EQUAL LABELINGS FOR P K -SIDED DICE

Transcription

1 EQUAL LABELINGS FOR P K -SIDED DICE A Thesis Presented to the Faculty of California State Polytechnic University, Pomona In Partial Fulfillment Of the Requirements for the Degree Master of Science In Mathematics By Omar Mendez 2015

2 SIGNATURE PAGE THESIS: EQUAL LABELINGS FOR P K -SIDED DICE AUTHOR: Omar Mendez DATE SUBMITTED: Spring 2015 Mathematics and Statistics Department Dr. Amber Rosin Thesis Committee Chair Mathematics & Statistics Dr. John Rock Mathematics & Statistics Dr. Robin Wilson Mathematics & Statistics ii

3 ACKNOWLEDGMENTS Graduate school has been the most fruitful experience of my life. Here, I learned to love mathematics through professors like Amber Rosin, Robin Wilson, and John Rock. I also met people in my math courses that I now consider lifelong friends. I would like to thank all these people for the love and support I have received these past few years. iii

4 ABSTRACT For m n-sided dice we call the possible sums m, m + 1, m + 2,..., mn the standard sums. We wish to relabel m n-sided dice in such a way that the standard sums have an equal likelihood of occurring. We call such a relabeling an equal relabeling. We will look at the case n = P k where P is prime and k is some positive integer. We will find the form of m for which m n-sided dice have an equal relabeling and we will provide the labeling. We also show that such equal relabelings require stupid dice and find formulas that characterize the least and greatest number of stupid dice for a given m. iv

5 Contents 1 Equal Labelings The Problem The Order of an Element Back to the Problem Stupid Dice The Inequality Lemmas Proving the Inequality Most/Least Stupid Dice The Case of 5 4-sided Dice The General Case Further Questions 20 Bibiliography 21 v

6 Chapter 1 Equal Labelings 1.1 The Problem A die is a solid that has markings on each of its faces. Dice are used for many activities, such as gambling or games involving probabilities. Because of this, dice are preferably constructed to have equal sides, equal angles, and have the same number of faces meet at each vertex. This is exactly the structure of the five platonic solids, and thus it is important to discuss dice that have the form of these solids. Recall that the five platonic solids are the tetrahedron (4-sided), cube (6-sided), octrahedron (8-sided),dodecahedron (12-sided), and icosahedron (20-sided). The form of these platonic solids can be further generalized. For example, the number of sides of a cube has the form pq where p and q are distinct primes. The number of sides of the docecahedron and icosahedron have the form 4p where p is a prime. Lastly, the number of sides of the tetrahedron and octrahedron have the form P k where P is prime and k is some positive integer. The case of dice with pq faces is examined 1

7 in [3]. We will focus on dice that have P k faces. Suppose we have a pair of four-sided dice where each die is labeled with 1 to 4 dots on each of the 4 faces. The standard sums for a pair of four-sided dice are 2, 3, 4, 5, 6, 7, and 8. When we look at the different ways to roll each sum we get probabilities ,,,,,, One may ask if it is possible to relabel a pair of four-sided dice in such a way that we have an equal likelihood for each of the standard sums to occur? The answer is no. If it were the case, for some integer t, we would have probabilities t t t t t t t,,,,,, In addition, the sum of these probabilities has to equal 1 and thus we obtain the 7t equation 16 = 1. This implies that 7 divides 16 which is not true. What if we consider more than 2 dice? Say 5? Consider the following labeling. D 1 : D 2 : D 3 : D 4 : D 5 : The first two dice lead us to sums 2, 3,..., 8, 9. Including the third dice yields sums 3, 4,..., 17, 18. Adding the last two dice gives us sums 5, 6,..., 19, 20. For probabilities, we see that we have two choices of which face to use for die 2 and die 3 and four choices on the last two dice. Hence each sum can be realized = 2 6 2

8 different ways. Thus the sums occur with equal probability. We just saw that a labeling was not possible for a pair of four-sided dice, but it was possible for five four-sided dice. One may now ask, for what m values do m P k -sided dice have an equal labeling? We will need the following definitions and lemmas to help us with this question. 1.2 The Order of an Element Definition. The order of a(mod b), denoted a b, is the least positive integer x such that a x (mod b) 1(mod b). As an example, consider the order of 2(mod 5). Notice that 2 1 2(mod 5) 2 2 4(mod 5) 2 3 3(mod 5) 2 4 1(mod 5). Therefore 4 is the least positive integer that works, and so the order of 2(mod 5) is 4. Lemma 1. If P k 1(mod b), then k = l P b for some l Z +. Proof. Say P k 1(mod b). Since P b is the smallest integer such that P P b 1 (mod b), k P b. Thus by the division algorihm, k = l P b + r where 3

