Matrix Games (or Two-Player Zero-sum Games)

Transcription

1 Lecture #2 Notes Summary Zero-Sum matrx games: Saddle ponts, Mnmax theorem Bmatrx games: Classcal examples, graphcal method to fnd mxed equlbra of 2 2 games A game n normal form wth two players s called a bmatrx game. The reason s that player when P 1 has m strateges and player P 2 has n strateges, the payoff functon π can be gven by a par of m n matrces. The game of matchng cons presented n the frst lecture P 1 /P ) 1 1, 1) 1, 1) 5 5, 5) 5, 5) ) 1 1 can be represented by a par of matrces A, B), where A = contans the payoffs π 1 of the frst 5 5 ) 1 1 player and B = contans the payoffs of the second player. 5 5 #1 Matrx Games or Two-Player Zero-sum Games) If A + B = 0 as ths s the case for the game of matchng cons), the game s called a zero-sum game. So we only need to gve the matrx A to defne the payoff functon π of ths game. Defnton 1 Matrx game). A matrx game wth the matrx A of sze m n s a zero-sum game wth two players, such that: the player P 1 has m strateges: Σ 1 = {S 1 1,..., S 1 m} the player P 2 has n strateges: Σ 2 = {S 2 1,..., S 2 n} the payoff functon s defned by π 1 S 1, S2 ) = A, and π 2 S 1, S2 ) = A,). We say that P 1 s the row player or the max player she selects a row of the matrx and wants to maxmze A, ), whle P 2 s the column player or the mn player she selects a column of the matrx and wants to mnmze A, ). Proposton 1. Let A be a matrx game. If player P 1 commts to the mxed strategy p m and player P 2 commts to the mxed strategy q n, then the expected payoff of player P 1 s π 1 p, q) = m =1 =1 n p q A, = p T Aq. Proof. Use the defnton of expected payoffs and rewrte the expresson wth matrx notaton. For matrx games, a pure Nash equlbrum s called a saddle pont of a matrx. Defnton 2 Saddle pont). Let A be a m n matrx. An entry A p,q of A s a saddle pont of A s A p,q s smultaneously a maxmum n the column q and a mnmum n the row p. Page 1 of 5

2 Saddle ponts of followng matrces are boxed: ) #1 Theorem 2. If M, and M k,l are saddle ponts of M, then M,l and M k, are also saddle ponts and M, = M k,l = M,l = M k,. Proof. M, s a maxmum of column, so M k, M,, and t s also a mnmum of row, so M, M,l. M k,l s a maxmum of column l, so M,l M k,l, and t s also a mnmum of row k, so M k,l M k,. Thus, we have M k, M, M,l M k,l M k,, whch shows that these 4 values are equal. In partcular, M,l = M, s a mnmum n row, and M,l = M k,l s a maxmum n column l, so M,l s a saddle pont. Smlarly we can see that M k, s a saddle pont. Defnton 3 maxmn values). The maxmn and mnmax values of M are defned respectvely as µ r M) = max mn M, A strategy S 1 A strategy S 2 µ c M) = mn max M, of the row player s called maxmn pure strategy f mn of the column player s called mnmax pure strategy f max M, = µ r M). M, = µ c M). Theorem 3. For any matrx M, µ r M) µ c M). Moreover µ r M) = µ c M) f and only f M has a saddle pont. Proof. For every l {1,..., n}, we have µ r M) = max mn M, max M,l. Hence, µ r M) mn l max M,l = µ c M). Now, let M p,q be a saddle pont. We have max M,q = M p,q and so µ c M) M p,q. Smlarly, mn M p, = M p,q mples µ r M) M p,q. So we have µ c M) µ r M), whch proves the frst sde of the equvalence. Conversely, assume that µ c M) = µ r M). Choose a maxmn pure strategy wth the ndex p and a mnmax pure strategy wth the ndex q. We have µ r M) = mn M p,, and let l be an ndex such that M p,l = µ r M) = µ c M). Snce the column q s mnmax, we have µ c M) = max M,q. Thus M p,l = max M,q M p,q, but l has be chosen so that M p,l s a mnmum n ts row, so M p,l = M p,q and M p,q s also a mnmum n ts row. Fnally, M p,q = M p,l = max M,q s a maxmum n ts column, and so M p,q s a saddle pont. We are now gong to defne the counterpart of maxmn values for mxed strateges: Page 2 of 5

