EQUAL RELABELINGS
FOR P Q-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
Alec Lewald
2017
SIGNATURE PAGE
THESIS:
EQUAL RELABELINGS
AUTHOR:
Alec Lewald
DATE SUBMITTED: Winter 2017
Department of Mathematics and Statistics
Dr. Amber Rosin
Thesis Committee Chair
Mathematics & Statistics
Dr. Berit Givens
Mathematics & Statistics
Dr. John Rock
Mathematics & Statistics
ii
ACKNOWLEDGMENTS
This thesis is dedicated to my family, friends, and professors who have supported
me throughout my education. A special thank you to my parents and Samier who
always pushed me even when it was abundantly clear that I had no intention of
working.
iii
ABSTRACT
For m n-sided dice, we will call the sums m, m + 1, m + 2, . . . , mn the standard
sums, and we will say that m n-sided dice have an equal relabeling if the dice can
be labeled with positive integers in such a way that the standard sums are equally
likely to occur. We will consider the case when n = pq for distinct primes p < q.
The m for which n = pq sided dice have an equal relabeling have previously been
characterized and it has been shown that all such equal relabelings require stupid
dice (dice with 1’s on every face). We find the minimum and maximum number of
stupid dice that are needed for an equal relabeling of pq-sided dice and show that
such a relabeling is possible for any number of stupid dice between the maximum
and minimum.
iv
Contents
Signature Page
ii
Acknowledgements
iii
Abstract
iv
1 Background
1
2 A Partial Proof of a Previously Unjustified Statement
5
3 The Greatest and Least Number of Stupid Dice
13
3.1
Stupid Dice Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
3.2
An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
3.3
An Equal Labeling . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
4 Further Questions
23
Bibiliography
24
v
Chapter 1
Background
Consider rolling a pair of 6-sided dice. The possible sums are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,
and 12 where the likelihood of rolling each of these sums is
and
1
36
1 2 3 4 5 6 5 4 3 2
, , , , , , , , , ,
36 36 36 36 36 36 36 36 36 36
repectively. Can we relabel the dice so that we still get the sums 2, 3, . . . , 12
but the sums have an equal chance of being rolled? The answer is no due to the
fact that there are 6 · 6 = 36 possible die rolls and 11 different sums. Since 11 does
not divide 36 it is impossible for each sum to have an equally likely chance of being
rolled.
Now consider rolling seven 6-sided dice. The possible sums are 7, 8, 9, . . . , 41, 42
where again the sums are not all equally likely. For example, to roll a 7 we would
need to roll a 1 on all seven dice, which has a
1
67
chance of occuring. Whereas
rolling an 8 would require that six of the dice result in a 1 but the last die results
in a 2, which has a
1
66
chance of occuring. Can we relabel the dice so the sums
7, 8, . . . , 42 are equally likely? Consider the following relabeling:
1
Die 1:
1
1
2
2
3
3
Die 2:
1
1
4
4
7
7
Die 3:
1
1
1
10
10
10
Die 4:
1
1
1
19
19
19
Die 5:
1
1
1
1
1
1
Die 6:
1
1
1
1
1
1
Die 7:
1
1
1
1
1
1
If we roll the first two dice, the sums will be 2, 3, 4, 5, 6, 7, 8, 9, 10 with each
combination having 4 ways of occuring because there are two choices on each die
of which face to use, so there are 2 · 2 = 4 ways of getting each sum. If we then
combine these with die 3 we get the sums 3, 4, . . . , 11 if we roll a 1, and we get
the sums 12, 13, . . . , 20 if we roll a 10. Each of these sums has 3 · 2 · 2 = 12 ways
of occuring. Further combining this with die 4 we get the sums 4, 5, . . . , 21 and
22, 23, . . . , 39 if we roll a 1 or 19 respectively. Each of these sums has 3 · 3 · 2 · 2 = 36
ways of occuring. Finally, when combined with dice 5, 6, and 7, each sum increases
by exactly 3. Therefore, this set of dice yield the sums 7, 8, 9, . . . , 42 where each
sum has 6 · 6 · 6 · 3 · 3 · 2 · 2 = 7776 ways of occuring. Since there are 67 possible
outcomes and each has 65 = 7776 ways of being rolled, there is a probability of
65
67
=
1
62
=
1
36
for each sum to be rolled.