9 0 r < P b. Now, 1 P k (mod b) P l P b+r (mod b) P l P b P r (mod b) 1 P r (mod b) P r (mod b) Since r < P b, it must be that r = 0. Therefore k = l P b. 1.3 Back to the Problem Suppose we have m many n = P k sided dice. Then dice rolling allows us to acquire the possible sums m, m + 1,..., mn. We can see that there are mn (m 1) many standard sums with a total of n m possible outcomes. In order to have an equal m labeling, mn (m 1) must divide n = (P k ) m and so mn (m 1) = P a for some a Z +. Further, leads to mn (m 1) = P a m(n 1) = P a 1 which yields P a 1 m =. n 1 Notice that m(n 1) = P a 1 tells us that P a 1(mod n 1). From the latter and the fact that P n 1 = k, lemma 1 tell us that we have a = kt for some t Z +. 4

10 Thus, P kt 1 n t 1 m = =. (1.1) P k 1 n 1 What we have shown is that for an equal relabeling of m n-sided dice to exist, m must have the form in (1.1). Two natural questions arise from this. If m has the form described above, will the number of standard sums always divide the number of outcomes? Also, is it guaranteed that such an m will always yield an equal relabeling of m n-sided dice? P kt 1 Let us explore the first question. Suppose m =. Then the number of P k 1 standard sums is m(n 1) + 1 = m(p k 1) + 1 = P kt. We know that the number of outcomes is n m = P km. Now, we need the number of standard sums to divide the number of outcomes. Thus P kt P km. Therefore we need the following inequality to hold: t < m. (1.2) In section 2.2, we will show that this inequality does indeed hold. As far as the second question is concerned, we will now see that if m has the form in (1.1), then m n-sided dice do have an equal labeling. Theorem 2. Let n = P k, where p is prime. Then there is an equal labeling of m n-sided dice if and only if m = n n t 1 1 where t 1. Proof. We already showed above that if we have an equal labeling, then m has the n t 1 n 1 1 form in (1.1). Now suppose m = where t 1. If t = 1, then m = = 1. n 1 n 1 This means that we have one die, and thus an equal labeling is 1, 2, 3,..., p k. 5

11 Let t > 1. Label the first t dice as follows: die 1 : 1, 2, 3,..., n die 2 : 1, 1 n + 1, 2 n + 1,..., (n 1) n + 1 die 3 : 1, 1 n 2 + 1, 2 n 2 + 1,..., (n 1) n die t : 1, 1 n t 1 + 1, 2 n t 1 + 1,..., (n 1) n t Since every integer can be uniquely expressed in base n, we get exactly one copy of the sums t, t + 1, t + 2,..., n t + t 1. Next, label the remaining m t dice with 1 ' s on every face. These m t dice yield m t m t n copies of the sum m t. Combining all dice, we get n copies of the sums t + (m t), t (m t), t (m t),..., n t + t 1 + (m t) which is equivalent to m, m + 1, m + 2,..., n t 1 + m. 6

12 Note that n t 1 n t 1 + m = n t 1 + n 1 (n t 1)(n 1) + n t 1 = n 1 n t+1 n = n 1 n(n t 1) = n 1 = nm. So we do, in fact, get the standard sums m, m + 1,..., nm, and they all occur n m t times, so they are equally likely. We just saw that our labeling depends on being able to label the first t dice. Since we have a total of m dice, in order for our labeling to make sense, it must be that P kt 1 t m =. (1.3) n 1 We will ultimately prove an inequality in section 2.2 that proves that (1.3) holds n t 1 for t 1. So if m has the form then there is an equal labeling. n Stupid Dice Something that can be noticed in our labeling is the use of dice with 1 ' s on every face. We will call such dice stupid dice. One may ask if it is possible to avoid stupid dice and if so, under what conditions? We will now explore this. Let us start by considering the sum m. In order to acquire the sum m from m dice we need each die to possess at least one face labeled 1. Now, let a i be the number 7

13 of 1 ' s on die i. Since each die has a i different ways of landing on a 1 and there are m dice, the product of these a i s yields the number of ways to obtain m. We also know that the number of ways to roll a sum of m is given by the total number of outcomes divided by the number of standard sums. That is, Thus, m n. m(n 1) + 1 n m a 1 a 2... a m 1 a m = m(n 1) + 1 n m = n t m t = n = P k(m t). If we want to avoid stupid dice, we need a i = n for each i. From this and the fact that a i P k(m t), each a i = P b i where 0 b i < k. Hence a i P k 1 for each i and so it follows that Thus for no stupid dice we would need Hence m a P m(k 1) i. i=1 m P k(m t) = a i P m(k 1). i=1 km kt = k(m t) m(k 1) = mk m, and so kt m. 8

