Coin ToGa: A Coin-Tossing Game

Transcription

1 Coin ToGa: A Coin-Tossing Game Osvaldo Marrero and Paul C Pasles Osvaldo Marrero OsvaldoMarrero@villanovaedu studied mathematics at the University of Miami, and biometry and statistics at Yale University After holding various aointments in academia and in industry, he is now a Professor at Villanova University He has ublished aers in eidemiology, mathematics, medicine, and statistics Other interests include listening to music, hysical exercises, and travel He is fluent in English, French, and Sanish, and he continues to imrove his knowledge of Dutch and German Paul C Pasles PaulPasles@villanovaedu is Assistant Professor of Mathematical Sciences at Villanova University He traces his ancestry to Chios, birthlace of Hiocrates the Pythagorean, not the hysician Paul s first book, with new revelations on Dr Franklin and his squares, is in rogress those who are first will robably be last Matthew 20:16, arahrased [1] Insiration For six decades, the Mathematical Association of America has sonsored the annual Putnam exam, challenging American and Canadian undergraduates to find novel solutions to imaginative mathematical questions There are cash incentives for the highestscoring individuals and teams, and it is erhas with such mercenary motives in mind that the 1997 Putnam roosed this question: Players 1, 2, 3,,n are seated around a table and each has a single enny Player 1 asses a enny to Player 2, who then asses two ennies to Player 3 Player 3 then asses one enny to Player 4, who asses two ennies to Player 5, and so on, layers alternately assing one enny or two to the next layer who still has some ennies A layer who runs out of ennies dros out of the game and leaves the table Find an infinite set of numbers for which some layer ends u with all n ennies [2] We examine a similar roblem with a robabilistic sin Let n be a ositive integer Suose that n eole are randomly assigned seats at a round table One of them is chosen again at random and called Player 1 Then the remaining articiants are numbered clockwise 2, 3,,n Player 1 tosses a coin, not necessarily fair If the coin comes u heads, then that layer is out of the game If it comes u tails, she stays in the game Then Player 2 tosses the same coin, etc After Player n s turn, the coin is VOL 34, NO 3, MAY 2003 THE COLLEGE MATHEMATICS JOURNAL 183

2 assed clockwise to the next remaining layer and the rocess continues When n 1 layers have been eliminated, then only the winner remains and the game terminates Let denote the robability that the coin comes u heads, and ut q 1 If 0, then the game almost surely does not terminate On the other hand, if 1, then the game almost surely terminates in n 1 stes, with Player n declared the winner Therefore, we will take 0 < < 1 This ensures that each layer has a ositive robability of winning, and also that the game terminates with robability 1 1 If 1/6, the robabilist might call this game Russian roulette, with reloading More generally, the game serves as a model for any simle rocess of successive elimination, from certain incarnations of the children s game of dodgeball to voting systems We will analyze this game to determine the advantage of each seat osition For brevity we will call this game Coin ToGa, for Coin Tossing Game We begin by looking at games that are not well-oulated n 2, 3 Then we discern some regular atterns by analyzing these secial cases Finally we derive a useful recursion for the general game with n layers and will use it to rove various theorems As a referee suggested to us, Coin ToGa could be used in courses or in-service workshos dealing with exerimental robability, or even at mathematical arties! The 2- and 3-erson games In the following, denotes the event that Player j wins For n 1, define the random variable A n by A n j if Player j wins, j 1, 2,,n, and ut P n PrA n j, the robability that the winner of an n-erson game is the jth layer The ossibilities for a two-erson game are shown in Figure 1 Figure 1 The tree diagram for n 2 The initial node 1 2 reflects the fact that both layers are still in the game and that Player 1 s turn is next, whereas the node label 2 1 shows that Player 2 s turn is next at that ste Player 2 can win in 1 ste robability, in 3 stes robability q 2, in 5 stes robability q 4, and so on That is because terminal node 2 is reached by first traversing the grah s single loo or cycle some nonnegative integer number of times, then moving to the left Therefore, the robability that Player 2 wins is P 2 X 2 m0 q2m / 2 1/1 + q It follows that P 2 X 1 1 1/1 + q q/1 + q <1/1 + q, so Player 2 is favored to win Figure 2 deicts the three-erson game As before, each node shows the survivors at a articular moment This tree contains cycles of length 2 and 3 Observe that the subtrees emanating from nodes 12, 23, and31 are isomorhic 1 Note added in roof: it has recently come to the authors attention that this game was studied by Blom et al, as General Russian Roulette, Mathematics Magazine #4 184 c THE MATHEMATICAL ASSOCIATION OF AMERICA

