An analysis of a war-like card game

AN ANAL YSIS OF A W AR-LIKE CARD GAME BORIS ALEXEEV AND JACOB TSI MERMAN Abstract. In his b ook “Mathematical Mind-Benders”, P eter Winkler p oses the following op en problem, originally due to the first author: “[In the game Peer Pressure,] tw o pla y ers are dealt some n umber of cards, initially fac e up , eac h card carrying a differen t integer. In eac h r ound, the play ers simultaneously play a card; the higher card is discarded and the low er card passed to the other play er. The play er who runs out of cards loses. As the num ber of cards dealt becomes larger, what is the limiting probability that one of the play ers wi l l hav e a wi nning strategy?” W e sho w that the answer to this question is zero, as Winkler suspected. M oreo v er, assume the cards are dealt so that one play er receiv es r ≥ 1 cards for every one card of the other. Then if r < ϕ = 1+ √ 5 2 , the limiting probability that either play er has a winning strategy is still zero, while if r > ϕ , it i s one. Introduction The card game “Peer Pressur e”, a v ar ia nt of “W a r ”, is played with a deck o f n cards, eac h car rying a distinct integer. The car ds a re initially dealt randomly to tw o play ers, with either exactly n / 2 car ds p er play e r o r each card r andomly going to one o f the play ers. In each round (“battle”) of the game, the play ers simult aneously play a card. The play er holding the higher card wins the round a nd receives the low er car d; how ever, the higher card is p ermanently discarded from the game. The play er who r uns out of car ds los es. W e assume tha t b oth players k now the origina l deck a nd th us ar e aware o f the conten ts of bo th play ers hands at all times. Recall that we say a play er has a winning str ate gy if she may a nnounce her strateg y b eforehand and still alwa ys win aga inst any strategy from her opp onent. A winning strategy may in genera l b e a mixed (probabilistic) stra tegy , but if one exists, there also exists a pure (deterministic) winning strategy . In Peer P ressure, if ther e are four or fewer cards, then one o f the players has a winning str a tegy . In particular, if one player has mor e ca rds, then she wins; if the play ers have an equa l num ber of cards, the play e r with the highest card wins . How ever, if there are fiv e cards, there is one p osition where neither player has a winning strategy: 1 , 2 , 4 versus 3 , 5. Suppo se our tw o play ers ar e named Alice and Bo b and hav e a and b car ds r esp ectively . W e pr ov e the following lemma that helps classify when a player has a winning strateg y : Main Lemm a. Le t ϕ = 1+ √ 5 2 ≈ 1 . 6180 3 b e the golden r atio, which notably satisfies 1 + ϕ = ϕ 2 . • If Ali c e has mor e than ϕ times as many c ar ds as Bob (that is, a > ϕb ), then Alic e has a winning str ate gy. • If Alic e has mor e t han 1 / ϕ times as many c ar ds as Bob (that is, b < ϕa ) and they ar e al l higher than Bob’s, then A lic e has a winning str ate gy. W e then use this lemma to prov e our main result: Date : Decem ber 2, 2009. 2010 Mathematics Subje ct Classific ation. 91A05. Key wor ds and phr ases. Two-person game, W ar (card game), winning strategy . 1 Definition. Say that a result holds generic al ly if it holds with pro babilit y approa ching o ne as the num b er of car ds, n , go es to infinit y . Main Theorem. In the original game with u nbiase d de aling, generic al ly neither player has a winning str ate gy. Mor e over, assume the c ar ds ar e de alt ra ndomly so that Alic e r e c eives r ≥ 1 c ar ds for every c ar d of Bob. If r < ϕ , generic al ly neither player has a winning str ate gy, while if r > ϕ , generic al ly Alic e has a winning str ate gy. This r esult deter mines the limiting pr obability that one o f the players