14 Therefore kt m. We will show that this inequality does not hold for t 2 by showing that m > kt holds for t 2 instead. This then tells us that stupid dice are unavoidable. 9

15 Chapter 2 The Inequality To show that stupid dice are necessary, we will need to show that P kt 1 m = > kt. (2.1) P k Lemmas We will eventually show that this inequality always holds for t 2. Note that if (2.1) holds then (1.2) and (1.3) hold as well. First we will prove a series of lemmas. P kt 1 P kt Lemma 3. For t 2, m = >. P k 1 P k Proof. For t 2, we know P k < P tk, and hence P k (P kt 1) = P kt+k P k > P kt+k P kt = P kt (P k 1). 10

16 Thus P kt 1 P k 1 P kt P k >. Lemma 4. For t 2, P kt k kt. Proof. We will prove this inequality via induction on t. For the base case t = 2, P 2k k = P k 2 k 2k. Suppose P kt k kt holds. Now, P k(t+1) k = P kt+k k = P kt k+k = P kt k P k ktp k kt 2 2tk (t + 1)k. Therefore for t 2, P kt k kt. 2.2 Proving the Inequality We are now in a position to prove the inequality in (2.1). Theorem 5. If m = P kt 1 P k 1 and t > 2, then m > kt. Proof. With the use of Lemma 3 and Lemma 4 we see that, P kt 1 m = > P k 1 P kt = P kt k kt. P k 11

17 We just showed that in order to create an equal labeling of m, n = P k -sided dice, stupid dice must be used. This inequality also allows us to show that (1.2) and (1.3) hold. Since m > kt, m > kt t and so m > t. Note that m t follows from m > t. We now know without question that if m has the form in (1.1), the number of standard sums divides the number of outcomes and that m having this form always allows an equal labeling. 12

18 Chapter 3 Most/Least Stupid Dice This unavoidability of stupid dice in equal labelings raises various questions. Is there a minimum amount of stupid dice that can be used in a labeling? Is there a maximum amount? Can we characterize these? We will entertain such questions with a concrete example. 3.1 The Case of 5 4-sided Dice Let n = 2 2 and m = 5. The first labeling is given by 13

19 D 1 : D 2 : Labeling 1 D 3 : D 4 : D 5 : Notice that the first two dice yield sums 2, 3, 4, 5. Including the third dice, we get sums 3, 4,..., 9, 10. Now, using the fourth die as well generates sums 4, 5,...,18, 19. The last die brings us to the sums 5, 6,..., 19, 20. Let us observe the probabilities associated with this labeling. For each sum there are two choices of which face to use on die 1 through die 4 and four choices on the last die. Therefore each sum can be realized = 2 6 different ways. We do indeed have an equal labeling. Since stupid dice are unavoidable and we only used one stupid die, this labeling exemplifies the case where the minimum amount of stupid dice are used. For the n = 2 2 and m = 5 case, we do have at least one other labeling. D 1 : D 2 : Labeling 2 D 3 : D 4 : D 5 :

20 Here, the first two dice give us possible sums 2, 3,..., 16, 17. Including the last three dice, we get the sums 5, 6,..., 19, 20. What about the probabilities? Since we have four choices of which face to use on die 3 through die 5, each sum can be realized 4 3 = 2 6 different ways. Again, an equal labeling. We will show that three is the maximum amount of stupid dice that can be used in a labeling by considering the product of a i s where a i is the number of 1 ' s on die i. We know that a 1 a 2 a 3 a 4 a 5 yields the number of ways to obtain m. The number of ways to get any sum is the number of outcomes divided by the number of sums, which is = = 2 6. Thus, a 1 a 2 a 3 a 4 a 5 = 2 6. Maximizing the amount of stupid dice used translates to choosing the maximum amount of a i s to equal 4 and still have this equation hold. This is done by choosing three a i s to equal 4 and the remaining a i s to equal 1. The product is now a 1 a 2 a 3 a 4 a 5 = = 2 6. Therefore three is the maximum amount of stupid dice that can be used in a labeling. Hence our labeling has the most amount of stupid dice that can be used. We just showed that for n = 2 2 and m = 5, one is the minimum amount of stupid dice that can be used in a labeling and three is the maximum amount. Is there a labeling which contains only two stupid dice? 15