This now generates the question of when such a relabeing is possible. More
generally we can look at m dice that have n sides each. The question of can m
n-sided dice be relabelled so that we still get the sums m, m + 1, . . . , mn with each
sum being equally likely was answered in [1] when n = pq where p and q are distinct
prime numbers with p < q. We have seen that for n = 6, m can equal 7 but not 2.
We will build on the results shown in [1].
2
When rolling m n-sided dice we call the sums m, m + 1, m + 2, m + 3, . . . , mn
the standard sums. This is due to the fact that if all m die were labeled with the
usual labels of 1, 2, . . . , (n − 1), n the possible sums would be m, m + 1, . . . , mn.
Note that the number of standard sums is mn − m + 1 = m(n − 1) + 1. We say
that m, n-sided dice have an equal relabeling if they can be relabeled so that the
standard sums are equally likely. For many equal relabelings, such as the one in
the example above, we will need dice that have a label of 1 on every face; we will
refer to these as stupid dice. Before continuing the work done in [1], we summarize
the main results in [1]. The principal result from the paper is the following:
Lemma 1. A set of m pq-sided dice have an equal relabeling if and only if
m=
nx pγy − 1
n−1
(*)
m=
nx q γy − 1
n−1
(**)
or
where n = pq for distinct primes p and q with p < q, x and y are nonnegative
integers such that (x, y) = (0, 1), and γ is the multiplicative order of p modulo
n − 1.
From Lemma 1, we see that the number of standard sums is
m(n − 1) + 1 = nx pγy = px+γy q x or nx q γy = px q x+γy . We will use this result later,
so we will record this as a lemma.
Lemma 2. For m pq-sided dice the number of standard sums is px+γy q x when
m=
nx pγy −1
n−1
and px q x+γy when m =
nx q γy −1
.
n−1
Lastly, it is shown that with the exception of a few possibilities for x and y, all
equal relabelings require stupid dice. In particular we have the following:
3
Lemma 3. So long as (x, y) = (0, 0), (0, 1) or (1, 0), stupid dice are necessary for
any equal relabeling of m n = pq sided dice.
This provides a great deal of information, but it does not address how many
(or how few) stupid dice are required. We will consider this question. There is
also an unjustified statement in [1], namely that an equal relabeling requires that
the number 1 appears on a given die 1, p, q, or pq times, and we will look at this
statement.
4
Chapter 2
A Partial Proof of a Previously
Unjustified Statement
As mentioned before, the proof of Lemma 3 given in [1] relies on the statement
that:
if m pq-sided dice have an equal relabeling, then the number of 1’s on any given
die must divide pq.
However, no justification for this statement is given. We attempt to justify it.
Since we need to have our m dice sum to m each die must have at least one
face labled as 1. We are going to look at two specific dice: one being a regular die
that we can choose from our given m dice and the other having faces that equal
the sums of the remaining dice. Die 1 will have pq sides. We need to prove that
each number that appears on die 1 appears 1, p, q, or pq times. For the following
proofs we will use the notation that ai and bi are the number of faces with value i
on dice one and two respectively. Furthermore, let s and t be the largest values to
appear on dice one and two respectively. Finally, we will denote k being on die 1
5
or die 2 as k ∈ D1 and k ∈ D2 respectively.
Lemma 4. If the smallest number to appear on die 1 is 1, each other number that
appears on die 1 appears less than or equal to the number of times that 1 appears
on die 1, and s appears the same number of times as 1 does on die 1.
Proof. Given our m dice we are required to find combinations that add up to the
standard sums of m, m+1, m+2, . . ., mn. Assume to the contrary that there exists
some die, among the original m dice, without a face of value 1. Then consider the
sum of the smallest faces on each of the die. Since m − 1 of the dice have at least
1 as their smallest value and one of the dice has at least 2 as its smallest value, we
have that the smallest sum we could get is
1 ∗ (m − 1) + 2 ∗ 1 = m − 1 + 2 = m + 1.
This means that the smallest sum that we can get is m + 1 which is a contradiction
due to the fact that we are required to get a summation of m. Therefore each of
our dice must have at least one face of value 1. Note that since each die has a 1,
then the smallest value that appears on die 2 is m − 1 since we have m − 1 dice
remaining after selecting a die for die 1.
