possibleprobabilityandirreducibilityofbalanced …...a nontransitive triple of dice consists of...

Research ArticlePossible Probability and Irreducibility of BalancedNontransitive Dice

Injo Hur and Yeansu Kim

Department of Mathematics Education Chonnam National University Gwangju City Republic of Korea

Correspondence should be addressed to Yeansu Kim ykimjnuackr

Received 13 November 2020 Revised 11 January 2021 Accepted 15 January 2021 Published 29 January 2021

Academic Editor Antonio Di Crescenzo

Copyright copy 2021 Injo Hur and Yeansu Kimis is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

We construct irreducible balanced nontransitive sets of n-sided dice for any positive integer n Onemain tool of the construction isto study so-called fair sets of dice Furthermore we also study the distribution of the probabilities of balanced nontransitive sets ofdice For a lower bound we show that the winning probability can be arbitrarily close to 12 We hypothesize that the winningprobability cannot be more than (12) + (19) and we construct a balanced nontransitive set of dice whose probabilityis (12) + (13 minus

153

radic24) asymp (12) + (1912)

1 Introduction

A nontransitive triple of dice consists of three dice la-beled A B and C with the property that all threeprobabilities that A rolls higher than B that B rolls higherthan C and that C rolls higher than A are greater than 12We write this as P(AgtB)gt (12) P(BgtC)gt (12) andP(CgtA)gt (12) It is called nontransitive because if wedefine the relation X≻Y as P(XgtY)gt (12) forX Y isin A B C then the relation ≻ is not transitive Forexample the following set of 6-sided dice is nontransitivesince P(AgtB) P(BgtC) P(CgtA) (1936)

1813 10 7

54A

1714 12 9

32B

1615 11 8

61C

Note that nontransitive sets of dice are first introducedby Gardner [1] further studied in [2 3] and have beengeneralized in several directions (see [4ndash14]) In [14]Schaefer and Schweig constructed balanced nontransitivesets of n-sided dice for any positive integer nge 3 (see below

for the definition of ldquobalancedrdquo and other terms)Main idea ofthe construction in [14] is to combine several balancednontransitive sets of dice erefore the sets of dice that areconstructed in [14] are reducible and Schaefer and Schweig[14] question whether there exist balanced irreducible non-transitive sets of n-sided dice for all n One main purpose ofthe paper is to construct irreducible nontransitive sets ofn-sided dice for any positive integer n Our main idea of theconstruction is to use so-called fair sets of 2-sided dice Herebeing fair means that probabilities P(AgtB) P(BgtC) andP(CgtA) are all 12 Although we used 2-sided dice toconstruct new sets of dice understanding fair sets of n-sideddice for any positive integer n seems to be an important step tounderstand all irreducible balanced nontransitive dice Wealso study fair sets of n-sided dice for any positive integer n

e second purpose of the paper is to study possibleprobabilities of balanced nontransitive sets of n-sided diceie possible value of P(AgtB) As far as we know previousknown constructions of balanced nontransitive sets of dicehave probability (12)ltP(AgtB)lt (35) (12) + (110)We first show that there exist balanced nontransitive sets ofdice such that P(AgtB) minus (12)gt 0 is arbitrarily small ie12 is a sharp lower bound for balanced nontransitive set of

HindawiJournal of MathematicsVolume 2021 Article ID 6648248 8 pageshttpsdoiorg10115520216648248

dice We conjecture that in general(12)ltP(AgtB)lt (12) + (19) To support the conjecturewe first explicitly calculate and prove that both the proba-bility P(AgtB) in our paper and the one in [14] are less than(12) + (19) We also provide another new construction ofa balanced nontransitive set of dice such thatP(AgtB) asymp (12) + (1912) which is less than(12) + (19) and is as far as we know the maximumamong known constructions of balanced nontransitive setsof dice

Let us now describe the content of our paper After weintroduce notation and preliminaries in Section 2 we studyfair sets of dice in Section 3 In Section 4 we provideconstruction of irreducible balanced sets of n-sided dice forany positive integer n using fair sets of 2-sided dice InSection 5 we calculate and study possible probabilityP(AgtB) for balanced nontransitive sets of dice

2 Notation and Preliminaries

We briefly recall definitions and notations in [14] that we aregoing to use

Definition 1 Fix an integer ngt 0 A set of n-sided dice is acollection of three pairwise-disjoint sets A B and C with|A| |B| |C| n and AcupBcupC 1 3n We think ofdice A B and C as being labeled with the elements of A Band C respectively and we assume that each die is fair (iethe probability of rolling any one of its numbers is 1n)

Definition 2 A set of dice is called as follows

(i) Balanced if P(AgtB) P(BgtC) P(CgtA)

(ii) Nontransitive if each of P(AgtB) P(BgtC) andP(CgtA) exceeds 12

(iii) Fair if P(AgtB) P(BgtC) P(CgtA) (12)

Definition 3 If D (A B C) is a set of n-sided dice define aword σ(D) by the following rule the ith letter of σ(D)

corresponds to the die on which the number i labels a side

For a word σ the number of letters in the word which wecall the length of σ is denoted by |σ| A number of A (resp BC) in σ is denoted by |A|σ (resp |B|σ |C|σ) For example if σcorresponds to a set of n-sided dice |A|σ |B|σ |C|σ n

Let D (A B C) be a set of n-sided dice and σ be itscorresponding word (Definition 3) en we denote byPD(AgtB) or Pσ(AgtB) the probability that the numberrolled on A is greater than the number rolled on B when weroll A and B Similarly we definePD(BgtC) Pσ(BgtC) PD(CgtA) and Pσ(CgtA)

Definition 4 For a set of n-sided dice D (and its corre-sponding word σ with length 3n) we letND(AgtB) ≔ n2PD(AgtB) ND(BgtC) ≔ n2PD(BgtC) andND(CgtA) ≔ n2PD(CgtA) Similarly we also defineNσ(AgtB) Nσ(BgtC) and Nσ(CgtA)

It is by definition that D (A B C) is balanced ifND(AgtB) ND(BgtC) ND(CgtA) and D (A B C)

is nontransitive if ND(AgtB) ND(BgtC) ND(CgtA)gt(n22)

Remark 1 ND(AgtB) in Definition 4 represents thenumber of consequences that a die A beats a die B when weconsider all possible outcomes when we roll A and B Notethat ND(AgtB) is the same as the notation 1113936siAq+

σ(D)(si) in[14]

Example 1 Let D be the following set of 3-sided dice

A 9 5 1

B 8 4 3

C 7 6 2

(1)

en σ(D) ACBBACCBA and P(AgtB) P(BgtC)