has a winning str ategy , an op en problem p osed by Peter Winkler in his b o ok “Mathematical Mind-Benders” , wher e it is attributed to the first author. [Win07] Proofs One interesting quirk in Peer Pressure is that it is no t immediately obvious that having better cards is necessarily a dv antageous. O f cour se, a b etter car d will be mor e likely to win in any given ro und, but the card may also end up in the hands of the opp onent, who may then use it to his adv antage. W e b egin by proving that this is not a problem. This lemma is not es s ent ial to the later results, but it do es simplify their pro ofs. Definition. W e s ay a collection o f car ds C is at le ast as go o d as ano ther collection C ′ if for all p ositive int egers k , either the k th highest car d in C is a t least as high as the k th highes t card in C ′ or C ′ has less than k cards . Monotonicity Lemma. Having b etter c ar ds do esn ’t hurt. That is, if in a c ertain p osition Alic e’s c ar ds ar e r eplac e d with b etter c ar ds and/or Bob’s c ar ds ar e r eplac e d with worse c ar ds, then Alic e is no less likely to win. In p articular, if Alic e had a winning str ate gy b efor e the r eplac ement, she stil l do es afterwa r d. Pr o of. W e prov e the r esult step by step. Extra cards do n’t h urt. Supp ose Alice r eceives extra ca rds, but no other change o cc ur s. Then clea rly she is no worse off b ecause she can play the same strategy as befor e, ig noring her extra cards . If she won befo re, she still wins. (If she lost befor e, she now has extra ca rds that may or may not help in the end.) Losing cards do esn’t help. By symmetry , if Bob has c a rds taken aw ay , but no other change o ccurs, he is no better off. Receiving a card from the opp onen t do esn’t hurt . Supp ose Bob g ives a card to Alice, but no o ther change occur s. The n Alice is no w orse off because this is equiv alent to Alice ga ining a card and Bob losing a card. Slightly impro ving one card do esn’t hurt. Supp ose Alice has a card A and Bob has a card B s uc h that A < B before the replacement and A > B after the replacement, but no other change o ccurs. (In particular, there a re no car ds of ra nk b etw een A and B .) Then let Alice play exactly as befo r e u ntil one of these cards is pla yed by either player. If b oth cards a re play ed simult aneously , then Alice is no w orse off bec ause s he wins a card instea d of Bob. Indeed, this is equiv alent to Bob winning the battle (a s b efore the replacement), follow ed by Bob giving Alice a card. If o nly one card is play ed and it wins, then it is removed from pla y and so the rela tiv e ra nking o f A and B do esn’t matter anywa y . I f only one card is play ed and it loses, then one of the play er s will end up ho lding both A a nd B . Again, the relative ra nk ing does n’t ma tter bec ause Alice c an pretend to switc h the ca rds. 2 Ha ving b etter cards do e sn’t hurt. The g eneral case consists of per fo rming the ab ov e mo difications one b y one. This may be a ccomplished, for example, by first improving Alice’s best ca r d, then her next b est, and so on, and afterward, giving Alice extr a cards and taking cards aw ay from Bo b.  W e ma y no w prov e o ur main lemma. Main Lemm a (redux) . L et ϕ = 1+ √ 5 2 ≈ 1 . 