3 Each node of the tree lists layers in order by whose turn is next, second from next, etc This has the advantage that each nonterminal node is uniquely determined by its labelling when n 3, and it is always clear whose turn is next As in the revious section, we will use the cycling roerties of the tree to avoid needing an infinite sheet of aer Figure 2 Thetreediagramforn 3 We take care to identify two nodes based not only on the set of remaining layers, eg, {1, 2}, but also on whose turn is next That is, we use ordered sets: node 1 2 node 21 Consider Player 1 There are two terminal nodes 1 in Figure 2 The first is reached directly by tossing THH; the second, by tossing TTHTH The accomanying robabilities are 2 q and 2 q 3 However, these account only for direct routes with no cycles Player 1 can also win as a result of several 3-cycles followed by 2-cycles Thus, P 3 X 1 2 q + 2 q q 3 + q q 2 + q q1 + q The factors 1 + q 3 + q 6 + and 1 + q 2 + q 4 + corresond to the 3- and 2-cycles Proceeding similarly for the other layers we obtain P 3 X 2 q2 1 + q 2 3 and P 3 X q Comare the trees for n 2, 3 It aears that the nth tree has exactly n! terminal nodes, with each layer aearing an equal number of times among them This can be roved inductively, but it will be clear anyway as a consequence of the roof of our recursion VOL 34, NO 3, MAY 2003 THE COLLEGE MATHEMATICS JOURNAL 185

4 Some observations By summarizing what we know so far, we will be able to see some atterns immediately Proofs are deferred until later The 1-erson game is straightforward, and we mention only for reference that P 1 X 1 1 The robabilities for the 2-erson game are best viewed in unreduced form: P 2 X 1 q 2, P 2 X 2 2 We begin to see a attern with the 3-erson game: P 3 X 1 q2 1 + q 2 2 3, P 3 X 2 P 3 X q q q 2 3, Similarly, a tree can be used to find the 4-erson values: P 4 X 1 q q 2 + 2q 3 + q , P 4 X 2 q3 1 + q + q 2 + 2q 3 + q , P 4 X 3 q q + q 2 + q 3 + q , P 4 X q 2 + 2q 3 + q However, the trees quickly become unwieldy Fortunately, the recurrence relation that we develo later allows a calculation of P n for any n, j For examle, using that recursion formula, we can find that P 5 X 2 4 q 1 + q + 2q 2 + 5q 3 + 5q 4 + 3q 5 + 3q 6 + 3q 7 + q Based on these calculations, we make several observations Theorem 1 For n 2,qP n X n P n X 1 This is surrising; although Coin ToGa is a circular game and Players n and 1 are adjacent layers, their robabilities of winning are quite closely related The theorem imlies that Player n is in a much better osition to win than Player 1 That can be extended into a generalized order relation which constitutes the next theorem As further motivation, Theorem 1 will be useful in the roof of Theorem 2 Theorem 2 For fixed n 1, P n is strictly monotone in j: P n X n >P n X n 1 > > P n X 1 That is, your chances of winning are better if your first turn comes later in the game This corresonds to the intuition that the more layers who are in front of you, the greater your chance of being the sole survivor This should hold true even as the game 186 c THE MATHEMATICAL ASSOCIATION OF AMERICA

