Konstantinos Drakakis. 1. Introduction

Transcription

1 FACTA UNIVERSITATIS (NIŠ) Ser. Math. Infor. Vol. 22, No. 1 (2007), pp A NOTE ON THE APPEARANCE OF CONSECUTIVE NUMBERS AMONGST THE SET OF WINNING NUMBERS IN LOTTERY Konstantinos Drakakis Abstract. We copute the nuber of ways in which integers can be chosen aong the integers 1,..., n, so that any 2 consecutive choices are at least k integers apart. Subsequently, we prove an interesting fact about Lottery: the winning 6 nubers (out of 49) in the gae of the Lottery contain 2 consecutive nubers with a surprisingly high probability (alost 50%). We also propose an alost fair casino gae based on this fact. 1. Introduction The gae of lottery exists and has been run in any countries (such as the UK, the US, Gerany, France, Ireland, Australia, Austria, Greece, Spain, etc.) for a nuber of years. In this gae, the player chooses nubers fro aong the nubers 1,..., n >, the order of the choice being uniportant and the values of n and varying fro country to country; the lottery organizers choose publicly nubers in the sae way, and if they are the sae as the ones the player chose, the player wins. Newspapers usually publish the winning set of nubers, along with (siple) statistics on the nuber of ties each particular nuber fro 1 to n has appeared in the winning set, while very sophisticated statistical studies of the lottery outcoes have already appeared in various papers in the atheatical literature [1, 2, 3]. It is however a slightly different and ore elusive statistical observation that will be of interest to us here. It has been noticed that, in the usual case Received January 30, Matheatics Subject Classification. Priary 05A05; Secondary 91A60. 1

2 2 K. Drakakis = 6 and n = 49, at least 2 of the winning nubers are very often close to each other. As 6 out of 49 is not really any, this sees at first to be paradoxical, if not altogether wrong, and ay reind us strongly of another very siilar faous paradox, the Birthday Paradox. In this work we will prove that this observation is well founded, even if we adopt the strictest interpretation of nubers being close, i.e. that they be consecutive. Our proble to solve then will be the following: What is the probability that, out of > 0 nubers drawn uniforly randoly fro the range 1,..., n, where n >, at least 2 are consecutive, or, ore generally, that no 2 nubers are closer than k integers apart? We will calculate this probability in 2 ways below: one quite echanical, by finding a recursion and then solving it by eans of generating functions, and one cobinatorial, which will actually yield a ore general result. We will also see that this proble, at least for the usual values = 6 and n = 49, leads to a novel and unexpected gabling application. 2. First Solution Let f(n, ) be the nuber of ways in which nubers can be chosen out of 1,..., n so that no two are consecutive. For any particular choice, one of the following will hold: Neither 1 nor n is chosen: we have to choose nubers aong 2,..., n 1 and the nuber of ways this can be accoplished in is f(n 2, ). 1 and/or n is chosen: the nuber of ways this can be accoplished in is, according to the inclusion-exclusion principle, the su of the nuber of ways of choosing 1 and choosing n inus the nuber of ways in choosing both. Observe now that 2 cannot be chosen if 1 is, and that n 1 cannot be chosen if n is. Then, in the first two cases the nuber of choices is f(n 2, 1), and in the last one f(n 4, 2), so that the total nuber of choices if 1 and/or n is chosen is 2f(n 2, 1) + f(n 4, 2). Accordingly, suing both cases: f(n, ) = f(n 2, ) + 2f(n 2, 1) f(n 4, 2). In addition to the recursive forula above, we need soe boundary conditions as well, corresponding to n = 0, 1, 2, 3 and = 0, 1. They are provided by the following:

3 A Note on the Appearance of Consecutive Nubers... 3 We can choose no nubers in only one way: ( n 0) f(n, 0) = 1. We can choose one nuber in n ways: ( n 0) f(n, 1) = n. f(3, 2) = 1. Let us now write down the generating function for f(n, ): F (z, w) = n=4 =2 f(n, )z n w. The upper boundary for is deterined by the fact that f(n, ) = 0 if + 1. By ultiplying the recursion forula by z n w, and applying the operator, we get: n=4 =2 where F (n, ) = F 1 (n, ) + 2F 2 (n, ) F 3 (n, ), F 1 (n, ) = F 2 (n, ) = F 3 (n, ) = n=4 =2 n=4 =2 n=4 =2 f(n 2, )z n w, f(n 2, 1)z n w, f(n 4, 2)z n w. For each of the three functions, we get F 1 (n, ) = +1 =2 = z 2 n=4 =2 = z 2 [ F (z, w) + z 3 w 2], f(n, )z n+2 w = z 2 =2 f(n, )z n w + f(3, 2)z 3 w 2 f(n, )z n w

