Possible Probability and Irreducibility of Balanced Nontransitive Dice

Hur, Injo; Kim, Yeansu

doi:https://doi.org/10.1155/2021/6648248

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Abstract Introduction Notation and Preliminaries Data Availability Conflicts of Interest Acknowledgments References Copyright Related Articles

Research Article | Open Access

Volume 2021 | Article ID 6648248 | https://doi.org/10.1155/2021/6648248

Possible Probability and Irreducibility of Balanced Nontransitive Dice

Injo Hur¹and Yeansu Kim¹

Academic Editor: Antonio Di Crescenzo

Received13 Nov 2020

Revised11 Jan 2021

Accepted15 Jan 2021

Published29 Jan 2021

Abstract

We construct irreducible balanced nontransitive sets of -sided dice for any positive integer . One main tool of the construction is to study so-called fair sets of dice. Furthermore, we also study the distribution of the probabilities of balanced nontransitive sets of dice. For a lower bound, we show that the winning probability can be arbitrarily close to . We hypothesize that the winning probability cannot be more than , and we construct a balanced nontransitive set of dice whose probability is .

1. Introduction

A nontransitive triple of dice consists of three dice, labeled , and , with the property that all three probabilities that rolls higher than , that rolls higher than , and that rolls higher than are greater than . We write this as , and . It is called nontransitive because if we define the relation as for , then the relation is not transitive. For example, the following set of 6-sided dice is nontransitive since :

Note that nontransitive sets of dice are first introduced by Gardner [1], further studied in [2, 3], and have been generalized in several directions (see [4–14]). In [14], Schaefer and Schweig constructed balanced nontransitive sets of -sided dice for any positive integer (see below for the definition of “balanced” and other terms). Main idea of the construction in [14] is to combine several balanced nontransitive sets of dice. Therefore, the sets of dice that are constructed in [14] are reducible, and Schaefer and Schweig [14] question whether there exist balanced irreducible nontransitive sets of -sided dice for all . One main purpose of the paper is to construct irreducible nontransitive sets of -sided dice for any positive integer . Our main idea of the construction is to use so-called fair sets of 2-sided dice. Here, being fair means that probabilities , and are all . Although we used 2-sided dice to construct new sets of dice, understanding fair sets of -sided dice for any positive integer seems to be an important step to understand all irreducible balanced nontransitive dice. We also study fair sets of -sided dice for any positive integer .

The second purpose of the paper is to study possible probabilities of balanced nontransitive sets of -sided dice, i.e., possible value of . As far as we know, previous known constructions of balanced nontransitive sets of dice have probability . We first show that there exist balanced nontransitive sets of dice such that is arbitrarily small, i.e., is a sharp lower bound for balanced nontransitive set of dice. We conjecture that, in general, . To support the conjecture, we first explicitly calculate and prove that both the probability in our paper and the one in [14] are less than . We also provide another new construction of a balanced nontransitive set of dice such that , which is less than , and is, as far as we know, the maximum among known constructions of balanced nontransitive sets of dice.

Let us now describe the content of our paper. After we introduce notation and preliminaries in Section 2, we study fair sets of dice in Section 3. In Section 4, we provide construction of irreducible balanced sets of -sided dice for any positive integer using fair sets of 2-sided dice. In Section 5, we calculate and study possible probability for balanced nontransitive sets of dice.

2. Notation and Preliminaries

We briefly recall definitions and notations in [14] that we are going to use.

Definition 1. Fix an integer . A set of -sided dice is a collection of three pairwise-disjoint sets , and with and . We think of dice , and as being labeled with the elements of , and , respectively, and we assume that each die is fair (i.e., the probability of rolling any one of its numbers is ).

Definition 2. A set of dice is called as follows:(i)Balanced if (ii)Nontransitive if each of , and exceeds (iii)Fair if

Definition 3. If is a set of -sided dice, define a word by the following rule: the letter of corresponds to the die on which the number labels a side.