P(CgtA) (59) erefore this set of dice is balancedand nontransitive

Definition 5 e concatenation of two words σ and τ issimply the word σ followed by τ denoted by στ

We recall a recursive relation of N(AgtB) N(BgtC)and N(CgtA) e following is in the proof of Lemma 24 in[14]

Lemma 1 (see (1) in [14]) Let σ and τ be two words thatcorrespond to two sets of dice respectively Let |σ| 3m and|τ| 3n (ie corresponding dice are m-sided and n-sidedrespectively) 1en we have

Nστ(AgtB) Nσ(AgtB) + Nτ(AgtB) + mn (2)

3 Fair Sets of Dice

In the following two lemmas ldquoprobabilitiesrdquo mean P(AgtB)P(BgtC) and P(CgtA) and x y or z is one of A B or C

Lemma 2 Assume that σ(D) has xy and yx ieσ(D) middot middot middot xy middot middot middot yx middot middot middot 1en exchanging the orders of xy

and yx at the same time ie σ( 1113957D) middot middot middot yx middot middot middot xy middot middot middot does notchange the ldquoprobabilitiesrdquo

Lemma 3 Assume that σ(D) is a word having three differentletters say x y and z 1en xyzσ(D) and σ(D)xyz havethe same ldquoprobabilitiesrdquo if and only if σ(D) has the samenumbers of x y and z

Definition 6 Two words are called similar σ(D1) sim σ(D2)if σ(D1) can be rewritten to σ(D2) by exchanging the ordersas allowed by virtue of Lemmas 2 and 3

Note that if σ(D1) sim σ(D2) then two words have thesame length and PD1

(AgtB) PD2(AgtB)

PD1(BgtC) PD2

(BgtC) and PD1(CgtA) PD2

(CgtA)

2 Journal of Mathematics

Example 2 Lemmas 2 and 3 indicate thatmiddot middot middot xy middot middot middot yx middot middot middot sim middot middot middot yx middot middot middot xy middot middot middot and xyzσ(D) sim σ(D)xy

z if and only if |x|σ(D) |y|σ(D) |z|σ(D)

Conjecture 1 (fair dice) If D is fair then the length of D is amultiple of 6 and it is similar to an iterated concatenation ofxyzzyx

Note that τ ≔ ABCCBA is a fair set of 2-sided dicewhich will be used in Section 4

Example 3 Let σ(D) be AABBCCCCBBAA It is fair and

AABBCCCCBBAA sim ABABCCCCBABA sim ABACBCCBCABA sim ABCABCCBACBA sim ABCCBAABCCBA (AB

CCBA)2

To support Conjecture 1 let us consider only two-diecase ie a fair word with only two letters A and B In thiscase a set of dice and a corresponding word are called fair ifP(AgtB) P(BgtA) (12)

Theorem 1 Assume that a given fair word σf has 4m letterswith |A| |B| 2m 1en σf sim (ABBA)m

Proof We may assume that σf starts with A (if not weexchange the notations of A and B) en let us claim thatthe factor ABBA can be extracted to the front withoutchanging probabilities ie σf sim ABBAωf where ωf is a fairword having 4(m minus 1) letters with |A| |B| 2(m minus 1)Mathematical induction with this fact leads us to the resultthat σf sim (ABBA)m

Observe first that σf sim ABω0 whereω0 is a word such thatABω0 is fair If σf ABω0 then we are done Suppose not ieσf A middot middot middot ABω1 We would like to see that ω1 contains asubword BA If ω1 would not have BA then the given fair diceshould be expressed by σf A middot middot middot ABA middot middot middot AB middot middot middot B which isnot fair at all Due to having BA in ω1 exchanging AB and BA

together will hold probabilities ie σf sim A middot middot middot BAω2 whereω2is the word obtained fromω1 by replacing BA byAB A similarargument works to reveal that ω2 has BA and thereforeσf sim A middot middot middot BAAω3 Keep doing this procedure untilσf sim ABω0

Next see that σf sim ABBAωf as claimed If ω0 starts withBA then we are done If not there are three cases that is ω0is one of ABω4 A middot middot middot ABω5 B middot middot middot BAω6 Since a similarargument can be applied let us consider the first case whenω0 ABω4 If ω4 would not have BA then it should beexpressed by A middot middot middot AB middot middot middot B is means thatσf sim ABABA middot middot middot AB middot middot middot B which is impossible to be fair Soω4 contains a subword BA Exchanging BA (in ω4) for AB

(in front of ω4 ) shows that σf sim ABBAωf As mentioned inthe other two cases a similar argument reveals thatσf sim ABBAωf as claimed

4 Irreducible Sets of Dice

In this section we construct irreducible balanced non-transitive sets of n-sided dice for any positive integer n Notethat this answers Question 52 in [14]

Definition 7 A balanced nontransitive word (and a corre-sponding set of dice) is called irreducible if there do not existbalanced nontransitive words σ1 and σ2 (both nonempty)such that σ σ1σ2

Lemma 4 Let σ be a balanced nontransitive word andτ ABCCBA 1en τσ is balanced and nontransitive

Proof τ is balanced but not nontransitive sinceP(AgtB) P(BgtC) P(CgtA) (12) Lemma 24 in[14] implies that τσ is balanced It remains to prove that it isnontransitive Let |σ| 3n Equation (2) implies

Nτσ(AgtB) 2 + Nσ(AgtB) + 2ngt 2 +n2

2+ 2n

(n + 2)2

2

(4)

Similarly we have Nτσ(BgtC) Nτσ(CgtA)gt ((n + 2)22) erefore τσ is also nontransitive (Definition 4)

Lemma 5 Let σ be an irreducible balanced nontransitiveword that corresponds to a set of either 3-sided or 4-sided diceand τ ABCCBA 1en (τ)kσ is an irreducible balancednontransitive word

Proof Applying Lemma 4 k times we conclude that (τ)kσis balanced and nontransitive It remains to prove that it isirreducible Suppose that (τ)kσ is reducible en we canwrite (τ)kσ as ππprime where π is an irreducible balancednontransitive word and πprime is a balanced nontransitiveword Since (τ)n is fair (so not nontransitive) and (τ)nABC

is not balanced for any nonnegative integer n π should beof the form (τ)kπPrime (πPrime is not empty) We write (τ)kσ as(τ)kπPrimeπ Since |π|lt 12 π is irreducible is impliesthat σ πPrimeπ corresponds to a balanced nontransitive setof 4-sided dice and it contains a balanced nontransitiveword π of length less than 12 which contradicts that σ isirreducible