5 enters the second round, third round, and so on If we assign the layers a ranking, Player 1 comes in last; hence the biblical quote that began this article It also aears that for fixed j 1, P n is strictly monotone in n: P j > Pj+1 > Pj+2 >, and so Player j s chances worsen each time the number of layers increases If true in general, this would be a useful lemma; it imlies that lim n P n exists Fortunately, we will be able to rove an even stronger limit result lim n P n 0 by other methods Excursion into number theory It is interesting to observe that our next theorem relates a robability question to some well-known and dee objects in number theory the modular forms and the cyclotomic olynomials Theorem 3 a For n 2, P n q l n 1 R n, j q { 2 3 n }, where l 1 if 1 j < n, and l 0 if j n, and R n, j is a monic olynomial with nonnegative coefficients In fact, these coefficients are ositive, excet for the linear and next-to-highest ower terms when j 1 or n Moreover, the sum of the coefficients R n, j 1 is indeendent of j: R n, j 1 n 1!, while R n, j 0 1, and for n > 3, R n, j 1 0 b R n, j 1/q q k R n,n j+1 q, wherek deg R n, j is described exlicitly below This looks remarkably like the two-function transformation law in the theory of modular forms! [See, for examle, [3, 145, eq 12]] However, as R n, j is neither eriodic nor transcendental, it holds no obvious allure for the ractical number theorist c For n 2, { 1 2 n 2 n 4, if 1 < j < n, deg R n, j n 2 n 2, if j 1, n 1 2 Since q 1, we can also write P n in terms of the olynomials f s q 1 + q + q 2 + +q s 1, so that P n q l R n, j q n s2 f s q 1 d The olynomial f s is always divisible by the cyclotomic olynomial s q r q ζ r, where this roduct runs through all rimitive sth roots of unity ζ r ; and in fact f s s whenever s is rime For any ositive integer s, a standard VOL 34, NO 3, MAY 2003 THE COLLEGE MATHEMATICS JOURNAL 187

6 result states that f s factors into irreducibles, each of which is cyclotomic Thus, we may write P n q l R n, j q n s2 d s d>1 d q 1 We omit the tedious details of the roof of Theorem 3; we merely note that it can be roved directly using the recursion on R n, j imosed by the one on P n Weuse R n, j in the closing section to rovide a simle roof of a robabilistic limit theorem The recursion We will obtain the robabilities for the N + 1-erson game in terms of the N-erson game Figure 3 The tree diagram for n N + 1 In Figure 3, the first subtree, which begins with node N + 1, is isomorhic to the tree for N layers That is, it is identical to the N-tree, excet that we have to subtract 1 from each node entry Thus, in this subtree, j occurs as a terminal node with robability P N 1 /, j 2 The factor occurs because we are conditioning on the event that the first coin toss before getting to the subtree is heads; the factor 1/ is necessary to account for the N + 1-length loos that are now ossible The second subtree is similar It begins with node 3 4 N + 11Tomathisto the N-tree, which begins with node N, we can subtract 2 from each node entry and then identify those entries mod N + 1 If we make this identification, then 188 c THE MATHEMATICAL ASSOCIATION OF AMERICA