For a word , the number of letters in the word, which we call the length of , is denoted by . A number of (resp. , ) in is denoted by (resp. , ). For example, if corresponds to a set of -sided dice, .

Let be a set of -sided dice and be its corresponding word (Definition 3). Then, we denote by or the probability that the number rolled on A is greater than the number rolled on B when we roll and . Similarly, we define , and .

Definition 4. For a set of -sided dice (and its corresponding word with length ), we let and . Similarly, we also define , , and .

It is by definition that is balanced if , and is nontransitive if .

Remark 1. in Definition 4 represents the number of consequences that a die beats a die when we consider all possible outcomes when we roll and . Note that is the same as the notation in [14].

Example 1. Let be the following set of 3-sided dice:Then, and . Therefore, this set of dice is balanced and nontransitive.

Definition 5. The concatenation of two words and is simply the word followed by , denoted by .

We recall a recursive relation of , and . The following is in the proof of Lemma 2.4 in [14].

Lemma 1. (see (1) in [14]). Let and be two words that correspond to two sets of dice, respectively. Let and (i.e., corresponding dice are -sided and -sided, respectively). Then, we have

3. Fair Sets of Dice

In the following two lemmas “probabilities” mean , , and and , , or is one of , , or .

Lemma 2. Assume that has and , i.e., . Then, exchanging the orders of and at the same time, i.e., does not change the “probabilities.”

Lemma 3. Assume that is a word having three different letters, say , , and . Then, and have the same “probabilities” if and only if has the same numbers of , , and .

Definition 6. Two words are called similar, , if can be rewritten to by exchanging the orders as allowed by virtue of Lemmas 2 and 3.

Note that, if , then two words have the same length and , and .

Example 2. Lemmas 2 and 3 indicate that and if and only if .

Conjecture 1. (fair dice). If is fair, then the length of is a multiple of 6 and it is similar to an iterated concatenation of .

Note that is a fair set of 2-sided dice, which will be used in Section 4.

Example 3. Let be . It is fair and

To support Conjecture 1, let us consider only two-die case, i.e., a fair word with only two letters and . In this case, a set of dice and a corresponding word are called fair if .

Theorem 1. Assume that a given fair word has letters with . Then, .

Proof. We may assume that starts with (if not, we exchange the notations of and ). Then, let us claim that the factor can be extracted to the front without changing probabilities, i.e., , where is a fair word having letters with . Mathematical induction with this fact leads us to the result that .
Observe first that , where is a word such that is fair. If , then we are done. Suppose not, i.e., . We would like to see that contains a subword . If would not have , then the given fair dice should be expressed by , which is not fair at all. Due to having in , exchanging and together will hold probabilities, i.e., , where is the word obtained from by replacing by . A similar argument works to reveal that has and therefore . Keep doing this procedure until .
Next see that as claimed. If starts with , then we are done. If not, there are three cases, that is, is one of , , . Since a similar argument can be applied, let us consider the first case when . If would not have , then it should be expressed by . This means that , which is impossible to be fair. So, contains a subword . Exchanging (in ) for (in front of ) shows that . As mentioned, in the other two cases, a similar argument reveals that as claimed.

4. Irreducible Sets of Dice

In this section, we construct irreducible balanced nontransitive sets of -sided dice for any positive integer . Note that this answers Question 5.2 in [14].

Definition 7. A balanced nontransitive word (and a corresponding set of dice) is called irreducible if there do not exist balanced nontransitive words and (both nonempty) such that .

Lemma 4. Let be a balanced nontransitive word and . Then, is balanced and nontransitive.

Proof. is balanced but not nontransitive since . Lemma 2.4 in [14] implies that is balanced. It remains to prove that it is nontransitive. Let . Equation (2) impliesSimilarly, we have . Therefore, is also nontransitive (Definition 4).