We are now ready to construct an irreducible balancednontransitive set of n-sided dice

Theorem 2 For any nge 3 there exists an irreducible bal-anced nontransitive set of n-sided dice

Proof We first consider the following two sets of dice

Journal of Mathematics 3

A3 9 5 1

B3 8 4 3

C3 7 6 2

A4 10 7 5 4

B4 12 9 3 2

C4 11 8 6 1

(5)

Both D3 (A3 B3 C3) and D4 (A4 B4 C4) are irre-ducible balanced and nontransitive As before we considerτ ABCCBA We now construct an irreducible balancednontransitive set of n-sided dice for any nge 3 We alreadyconstructed such a set of dice when n 3 and n 4 Whennge 5 is odd we write n 3 + 2k kge 1en (τ)kσ(D3) is anirreducible balanced nontransitive set of n-sided dice due toLemma 5 When nge 5 is even we write n 4 + 2k kge 1en (τ)kσ(D4) is an irreducible balanced nontransitive setof n-sided dice due to Lemma 5

5 Possible Probability

Let (A B C) be a balanced nontransitive set of n-sided dicewith P(AgtB) P(BgtC) P(CgtA)gt (12) In this sec-tion our interest is in a possible probability P(AgtB) Let usstate what is on our mind

Conjecture 2 If (A B C) is a balanced nontransitive set ofn-sided dice such that P(AgtB)gt (12) then

12lt1113874 1113875P(AgtB)lt

12

+19 (6)

51 Probabilities of Our Construction in 1eorem 2 Tosupport Conjecture 2 we first calculate all possible proba-bilities of the set of dice which are constructed in Section 4

Lemma 6 Let σ (resp τ) be a word of length 3m (resp 3n)that corresponds to a set of dice 1en

Pστ(AgtB) 12

+Nσ(AgtB) minus m

221113872 11138731113872 1113873 + Nτ(AgtB) minus n221113872 11138731113872 1113873

(m + n)2

(7)

Similarly we have

Pστ(BgtC) 12

+Nσ(BgtC) minus m

221113872 11138731113872 1113873 + Nτ(BgtC) minus n221113872 11138731113872 1113873

(m + n)2

Pστ(CgtA) 12

+Nσ(CgtA) minus m

221113872 11138731113872 1113873 + Nτ(CgtA) minus n221113872 11138731113872 1113873

(m + n)2

(8)

Proof It is an easy consequence of equation (2) sinceNστ(AgtB) (m + n)2Pστ(AgtB)

Remark 2 Note that Nσ(AgtB) represents the number ofconsequences that a die A beats a die B erefore(Nσ(AgtB) minus (m22)) in Lemma 52 represents how far σ isfrom being fair For example if Nσ(AgtB) minus (m22) 0then Pσ(AgtB) (12)

Lemma 7 Let σ and τ be as in Lemma 6 Assume that(12)ltPσ(AgtB)lt a and (12)ltPτ(AgtB)lt b 1en(12)ltPστ(AgtB)ltmax a b

Proof Equation (2) with Nστ(AgtB) (m + n)2Pστ(AgtB)

Nσ(AgtB) m2Pσ(AgtB) and Nτ(AgtB) n2Pτ(AgtB)

implies

(m + n)2Pστ(AgtB) m

2Pσ(AgtB) + n

2Pτ(AgtB) + mn

ltmax a b m2

+ n2

+1

max a b mn1113888 1113889ltmax a b (m + n)

2

(9)

since (12)ltmax a b

We are now ready to calculate the probability P(AgtB)Let σ be a word that is constructed in eorem 2 en σ iseither (τ)kτ1 or (τ)kτ2 for some nonnegative integers kwhere

D τ1( 1113857

9 5 1

8 4 3

7 6 2

⎧⎪⎪⎨

⎪⎪⎩

D τ2( 1113857

10 7 5 4

12 9 3 2

11 8 6 1

⎧⎪⎪⎨

⎪⎪⎩

D(τ)

6 1

5 2

4 3

⎧⎪⎪⎨

⎪⎪⎩

(10)

Note that (τ)k is a fair word (ie Pτk (AgtB) (12))For i 1 2 Lemma 21 implies that

2k + ni( 11138572Pσ(AgtB) (2k)

2Pτk (AgtB) + ni( 1113857

2Pτi

(AgtB) + 2kni

(2k)

2

2+

n2i + 22

1113890 1113891 + 2kni ni + 2k( 1113857

2+ 2

21113890 1113891

(11)

which is the closest integer greater than ((ni + 2k)22) Here3ni is the length of τi and therefore ni i + 2

erefore Conjecture 2 is true in this case

Remark 3 Similarly we can also calculate possible proba-bilities of the set of dice that is constructed in [14] Let σ be aword that is constructed in eorem 21 of [14] en σ is aproduct of several τ1 τ2 and τ3 where


D τ1( 1113857

9 5 18 4 37 6 2

⎧⎪⎨

⎪⎩

D τ2( 1113857

12 10 3 19 8 7 211 6 5 4

⎧⎪⎨

⎪⎩

D τ3( 1113857

15 11 7 4 314 10 9 5 213 12 8 6 1

⎧⎪⎨

⎪⎩

(12)

Pτi(AgtB) ([n2

i + 22]n2i ) for i 1 2 3 whereni i + 2 erefore Lemma 7 implies that Pσ(AgtB)le(59) (12) + (118) erefore Conjecture 2 is true inthis case

52 Best Bound for Probability We first show that (12) isthe greatest lower bound of P(AgtB) which means that theinequality P(AgtB)gt (12) is optimal

Theorem 3 1ere exists a balanced nontransitive set of dicesuch that P(AgtB) minus (12)gt 0 becomes arbitrary small

Proof Let n 2m + 1(m isin N) en we construct a word σ(or a corresponding set of dice) as follows First consider theword (ABCCBA)m(BAC) which is not balanced since thenumber of the events for C to beat A or B is one more thanthe ones of the events for A or B to beat C (therefore C

always beats both A and B at least in probability) To changethis word to a balanced one let us replace BC by CB on thefirst factor ABCCBA In all the word σ obtained by

σ ≔ (ACBCBA)(ABCCBA)mminus 1

(BAC) (13)

is balanced A direct computation of P(AgtB) on σ or moreprecisely

Nσ(D)(AgtB) 0 + 2 + (2 + 4) + middot middot middot +(2(m minus 1) + 2m)

+(2m + 1) 2m2

+ 2m + 1(14)

shows that

P(AgtB) 2m

2+ 2m + 1

(2m + 1)2

12

+05n2 (15)

which goes to (12) as n⟶infin is indicates that (12) isthe greatest lower bound of P(AgtB) and the inequalityP(AgtB)gt (12) could not get better