To generate equally likely standard sums with our dice we must have that each
sum appears the same number of times. By way of contradiction let k ∈ D1 with
ak > a1 . Then we know that k + m − 1 appears at least ak ∗ bm−1 times since m − 1
appears bm−1 times on die 2. Next consider that the only values we can use that
sum to m are 1 ∈ D1 and m − 1 ∈ D2 since these are the minimum values on each
die. Therefore the number of times that the sum m appears is exactly a1 ∗ bm−1 .
Since a1 ∗ bm−1 < ak ∗ bm−1 the sum k + m − 1 is more likely to occur than the sum
m, which is a contradiction. Therefore we must have that for all k ∈ D1 , ak ≤ a1 .
6
Note that the same reasoning applied to die 2 implies that bj ≤ bm−1 for all j ∈ D2 .
Finally, we will consider the sum s + t. If we use a value from die 1 that is less
than s, say s − j, then the value we would need from die 2 is s + t − (s − j) = t + j.
Since t is the max value on die 2 it must be that j ≤ 0 making our value for die
1 at least s. Similarly, if we used t − i for our value from die 2 we would need to
use s + i from die 1 making i ≤ 0. Since s is our largest value on die 1 we have
that i ≤ 0 making our value for die 2 at least t. Since our value for die 1 is at least
s and our value for die 2 is at least t we have that they must be exactly s and t.
Therefore the only values that sum to s + t are s on die 1 and t on die 2. Since
all sums appear an equal number of times, we have that a1 ∗ b1 = as ∗ bt . Using
our above facts we have that as ≤ a1 and bt ≤ b1 . If as < a1 or bt < b1 then we
would have that as ∗ bt < a1 ∗ b1 which would contradict our previous statement.
Therefore, as = a1 and bt = b1 .
It is important to note that since die 1 was arbitrary we have that all of these
results hold no matter which die is choosen to be die 1.
For the rest of the paper we will need terminology to describe the cases ai = a1
or bj = bm−1 . If this happens, then we refer to i and j as being f ull. Likewise,
if ai = 0 or bj = 0 then the value i or j is empty. Furthermore, if a sum, say d,
has the same number of ways of being obtained as m then we will say d is maxed.
These will be important due to the fact that if a value is full then the die which it
is on can no longer have any more faces with said value. Similarly, if a combination
is maxed then if there are any other two values that make that combination which
have not yet been acounted for, one must be empty. Finally we will refer to values
of our dice as mirroring if ai = as−(i−1) and bj = bt−(j−(m−1)) .
7
Lemma 5. Die 1 is mirrored from 1 to 1+k1 where 1+k1 is the smallest non-empty
number on die 1 other than 1; furthermore, 1 + k1 ∈ D1 is full. We will also show
that die 2 is mirrored from m − 1 to m − 1 + k1 with m − 1 through m − 2 + k1
being full while m − 1 + k1 is empty.
Proof. Let 1 + k1 be the smallest non-empty number on die 1 other than 1. Then
we know that to get the sums of m, m + 1, . . . , m − 1 + k1 we must use the value
1 ∈ D1 since (1 + k1 ) + (m − 1) is larger than any of these values. Then for any
value we just listed, say x, we know that x − 1 ∈ D2 and x − 1 must be full
since there are no other combinations that can result in x other than 1 + x − 1.
Therefore the numbers m − 1, m, m + 1, . . . , m − 2 + k1 on die 2 are full. Since
s + m − 1 is a maxed combination, due to s and m − 1 being full, we have that any
number smaller than s on die 1 that sums to s + m − 1 which uses the numbers
m, m + 1, m + 2, . . . , m − 2 + k1 as summands from die 2 is empty. This list of empty
numbers on die 1 is s − 1, s − 2, . . . , s − k1 + 1.
Similarly to the above s − k1 is the largest non-empty number on die 1 other
than s. Therefore to get the sums of s + t − k1 + 1, s + t − k1 + 2, . . . , s + t − 1
we must use s ∈ D1 since s − k1 + t = s + t − k1 is too small. Then for any
value we just listed, say y, we know that y − s ∈ D2 and y − s must be full since
there are no other combinations that can result in y other than s + y − s. Therefore
t−k1 +1, t−k1 +2, . . . , t−1 on die 2 are all full. Since 1+t is a maxed combination,
due to 1 and t being full, we have that any number larger than 1 on die 1 that sums
to t + 1 which uses the numbers t − k1 + 2, t − k1 + 3, . . . , t − 1 as the summands
from die 2 is empty. This list of empty numbers on die 1 is 2, 3, . . . , k1 − 1, k1 .
