Comments on "Reverse auction: the lowest positive integer game"

Fluctuation and Noise Letters V ol. 0, No. 0 (2008) 000–000 c  W orld Scient ific Publishing Company COMMENTS ON ‘REVERSE A UCTION: THE LO WEST UNIQUE POSITIVE INTEGER GAME’ A. P . FLITNEY Scho ol of Physics, University of Melb ourne Parkvil le, VIC 3010, Austr alia. aflitney@unimelb.e du.au Receiv ed (r eceived date) Revised (revised date) Accepted (accepted date) In Zeng et al. [Fluct. Noise Lett. 7 (2007) L439–L447] the analysis of the low est unique posi tive inte ger game is simplified by some reasonable assumptions that m ake the prob- lem tractab le for ar bi trary n umbers of pl ay ers. How ev er, here we sho w that the solution obtained for rational play ers is not a N ash equilibri um and that a rational utility max- imizer with full computational capability would arrive at a solution with a superior expected pay off. An exact solution is presented for the three- and four-play er cases and an approximate solution for an arbitrary num ber of play ers. Keywor ds : rev erse auction, game theory , m i norit y game, rational choice , LUPI 1. In trod uct ion The low est unique p ositiv e in teger game can b e briefly desc r ibed as follows: Each of n pla yers secretly selects an int eger x in the rang e [1 , n ] with the pla yer selecting the smallest unique integer receiving a utility of one, while the other play ers score nothing. If there is no low est unique in teger then all play ers scor e zero . In Zeng et al. [1 ] the assumption is made that “...a player is indifferent b et w een t wo strategies conditioned on the other play ers’ choices and a play er will alwa ys pick the low est num b er. Arguably this might b e to o strong.” In the following section we show the latter assumption is indeed to o strong. W e make us e of the following g ame-theoretic concepts: A str ate gy pr ofile is a set of str ategies, one for each play er; a Nash e qu il ibrium (NE) is a s trategy profile from w hich no pla yer c a n improve their pay off by a unilateral change in strategy; a Par eto optimal (PO) strategy pro file is one fro m whic h no pla yer can impro ve their result without so meone else b eing w orse off. R1 R2 Comments on R everse Auction 2. Nash equilibrium fo r the n = 3 , 4 pla y e r cases It is ea sy to show tha t the so lution o f Zeng et al is not a NE for small n . I n the three play er ca s e, if Bob and Charles adopt the s trategy ( 1 2 , 1 2 , 0), where the i th nu mber in the parentheses is the probability of selec ting the integer i , Alice can maximize her pa yoff by selec ting the strateg y (0 , 0 , 1). In this ca se Alice wins when ever the other t w o choo s e the same in teger. Hence her exp ected pay off is 1 2 , double that obta ined b y selecting the stra teg y ( 1 2 , 1 2 , 0). Bob and Charles win in only 1 4 of the cases. This result is an (a symmetric) NE. Given that $ A + $ B + $ C = 1 the result is a lso PO : the sum o f the payoffs is maximal so no other strategy pro file can give one player a higher pay off w itho ut so meone else b eing w orse off. F or n = 4 there is an analogo us solution. If Bob, Char les and Debra play the strategy ( 1 2 , 1 2 , 0 , 0), Alice’s optimal play is to select ‘3’ with proba bilit y one . Then she wins if the others hav e all s e le cted ‘1 ’ o r all selected ‘2’. The exp ected payoff to all play ers is 1 4 and so the equilibrium is fair to all play ers. Again this so lution is a NE and is PO with the maximum p ossible sum of payoffs (one). F or n > 4, the strategy “always choos e ‘3 ’ ” is no lo nger optimal against a group o f play ers choosing a mixed ‘1’ or ‘2’ strategy and there is no simple analo gue to the ab o ve NE strateg y