For the upper bound of P(AgtB) recall that Conjecture2 states that

P(AgtB)lt12

+19 (16)

To support this inequality we provide three new con-structions of sets of dice of which probability is close to(12) + (19) For simplicity assume that n 6p (later wediscuss when n 6p + 2 or n 6p + 4 which is similar to the

present case) Let us start with the most unmixed fair n-sideddice σf

σf ≔ A middot middot middot AB middot middot middot BC middot middot middot CC middot middot middot CB middot middot middot BA middot middot middot A (17)

where ldquomiddot middot middotrdquo means that all the letters are the same as theboundary letters and the numbers of letters in ldquomiddot middot middotrdquo are thesame ie |A middot middot middot A| |B middot middot middot B| |C middot middot middot C| 3p

We apply the following algorithm to construct the largestprobability P(AgtB)

Algorithm 1

Step 1 if there is no triple AB BC CA in a word thenreplace xy and yx in a word for x y isin A B C so thatthe resulting word contains AB BC CAStep 2 if you find AB BC CA then replace AB BC CA

by BA CB AC respectively

Remark 4

Replacement in Step 1 does not change the probabilitydue to Lemma 2Replacement in Step 2 increases the number of eventsfor A (resp B or C) to beat B (resp C or A) by 1 Inparticular doing this replacement keeps the wordsbalanced and nontransitive erefore we may expectthat keeping replacing in this way would lead the largestprobability of P(AgtB)

Since there is no CA in σf we first apply Step 1 (ex-change of BC andCB) several times to σf in order to obtain aword σ1 which is still fair but having sequels CA by

σ1 ≔ A middot middot middot AA middot middot middot AA middot middot middot AB middot middot middot BB middot middot middot BC middot middot middot C

C middot middot middot CC middot middot middot CB middot middot middotBC middot middot middot CC middot middot middot CB middot middot middot B

B middot middot middot BB middot middot middot BC middot middot middotCA middot middot middot AA middot middot middot AA middot middot middot A

(18)

(ie the last one-third of the first B middot middot middot B in (17) moves to themiddle and the last one-third of the second C middot middot middot C in (17)does to the front of the second A middot middot middot A (the moved ones are inbold) Here |A middot middot middot A| |B middot middot middot B| |C middot middot middot C| p in (18)erefore the numbers of replacing BC and CB by CB andBC respectively are the same as 3p2)

We now apply Step 2 3p2 times to obtain a balancednontransitive word σ2 by

σ2 ≔ B middot middot middotBA middot middot middot AA middot middot middot AA middot middot middot AC middot middot middot CC middot middot middot C

C middot middot middot CB middot middot middotBB middot middot middot BC middot middot middot CC middot middot middot CB middot middot middot B

B middot middot middot BB middot middot middot BA middot middot middot AA middot middot middot AA middot middot middot AC middot middot middotC

(19)

(ie in (18) the first B middot middot middot B moves in front the second B middot middot middot B

does to almost middle and the last C middot middot middot C does to the ende moved ones are in bold as before) Here the probabilityP(AgtB) of σ2 increases to (12) + (3p2(6p)2) (712)

Since there is no AB and CA in σ2 we apply Step 1 6p

times as follows we move B and C in σ2 to the ones in σ3 (all


of which are in bold) in order to produce the products AB

and CA

σ2 ≔ B middot middot middot BA middot middot middot AA middot middot middot AA middot middot middot AC middot middot middot CC middot middot middot C

C middot middot middot CB middot middot middot BB middot middot middot BC middot middot middot CC middot middot middotCB middot middot middot B

B middot middot middot BB middot middot middot BA middot middot middot AA middot middot middot AA middot middot middot AC middot middot middot C

σ3 ≔ B middot middot middot BA middot middot middot AA middot middot middot AA middot middot middot ABC middot middot middot CC middot middot middot C

C middot middot middot C _B middot middot middot euroBeuroB middot middot middot euroBeuroC middot middot middot euroCeuroC middot middot middot _CB middot middot middot B

B middot middot middot BB middot middot middot BCA middot middot middot AA middot middot middot AA middot middot middot AC middot middot middot C

( _B and _C have beenmoved toB andC respectively)

(20)

is means that we need to exchange BC to CB in themiddle part (which is blue and has two dots above thecharacter on σ3) 6p times (3p to move B and another 3p tobeat C) en to increase P(AgtB) one more exchangefrom BC to CB in the middle part is necessary

In addition to this observation we would like to measurehow many times we can apply Step 2 to σ3 in terms of p (thisis because the denominator of P(AgtB) is (6p)2 so wewould better express in p in order to examine asymptoticbehavior of P(AgtB) ) Put by m the number of furtherreplacement such thatm-many Brsquos move betweenB middot middot middot B andA middot middot middot A in front For this we need extra 3pm-many ex-changes of BC and CB in the middle part Since theremaining of each Brsquos or Crsquos in the middle is 2p minus m thepossible exchange of BC toCB is (2p minus m)2 which should beat least 9pm( 3pm + 3pm + 3pm) (the last 3pm is due tomove m-many Brsquos in the further replacement) As a sum-mary we have that

(2p minus m)(2p minus m)ge 9pm(mle 2p)

⟹mle13 minus

154

radic

2p

(21)

erefore the largest probability on P(AgtB) in theargument above is

12

+3p

2+(13 minus

154

radic2)p middot 3p

(6p)2

12

+15 minus

154

radic

24asymp12

+1

925lt12

+19

(22)

Note that the maximum m (in N) means that since allBC are exhausted in the last word the probability P(AgtB)

is not able to be larger in this constructione cases when n 6p + 2 and n 6p + 4 can be in-

vestigated by a similar procedure By applying Step 1 (orexchanging BC by CB) and Step 2 (or replacing AB BC andCB by BA CB and BC) the probability P(AgtB) increases by

p(3p + 1)

[2(3p + 1)]2 (when n 6p + 2)

orp(3p + 2)

[2(3p + 2)]2 (when n 6p + 4)

(23)

(which is similar to the increase p(3p)(6p)2 when n 6p)In these cases after doing this procedure new words areexpressed by

n 6p + 2 σ2 B middot middot middot BA middot middot middot AA middot middot middot AA middot middot middot AABC middot middot middot CC middot middot middot C

C middot middot middot CCB middot middot middot BB middot middot middot BC middot middot middot CC middot middot middot CC middot middot middot B

