Game Theory. Lecture Notes By Y. Narahari. Department of Computer Science and Automation Indian Institute of Science Bangalore, India February 2008

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1 Game Theory Lecture Notes By Y. Narahar Department of Computer Scence and Automaton Indan Insttute of Scence Bangalore, Inda February 2008 Chapter 10: Two Person Zero Sum Games Note: Ths s a only a draft verson, so there could be flaws. If you fnd any errors, please do send emal to har@csa.sc.ernet.n. A more thorough verson would be avalable soon n ths space. A two person zerosum game s of the form {1,2},S 1,S 2,u 1, u 1. Note that when a player tres to mze her payoff, she s also smultaneously mzng payoff of the other player. For ths reason, these games are also called strctly compettve games. Player 1 s usually called the row player and player 2 s called the column player. Let S 1 = {s 11,s 12,...,s 1m } and S 2 = {s 21,s 22,...,s 2n }. Wthout any confuson, we wll assume from now on that S 1 = {1,2,...,m} and S 2 = {1,2,...,n}. Example 1: Matchng Pennes Consder the standard matchng pennes game, whose payoff matrx s gven by: 2 1 H T H 1, 1 1,+1 T 1, +1 1, 1 Snce u 2 (s 1,s 2 ) = u 1 (s 1,s 2 ) s 1 S 1, s 2 S 2, such a payoff matrx can also be specfed by a smpler matrx A where a = u 1 (,). For example, the matchng pennes game can be represented as [ ] 1 1 A = 1 1 Notes Note 1: Snce the payoffs n a fnte two person zerosum game can be completely descrbed by a sngle matrx, namely the matrx that represents u 1 (,), such a game s aptly called a matrx game. Note 2: An mmedate generalzaton of a zerosum game s a constant sum game: ({1,2},S 1,S 2,u 1,u 2 ) such that u 1 (s 1,s 2 ) + u 2 (s 1,s 2 ) = C, s 1 S 1 ;s 2 S 2 wth C a gven constant. Most results that 1

2 hold for zerosum games also hold for constant sum games. Note 3: The above game does not have a pure strategy Nash equlbrum. Example 2: A Zerosum Game wth a Pure Strategy Nash Equlbrum Consder the followng zerosum game. 1 \ In ths case, t s easy to see that the profle (1, 1) s a pure strategy Nash equlbrum. Defnton: Saddle Pont of a Matrx Gven a matrx A = [a ], the element a s called a saddle pont of A f a a l l = 1,...,n a a k k = 1,...,m That s, the element a s smultaneously a mum n ts row and a mum n ts column. Proposton: For a matrx game wth payoff matrx A, a s a saddle pont f and only f the outcome (, ) s a pure strategy Nash equlbrum. Proof: Let a be a saddle pont a s a row mum and a s a column mum a s a row mum and +a s a column mum The column player s playng a best response w.r.t. strategy of the row player and the row player s playng a best response w.r.t. strategy of the column player. (,) s a Nash equlbrum. The followng theorem gves a necessary and suffcent condton for the exstence of a pure strategy Nash equlbrum or saddle pont. Theorem: In a matrx A = [a ], let u R u C = = a Then the matrx A has a saddle pont f and only f u R = u C. The followng proposton gves a useful property of saddle ponts. a Proposton: If n a matrx A, the elements a and a hk are both saddle ponts, then a k and a h are also saddle ponts. 2