21 Let us look at the following labeling: D 1 : D 2 : Labeling 3 D 3 : D 4 : D 5 : Notice that two stupid dice are used in this labeling. Also, the first two dice lead us to sums 2, 3,..., 8, 9. Including the third dice yields sums 3, 4,..., 17, 18. Adding the last two dice gives us sums 5, 6,..., 19, 20. For probabilities, we see that we have two choices of which face to use for die 2 and die 3 and four choices on the last two dice. Hence each sum can be realized = 2 6 different ways. 3.2 The General Case n t 1 Let us now look at the general case where n = P k and m =. We want to n 1 find an equal labeling which contains the maximum amount of stupid dice. First, we will figure out how many stupid dice are needed for such a labeling. We know that the number of ways to get a sum of m is given by the number of outcomes divided by the number of standard sums. Therefore m n m n m t n t = = n = = yields the number of ways to mn m + 1 obtain the sum m. Let a i be the number of 1 s on die i. We know the product P k(m t) P km kt a 1 a 2... a m also gives the number of ways to get m. Thus a 1 a 2... a m 1 a m = 16

22 P k(m t). Getting the maximum amount of stupid dice translates to choosing the maximum amount of a i s to equal P k. This happens when m t of the a i s equal P k and the rest of the a i s equal 1. Therefore the maximum amount of stupid dice that can be used is m t. We actually already considered an equal labeling which uses m t stupid dice. The labeling was shown in the proof of Theorem 2. What is the least amount of stupid dice that can be used? Again, consider the product of a i s where a i is the number of 1 s on die i. In order to minimize the number of stupid dice that may arise from this product we will start by setting each a i to equal P k 1 and then increasing some a i s to get the desired number of ways to obtain m. If each a i = P k 1, then the product a 1 a 2... a m would be P m(k 1) = P mk m, but we want the product to be P km kt. So we need to distribute m kt more P ' s. The way to do this that makes the least number of a i s equal to P k is as follows: P p k P k..... P k P _ k P p k 1 P k P k 1 P k 1 _ = P k(m kt) P kt(k 1) m-kt times kt times = P km k2 t P k2 t kt = P km k2 t+k 2 t kt = P km kt. Thus the minimum amount of stupid dice that can be used in an equal labeling is m kt. Let us now consider a labeling with m kt stupid dice. First, for any integer z let z represent the sequence z, z,..., z, z. We will call the sequence z a z block. Consider p.. _ P k 1 times the following labeling. 17

23 Label the first kt dice die 1: 1, 2, 3, 4,..., P die 2: 1, P + 1, 2P + 1, 3P + 1,..., (P 1)P + 1 die 3: 1, P 2 + 1, 2P 2 + 1, 3P 2 + 1,..., (P 1)P die 4: 1, P 3 + 1, 2P 3 + 1, 3P 3 + 1,..., (P 1)P die kt: 1 P kt 1 + 1, 2P kt 1 + 1, 3P kt 1 + 1,..., (P 1)P kt Since every integer can be uniquely expressed in base P, each sum will use exactly one block from each die, so we get (P k 1 ) kt copies of the sums kt, kt + 1,..., P kt + kt 1. Next, label the remaining m kt dice with 1 s on every face. These m kt dice give us n m kt = P k(m kt) copies of the sum m kt. Combining all dice, we get (P k 1 ) kt (P k ) m kt = (P k2 t kt )(P km k2t ) = P k2 t kt+km k 2 t = P km kt copies of the sums m, m + 1,..., P kt + m 1. 18

24 Notice that P kt P kt 1 + m 1 = P kt + 1 P k 1 P kt (P k 1 ) + P kt 1 (P k 1) = P k 1 P k2 t P kt + P kt 1 P k + 1 = P k 1 P k2 t P k = P k 1 P k (P kt 1) = P k 1 = nm, so we do indeed get the standard sums. 19

25 Chapter 4 Further Questions In section 3.2 we found that the minimum amount of stupid dice that a labeling may contain when considering m many n = P k sided die is m kt. The maximum amount of stupid dice was shown to be m t. A natural question that arises is if we can always find a labeling with j stupid dice where m kt < j < m t? Similarly, given a particular number of stupid dice, how many different labelings are possible with that number of stupid dice? The labelings we have shown all involve (possibly repeated) arithmetic sequences; must all labelings look like this? Lastly, is it possible to extend the results for n = P k sided dice to any positive integer n? 20

26 Bibliography [1] J. Gallian, Contemporary Abstract Algebra, Houghton Mifflin Company, [2] A. Medina, Relabeling PQ-Sided Dice:An Equal Probability Problem, California State Polytechnic University Pomona, [3] F. Bermudez, A. Medina, A. Rosin, E. Scott. Are Stupid Dice Necessary?, The College Mathematics Journal, Vol. 44, No. 4, September