6180 3 b e the golden r atio, which notably satisfies 1 + ϕ = ϕ 2 . • If Ali c e has mor e than ϕ times as many c ar ds as Bob (that is, a > ϕb ), then Alic e has a winning str ate gy. • If Alic e has mor e t han 1 / ϕ times as many c ar ds as Bob (that is, b < ϕa ) and they ar e al l higher than Bob’s, then A lic e has a winning str ate gy. Pr o of. W e prov e the r esult by induction on the to ta l num b er of cards , a + b . Note that b oth results a re certainly true w he n a = 0 or b = 0. Alice has man y cards. Supp ose that a > ϕb > 0 . By the Monotonicit y Lemma, we may a ssume that Bob has the b highest cards, since this is the worst p o ssible s ituation for Alice. In order to win, Alice plays her low est current card until all of Bo b’s cards are less than hers. In any battle, Bob may either lose one o f his high c a rds and receive one of Alice’s low cards or Bo b may give Alice bac k one of her or iginal lo w car ds. In other words, Alice loses a card if and only if one of B o b’s high cards is discarded. Therefore, afterward, Bob has at mo st b cards, all of which a re lower than Alice’s a − b ca rds. The r esult holds by induction bec ause b < ϕ ( a − b ). Alice has eno ugh high cards. Suppo se that ϕa > b > 0 and all o f Alice’s cards are higher than Bob’s. In order to w in, Alice plays each of her a cards once. She will win every battle, so afterward she will still hav e a cards, while Bob will hav e b − a ca rds. The result holds by induction because a > ϕ ( b − a ).  Armed with the Main Lemma, w e prove the Main Theo rem. Rec all that a re s ult holds generic al ly if it holds with pr obability a pproaching one as the num b er of cards, n , g o es to infinity . Main Theorem (r edux) . In the original game with unbiase d de aling, generic al ly neither player has a winning str ate gy. Mor e over, assume the c ar ds ar e de alt r andomly so that Alic e r e c eives r ≥ 1 c ar ds for every c ar d of Bob. If r < ϕ , generic al ly neither player has a winning stra te gy, while if r > ϕ , generic al ly Ali c e has a winning str ate gy. Pr o of. First o f a ll, b eca us e only the r elative ordering of the cards matters, we as sume the cards are num bered 1 to n . Also, r ecall that if Alice has a mixe d (probabilis tic) winning strategy , then s he also has a pure (deterministic) winning strategy . W e b egin with the w eaker result with un biased dealing. Divide the ca rds into fiv e equally-sized in terv als: C 1 = (0 , n/ 5], C 2 = ( n/ 5 , 2 n/ 5] , . . . , C 5 = (4 n/ 5 , n ]. Suppose Alice reveals her pure stra tegy in adv ance to Bob. W e will show how Bo b ca n us e these in ter v als to defeat it. Sp ecifically , for 1 ≤ i < 5, he will use his cards in C i +1 to defeat Alice’s car ds in C i . Finally , he will use his leftov er cards to defeat Alice’s C 5 . See the Figure for a visual expla nation. W e exp ect each player to receive ha lf of the cards in each int erv al, so b y the la w of large num ber s, Bob will gener ically receive a t least . 099 n cards fr om each int erv al; Alice will receive a t most the remainder , . 101 n . No w supp ose that only Bob’s c a rds in C 2 and Alice’s cards in C 1 are under considera tion. By the Main Lemma, Bob may use . 063 n > . 101 n/ϕ car ds from C 2 to defeat Alice’s cards in C 1 (note that all of the ca rds in C 2 are higher than those in C 1 ), leaving at least (0 . 0 99 − 0 . 063) n = 0 . 03 6 n un used cards left 3 C 1 C 2 C 3 C 4 C 5 Alice Bob Figure. The unshaded interv als illustra te how Bob uses s lightly more tha n a 1 /ϕ - prop ortion of his cards in C i +1 to defeat Alice’s ca rds in C i . B o b’s leftov er c ards in all of his C i , repres e n ted by shaded interv als, ar e sufficient in num ber to ov er whelm Alice’s cards in C 5 . ov e r. If he do es similarly for his C 3 through C 5 , Bob will hav e at least 4( . 036 n ) + . 09 9 n = . 243 n car ds left ov e r. Again by the Ma in Lemma, Bob may us e these ca rds to defeat Alice’s C 5 bec ause . 243 /. 