3 Examples: Saddle Ponts For the matrx game (matchng pennes), A = [ ] u R u C = = For the matrx game wth payoff matrx A = a = { 1, 1} = 1 a = {+1,+1} = u R u C = a = {1, 1, 2} = 1 = a = {1,2,2} = 1 Therefore u R = u C wth a 11 as the saddle pont. Mxed Strateges n Matrx Games Let x = (x 1,x 2,...,x n ) and y = (y 1,...,y m ) be the mxed strateges of the row player and the column player respectvely. Note that a s player 1 s payoff when the row player chooses row and column player chooses column wth probablty 1. The correspondng payoff for the column player s a. The expected payoff to the row player wth the above mxed strateges x and y s gven by: = u 1 (x,y) = a x y =1 = xay where x = (x 1,...,x m ); y = (y 1,...,y n ) T ;A = [a ] The expected payoff to column player = xa y. When the row player plays x, he assures hmself of an expected payoff xay The row player should therefore look for a mxed strategy x that mzes the above. That s, an x such that xay 3

4 In other words, an optmal strategy for row player s to do mzaton. Note that the row player chooses a mxed strategy that s best for her on the assumpton that whatever she does, the column player wll choose an acton that wll hurt her (row player) as much as possble. Ths s a a drect consequence of ratonalty and the fact that the payoff for each player s the negatve of the other player s payoff. Smlarly, when the column player plays y, he assures hmself of a payoff That s, he assures hmself of losng no more than = xay = xay xay The column player s optmal strategy should be to mze ths loss: Ths s called mzaton. An Important Lemma xay Ths lemma asserts that when the row player plays x, among the most effectve reples y of the column player, there s always at least one pure strategy. Symbolcally, Proof: For a gven, the summaton xay = a x a x gves the payoff to the row player when she plays x = (x 1,...,x m ) and the column player player the pure strategy y. That s, a x = u 1 (x,y ) Therefore a x gves the mum payoff that the row player gets when she plays x and when the column player plays only pure strateges. Snce a pure strategy s a specal case of mxed strateges, we have a x xay (1) 4

5 On the other hand, ( m ) xay = y a x =1 ) r y ( a x =1 = a x snce y = 1 =1 Therefore, we have: From (1) and (2), we have, Smlarly, t can be shown that xay a x y (S 2 ); x (S 1 ) xay a x (2) xay = a x xay = a y =1 From the above lemma, we can descrbe the optmzaton problems of the row player and column players as follows. Row Player s Optmzaton Problem (Maxmzaton) subect to mze a x x = 1 x 0 = 1,2,...,m Call the above problem P 1. Note that ths s equvalent to xay 5

6 Column Player s Optmzaton Problem (Mnmzaton) subect to mze a y =1 y = 1 =1 y 0 = 1,...,n Call the above problem P 2. Note that ths s equvalent to xay We now show that the problems P 1 and P 2 are equvalent to approprate lnear programs. Proposton: The followng problems are equvalent. Maxmze m subect to x = 1 P 1 a x x 0 = 1,...,m Maxmze z subect to z a x 0 = 1,...,n x = 1 LP 1 x 0 = 1,...,m Proof: Note that P 1 s a mzaton problem and therefore by lookng at the constrants z a x 0 = 1,2,...,n any optmal soluton z wll satsfy the equalty n the above constrant. That s, z = a x for some {1,...,n} 6

7 Let be one such value of. Then z = a x Because z s a feasble soluton of LP 1, we have a x m a x = 1,...,n Ths means a x = a x If not, we have z < a x = 1,2,...,n If ths happens, we can fnd a feasble soluton ẑ such that ẑ > z. Such a ẑ s precsely the one for whch equalty wll hold. But snce z s a mal value, the exstence of ẑ > z s a contradcton! The summary so far s: The row player s optmal strategy s mzaton: Ths s equvalent to the followng problem: xay m mze a x subect to P 1 x = 1 x 0 = 1,2,...,n The above s equvalent to the followng LP: mze z subect to z a x 0 = 1,...,n LP 1 x = 1 x 0 The column player s optmal strategy s mzaton: Ths s equvalent to: xay mze a y =1 subect to P 2 y = 1 y 0 = 1,...,n 7