Since t − k1 + 1, t − k1 , . . . , t − 2, t − 1, t are full, we have the sums of t − k1 +
2, t − k1 + 3, . . . , t − 1, t, t + 1 being maxed due to the fact that we can add these
8
values to 1 ∈ D1 . Since we know that 1 + k1 ∈ D1 is non-empty but t + 1 is maxed,
we know that t − k1 ∈ D2 is empty or else the sum t + 1 would occur too many
times.
Consider the sum of s + t − k1 . The labels t − k1 + 1, t − k1 + 2, . . . , t − 1, t are
too large to sum with s to attain this, but any value on die 1 that is less than s − k1
would be too small to combine with t. Therefore we must use some value on die 1
that is greater than or equal to s − k1 but smaller than s. Since all of these values
except for s − k1 are empty we must only use s − k1 for this sum. Thus, s − k1 is
full or else s + t − k1 would not be maxed. Also notice that if we sum s − k1 ∈ D1
with m − 1 + k1 ∈ D2 we get s + m − 1 which we have already established is maxed,
leaving us with the conclusion that m − 1 + k1 ∈ D2 is empty.
This brings us to our final point. Consider the sum m + k1 . This can only be
achieved through using 1 or 1 + k1 ∈ D1 since any other number would be too big
or is empty. These reqiure the summands of m − 1 + k1 or m − 1 ∈ D2 respectively.
Since m − 1 + k1 ∈ D2 is empty we must have that 1 + k1 ∈ D1 is full to make
m + k1 maxed.
In conclusion, die 1 contains at least 1, 1+k1 , s−k1 , and s which are all full while
having 2, 3, . . . , k1 and s−k1 +1, s−k1 +2, . . . , s−1 be empty. Similarly we have that
die 2 contains at least m−1, m, m+1, . . . , m−2+k1 and t−k1 +1, t−k1 +2, . . . t−1, t
full while having m − 1 + k1 and t − k1 be empty. Using our definition, we have
that die 1 mirrors from at least 1 to 1 + k1 while die 2 mirrors from at least m − 1
to m − 1 + k1 .
Lemma 6. If there exists a non-empty element of a die that is also not full, then
there is at least one die that doesn’t mirror.
9
Proof. Let e and g be the smallest elements on D1 and D2 respecitvely such that
they are neither empty nor full. By way of contradiction assume that both dice
mirror.
We first show that e − 1 = g − (m − 1). Assume by way of contradiction that
e − 1 < g − (m − 1) making e + (m − 1) < g + 1. Since e is not full we have that
e + (m − 1) is not maxed from just e ∈ D1 and m − 1 ∈ D2 . Since m − 1 is the
smallest number on D2 we have that there exists some x ∈ D1 such that 1 ≤ x < e
and a y ∈ D2 such that m − 1 < y ≤ t where x + y = e + (m − 1). Due to the
fact that e + (m − 1) < g + 1 we have that m − 1 < y < g. Since x < e and y < g
we have that both are either full or empty. If one of them is empty then their sum
does not contribute to maxing e + (m − 1) but if both are full then e + (m − 1)
would occur too many times. Therefore e − 1 ≥ g − (m − 1). We can similarly show
that e − 1 ≤ g − (m − 1) making e − 1 = g − (m − 1).
Now let f and h be such that f = s − (e − 1) and h = t − (g − (m − 1)). Since
the dice mirror ae = af and bg = bh . By the above equations we get g − (m − 1) =
e − 1 = s − f which makes g + f − (m − 1) = s. Now if we consider the sum of
s ∈ D1 and (m − 1) ∈ D2 , we can see that both of the elements are full making
s + (m − 1) maxed. Using our above equation we get that
s + (m − 1) = g + f − (m − 1) + (m − 1) = g + f,
but since that sum is already maxed we must have that either g or f is empty.
By assumption g and e are nonempty and af = ae , therefore f must be nonempty,
which gives us a contradiction. Therefore, if there does exist a smallest such e ∈ D1
such that e is neither empty nor full, then D1 can’t mirror.