4 4 K. Drakakis F 2 (n, ) = +1 =2 = wz 2 =1 = wz 2 = z 2 w [ n=4 =2 f(n, 1)z n+2 w = f(n, )z n w =1 f(n, )z n w + f(3, 2)z 3 w 2 + F (z, w) + z 3 w 2 + w ] nz n, f(n, )z n+2 w +1 f(n, 1)z n w F 3 (n, ) = +1 =2 = w 2 z 4 =0 = w 2 z 4 f(n, 2)z n+4 w = n=4 =2 f(n, )z n w =0 f(n, )z n w + f(3, 2)z 3 w 2 + ] f(n, 1)z n w + f(n, 0)z n ] = z 4 w [F 2 (z, w) + z 3 w 2 + w nz n + z n. We still need three auxiliary coputations: ( ) z nz n = z nz n 1 2 = z = z 2 2 z 1 z (1 z) 2 ; z n = 1 1 z ; nz n = z ( ) 1 nz n 1 = z = 1 z z (1 z) 2. f(n, )z n+4 w +2

5 A Note on the Appearance of Consecutive Nubers... 5 Putting all of the above together, and after soe further algebraic siplifications, we find: F (z, w) = w 2 z z(z 3 + w(z 1)2 ) (z 1) 2 (1 z wz 2 ). Of course, this is not the full generating function, as the cases n = 1, 2, 3 and = 0, 1 are entirely issing; we oitted the in order to avoid to have to deal with weird boundary conditions such as f( 3, 1) etc. But now we can add the back. Reeber that f(n, 0) = 1, n 0 and f(n, 1) = n, n 1; but we have already carried out the relevant coputations as auxiliary coputations above. Therefore: F(z, w) = F (z, w) + z 3 w z + zw (1 z) 2, where the first fraction is the generating function for f(n, 0) and the second for f(n, 1). After soe algebraic siplifications, we find: 1 + zw F(z, w) = 1 z wz 2 = 1 + zw 1 z wz 2 = 1 z(1 + zw) z 1 z(1 + wz) = 1 [z(1 + wz)] n z n=1 n ( ) n ( ) n + 1 = z n+ 1 w = z n w, so that n=1 =0 =0 n= ( ) n + 1 f(n, ) =. If then we draw nubers fro the range 1,..., n, the probability no two are consecutive is: ) q(n, ) = ( n +1 ( n ), so that the solution to our original proble is: p(n, ) = 1 ( n +1 ) ( n ). We should note here that a proof of the forula for f(n, ) based on induction appears in [6].

6 6 K. Drakakis 3. Second Solution The second solution, cobinatorial in nature, allows us to solve a ore general proble: in how any ways f k (n, ) can we choose nubers aong the nubers 1,..., n so that the iniu distance between any two of our choices (which we will be calling the distance of our choice) is k > 0? There is a very siple forula for that. Iagine we have nubered n balls with the nubers 1,..., n, and that we have chosen the nubers 1 N 1 <... < N n. For every nuber chosen but the last one, reove the nubers of the k 1 balls iediately following it; as for the reaining balls, renuber the consecutively and in the order they are. We will end up with n (k 1)( 1) balls nubered consecutively fro 1 to n (k 1)( 1), and (k 1)( 1) blank ones. This final situation will not depend on the balls we chose originally, although the exact positioning of the blank balls aong the nubered ones will. Notice finally that the original nuber of every ball can be recovered: it is the nuber of balls preceding it, including itself! Any valid choice of nubers in the original nubering will correspond to a choice of nubers after renubering, and vice versa: after we choose nubers between 1 and n (k 1)( 1), we insert blanks as described above and renuber, getting a valid choice of nubers in the original nubering. This correspondence is obviously bijective. Therefore, ( ) n (k 1)( 1) f k (n, ) =, n > > 1, k 1. For k = 2 we recover the result of our first solution, and hence the sae probability p(n, ) of at least two choices being consecutive. We also obtain the ore general forula p k (n, ) = 1 ( n (k 1)( 1) ( n ) ) for the probability that at least two of the winning nubers have a distance less than k. 4. Application in Gabling The probability p(n, ) can actually be quite large, aybe unexpectedly large: for exaple, for the usual values n = 49 and = 6, we find p(49, 6)