8 The above s equvalent to the followng LP: mze w subect to w a x 0 = 1,...,m LP 2 =1 y = 1 y 0 = 1,...,n Mn Theorem Ths result s one of the mportant landmarks n the ntal decades of game theory. Ths result was proved by von Neumann n 1928 usng the Brower s fxed pont theorem. Later, he and Morgenstern provded an elegant proof of ths theorem usng LP dualty. The key mplcaton of the theorem s the exstence of a mxed strategy Nash equlbrum n any matrx game. Theorem: For every (m n) matrx A, there s a stochastc row vector x = (x 1,...,x m) and a stochastc column vector y = (y 1,...,y n) T such that x Ay = xay Proof: Gven a matrx A, we have derved lnear programs LP 1, LP 2. LP 1 represents the optmal strategy of row player whle LP 2 represents the optmal strategy of column player. Frst we make the observaton that the lnear program LP 2 s the dual of the lnear program LP 1. We now nvoke the strong dualty theorem whch says: If an LP has an optmal soluton, then ts dual also has an optmal soluton; moreover the optmal value of the dual s the same as the optmal value of the orgnal (prmal) LP. Please refer the appendx for a quck prmer on LP dualty. To apply the strong dualty theorem n the current context, we frst observe that the problem P 1 has an optmal soluton by the very nature of the problem. Snce LP 1 s equvalent to the problem P 1, the mmedate mplcaton s that LP 1 has an optmal soluton. Thus we have two LPs LP 1 and LP 2 whch are duals of each other and LP 1 has an optmal soluton. Then by the strong dualty theorem, LP 2 also has an optmal soluton and the optmal value of LP 2 s the same as the optmal value of LP 1. Let z,x 1,...,x m be an optmal soluton of LP 1. Then, we have z = a x By the feasblty of the optmal soluton n LP 1, we have for some {1,...,n} Ths mples that a x a x for = 1,...,n a x = a x 8

9 Thus = x Ay z = x Ay Smlarly, let w,y 1,...,y n be an optmal soluton of LP 2. Then (by the lemma) w = a y for some {1,...,m} =1 By the feasblty of the optmal soluton n LP 2, we have a y a y for = 1,2,...,m Therefore =1 a y =1 =1 = a y =1 = xay w = xay (by Lemma) By the strong dualty theorem, the optmal values of the prmal and the dual are the same and therefore z = w. Ths means x Ay = xay Ths proves the theorem. We now show that the mxed strategy profle (x,y ) s n fact a mxed strategy Nash equlbrum of the matrx game A. For ths, consder That s, x Ay xay x (S 1 ). Ths mples Further That s, x Ay x Ay y (S 2 ). Ths mples x Ay x Ay = xay xay x (S 1 ) u 1 (x,y ) u 1 (x,y ) x (S 1 ) x Ay xay = x (S 2 ) x Ay x Ay y (S 2 ) u 2 (x,y ) u 2 (x,y) y (S 2 ) Thus (x,y ) s a mxed strategy Nash equlbrum or a randomzed saddle pont. Ths means the theorem guarantees the exstence of a mxed strategy Nash equlbrum for any matrx game. 9

10 A Necessary and Suffcent Condton for a Nash Equlbrum We now state and prove a key theorem that provdes necessary and suffcent condtons for a mxed strategy profle to be a Nash equlbrum n matrx games. Theorem: Gven a two player zerosum game ({1,2},S 1,S 2,u 1, u 1 ) a mxed strategy profle (x,y ) s a Nash equlbrum f and only f and Furthermore x y arg x (S 1 ) arg y (S 2 ) xay xay u 1 (x,y ) = u 2 (x,y ) = xay = xay Proof: Frst we prove the necessty. Suppose (x,y ) s a Nash equlbrum. Then Also, note that u 1 (x,y ) u 1 (x,y ) x (S 1 ) u 1 (x,y ) = u 1(x,y ) (3) u 1 (x,y ) u 1(x,y) x (S 1 ) u 1(x,y ) { u 1(x,y) snce f(x) g(x) x x f(x) x g(x). From (3) and (4), we have u 1 (x,y ) On smlar lnes, usng u 1 (x,y ) = u 2 (x,y ),, we can show that We have u 1 (x,y ) u 1 (x,y ) = u 2 (x,y ) = { u 2(x,y)} } (4) u 1(x,y) (5) u 1(x,y) (6) 10