B middot middot middot BB middot middot middot BBA middot middot middot AA middot middot middot AA middot middot middot AAC middot middot middot C

n 6p + 4 σ2prime B middot middot middot BA middot middot middot AA middot middot middot AA middot middot middot AAABBC middot middot middot CC middot middot middot C

C middot middot middot CCCB middot middot middot BB middot middot middot BC middot middot middot CC middot middot middot CCC middot middot middot B

B middot middot middot BB middot middot middot BBBA middot middot middot AA middot middot middot AA middot middot middot AAAC middot middot middot C

(24)

Here each middot middot middot has exactly p minus 2 letters For example|A middot middot middot A| p minus 2 + 2 p

Next to make the largest P(AgtB) we do Step 2 furtheras in the previous case Shortly speaking when n 6p + 2we have that

(2p minus m)(2p + 1 minus m) minus (3p + 1)ge 3m(3p + 1)(mle 2p)

hArrm2

minus (13p + 4)m + 4p2

minus p minus 11113872 1113873ge 0

rArrmle13p + 4 minus

153p2

+ 108p + 201113969

2

(25)

Hence the increased probability on P(AgtB) in theargument (ie except 12) is

p + 13p + 4 minus

153p2

+ 108p + 201113969

1113874 111387521113874 11138751113874 1113875(3p + 1)

(6p + 2)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (26)


Note that since the numerator13p + 4 minus

153p2 + 108p + 20

1113968is increasing in p the last

inequality holds for all pSimilarly when n 6p + 4 we have that

(2p minus m)(2p + 2 minus m) minus 2(3p + 2)ge 3m(3p + 2)(mle 2p)

hArrm2

minus (13p + 8)m + 4p2

minus 2p minus 41113872 1113873ge 0


153p2

+ 216p + 801113969

2

(27)

Hence the increased probability on P(AgtB) for n

6p + 4 (again except 12) is

p + 13p + 8 minus

153p2

+ 216p + 801113969

21113874 11138751113874 1113875(3p + 2)

(6p + 4)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (28)

Note again that the numerator13p + 8 minus

153p2 + 216p + 80

1113968increases in p

Let us summarize the argument above Start with themost unmixed fair word σf (which is (17)) By replacing ABBC CA by BA CB AC ie (Step 2) as much as possible theprobability P(AgtB)( P(BgtC) P(CgtA)) increasesmost en the computation above tells us that

P(AgtB)lt12

+19 (29)

which upholds Conjecture 2

Remark 5

(i) Let us tell why we think the last constructed word inthis argument provides the largest probability onP(AgtB) First observe that we have considered theonly case when n was even (when n 6p 6p + 2 or6p + 4) We however believe that this would beenough due to the following observation A directcomputation shows that there is only one possibleprobability P(AgtB) when n 3 or equivalentlyany three balanced nontransitive dice with 3 sidescan be expressed by a balanced nontransitive wordwhich is similar to σ3 CBABACACB In this casePσ3(AgtB) (59) en add CBA at the end of σ3and then exchange CA (in σ3) to AC en theconstructed word σ4 is σ4 CBABAACCBCBAwhich becomes balanced and nontransitive For thisPσ4(AgtB) (916) which is greater thanPσ3(AgtB) (59) A similar computation can bedone for more general n erefore with this trick(however we do not know if we can do this trick atall times) the probability of P(AgtB) would getgreater for even nrsquos than odd ones

(ii) Second proving that the abovementioned proba-bility is the largest probability related to construction

of all balanced nontransitive sets of dice and it alsoseems to be related to property of fair sets of dice ieConjecture 1 and we leave this for future work

Remark 6 In [13] Schaefer further generalized the results in[14] to sets of n-sided m dice for any positive integer m Itwill be very interesting to generalize our results to sets of m

dice for mge 4 and we also leave this for future work

Data Availability

No data were used to support this study

Conflicts of Interest

e authors declare that they have no conflicts of interest

Acknowledgments

e first author was supported by Basic Science ResearchProgram through the National Research Foundation ofKorea (NRF) funded by the Korea government (NRF-2019R1F1A1061300) e second author was partiallysupported by Chonnam National University (Grant no2018-0978) and by the National Research Foundation ofKorea (NRF) grant funded by the Korea government (MSIP)(no 2017R1C1B2010081) e authors would like to thankthe referees for a number of corrections and very usefulsuggestions

References

[1] M Gardner ldquoMathematical gamesrdquo Scientific Americanvol 223 no 6 pp 110ndash115 1970


[3] R P Savage ldquoe paradox of nontransitive dicerdquo 1eAmerican Mathematical Monthly vol 101 no 5 pp 429ndash4361994


[4] L Angel andM Davis ldquoA direct construction of nontransitivedice setsrdquo Journal of Combinatorial Designs vol 25 no 11pp 523ndash529 2017

[5] J Buhler A Gamst R Graham and A Hales ldquoExplicit errorbounds for lattice edgeworth expansionsrdquo in Connections inDiscrete Mathematics A Celebration of the Work of RonGraham S Butler J Cooper and G Hurlbert Edspp 321ndash352 Cambridge University Press Cambridge UK2018

[6] J Buhler R Graham and A Hales ldquoMaximally nontransitivedicerdquo 1e American Mathematical Monthly vol 125 no 5pp 387ndash399 2018

[7] B Conrey J Gabbard K Grant A Liu and K E MorrisonldquoIntransitive dicerdquo Mathematics Magazine vol 89 no 2pp 133ndash143 2016

[8] C Cooley W Ella M Follett E Gilson and L Traldi ldquoTieddice II some asymptotic resultsrdquo Journal of CombinatorialMathematics and Combinatorial Computing vol 90pp 241ndash248 2014

[9] E Gilson C Cooley W Ella M Follett and L Traldi ldquoeefron dice voting systemrdquo Social Choice and Welfare vol 39no 4 pp 931ndash959 2012

[10] J Grime ldquoe bizarre world of nontransitive dice games fortwo or more playersrdquo 1e College Mathematics Journalvol 48 no 1 pp 2ndash9 2017

[11] B G Kronenthal and L Traldi ldquoTied dicerdquo Journal ofCombinatorial Mathematics and Combinatorial Computingvol 68 pp 85ndash96 2009

[12] C M Rump ldquoStrategies for rolling the efron dicerdquo Mathe-matics Magazine vol 74 no 3 pp 212ndash216 2001

[13] A Schaefer ldquoBalanced non-transitive dice II tournamentsrdquo2017 httpsarxivorgabs170608986