101 > ϕ . In the previous paragr aph, w e pretended that Bob ma y conside r the g ame a s the sum o f five indep endent games. How ever, this is justified b ecause Alice has r evealed her pure s trategy in adv ance . Becaus e Bob knows where Alice w ill play , he may use the appr opriate cards to defeat her. Note that B ob may choos e b efor ehand whic h cards a re allo cated where, so it does not matter in what o r der Alice pla ys; in particular, Bob can c hoos e his “lefto v er c a rds” beforeha nd, a s all that matters is their num ber. There fo re g e nerically , Alice has no winning strateg y and by symmetry , neither do es Bob. Now we prov e the stro nger result with dealing biased tow ards Alice. If r > ϕ , this is easy . Gener ically , Alice will ha ve more than ϕ times as many cards a s Bob and thus win b y the Main Lemma. Now supp ose that r < ϕ is fixed. W e follo w the same appr oach as b efore. Divide the c a rds into k equally-sized in terv als C i =  i − 1 k n, i k n  , where k will be c hosen la ter to dep end only o n r and not on n . In each interv al, we exp ect play ers to re c eive cards in an r : 1 prop ortion. By the law of large num b ers, for any co nstant δ > 0, Bob generically receives at leas t 1 − δ of the n um ber of car ds he exp ects in e ach o f the int erv als. (Note that we crucially use here that k doe s not dep end o n n .) By the Ma in Lemma, we may choose δ so small (but indepe nden t of k ) that B ob may use his ca rds in C i +1 to defeat Alice’s car ds in C i for all 1 ≤ i < k and still have a p ositive prop ortion of his cards left o ver in ea ch interv al. Now choose k so la rge (but indep endent o f n ) that Bob’s r emaining cards in C 1 and his leftov er cards from all of the o ther C i are more than ϕ times the num ber of Alice’s car ds in C k . This is p o ssible bec a use we insured that each C i has at least a fixed p ositive propo rtion o f cards left ov er, so Bob ma y overwhelm Alice with his extra cards . No w, a s b efor e, B o b’s leftov er cards defeat Alice ’s C k by the Main Lemma. (Again, Bob’s car ds may be allo cated before a n y ac tua l play .) Finally , if Alice reveals her s trategy in adv ance, Bob may co m bine his s tr ategies on all o f the interv als C i to defeat her. Therefore, if r < ϕ , Alice gener ically do es not hav e a winning strateg y . B o b g enerically do esn’t hav e a winning str a tegy either, b ecause his cards ar e even worse than in the unbiased case.  Fur ther directions The results in this pap er may be contin ued in a few natura l directions. F or example, by using techniques similar to those presented a bove, Ja cob F ox has determined the threshold for the num ber o f battles a play er ca n gua rantee winning in the unbiased mo del. [F o x08] In particular , if f ( n ) is a function that grows slow er than √ n (using L andau’s asympto tic notation, o ( √ n )), then generically bo th o f the play ers may gua rantee winning a t least f ( n ) of the ba ttles. Ho w ever, if f ( n ) is a function that 4 grows faster than √ n (using Landau’s asymptotic no tation, ω ( √ n )), then gener ically neither play er may guarantee winning at least f ( n ) of the battles. The threshold √ n comes fro m the ce n tral limit theo rem. In another direction, no te that the Main Lemma classifies some of the hands where Alice has a winning strategy , and it can also b e used to classify hands wher e neither play e r has a winning strategy (as in the Main Theorem). W e leave as an o pen problem whether or not one ma y prov e stronger results ab out winning strategies: Problem. Classify the situations when a giv en play er ha s a winning strategy . Ackno wledgments The authors wis h to thank Dan Cranston a nd J acob F ox for helpful discussions. References [F o x08] Jacob F o x, pers onal comm unication, No v em ber 2008. [Win07] Pe ter Winkler, Mathematic al mind-b enders , c h. 11, p. 134, A K P eters Ltd., W ellesley , MA, 2007. Dep ar tment of M a themati cs, Princeton University, Fine Hall, W ashing ton Road, Princeton, NJ 08544-100 0 E-mail addr e ss : balexeev@m ath.princ eton.edu, jtsime rm@math.p rinceton.edu 5