profiles. Asymmetric stra tegy profiles s uch as those g iv en ab ov e ar e difficult to realize in practice since in the absence of communication it is not p ossible to decide o n who plays the o dd strategy . W e will now search for a symmetric NE s trategy pro file where all the play ers choo s e the same (mixed) strateg y . Suppose all players but Alice choose the strateg y ( p 1 , p 2 , . . . , p n ), while Alice plays ( π 1 , π 2 , . . . , π n ), with the normalizatio n conditions 1 Σ p i = Σ π i = 1. In the end we will set π i = p i ∀ i to give a symmetric str ategy profile. F or Alice’s stra tegy to yield her maximum pay off (given the others’ strateg ies) it is necessa ry , though no t s ufficien t, for d $ A /dπ i = 0 , ∀ i . Using the normaliza tion co nditions to substitute for p n and π n , we can write Alice’s exp ected winnings as $ A = π 1 (1 − p 1 ) n − 1 + 1 − n − 1 X k =1 π k ! n − 1 X j =1 p n − 1 j + n − 1 X i =2 π i   (1 − i X j =1 p j ) n − 1 + i − 1 X j =1 p n − 1 j   . (1) By differentiating with r espect to ea ch o f the π i and setting the res ult equal to zer o n − 1 no n-linear coupled equations in the n − 1 v ar iables p 1 , . . . p n − 1 are obtained. Amongst the simultaneous solutions of these eq uations will be one that is maximal for Alice. B y se tting π i = p i ∀ i w e obtain a s trategy that is maximal for a ll play ers and is thus a NE. W e note that the der iv a tiv es of $ A do not inv olv e the π i . F or the ca s e of n = 3 we have $ = π 1 (1 − p 1 ) 2 + π 2 ( p 2 1 + (1 − p 1 − p 2 ) 2 ) + (1 − π 1 − π 2 )( p 2 1 + p 2 2 ) , (2) 1 In addition, i f Al ice picks either n or n − 1 she can only win if al l the other pla y ers hav e chose n the s ame int eger. This wi ll mean that f or the N E strategy π n − 1 = π n . Ho we ve r, in the following analysis w e shall not m ake use of this relation. A. P. Flitney R3 (where the subscr ipt A has b een dropp ed for simplicity) resulting in d $ dπ 1 = 1 − 2 p 1 − p 2 2 (3a) d $ dπ 2 = 1 − 2 p 1 + p 2 1 − 2 p 2 + 2 p 1 p 2 . (3b) This has the unique (for the physical range of p 1 , p 2 ) solution p 1 = 2 √ 3 − 3 , p 2 = 2 − √ 3 . (4) W e note that p 2 = 1 − p 1 − p 2 , as o bserved in the ea rlier foo tnote. When Bob and Charles play the stra tegy (4), that is, when they select ‘1’ with pr obabilit y 2 √ 3 − 3 ≈ 0 . 4 6 4 and ‘2’ or ‘3’ each with proba bilit y 2 − √ 3 ≈ 0 . 268, Alice’s pay off is indep enden t of her s trategy . The game being symmetric, the same is true for any of the play ers whe n the other tw o c ho ose (4). Thus, no player can improve their strategy by a unilater al change in strategy , demonstrating tha t (4) is a NE. When all players select this s trategy , the exp ected pay off to ea c h is 4(7 − 4 √ 3) ≈ 0 . 2 87, which is higher than the pay off of 0 . 25 that results when ea ch play er s e le cts only betw een ‘1’ o r ‘2’ with equal pro babilit y , the “rational” play er result of Ref. [1]. It is interesting, and some what an ti-intuit ive, that the solution inv olves a non-zero v alue for p 3 = 1 − p 1 − p 2 since ‘3’ can never b e the lowest integer, though it ca n be the only unique integer. Pro ceeding in the same manner for n = 4, (1) reduces to $ = π 1 (1 − p 1 ) 3 + π 2 [ p 3 1 + (1 − p 1 − p 2 ) 3 ] + π 3 [ p 3 1 + p 3 2 + (1 − p 1 − p 2 − p 3 ) 3 ] + (1 − π 1 − π 2 − π 3 )( p 3 1 + p 3 2 + p 3 3 ) . (5) Different iating with resp ect to each of π 1 , π 2 , and π 3 and setting the results equal to zero gives the unique (ph ysical) solution p 1 ≈ 0 . 488 , p 2 ≈ 0 . 2 50 , p 3 ≈ 0 . 1 31 , (6) again with the rela tio nship p 3 = 1 − p 1 − p 2 − p 3 . The exa ct v alues for the p i are complicated and unilluminating. The pay off to each player when they all choos e the strategy (6), that is, when eac h player s elects ‘1’ with probability ≈ 0 . 