7 j occurs as a terminal node in this subtree with robability qp N 2 /, j 2 Now consider the third subtree For j 3, j occurs as a terminal node in this subtree with robability q 2 P N 3 / For instance, 2 occurs in this subtree with robability q 2 P N X 1 / q 2 P N X N /,since again we interret layers modn + 1 In the last, or N + 1st, subtree, Player j wins with robability q N P N N 1, for any j N + 1 Sans the denominator, these terms are organized in Table 1 Table 1 Numerators of the recursive robabilities 1st subtree 2nd subtree 3rd subtree Nth subtree N + 1st subtree X 1 0 qp N X N q 2 P N X N 1 q N 1 P N X 2 q N P N X 1 X 2 P N X 1 0 q 2 P N X N q N 1 P N X 3 q N P N X 2 X 3 P N X 2 qp N X 1 0 q N 1 P N X 4 q N P N X 3 X N P N X N 1 qp N X N 2 q 2 P N X N 3 0 q N P N X N X N+1 P N X N qp N X N 1 q 2 P N X N 2 q N 1 P N X 1 0 We can calculate P N+1 X 1 by summing the robabilities of 1 aearing as a terminal node in all of the aforementioned N-subtrees: P N+1 X 1 1 { 0 + qpn X N + q 2 P N X N 1 + +q N P N X 1 } q N t+1 P N X t } t1 For the other layers we can use the same rocedure: P N+1 X 2 1 { PN X q 2 P N X N + q 3 P N X N q N P N X 2 } { } P N X 1 + q N t+2 P N X t, t2 1 { P N+1 X 3 PN X 2 + qp N X q 3 P N X N + + q N P N X 3 } { } P N X 2 + qp N X 1 + q N t+3 P N X t, P N+1 X N t3 1 { PN X N 1 + qp N X N 2 + VOL 34, NO 3, MAY 2003 THE COLLEGE MATHEMATICS JOURNAL 189

8 P N+1 X N+1 Therefore, in general, + q N 2 P N X q N P N X N } { PN X N 1 + qp N X N q N 2 P N X 1 + q N P N X N }, 1 { PN X N + qp N X N 1 + q 2 P N X N q N 1 P N X } { PN X N + qp N X N 1 + q 2 P N X N q N 1 P N X 1 } P N+1 { j 1 q j t 1 P N X t + t1 } q j t+n P N X t for 1 j N + 1andN 4 In fact, it can be verified quickly that this last equation holds for N 1 It is imortant to note that we have roved the recursion based on Figure 3 and not on any of the as-yet unroved theorems already stated Quite to the contrary, we will use our recurrence relation to rove those other theorems Various equivalent formulations of the recursion relation are given next Some of these will be crucial in the roofs of Theorems 1 and 2 Recursion For 1 j N + 1andN 1, 1 P N+1 { j 1 q j t 1 P N X t + t1 t j } q j t+n P N X t 2 P N+1 q j t+n+n+1 t j/n P N X t t1 { j 2 } 3 P N+1 q m P N m 1 + q m P N m+n 4 P N+1 m0 m0 m j 1 t j m j q m P N m 1+N+1 m+n j+2/n+1 Probabilities Using the recursion, we can rove Theorems 1 and 2, reviously stated Proof of Theorem 1 Put N n 1 We will use version 2 of the recursion For 1 t N, t 1 t N and 1 N N 190 c THE MATHEMATICAL ASSOCIATION OF AMERICA

9 Thus, P N+1 X 1 This roves Theorem 1 q q 1 t+n+n+1 t 1/N P N X t t1 q 1 t+n P N X t t1 q N+1 t 1 P N X t t1 q q N+1 t+n+n+1 t N 1/N P N X t t1 qp N+1 X N+1, for N 1 One can establish a bijection between sequences of coin flis that win for Player 1 and those that win for Player n Thus, an alternate roof of the receding theorem is ossible Proof of Theorem 2 Induction on n Clearly the theorem holds for n 1, 2 Now let n 2 and suose P m X i <P m X i+1 for all m n and i {1, 2,,n 1} By version 3 of the recursion, P n+1 +1 Pn+1 n+1 { j 1 q m P n m + m0 m j+1 n q m P n +1 m+n j 2 n } q m P n m m P n m+n m0 n+1 { j 2 m j q [ ] m P n m Pn m 1 m0 n + q m[ ] P n +1 m+n Pn m+n m j+1 } q j P n X n + q j 1 P n X 1 By Theorem 1, q j P n X n q j 1 P n X 1,so P n+1 +1 Pn+1 n+1 { j 2 q [ ] m P n m Pn m 1 m0 n + q [ ] } m P n +1 m+n Pn m+n m j+1 VOL 34, NO 3, MAY 2003 THE COLLEGE MATHEMATICS JOURNAL 191

