoff matrx to player I and A T s the payoff to player II. So, f we want the payoffs to player II to be the same as the payoffs to player I, then we must have the same game matrces for each player and so A = A T. If ths s the case, the matrx must be square, a j = a j, and the dagonal elements of A, namely, a, must be 0. We can say more. In what follows t s helpful to keep n mnd that for any approprate sze matrces (AB) T = B T A T. Theorem 2.3. For any skew symmetrc game v(a) = 0 and f X s optmal for player I, then t s also optmal for player II. Proof. Let X be any strategy for I. Then E(X, X) = X A X T = X A T X T = (X A T X T ) T = X A X T = E(X, X). Therefore E(X, X) = 0 and any strategy played aganst tself has zero payoff. Let (X, Y ) be a saddle pont for the game so that E(X, Y ) E(X, Y ) E(X, Y ), for all strateges (X, Y ). Then for any (X, Y ), we have E(X, Y ) = X A Y T = X A T Y T = (X A T Y T ) T = Y A X T = E(Y, X). Hence, from the saddle pont defnton, we obtan E(X, Y ) = E(Y, X) E(X, Y ) = E(Y, X ) E(X, Y ) = E(Y, X ). Then E(Y, X) E(Y, X ) E(Y, X ) = E(Y, X) E(Y, X ) E(Y, X ).

28 SYMMETRIC GAMES 69 But ths says that Y s optmal for player I and X s optmal for player II and also that E(X, Y ) = E(Y, X ) = v(a) = 0. EXAMPLE 2.5 General Soluton of 3 3 Symmetrc Games. For any 3 3 symmetrc game we must have A = 0 a a 0 b c. b c 0 Any of the followng condtons gves a pure saddle pont:. a 0, b 0 = saddle at (, ) poston, 2. a 0, c 0 = saddle at (2, 2) poston, 3. b 0, c 0 = saddle at (3, 3) poston. Here s why. Let s assume that a 0, c 0. In ths case f b 0 we get v = max{mn{a, b}, 0, c} = 0 and v + = mn{max{ a, b}, 0, c} = 0, so there s a saddle n pure strateges at (2, 2). All cases are treated smlarly. To have a mxed strategy, all three of these must fal. We next solve the case a > 0, b < 0, c > 0 so there s no pure saddle and we look for the mxed strateges. Let player I s optmal strategy be X = (x, x 2, x 3 ). Then E(X, ) = ax 2 bx 3 0 = v(a) E(X, 2) = ax cx 3 0 E(X, 3) = bx + cx 2 0 Each one s nonnegatve snce E(X, Y ) 0 = v(a), for all Y. Now, snce a > 0, b < 0, c > 0 we get x 3 a x 2 b, x c x 3 a, x 2 b x c so x 3 a x 2 b x c x 3 a, and we must have equalty throughout. Thus, each fracton must be some scalar λ, and so x 3 = a λ, x 2 = b λ, x = c λ. Snce they must sum to one, λ = /(a b + c). We have found the optmal strateges X = Y = (c λ, b λ, a λ). The value of the game, of course s zero. For example, the matrx A =

29 8 MATRIX TWO-PERSON GAMES tree, whch s nothng more than a pcture of what happens at each stage of the game where a decson has to be made. The numbers at the end of the branches are the payoffs to player I. The number 2, for example, means that the net gan to player I s $500 because player II had to pay $000 for the ablty to pass and they splt the pot n ths case. The crcled nodes are spots at whch the next node s decded by chance. You could even consder Nature as another player. We analyze the game by frst convertng the tree to a game matrx whch, n ths example becomes I I/II II II2 II3 II I To see how the numbers n the matrx are obtaned, we frst need to know what the pure strateges are for each player. For player I, ths s easy because she makes only one choce and that s pass (I2) or spn (I). For player II, II s the strategy; f I passes, then spn, but f I spns and survves, then pass. So, the expected payoff 2 to I s I aganst II : 5 ( ) ( ) = 4, and I2 aganst II : 5 6 ( 2) + 6 () = 3 2. Strategy II3 says the followng: If I spns and survves, then spn, but f I passes, then spn and fre. The expected payoff to I s I aganst II3 : 5 ( (0) + ) 6 () + 6 ( ) = 36, and I2 aganst II3 : 5 6 ( 2) + 6 () = 3 2. The remanng entres are left for the reader. The pure strateges for player II are summarzed n the followng table. II If I2, then S; If I, then P. II2 If I2, then P; If I, then P. II3 If I, then S; If I2, then S. II4 If I, then S; If I2, then P. 2 Ths uses the fact that f X s a random varable takng values x, x 2,..., x n wth probabltes p, p 2,..., p n, respectvely, then EX = n = x p. In I aganst II, X s 2 wth probablty 5 6 and wth probablty. See the appendx for more. 6

30 THE BASICS 9 Ths s actually a smple game to analyze because we see that player II wll never play II, II2, or II4 because there s always a strategy for player II n whch II can do better. Ths s strategy II3, whch stpulates that f I spns and survves the shot, then II should spn, whle f I passes, then II should spn and shoot. If I passes, II gets 36 and I loses 36. If I spns and shoots, then II gets 3 2 and I loses 3 2. The larger of these two numbers s 36, and so player I should always spn and shoot. Consequently, player II wll also spn and shoot. The dotted lne n Fgure.3 ndcates the optmal strateges. The key to these strateges s that no sgnfcant value s placed on survvng. Remark. Extensve form games can take nto account nformaton that s avalable to a player at each decson node. Ths s an mportant generalzaton. Extensve form games are a topc n sequental decson theory, a second course n game theory. Fnally, we present an example n whch s t clear that randomzaton of strateges must be ncluded as an essental element of games. EXAMPLE.6 Evens or Odds. In ths game, each player decdes to show one, two, or three fngers. If the total number of fngers shown s even, player I wns + and player II loses. If the total number of fngers s odd, player I loses, and player II wns +. The strateges n ths game are smple: decdng how many fngers to show. We may represent the payoff matrx as follows: Evens Odds I/II The row player here and throughout ths book wll always want to maxmze hs payoff, whle the column player wants to mnmze the payoff to the row player, so that her own payoff s maxmzed (because t s a zero sum game). The rows are called the pure strateges for player I, and the columns are called the pure strateges for player II. The followng queston arses: How should each player decde what number of fngers to show? If the row player always chooses the same row, say, one fnger, then player II can always wn by showng two fngers. No one would be stupd enough to play lke that. So what do we do? In contrast to 2 2 Nm or Russan roulette, there s no obvous strategy that wll always guarantee a wn for ether player. Even n ths smple game we have dscovered a problem. If a player always plays the same strategy, the opposng player can wn the game. It seems