11 = 2(x,y)} = 1(x,y) u 1 (x,y ) = 1(x,y) We know that u 1(x,y) u 1(x,y) = u 1 (x,y ) by (5) Smlarly we know that (3) and (6) mply that (4) and (7) mply that From the above two expressons, we have u 1(x,y) u 1 (x,y ) = u 1 (x,y ) = u 1(x,y ) = u 1 (x,y ) u 1(x,y) u 1(x,y) x y arg x (S 1 ) arg y (S 2 ) u 1(x,y) u 1(x,y) Ths completes the necessty part of the proof. To prove the suffcency, we are gven that (8) and (9) are satsfed and we have to show that (x,y ) s a Nash equlbrum. Ths s left to the reader to prove. The crucal aspect whch s requred for provng the suffcency s the exstence of a mxed strategy Nash equlbrum, whch s guaranteed by the theorem. Appendx: A quck Prmer on LP Dualty Frst we consder an example of an LP n canoncal form: subect to The dual of ths s the LP s gven by mze 6x 1 + 8x 2 10x 3 3x 1 + x 2 x 3 4 5x 1 + 2x 2 7x 3 7 x 1,x 2,x 3 0 mze 4w 1 + 7w 2 11

12 subect to 3w 1 + 5w 2 6 w 1 + 2w 2 8 w 1 7w 2 10 w 1,w 2 0 In general, gven the prmal LP n canoncal form s: The dual of the above prmal s gven by c = [c 1...c n ] x = [x 1 x n ] T A = [a ] m n b = [b 1 b m ] T w = [w 1 w m ] mze cx subect to Ax b x 0. mze wb subect to wa c w 0. A prmal LP n standard form s The dual of the above prmal s: mze cx subect to Ax = b x 0. mze wb subect to wa c w unrestrcted If we consder a mzaton problem, then correspondng to the prmal: we have the dual gven by mze cx subect to Ax b x 0. mze wb subect to wa c w 0 It s a smple matter to show that the dual of the dual of a (prmal) problem s the orgnal (prmal) problem tself. We now state a few mportant results concernng dualty, whch are relevant to the current context. 12

13 Weak Dualty Theorem: If the prmal s a mzaton problem, then the value of any feasble prmal soluton s greater than or equal to the value of any feasble dual soluton. If the prmal s a mzaton problem, then the value of any feasble prmal soluton s less than or equal to the value of any feasble dual soluton. If x 0 s a feasble prmal soluton and w 0 s a feasble dual soluton, and cx 0 = w 0 b, then x 0 s an optmal soluton of the prmal problem and w 0 s an optmal soluton of the dual problem. Strong Dualty Theorem: Between a prmal and ts dual, f one of them has an optmal soluton then the other also has an optmal soluton and the values of the optmal solutons are the same. Note that ths s the key result whch was used n provng the theorem. Fundamental Theorem of Dualty: Gven a prmal and ts dual, exactly one of the followng statements s true. Problems 1. Both possess optmal soluton x and w wth cx = w b. 2. One problem has unbounded obectve value n whch case the other must be nfeasble. 3. Both problems are nfeasble. 1. (Problem taken from the book by Jones [1]). Construct a two player zero sum game wth S 1 = {A,B,C}, S 2 = {X,Y,Z} wth value = 1 2 and such that the set of optmal strateges for the row player s exactly the set { 3 (α,1 α,0); 8 α 5 } 8 2. (Problem taken from the book by Osborne and Rubnsten [2]). Let G be a two player zero sum game that has a pure strategy Nash equlbrum. (a) Show that f some of the player 1 s payoffs n G are ncreased n such a way that the resultng game G s strctly compettve then G has no equlbrum n whch player 1 s worse off than she was n an equlbrum of G. (Note that G may have no equlbrum at all.) (b) Show that the game that results f player 1 s prohbted from usng one of her actons n G does not have an equlbrum n whch player 1 s payoffs s hgher than t s n an equlbrum of G. (c) Gve examples to show that nether of the above propertes necessarly holds for a game that s not strctly compettve. 3. (Problem taken from the book by Osborne and Rubnsten [2]). Army A has a sngle plane wth whch t can strke one of three possble targets. Army B has one ant-arcraft gun that can be assgned to one of the targets. The value of target k s v k, wth v 1 > v 2 > v 3 > 0. Army A can destroy a target only f the target s undefended and A attacks t. Army A wshes to mze the expected value of the damage and army B wshes to mze t. Formulate the stuaton as a (strctly compettve) strategc game and fnd ts mxed strategy Nash equlbra. 4. For the followng two player zero sum game, wrte down the prmal and dual LPs and compute all Nash equlbra. 13