10 At least one of these two sums is nonemty, and by the induction hyothesis all of its terms are ositive Thus P n+1 +1 P n+1 >0 Limits We conclude with some simly-roved results that have a nice interretation for Coin ToGa Theorem 4 For j Z +, lim n P n 0 Proof Induction on j Forn Z +, Theorem 2 says P n X 1 <P n X 2 < < P n X n, and we know that n t1 P nx t 1sinceq 1 Thus P n X 1 <1/n, so lim n P n X 1 0 Now, suose the theorem holds for the first j ositive integers: lim n P n X 1 lim n P n X 2 lim n P n 0 We will show that as a consequence lim n P n +1 0 For n j, n t j+1 P nx t 1 j t1 P nx t,andthen j terms on the left-hand side are ordered P n +1 <P n +2 < < P n X n,by Theorem 2 This imlies that 1 j P n +1 < t1 P n X t 0 n j as n j fixed, by the induction hyothesis Therefore lim n P n +1 0, comleting the induction This result says that if you remain at a fixed osition eg, you are always in the role of Player 4 then as more layers enter the game, your chances of winning decrease to zero Intuitively obvious, erhas, but it is quite useful in roving the next result Theorem 5 For n Z +, lim n P n X n 0 Proof It is a subtle oint, but this theorem is not contained in the revious one Rather, it is a consequence of Theorems 1 and 4 together: lim P n X n lim qp n X 1 q lim P n X 1 0 n n n We can interret Theorem 5 as follows Each day, a new game is held Let s assume it is not Russian roulette, so you are ermitted to attend on many consecutive days More layers enter the game each day Given a choice, you should hold on to the last seat assuming you are given a choice and the seats are not assigned randomly; that s what Theorem 2 told you Even so, Theorem 5 says that your robability of winning still declines toward zero as the number of layers increases Recall from the first section that if 0, then almost surely the game will cycle indefinitely and there will be no winner That is, lim q 1 P n 0for1 j n, which makes the next theorem all the more interesting Theorem 6 lim q 1 P n X i 1 for 1 i, j < n P n 192 c THE MATHEMATICAL ASSOCIATION OF AMERICA

11 Proof By Theorem 3 art a, lim q 1 P n X i R n,i q lim P n q 1 R n, j q R n,i 1 n 1! R n, j 1 n 1! 1 for 1 i, j < n An alternate roof would use Theorems 1 and 2 in tandem If P n X i /P n is close to 1, then Players i and j have a nearly equal chance of winning; thus, Theorem 6 is concerned with the fairness of the game If n layers want to make the game more fair, they will choose a coin that is biased more heavily towards tails As long as q > 1/2, a more biased coin actually results in a fairer game! However, there is a trade-off involved: the increased bias of the coin will lengthen the game, so that the erfectly fair game almost surely never terminates References 1 Good News Bible, American Bible Society, New York, Problem A-2, The Fifty-Eighth William Lowell Putnam Exam Mathematical Cometition, 1997 Some solutions to this exercise can be found in Mathematics Magazine , and American Mathematical Monthly E Hecke, Neuere Fortschritte in der Theorie der ellitischen Modulfunktionen, Comtes rendus du congrès international des mathématiciens Oslo Mathematics Without Words Is there any trigonometric identity that can t be established with a wordless icture? Here is Roger Nelsen Lewis & Clark College, nelsen@lclarkedu with an examle that is out of the ordinary tanπ/4 + α tanπ/4 α 1 VOL 34, NO 3, MAY 2003 THE COLLEGE MATHEMATICS JOURNAL 193