3 Defnton 4 row and column values). The row and column values of a matrx M R m n are defned respectvely as v r M) = max mn p T Mq = max mn p T M ) v c M) = mn max p T ) Mq = mn max Mq A mxed strategy p of the row player s called optmal f mn p T M ) = v rm). ) A mxed strategy q of the column player s called optmal f max Mq = v cm). Theorem 4. Let p and q be optmal strateges of the row and column players for a matrx game M. Then, µ r M) v r M) p T M q v c M) µ c M). Proof. cf. Exercse 2 of the Worksheet #2. Theorem 5 mnmax theorem). Let M be a m n matrx game. Then the row and the column players have optmal mxed strateges p and q, and the row and column values of the game are equal: v r M) = p T M q = v c M). The common row and column values defne the value of the game, whch we denote by vm). Proof. We know from the exstence theorem of Nash cf. frst Lecture) that a Nash equlbrum p, q ) exsts. So q s a best response to p,.e. p T Mq = mn q n p T Mq. Combnng ths equalty wth mn p T Mq max mn p T Mq = v r M) we obtan p T Mq v r M). Smlarly, p s a best response to q yelds p T Mq v c M). Now, note that optmal strateges p and q for the row and column player exst optmzaton of a contnuous functon over a compact set). So we can apply Theorem 4, and we obtan: whch proves that all these values are equal. v r M) p T M q v c M) p T Mq v r M), Corollary 6. Let M be a m n matrx game. The followng statements are equvalent. ) p, q ) s a par of optmal strateges; ) p, q ) s a Nash equlbrum. Proof. ) ) : If p, q) are optmal, then mn p T Mq = v r M) = p T M q, so q s a mnmzer of p T Mq over n,.e. q BR 2 p). Smlarly, max p T M q = v c M) = p T M q tells us that p BR 1 q). ) ) : If p, q ) s a Nash equlbrum, then p BR 1 q ) mples p T Mq = max p T Mq, Page 3 of 5

4 and we know that p T Mq = vm) = v c M). So q s an optmal strategy of the column player. Smlarly, q BR 2 p ) mples v r M) = vm) = p T Mq = mn q n p T Mq, whch tells that p s an optmal strategy for the row player. We also have a result whch s the counterpart of Theorem 2 for mxed strateges: Corollary 7. If p and r are both optmal strateges of the row player, and q and s are both optmal for the column player, then p T Mq = p T Ms = r T Mq = r T Ms. Proof. Clear from the mnmax theorem, and all these payoffs must be equal to vm). We next defne symmetrc games, whch are games n whch both players are ndstngushable. Defnton 5 symmetrc game). A matrx game s called symmetrc f the matrx M of the game s skew-symmetrc M = M T ). Rock Paper Scssors P 1 / P 2 R P S R P S #2 Theorem 8. The value of a symmetrc game s zero. Moreover f p s optmal for the frst player, then p s also an optmal strategy for the column player Proof. Let p be optmal for the row player,.e. p maxmzes mnp T M) = mn M T p) = mn Mp) = maxmp), so p mnmzes max Mp),.e. p s optmal for the column player. The value of the game s thus p T Mp =, p p M, = < p p M, + < p p M, }{{} = < ppm, = 0. Bmatrx Games or Two-Player Non-Zero-Sum Games) In non-zero-sum games, the stuaton s not so clear anymore. We wll see that Nash equlbra mght lead to stuatons that are bad for both players. Moreover, usng Nash equlbra to predct the outcome of the game s perlous, snce several equlbra mght coexst and the problem of selectng a Nash equlbrum s challengng. Prsoner s Dlemma: P 1 /P 2 plead gulty accuse the other one ) #1 plead gulty 2, 2) 10, 1) accuse the other one 1, 10) 5, 5) Page 4 of 5

5 Chcken Game: P 1 /P 2 turn don t turn turn 0, 0) 2, 5) don t turn 5, 2) 10, 10) #2 Battle of the sex: / 10, 3) 2, 2) 0, 0) 3, 10) #3 Graphcal method to fnd the Nash equlbra of 2 2 bmatrx games Defne the sets S 1 = {p, q) m n : p BR 1 q)} S 2 = {p, q) m n : q BR 2 p)} By defnton, the set of Nash equlbra s S 1 S 2. For 2 2 bmatrx games, these sets can be plotted n the square [0, 1] [0, 1]. Consder the game of Battle of Sex Example #3). We wrte p = [p, 1 p] T 2 and q = [q, 1 q] T 2. For the male player, the strategy s a best response to q ff 10q + 21 q) 31 q) q Smlarly s a best response to q ff q Now for the female player, the strategy s a best response to p ff 3p 2p p) p Smlarly s a best response to p ff p We know that the set of best responses are the convex hull of the pure best responses. So f q 1 11, every p 2 s a best response to q. Smlarly BR 2 [ 10 11, ) 1 11 ]T = 2. The best responses sets are dsplayed on the followng plot: 1 Nash equlbrum #4 0 1 Ths shows that ths b-matrx game has two pure equlbra, ),, ) and one mxed equlbrum [ 10 11, 1 11 ]T, [ 1 11, ) ]T. Page 5 of 5