Conjecture 1. Dice 1 and 2 are mirrored with each element being either full or
empty. (not finished proving)
10
Proof. (partial ) We will induct on the number of gaps on D1 . Let lemma 5 be our
base case. For our inductive case let the elements 1, 1 + k1 , 1 + k1 + k2 , . . . , 1 + k1 +
. . . + kn ∈ D1 be full where each ki represents a gap between each face labling value,
i.e. all values between each 1 + k1 + k2 + . . . + ki and 1 + k1 + k2 + . . . + ki + ki+1 are
empty. Furthermore let 1 + k1 + . . . + kn + kn+1 ∈ D1 be non-empty and all other
values on D1 between 1 + k1 + . . . + kn and 1 + k1 + . . . + kn + kn+1 be empty. Next,
let all elements, m − 1 ≤ x ≤ m − 1 + kn on D2 , be full or empty. Finally, we let
all elements of D1 between 1 and 1 + k1 + . . . , +kn and all elements of D2 between
m − 1 and m − 1 + kn mirror.
We then let m − 1 + kn < v ≤ m − 1 + kn+1 with v ∈ D2 such that j is the
smallest such element where it is neither full nor empty. Since it is neither full nor
empty the sum of 1+v is not maxed from just adding 1 ∈ D1 and v ∈ D2 . Therefore
there exists some non-empty x ∈ D2 such that 1 + k1 + . . . + kj + x = 1 + v for some
1 ≤ j ≤ n. The reason that we have this restriction on j is due to the fact that if
j = n + 1 then we would be using 1 + k1 + . . . , +kn + kn+1 ∈ D1 but the smallest
sum we could attain from this would be (1 + k1 + . . . , +kn + kn+1 ) + (m − 1) =
m + k1 + . . . , +kn + kn+1 but the max that 1 + v could be is m + kn+1 .
If we use our formula of 1+k1 +. . . , +kj +x = 1+v we get x = v−k1 −. . .−kj < v.
Since x < v we have that x is either full or empty from or hypothesis. If x is full
then v is empty, otherwise 1 + v would be over our max. However, if x is empty
then contradict our requirement that x is non-empty. Since we assumed that v was
the smallest non-empty or full element between m − 1 + kn and m − 1 + kn+1 , we
have that all elements on D2 between m − 1 and m − 1 + kn+1 are either full or
empty.
11
So far we have shown that 1 appears on all m dice and appears the most often,
each die mirrors from 1 to at least 1 + k1 where 1 + k1 is the smallest non-empty
number on the die, and if there exists a non-empty element of a die that is also not
full, then the die does not mirror. The end goal of these Lemmas was to show that
each non-empty ai was equal. To complete the proof, we would need to show that
the dice mirror completely. We have not yet been able to show this, so for now we
state the following conjecture:
Conjecture 2. If m pq-sided dice have an equal relabeling, then the number of 1’s
on any given die must divide pq.
We will assume Conjecture 2 is true in the next chapter.
12
Chapter 3
The Greatest and Least Number
of Stupid Dice
3.1
Stupid Dice Facts
By Lemma 3 we know that equal relabelings require stupid dice, but how many
stupid dice are needed? Suppose m pq-sided dice have an equal relabeling. Then
the number of standard sum is equal to either px+γy q x or px q x+γy by Lemma 2.
The number of rolls that sum to each standard sums is calculated by dividing the
total number of outcomes by the number of standard sums. In either case the total
number of sums is nm = pm q m . In the first case the number of times each sum
appears is
pm q m
px+γy q x
sum appears is
= pm−x−γy q m−x . In the second case, the number of times each
pm q m
px q x+γy
= pm−x q m−x−γy . We will consider the case where we have
px+γy q x standard sums, the other case is very similar. The least standard sum is
m, and if we only label faces with positive integers, then in order to achieve a sum
of m using m dice, every die must have at least one face labeled as 1. Let ci be
13
the number of times 1 appears on die i. Then each ci must equal 1, p, q, or pq from
Conjecture 2. Since the sum m can only be achieved by using a 1 from each die,
the number of times the sum m appears is
c1 c2 . . . cm = pm−x−γy q m−x .
For a die to be a stupid die ci = n = pq .
To find the maximum and minimum numbers of stupid dice that are possible
in an equal relabeling we need to find the least and greatest possible amount of ci ’s
that equal pq. In all cases to get c1 , c2 , . . . , cm = pm−x−γy q m−x , m − x − γy of the
ci ’s must have a factor of p, and m − x ci ’s must have a factor q. To find the largest
amount of stupid dice, we want as much overlap as possible, so of the m − x ci ’s
that have a factor of q, assume that m − x − γy of them also have p as a factor.