14 A B A 2, -2 3,-3 B 4,-4 1, For the followng two player zero sum game, wrte down the prmal and dual LPs and compute all Nash equlbra. A B C A 2, -2 3,-3 1,-1 B 4,-4 1, -1 2,-2 C 4,-4 1, -1 3,-3 6. Gven a two player zero sum game wth 3 pure strateges for each player, whch numbers among {0, 1,..., 9} cannot be the total number of pure strategy Nash equlbra for the game? Justfy your answer. 7. In a matrx A = [a ], f two elements a and a hk are saddle ponts, then show that a k and a h are also saddle ponts. 8. Gven a matrx A = [a ], defne u R = u C = Show that A has a saddle pont f and only f u R = u R. 9. For the followng matrx game, formulate an approprate LP and compute all mxed strategy equlbra A = Show that the followng holds for any two player game. x (s 1 ) xay y (s 2 ) a a y (s 2 ) xay x (s 1 ) 11. Show that the payoffs n Nash equlbrum of a symmetrc matrx game ( matrx game wth symmetrc payoff matrx) wll be equal to zero for each player. 12. Complete the suffcency part of the theorem that provdes a necessary and suffcent condton for a mxed strategy profle (x,y ) to be a Nash equlbrum n a matrx game. 14

15 To Probe Further Two person zerosum games provde, perhaps, the smplest class of games whch were studed durng the ntal years of game theory. John von Neumann s credted wth the theorem, whch he proved n 1928 [3] by nvokng the Brower s fxed pont theorem. The classc book by Neumann and Morgenstern [4] contaned a detaled exposton of matrx games, ncludng the LP dualty based approach to the theorem. The book by Myerson [5] and the book on lnear programg by Chavatal [6] have nspred the exposton n ths chapter. Other books whch can be consulted are the ones by Osborne [7], by Rapoport [8], and by Straffn [9]. References [1] Jones. Game Theory. John Wley & Sons, [2] Martn J. Osborne and Arel Rubnsten. A Course n Game Theory. Oxford Unversty Press, [3] John von Neumann. Zur theore der gesellschaftsspele. Annals of Mathematcs, 100: , [4] John von Neumann and Oskar Morgenstern. Theory of Games and Economc Behavor. Prnceton Unversty Press, [5] Roger B. Myerson. Game Theory: Analyss of Conflct. Harvard Unversty Press, [6] Vasek Chvatal. Lnear Programg. W.H. Freeman & Company, [7] Martn J. Osborne. An Introducton to Game Theory. The MIT Press, [8] Anatol Rapoport. Two Person Game Theory. Dover Publcatons, Inc., New York, USA, [9] Phlp D. Straffn Jr. Game Theory and Strategy. The Mathematcal Assocaton of Amerca,