[14] A Schaefer and J Schweig ldquoBalanced nontransitive dicerdquo1eCollege Mathematics Journal vol 48 no 1 pp 10ndash16 2017


dice We conjecture that in general(12)ltP(AgtB)lt (12) + (19) To support the conjecturewe first explicitly calculate and prove that both the proba-bility P(AgtB) in our paper and the one in [14] are less than(12) + (19) We also provide another new construction ofa balanced nontransitive set of dice such thatP(AgtB) asymp (12) + (1912) which is less than(12) + (19) and is as far as we know the maximumamong known constructions of balanced nontransitive setsof dice

Let us now describe the content of our paper After weintroduce notation and preliminaries in Section 2 we studyfair sets of dice in Section 3 In Section 4 we provideconstruction of irreducible balanced sets of n-sided dice forany positive integer n using fair sets of 2-sided dice InSection 5 we calculate and study possible probabilityP(AgtB) for balanced nontransitive sets of dice

2 Notation and Preliminaries

We briefly recall definitions and notations in [14] that we aregoing to use

Definition 1 Fix an integer ngt 0 A set of n-sided dice is acollection of three pairwise-disjoint sets A B and C with|A| |B| |C| n and AcupBcupC 1 3n We think ofdice A B and C as being labeled with the elements of A Band C respectively and we assume that each die is fair (iethe probability of rolling any one of its numbers is 1n)

Definition 2 A set of dice is called as follows

(i) Balanced if P(AgtB) P(BgtC) P(CgtA)

(ii) Nontransitive if each of P(AgtB) P(BgtC) andP(CgtA) exceeds 12

(iii) Fair if P(AgtB) P(BgtC) P(CgtA) (12)

Definition 3 If D (A B C) is a set of n-sided dice define aword σ(D) by the following rule the ith letter of σ(D)

corresponds to the die on which the number i labels a side

For a word σ the number of letters in the word which wecall the length of σ is denoted by |σ| A number of A (resp BC) in σ is denoted by |A|σ (resp |B|σ |C|σ) For example if σcorresponds to a set of n-sided dice |A|σ |B|σ |C|σ n

Let D (A B C) be a set of n-sided dice and σ be itscorresponding word (Definition 3) en we denote byPD(AgtB) or Pσ(AgtB) the probability that the numberrolled on A is greater than the number rolled on B when weroll A and B Similarly we definePD(BgtC) Pσ(BgtC) PD(CgtA) and Pσ(CgtA)

Definition 4 For a set of n-sided dice D (and its corre-sponding word σ with length 3n) we letND(AgtB) ≔ n2PD(AgtB) ND(BgtC) ≔ n2PD(BgtC) andND(CgtA) ≔ n2PD(CgtA) Similarly we also defineNσ(AgtB) Nσ(BgtC) and Nσ(CgtA)

It is by definition that D (A B C) is balanced ifND(AgtB) ND(BgtC) ND(CgtA) and D (A B C)

is nontransitive if ND(AgtB) ND(BgtC) ND(CgtA)gt(n22)

Remark 1 ND(AgtB) in Definition 4 represents thenumber of consequences that a die A beats a die B when weconsider all possible outcomes when we roll A and B Notethat ND(AgtB) is the same as the notation 1113936siAq+

σ(D)(si) in[14]

Example 1 Let D be the following set of 3-sided dice

A 9 5 1

B 8 4 3

C 7 6 2

(1)

en σ(D) ACBBACCBA and P(AgtB) P(BgtC)

P(CgtA) (59) erefore this set of dice is balancedand nontransitive

Definition 5 e concatenation of two words σ and τ issimply the word σ followed by τ denoted by στ

We recall a recursive relation of N(AgtB) N(BgtC)and N(CgtA) e following is in the proof of Lemma 24 in[14]

Lemma 1 (see (1) in [14]) Let σ and τ be two words thatcorrespond to two sets of dice respectively Let |σ| 3m and|τ| 3n (ie corresponding dice are m-sided and n-sidedrespectively) 1en we have

Nστ(AgtB) Nσ(AgtB) + Nτ(AgtB) + mn (2)

3 Fair Sets of Dice

In the following two lemmas ldquoprobabilitiesrdquo mean P(AgtB)P(BgtC) and P(CgtA) and x y or z is one of A B or C

Lemma 2 Assume that σ(D) has xy and yx ieσ(D) middot middot middot xy middot middot middot yx middot middot middot 1en exchanging the orders of xy

and yx at the same time ie σ( 1113957D) middot middot middot yx middot middot middot xy middot middot middot does notchange the ldquoprobabilitiesrdquo

Lemma 3 Assume that σ(D) is a word having three differentletters say x y and z 1en xyzσ(D) and σ(D)xyz havethe same ldquoprobabilitiesrdquo if and only if σ(D) has the samenumbers of x y and z

Definition 6 Two words are called similar σ(D1) sim σ(D2)if σ(D1) can be rewritten to σ(D2) by exchanging the ordersas allowed by virtue of Lemmas 2 and 3

Note that if σ(D1) sim σ(D2) then two words have thesame length and PD1

(AgtB) PD2(AgtB)

PD1(BgtC) PD2

(BgtC) and PD1(CgtA) PD2

(CgtA)








CCBA)2














2+ 2n

(n + 2)2

2

(4)









A3 9 5 1

B3 8 4 3

C3 7 6 2

A4 10 7 5 4

B4 12 9 3 2

C4 11 8 6 1

(5)





12lt1113874 1113875P(AgtB)lt

12

+19 (6)



Pστ(AgtB) 12

+Nσ(AgtB) minus m

221113872 11138731113872 1113873 + Nτ(AgtB) minus n221113872 11138731113872 1113873

(m + n)2

(7)

Similarly we have

Pστ(BgtC) 12

+Nσ(BgtC) minus m

221113872 11138731113872 1113873 + Nτ(BgtC) minus n221113872 11138731113872 1113873

(m + n)2

Pστ(CgtA) 12

+Nσ(CgtA) minus m

221113872 11138731113872 1113873 + Nτ(CgtA) minus n221113872 11138731113872 1113873

(m + n)2

(8)






implies


2Pσ(AgtB) + n

2Pτ(AgtB) + mn

ltmax a b m2

+ n2

+1


2

(9)

since (12)ltmax a b


D τ1( 1113857

9 5 1

8 4 3

7 6 2

⎧⎪⎪⎨

⎪⎪⎩

D τ2( 1113857

10 7 5 4

12 9 3 2

11 8 6 1

⎧⎪⎪⎨

⎪⎪⎩

D(τ)

6 1

5 2

4 3

⎧⎪⎪⎨

⎪⎪⎩

(10)


2k + ni( 11138572Pσ(AgtB) (2k)