7 A Note on the Appearance of Consecutive Nubers Therefore, the observation that the winning six nubers of the lottery often contain two that are very close is well founded; in alost one gae out of two the winning set of nubers contains 2 consecutive ones! Moreover, as p(49, 6) is very close to 0.5, the proble we just studied can be turned into a successful casino gae: the player bets ee that 6 nubers randoly chosen aong 1,..., 49 will contain at least two consecutive ones. If this happens, the player gets ee fro the house, otherwise the house wins the player s oney. This gae is alost fair, as the player has an alost 50% chance to win; but (s)he actually has slightly less than that, and this gives the house a profitable advantage! Indeed, the house wins oney at the approxiate rate of e 1 2p(49, 6) cent per every e placed as a bet. 5. A Slight Variant What would happen, though, if the player suggests that nubers 1 and n be treated as consecutive as well, naely if we order the nubers on a ring instead of a line? There should now be fewer possible choices for nonconsecutive nubers. Indeed, let now g k (n, ) be the nuber of possible choices of n aong n > 0 nubers so that the iniu distance between any two of the chosen ones is k; in other words, aong any two chosen nubers, with the property that no nuber between the is chosen, there are at least k 1 nubers lying between the. Then, we can split the choices into those in which one nuber aong 1,..., k 1 is chosen, and those in which this is not the case: If one ball aong 1,..., k 1 is chosen, then the reaining 1 balls can be chosen aong n 2k + 1 balls (we exclude the chosen ball and the k 1 adjacent balls on either side); but now, by reoving a block of 2k 1 balls fro the circle, we turn it into a line, so the total nuber of choices, for a fixed choice within 1,..., k, is f k (n 2k + 1, 1); and since every different choice within 1,..., k leads to different possible choices, the total nuber of choices in this category is (k 1)f k (n 2k + 1, 1). If no ball is chosen aong 1,..., k 1, then we can just reove the, turn the circle into a line, and renuber: we need to choose balls aong the reaining n k + 1, obeying the distance restrictions, and this can happen in f k (n k + 1, ) ways. Therefore,

8 8 K. Drakakis g k (n, ) = (k 1)f k (n 2k+1, 1)+f k (n k+1, ), n > > 1, k 0. If we define now p k (n, ) = 1 g k(n, ) ( n ) = 1 ( n k+1 (k 1)( 1) ) ( + n 2k+1 (k 1)( 2) ) we find that p 2 (49, 6) = p(49, 6) Therefore, if soe casino agreed to play this variant of the gae with a player, the player would have a slight advantage over the house, and the latter would loose oney, at a rate of e 2p(49, 6) per e placed as a bet! ( n ) 1 Table 5.1 gives the values of p k (49, 6) and p k (49, 6) for k N. Table 5.1: The probabilities that the winning set of nubers of the standard Lottery has a iniu distance less than k. k p k (49, 6) p k (49, 6) Exaples We now offer 2 exaples fro actual lotteries, the Austrian Lottery (n = 45, = 6) and the Greek Lottery (n = 49, = 6). The results can be viewed in Table 6., and it is clear that the fit is very good in both cases. The observed frequencies of the Austrian lottery coe fro ore than 10 years of data (1400 entries in total) [4], while the observed frequencies of the Greek lottery coe fro the coplete results of the years fro 1990 to 2006 (1527 entries in total) [5].

9 A Note on the Appearance of Consecutive Nubers... 9 Table 6.2: A coparison between the observed and the expected frequencies of occurrence of a given inial distance in the winning set of lottery nubers for the Austrian (left) and the Greek (right) lottery, according to data sets spanning ore than a decade Distance Observed Expected Distance Observed Expected Suary and Conclusion We counted the nuber of ways in which integers can be chosen aong the integers 1,..., n, so that any 2 consecutive choices are at least k integers apart, in the 2 cases where 1 and n are considered non-consecutive (linear order) or consecutive (ring order). To do so, we followed 2 different ethods: the first was based on generating functions, copletely echanical (what one ight label as technique before insight ), but restricted to the case k = 2 only; the second solved the general proble through a cobinatorial arguent, and is uch shorter and, in our opinion, uch clearer. A natural odel for this proble is the winning set of nubers in lottery. For the particular case n = 49, = 6, k = 2, we discovered the (probably surprising and counterintuitive, as in the case of the Birthday Paradox) result that the probability of finding such a set of nubers is just below 50% if 1 and 49 are considered non-consecutive, and just above 50% if they are considered consecutive. This leads to the design of an alost fair casino gae, where, depending on whether the forer or the latter case holds, either the house or the player gain a very slight but systeatic edge over the other, respectively. 8. Acknowledgeents The author would like to thank an anonyous student of his for counicating to the author his observation about the frequency of appearance of

10 10 K. Drakakis consecutive nubers in the set of the Lottery winning nubers, and thus stiulating hi to write this article. He would also like to thank E. Neuwirth at the Didactic Center for Coputer Science, University of Vienna, for providing the data on the Austrian Lottery. Finally, he would like to thank the anonyous reviewer whose coents resulted to a title describing the content of the paper uch ore accurately. R E F E R E N C E S 1. N. Henze: The distribution of spaces on lottery tickets. Fibonacci Quarterly 33 (1995), N. Henze and H. Riedwyl: How to win ore: Strategies for increasing a lottery win. A.K. Peters, Natick, Massachusetts, L. Holst: On discrete spacings and the Bose-Einstein distribution. Contributions to probability and statistics, Hon. G. Blo, 1985, E. Neuwirth: Personal counication H. Ryser: Cobinatorial Matheatics. Carus Matheatical Monographs, Claude Shannon Institute Ireland Konstantinos.Drakakis@ucd.ie University College Dublin Electronic & Electrical Engineering Belfield, Dublin 4 Ireland