Robert E. Kennedy Curtis N. Cooper. Central Missouri State University Warrensburg, Missouri
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1 THE STATISTICS OF THE SMALLEST SPACE ON A LOTTERY TICKET by Robert E Kennedy Curtis N Cooper Central Missouri State University Warrensburg Missouri Introduction A day hardly goes by without hearing that some lucky person has become an instant millionaire by winning a lottery Recently one of the authors was visiting relatives in Florida when a sequence of winning lottery numbers was announced (In the Florida state lottery one chooses six distinct integers from 1 to 9 Someone suggested that a person might just as well choose and as any other sequence In fact why not choose any six consecutive integers what difference does it make? The chances are the same as any other sequence of six distinct integers! This led to the following analysis of the least interval between consecutive members of a sequence of six integers Here we are concerned with the set of possible lottery tickets for the Florida state lottery That is the set of all possible six distinct integers from 1 to 9 The calculation given below can be generalized to r integers from 1 to n are chosen The generalization will be given at the end of this article For clarity however we will use Florida s lottery as an example of the technique involved In what follows we let L be the set of all possible Florida lottery tickets That is L { ( t 1 t 2 t 3 t t 5 t : 1 t 1 < t 2 < t 3 < t < t 5 < t 9 } 1
2 We also define the function f on L by: f ( t 1 t 2 t 3 t t 5 t min { t i+1 t i : i } Thus if t L we can think of f(t as the smallest space on the ticket t Our purpose is to determine the mean smallest space with respect to the members of L That is t L f(t ( 9 will be determined Determination of the Mean of the Smallest Spaces of L Consider the set of 5-tuples D { ( d 1 d 2 d 3 d d 5 : 5 d 1 + d 2 + d 3 + d + d 5 8 ; d i 1 } and the function F : L D defined by F ( (t 1 t 2 t 3 t t 5 t ( t 2 t 1 t 3 t 2 t t 3 t 5 t t t 5 It is clear that F is a function from L onto D This will enable us to efficiently determine f(t t L by use of a particular partition of D If d D we will denote { t L : F (t d } by F 1 (d and the number of elements of F 1 (d by #F 1 (d It is immediate that #F 1 (d 9 s 2
3 where d ( d 1 d 2 d 3 d d 5 and s d 1 + d 2 + d 3 + d + d 5 For d ( d 1 d 2 d 3 d d 5 D we define s(d d 1 + d 2 + d 3 + d + d 5 a(d min { d 1 d 2 d 3 d d 5 } m(d # { i : d i a(d } When the context is clear we will just write s a or m We see then that f(t a ( F (t and that 5 s 8 1 a 9 and 1 m 5 For each triple ( i j k with 5 i 8 1 j 9 and 1 m 5 we define D ijk { d D : s(d i; a(d j; m(d k } and note that D { D ijk : 5 i 8; 1 j 9; 1 k 5 } is a partition of D Since ( f(t (9 i j (#D ijk t D D ijk D we proceed to determine the right side of (* by first considering each k and 5 For this we use the following theorem Its statement and proof are found in [1: Theorem 23; pp 15 1] 3
4 T heorem For integers r r 1 r 2 r n the number of solutions to x 1 + x x n r x i r i for i 1 2 n is ( n 1 + r r1 r 2 r n n 1 Thus if we let s and a be given we use the above theorem to find the number of solutions to d 1 + d 2 + d 3 + d + d 5 s d i r i ; i for m and 5 For example if m 1 d i a for some i and d j a + 1 for j i Since there are ( 5 1 ways to choose the di and by the theorem x 1 + x 2 + x 3 + x s a x i a + 1; i has ( s 5a 1 3 solutions it follows that ( s 5a 1 #D sa1 5 3 Similarly we obtain ( ( ( 5 s 5a 1 s 5a 1 #D sa ( ( ( 5 s 5a 1 s 5a 1 #D sa ( ( ( 5 s 5a 1 s 5a 1 #D sa and #D sa5 ( ( 5 s 5a ( s 5a 1 1
5 It should be noted here that we will use the convention ( { n 1 if n 1; 1 0 otherwise Thus for fixed s and a 5 #D sak k1 5 k1 ( ( 5 s 5a 1 k k and by [2: page 58; formula 21] we have 5 ( ( s 5a + s 5a 1 #D sak k1 Hence which is by the above f(t 8 t L i5 j1 k (9 i j (#D ijk 9 (9 i j 5 i5 j1 k1 #D ijk 8 i5 9 [( ( ] i 5j + i 5j 1 (9 i j j1 and by telescoping the inner sum 8 i5 j1 9 9 j1 i0 9 ( i 5j + (9 i ( ( 9 i i 5j + 9 ( 5 5j j1 1 by [2; page 58; formula 25] We have then that the mean smallest space on a (Florida lottery ticket is ( 55j t L ( f(t 9 j1 9 ( 9 which is approximately 188 5
6 Distribution Of interest also would be a list of how lottery tickets are distributed with respect to the smallest space concept For example how many of the ( 9 Florida state lottery tickets have a smallest space of 3? This can be readily answered by noting that by omitting the j factor in the summand of (* and summing with a fixed j we have that the number of Florida lottery tickets with a smallest space of a is D ijk D ja (9 i #D ijk which simplifies to ( 5 5a ( 9 5a Hence we can construct the following list of how the Florida lottery tickets are distributed with respect to the smallest space idea smallest space number of such tickets Thus it can be observed that close to 91% of all possible Florida state lottery tickets have a smallest space of 1 2 or 3 It seems then that it might be wise to choose a lottery ticket that has a smallest space of 1 2 or 3 and avoid those with a smallest space greater than 3 Conclusion As stated earlier the above could be generalized to a lottery where r
7 numbers from the sequence n are chosen Using the same technique as before it is easily shown that the mean of the smallest space of all possible lottery tickets where r numbers are chosen from n is [ ] n1 r1 j1 ( n(r1(j1 r ( n r and that the number of such lottery tickets with a least space of a is ( n (r 1(a 1 r ( n (r 1a r Of course another approach in investigating lottery tickets might be to analyze the collection of lottery tickets with respect to the largest space on a ticket This should also be of interest and we encourage the reader to make such an analysis Ref erences [1] JL Mott A Kandel TP Baker Discrete Mathematics for Computer Scientists Reston Publishing Co (1983 [2] DE Knuth The Art of Computer Programming Addison-Wesley Publishing Co (199 vol 1 7
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Curtis Cooper f Robert E* Kennedy, and Milo Renberg Dept. of Mathematics, Central Missouri State University, Warrensburg, MO 64093-5045 (Submitted January 1997-Final Revision June 1997) 0. INTRODUCTION
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