2Pτk (AgtB) + ni( 1113857

2Pτi

(AgtB) + 2kni

(2k)

2

2+

n2i + 22

1113890 1113891 + 2kni ni + 2k( 1113857

2+ 2

21113890 1113891

(11)





D τ1( 1113857

9 5 18 4 37 6 2

⎧⎪⎨

⎪⎩

D τ2( 1113857

12 10 3 19 8 7 211 6 5 4

⎧⎪⎨

⎪⎩

D τ3( 1113857

15 11 7 4 314 10 9 5 213 12 8 6 1

⎧⎪⎨

⎪⎩

(12)

Pτi(AgtB) ([n2







(BAC) (13)



+(2m + 1) 2m2

+ 2m + 1(14)

shows that

P(AgtB) 2m

2+ 2m + 1

(2m + 1)2

12

+05n2 (15)



P(AgtB)lt12

+19 (16)






Algorithm 1



Remark 4






(18)






(19)







and CA








(20)




⟹mle13 minus

154

radic

2p

(21)


12

+3p

2+(13 minus

154

radic2)p middot 3p

(6p)2

12

+15 minus

154

radic

24asymp12

+1

925lt12

+19

(22)




p(3p + 1)

[2(3p + 1)]2 (when n 6p + 2)

orp(3p + 2)

[2(3p + 2)]2 (when n 6p + 4)

(23)








(24)




hArrm2

minus (13p + 4)m + 4p2

minus p minus 11113872 1113873ge 0


153p2

+ 108p + 201113969

2

(25)


p + 13p + 4 minus

153p2

+ 108p + 201113969

1113874 111387521113874 11138751113874 1113875(3p + 1)

(6p + 2)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (26)



153p2 + 108p + 20




hArrm2

minus (13p + 8)m + 4p2

minus 2p minus 41113872 1113873ge 0


153p2

+ 216p + 801113969

2

(27)



p + 13p + 8 minus

153p2

+ 216p + 801113969

21113874 11138751113874 1113875(3p + 2)

(6p + 4)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (28)


153p2 + 216p + 80



P(AgtB)lt12

+19 (29)


Remark 5






Data Availability




Acknowledgments


References























CCBA)2














2+ 2n

(n + 2)2

2

(4)









A3 9 5 1

B3 8 4 3

C3 7 6 2

A4 10 7 5 4

B4 12 9 3 2

C4 11 8 6 1

(5)





12lt1113874 1113875P(AgtB)lt

12

+19 (6)



Pστ(AgtB) 12

+Nσ(AgtB) minus m

221113872 11138731113872 1113873 + Nτ(AgtB) minus n221113872 11138731113872 1113873

(m + n)2

(7)

Similarly we have

Pστ(BgtC) 12

+Nσ(BgtC) minus m

221113872 11138731113872 1113873 + Nτ(BgtC) minus n221113872 11138731113872 1113873

(m + n)2

Pστ(CgtA) 12

+Nσ(CgtA) minus m

221113872 11138731113872 1113873 + Nτ(CgtA) minus n221113872 11138731113872 1113873

(m + n)2

(8)






implies


2Pσ(AgtB) + n

2Pτ(AgtB) + mn

ltmax a b m2

+ n2

+1


2

(9)

since (12)ltmax a b


D τ1( 1113857

9 5 1

8 4 3

7 6 2

⎧⎪⎪⎨

⎪⎪⎩

D τ2( 1113857

10 7 5 4

12 9 3 2

11 8 6 1

⎧⎪⎪⎨

⎪⎪⎩

D(τ)

6 1

5 2

4 3

⎧⎪⎪⎨

⎪⎪⎩

(10)


2k + ni( 11138572Pσ(AgtB) (2k)

2Pτk (AgtB) + ni( 1113857

2Pτi

(AgtB) + 2kni

(2k)

2

2+

n2i + 22

1113890 1113891 + 2kni ni + 2k( 1113857

2+ 2

21113890 1113891

(11)





D τ1( 1113857

9 5 18 4 37 6 2

⎧⎪⎨

⎪⎩

D τ2( 1113857

12 10 3 19 8 7 211 6 5 4

⎧⎪⎨

⎪⎩

D τ3( 1113857

15 11 7 4 314 10 9 5 213 12 8 6 1

⎧⎪⎨

⎪⎩

(12)

Pτi(AgtB) ([n2







(BAC) (13)



+(2m + 1) 2m2

+ 2m + 1(14)

shows that

P(AgtB) 2m

2+ 2m + 1

(2m + 1)2

12

+05n2 (15)



P(AgtB)lt12

+19 (16)






Algorithm 1



Remark 4






(18)






(19)







and CA








(20)




⟹mle13 minus

154

radic

2p

(21)


12

+3p

2+(13 minus

154

radic2)p middot 3p

(6p)2

12

+15 minus

154

radic

24asymp12

+1

925lt12

+19

(22)




p(3p + 1)

[2(3p + 1)]2 (when n 6p + 2)

orp(3p + 2)

[2(3p + 2)]2 (when n 6p + 4)

(23)








(24)




hArrm2

minus (13p + 4)m + 4p2

minus p minus 11113872 1113873ge 0


153p2

+ 108p + 201113969

2

(25)


p + 13p + 4 minus

153p2

+ 108p + 201113969

1113874 111387521113874 11138751113874 1113875(3p + 1)

(6p + 2)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (26)



153p2 + 108p + 20




hArrm2

minus (13p + 8)m + 4p2

minus 2p minus 41113872 1113873ge 0


153p2

+ 216p + 801113969

2

(27)



p + 13p + 8 minus

153p2

+ 216p + 801113969

21113874 11138751113874 1113875(3p + 2)

(6p + 4)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (28)


153p2 + 216p + 80



P(AgtB)lt12

+19 (29)


Remark 5






Data Availability




Acknowledgments


References

















A3 9 5 1

B3 8 4 3

C3 7 6 2

A4 10 7 5 4

B4 12 9 3 2

C4 11 8 6 1

(5)





12lt1113874 1113875P(AgtB)lt

12

+19 (6)



Pστ(AgtB) 12

+Nσ(AgtB) minus m

221113872 11138731113872 1113873 + Nτ(AgtB) minus n221113872 11138731113872 1113873

(m + n)2

(7)

Similarly we have

Pστ(BgtC) 12

+Nσ(BgtC) minus m

221113872 11138731113872 1113873 + Nτ(BgtC) minus n221113872 11138731113872 1113873

(m + n)2

Pστ(CgtA) 12

+Nσ(CgtA) minus m

221113872 11138731113872 1113873 + Nτ(CgtA) minus n221113872 11138731113872 1113873