488 , ‘2’ with probability ≈ 0 . 250 and ‘3’ or ‘4’ each with probability ≈ 0 . 131, is approximately 0.134. This is higher than that obtaina ble if all the players simply s elect b et w een ‘1’ and ‘2’ (0.12 5). Again, when three play ers cho ose (6) the pay off to the four th play er is independent o f their strategy , demo ns trating that the str ategy pr o file is a NE. Note the symmetric mixed strateg y NE pr ofiles hav e lower average pay offs than the asymmetric ones found e a rlier. 3. Appro ximate so lution for an arbitrary num ber of pla y ers In general, since we hav e n − 1 coupled equations of deg ree n − 1, for n > 5 no analytic solution will b e po ssible, and fo r n = 5 the solution will b e pro blematic. By insp ection of (4) and (6) the mixed stra teg y with π i = 1 2 i for i < n, π n = π n − 1 = 1 2 n − 1 , (7) R4 Comments on R everse Auction is an approximation to the symmetric NE so lutio ns for n = 3 , 4. E quation (7) is in keeping with our intuition by giving higher weigh ts to the selection of smaller int egers. The pay off to ea c h play er for n > 2 if all select (7) is $ = n − 1 X k =1   1 2 k k X j =1  1 2 j  n − 1   + 1 2 n − 1 n − 1 X j =1  1 2 j  n − 1 . (8) F or n = 3 the pay off is 9 32 ≈ 0 . 2 81 and for n = 4 it is ≈ 0 . 133, bo th very clo se to the v alues for the exact symmetric NE given in the previous section. The payoff (8) as a function of n is shown in T able 1, a long with the payoffs from Ref. [1] of 1 / 2 n − 1 and the exa ct solutions for the n = 3 and 4 ca ses. The payoff given in Ref. [1] is slightly smaller than the payoff g iv en by (8) but will asymptote to it as n increases. 4. Conclusion W e have found b oth a symmetric and symmetric NE strateg y profiles for a thre e- and four-player low est unique p ositive integer g ame with pay offs sup erior to that resulting from the simplifying assumption of Ref. [1]. In particular the assumption that a play er w ill alwa ys cho ose the low est integer in a situation where they hav e a choice results in a strategy that is not a NE. The asymmetric NE a re also PO, and in the case of n = 4, is fair to all players. The symmetric solutions are unique amongst sy mmetric strategy profiles but yield a low er pa yoff than the asymmetric solutions. Anti-in tuitiv ely , the NE strategy profiles includes a no n-zero probability for selecting the la r gest integer since this may b e the only unique integer. F or ar bitrary n , w e prop ose a simple sy mmetr ic stra tegy pro file with geometri- cally decr easing probabilities of selecting hig he r integers. This gives very close to the pay offs of the exa ct symmetric NE solutio ns for the tw o ca ses for which ex act solutions were obtained. The r a tional player solution of Ref. [1 ] is simpler than our s but gives payoffs slightly sma lle r. Ac kno wl edgmen t F unding was provided by the Austr alian Resea rc h Council grant num ber DP0 559273. References [1] Q. Zeng, B. R. Davis, and D. A bbott, R everse auction game: the l owes t uni que p ositive inte ger game, Fluct. Noise L ett. 7 ( 20 07) L439– L447. n 3 4 5 6 7 8 Equation (8) 0.281 0.133 0.064 5 0.0317 0.0157 0.007 84 Reference [1] 0.25 0.125 0.0625 0.0313 0.0156 0.0078 1 Exact 0.287 0.134 T able 1. The pa yo ff for the approximate symmetric Nash equilibr ium solution of strategy (7) along with the r at ional pla y er pay offs from Ref. [1] and the exact symmetric Nash equilibrium pay offs of strategies (4) and (6), for the three- and four- pla y er cases, resp ec tiv ely . Exact solutions for the other cases hav e not be calculated. Pa yoffs hav e b een r ound ed to three significan t figures.