Lemma 5. Let be an irreducible balanced nontransitive word that corresponds to a set of either 3-sided or 4-sided dice and . Then, is an irreducible balanced nontransitive word.

Proof. Applying Lemma 4 times, we conclude that is balanced and nontransitive. It remains to prove that it is irreducible. Suppose that is reducible. Then, we can write as where is an irreducible balanced nontransitive word and is a balanced nontransitive word. Since is fair (so not nontransitive) and is not balanced for any nonnegative integer , should be of the form ( is not empty). We write as . Since , is irreducible. This implies that corresponds to a balanced nontransitive set of 4-sided dice and it contains a balanced nontransitive word of length less than 12, which contradicts that is irreducible.

We are now ready to construct an irreducible balanced nontransitive set of -sided dice.

Theorem 2. For any , there exists an irreducible balanced nontransitive set of -sided dice.

Proof. We first consider the following two sets of dice:Both and are irreducible, balanced, and nontransitive. As before, we consider . We now construct an irreducible balanced nontransitive set of -sided dice for any . We already constructed such a set of dice when and . When is odd, we write . Then, is an irreducible balanced nontransitive set of -sided dice due to Lemma 5. When is even, we write . Then, is an irreducible balanced nontransitive set of -sided dice due to Lemma 5.

5. Possible Probability

Let be a balanced nontransitive set of -sided dice with . In this section, our interest is in a possible probability . Let us state what is on our mind.

Conjecture 2. If is a balanced nontransitive set of -sided dice such that , then

5.1. Probabilities of Our Construction in Theorem 2

To support Conjecture 2, we first calculate all possible probabilities of the set of dice which are constructed in Section 4.

Lemma 6. Let (resp. ) be a word of length (resp. ) that corresponds to a set of dice. Then,Similarly, we have

Proof. It is an easy consequence of equation (2) since .

Remark 2. Note that represents the number of consequences that a die beats a die . Therefore, in Lemma 5.2 represents how far is from being fair. For example, if , then .

Lemma 7. Let and be as in Lemma 6. Assume that and . Then, .

Proof. Equation (2) with and impliessince .

We are now ready to calculate the probability . Let be a word that is constructed in Theorem 2. Then, is either or for some nonnegative integers , where

Note that is a fair word (i.e., ). For , Lemma 2.1 implies thatwhich is the closest integer greater than . Here, is the length of and, therefore, .

Therefore, Conjecture 2 is true in this case.

Remark 3. Similarly, we can also calculate possible probabilities of the set of dice that is constructed in [14]. Let be a word that is constructed in Theorem 2.1 of [14]. Then, is a product of several , and , where for , where . Therefore, Lemma 7 implies that . Therefore, Conjecture 2 is true in this case.

5.2. Best Bound for Probability

We first show that is the greatest lower bound of , which means that the inequality is optimal.

Theorem 3. There exists a balanced nontransitive set of dice such that becomes arbitrary small.

Proof. Let . Then, we construct a word (or a corresponding set of dice) as follows. First, consider the word which is not balanced, since the number of the events for to beat or is one more than the ones of the events for or to beat (therefore, always beats both and at least in probability). To change this word to a balanced one, let us replace by on the first factor . In all, the word obtained byis balanced. A direct computation of on , or more precisely,shows thatwhich goes to as . This indicates that is the greatest lower bound of and the inequality could not get better.

For the upper bound of , recall that Conjecture 2 states that

To support this inequality, we provide three new constructions of sets of dice of which probability is close to . For simplicity, assume that (later, we discuss when or which is similar to the present case). Let us start with the most unmixed fair -sided dice ,where “” means that all the letters are the same as the boundary letters and the numbers of letters in “” are the same, i.e., .

We apply the following algorithm to construct the largest probability .

Algorithm 1. Step 1: if there is no triple in a word, then replace and in a word for so that the resulting word contains . Step 2: if you find , then replace by , respectively.