(m + n)2

(8)






implies


2Pσ(AgtB) + n

2Pτ(AgtB) + mn

ltmax a b m2

+ n2

+1


2

(9)

since (12)ltmax a b


D τ1( 1113857

9 5 1

8 4 3

7 6 2

⎧⎪⎪⎨

⎪⎪⎩

D τ2( 1113857

10 7 5 4

12 9 3 2

11 8 6 1

⎧⎪⎪⎨

⎪⎪⎩

D(τ)

6 1

5 2

4 3

⎧⎪⎪⎨

⎪⎪⎩

(10)


2k + ni( 11138572Pσ(AgtB) (2k)

2Pτk (AgtB) + ni( 1113857

2Pτi

(AgtB) + 2kni

(2k)

2

2+

n2i + 22

1113890 1113891 + 2kni ni + 2k( 1113857

2+ 2

21113890 1113891

(11)





D τ1( 1113857

9 5 18 4 37 6 2

⎧⎪⎨

⎪⎩

D τ2( 1113857

12 10 3 19 8 7 211 6 5 4

⎧⎪⎨

⎪⎩

D τ3( 1113857

15 11 7 4 314 10 9 5 213 12 8 6 1

⎧⎪⎨

⎪⎩

(12)

Pτi(AgtB) ([n2







(BAC) (13)



+(2m + 1) 2m2

+ 2m + 1(14)

shows that

P(AgtB) 2m

2+ 2m + 1

(2m + 1)2

12

+05n2 (15)



P(AgtB)lt12

+19 (16)






Algorithm 1



Remark 4






(18)






(19)







and CA








(20)




⟹mle13 minus

154

radic

2p

(21)


12

+3p

2+(13 minus

154

radic2)p middot 3p

(6p)2

12

+15 minus

154

radic

24asymp12

+1

925lt12

+19

(22)




p(3p + 1)

[2(3p + 1)]2 (when n 6p + 2)

orp(3p + 2)

[2(3p + 2)]2 (when n 6p + 4)

(23)








(24)




hArrm2

minus (13p + 4)m + 4p2

minus p minus 11113872 1113873ge 0


153p2

+ 108p + 201113969

2

(25)


p + 13p + 4 minus

153p2

+ 108p + 201113969

1113874 111387521113874 11138751113874 1113875(3p + 1)

(6p + 2)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (26)



153p2 + 108p + 20




hArrm2

minus (13p + 8)m + 4p2

minus 2p minus 41113872 1113873ge 0


153p2

+ 216p + 801113969

2

(27)



p + 13p + 8 minus

153p2

+ 216p + 801113969

21113874 11138751113874 1113875(3p + 2)

(6p + 4)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (28)


153p2 + 216p + 80



P(AgtB)lt12

+19 (29)


Remark 5






Data Availability




Acknowledgments


References

















D τ1( 1113857

9 5 18 4 37 6 2

⎧⎪⎨

⎪⎩

D τ2( 1113857

12 10 3 19 8 7 211 6 5 4

⎧⎪⎨

⎪⎩

D τ3( 1113857

15 11 7 4 314 10 9 5 213 12 8 6 1

⎧⎪⎨

⎪⎩

(12)

Pτi(AgtB) ([n2







(BAC) (13)



+(2m + 1) 2m2

+ 2m + 1(14)

shows that

P(AgtB) 2m

2+ 2m + 1

(2m + 1)2

12

+05n2 (15)



P(AgtB)lt12

+19 (16)






Algorithm 1



Remark 4






(18)






(19)







and CA








(20)




⟹mle13 minus

154

radic

2p

(21)


12

+3p

2+(13 minus

154

radic2)p middot 3p

(6p)2

12

+15 minus

154

radic

24asymp12

+1

925lt12

+19

(22)




p(3p + 1)

[2(3p + 1)]2 (when n 6p + 2)

orp(3p + 2)

[2(3p + 2)]2 (when n 6p + 4)

(23)








(24)




hArrm2

minus (13p + 4)m + 4p2

minus p minus 11113872 1113873ge 0


153p2

+ 108p + 201113969

2

(25)


p + 13p + 4 minus

153p2

+ 108p + 201113969

1113874 111387521113874 11138751113874 1113875(3p + 1)

(6p + 2)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (26)



153p2 + 108p + 20




hArrm2

minus (13p + 8)m + 4p2

minus 2p minus 41113872 1113873ge 0


153p2

+ 216p + 801113969

2

(27)



p + 13p + 8 minus

153p2

+ 216p + 801113969

21113874 11138751113874 1113875(3p + 2)

(6p + 4)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (28)


153p2 + 216p + 80



P(AgtB)lt12

+19 (29)


Remark 5






Data Availability




Acknowledgments


References


















and CA








(20)




⟹mle13 minus

154

radic

2p

(21)


12

+3p

2+(13 minus

154

radic2)p middot 3p

(6p)2

12

+15 minus

154

radic

24asymp12

+1

925lt12

+19

(22)




p(3p + 1)

[2(3p + 1)]2 (when n 6p + 2)

orp(3p + 2)

[2(3p + 2)]2 (when n 6p + 4)

(23)








(24)




hArrm2

minus (13p + 4)m + 4p2

minus p minus 11113872 1113873ge 0


153p2

+ 108p + 201113969

2

(25)


p + 13p + 4 minus

153p2

+ 108p + 201113969

1113874 111387521113874 11138751113874 1113875(3p + 1)

(6p + 2)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (26)



153p2 + 108p + 20




hArrm2

minus (13p + 8)m + 4p2

minus 2p minus 41113872 1113873ge 0


153p2

+ 216p + 801113969

2

(27)



p + 13p + 8 minus

153p2

+ 216p + 801113969

21113874 11138751113874 1113875(3p + 2)

(6p + 4)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (28)


153p2 + 216p + 80



P(AgtB)lt12

+19 (29)


Remark 5






Data Availability




Acknowledgments


References


















153p2 + 108p + 20




hArrm2

minus (13p + 8)m + 4p2

minus 2p minus 41113872 1113873ge 0


153p2

+ 216p + 801113969

2

(27)



p + 13p + 8 minus

153p2

+ 216p + 801113969

21113874 11138751113874 1113875(3p + 2)

(6p + 4)2 ⟶

p⟶infin

112

+13 minus

153

radic

24lt

1912lt19 (28)


153p2 + 216p + 80



P(AgtB)lt12

+19 (29)


Remark 5






Data Availability




Acknowledgments


References

















possibleprobabilityandirreducibilityofbalanced …...a nontransitive triple of dice consists of...

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