Remark 4. Replacement in Step 1 does not change the probability due to Lemma 2. Replacement in Step 2 increases the number of events for (resp. or ) to beat (resp. or ) by 1. In particular, doing this replacement keeps the words balanced and nontransitive. Therefore we may expect that keeping replacing in this way would lead the largest probability of .

Since there is no in , we first apply Step 1 (exchange of and ) several times to in order to obtain a word which is still fair but having sequels by(i.e., the last one-third of the first in (17) moves to the middle and the last one-third of the second in (17) does to the front of the second (the moved ones are in bold). Here, in (18). Therefore, the numbers of replacing and by and , respectively, are the same as .)

We now apply Step 2 times to obtain a balanced nontransitive word by(i.e., in (18), the first moves in front, the second does to almost middle, and the last does to the end. The moved ones are in bold as before). Here, the probability of increases to .

Since there is no and in , we apply Step 1 times as follows; we move and in to the ones in (all of which are in bold) in order to produce the products and .

This means that we need to exchange to in the middle part (which is blue and has two dots above the character on ) times ( to move and another to beat ). Then, to increase , one more exchange from to in the middle part is necessary.

In addition to this observation, we would like to measure how many times we can apply Step 2 to in terms of (this is because the denominator of is , so we would better express in in order to examine asymptotic behavior of ). Put by the number of further replacement such that -many ’s move between and in front. For this, we need extra -many exchanges of and in the middle part. Since the remaining of each ’s or ’s in the middle is , the possible exchange of to is , which should be at least (the last is due to move -many ’s in the further replacement). As a summary, we have that

Therefore, the largest probability on in the argument above is

Note that the maximum (in ) means that, since all are exhausted in the last word, the probability is not able to be larger in this construction.

The cases when and can be investigated by a similar procedure. By applying Step 1 (or exchanging by ) and Step 2 (or replacing , , and by , , and ), the probability increases by(which is similar to the increase when ). In these cases, after doing this procedure, new words are expressed by

Here, each has exactly letters. For example, .

Next, to make the largest , we do Step 2 further as in the previous case. Shortly speaking, when , we have that

Hence, the increased probability on in the argument (i.e., except ) is

Note that, since the numerator is increasing in , the last inequality holds for all .

Similarly, when , we have that

Hence, the increased probability on for (again except ) is

Note again that the numerator increases in .

Let us summarize the argument above. Start with the most unmixed fair word (which is (17)). By replacing , , by , , , i.e., (Step 2) as much as possible, the probability increases most. Then, the computation above tells us thatwhich upholds Conjecture 2.

Remark 5. (i)Let us tell why we think the last constructed word in this argument provides the largest probability on . First, observe that we have considered the only case when was even (when , , or ). We, however, believe that this would be enough due to the following observation. A direct computation shows that there is only one possible probability when , or equivalently, any three balanced, nontransitive dice with 3 sides can be expressed by a balanced, nontransitive word which is similar to . In this case, . Then, add at the end of and then exchange (in ) to . Then, the constructed word is , which becomes balanced and nontransitive. For this, , which is greater than . A similar computation can be done for more general . Therefore, with this trick (however, we do not know if we can do this trick at all times), the probability of would get greater for even ’s than odd ones.(ii)Second, proving that the abovementioned probability is the largest probability related to construction of all balanced nontransitive sets of dice and it also seems to be related to property of fair sets of dice, i.e., Conjecture 1, and we leave this for future work.

Remark 6. In [13], Schaefer further generalized the results in [14] to sets of -sided dice for any positive integer . It will be very interesting to generalize our results to sets of dice for and we also leave this for future work.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The first author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Korea government (NRF-2019R1F1A1061300). The second author was partially supported by Chonnam National University (Grant no. 2018-0978) and by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIP) (no. 2017R1C1B2010081). The authors would like to thank the referees for a number of corrections and very useful suggestions.

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Copyright © 2021 Injo Hur and Yeansu Kim. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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