Therefore we have have m − x − γy dice where ci = pq, m − x − (m − x − γy) = γy
dice where ci = q, and m − (m − x − γy) − γy = x dice with ci = 1, making the
making the maximum amount of stupid dice m − x − γy.
To find our least number of stupid dice we want as little overlap as possible.
Since m − x − γy ci ’s have p as a factor, there are m − (m − x − γy) = x + γy ci ’s
that don’t have p as a factor. If these x + γy ci ’s have q’s as factors then there are
still (m − x) − (x + γy) = m − 2x − γy factors of q remaining. Since each ci can
only have at most 1 factor of q, the remaining factors of q must be factors of ci ’s
which also have p as a factor. Therefore, there are at least m − 2x − γy dice whose
ci = pq, making the minimum amount of stupid dice m − 2x − γy.
Thus, the minimum number of stupid dice is m − 2x − γy and the maximum
number of stupid dice is m − x − γy.
14
3.2
An Example
Before tryinig to find a general relabeling, we consider a specific example. Let
p = 3, q = 5, x = 4, and y = 2. To find γ, consider
31 ≡ 3 (mod 14)
32 ≡ −5 (mod 14)
33 ≡ −1 (mod 14)
34 ≡ −3 (mod 14)
35 ≡ 5 (mod 14)
36 ≡ 1 (mod 14),
thus γ = 6. This makes n = 15 and
m=
154 36·2 − 1
= 1, 921, 728, 616.
14
The largest amount of stupid dice we could use is m − x − γy = 1, 921, 728, 600
and the least amount is m − 2x − γy = 1, 921, 728, 596. We will show a labeling for
1, 921, 728, 599 stupid dice. Consider the following labeling:
The first 14 dice are labeled as
15
Die 1:
1,2,3,4,5
Each value appears 3 times
Die 2:
1,6,11
Each value appears 5 times
Die 3:
1,16, 31
Each value appears 5 times
Die 4:
1, 46, 91
Each value appears 5 times
Die 5:
1, 136, 271
Each value appears 5 times
Die 6:
1, 406, 811
Each value appears 5 times
Die 7:
1, 1216, 2431
Each value appears 5 times
Die 8:
1, 3646, 7291
Each value appears 5 times
Die 9:
1, 10936, 21871
Each value appears 5 times
Die 10:
1, 32806, 65611
Each value appears 5 times
Die 11:
1, 98416, 196831
Each value appears 5 times
Die 12:
1, 295246, 590491
Each value appears 5 times
Die 13:
1, 885736, 1771471
Each value appears 5 times
Die 14:
1, 2657206, 5314411
Each value appears 5 times
Combining dice 1 and 2 we can see that the labelings from die 1 fill in the gaps
of die 2 creating the sums 2, 3, . . . , 16 with each sum appearing 15 times. Consider
the fact that the gaps between the faces on die 3 are each of size 15 and dice 1
and 2 combine for 15 consecutive sums. This means that when we combine die 3
with our previous combination the gaps will get filled in and the new sums will
be 3, 4, . . . , 47. We can continue this process for all 14 dice and get that the sums
14, 15, 16, . . . , 7971628 occur 3 · 513 = 3662109375 times.
The values on the next 3 dice appear exactly once each. The 15th die is labeled
as
1, 7971616, 15943231, 23914846, 31886461, 39858076, 47829691, 55801306, 63772921, 71744536,
79716151, 87687766, 95659381, 103630996, 111602611,
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the 16th die is labeled as
1, 119574226, 239148451, 358722676, 478296901, 597871126, 717445351, 837019576, 956593801,
1076168026, 1195742251, 1315316476, 1434890701, 1554464926, 1674039151,
and the 17th die is labeled as
1, 1793613376, 3587226751, 5380840126, 7174453501, 8968066876, 10761680251, 12555293626,
14348907001, 16142520376, 17936133751, 19729747126, 21523360501, 23316973876, 25110587251.
Similar to before, the gaps between the faces on die 15 are of size 7971616.
When we combine this with our 7071615 consecutive sums from the first 14 dice we
get the sums of 15, 16, 17, . . . , 119574239 where each sum still appears 3662109375
times.
Repeating this with dice 16 and 17 makes the sums 17, 18, 19, . . . , 26904200641
where each sum appears 3662109375 times. The remaining 1, 921, 728, 599 dice are
stupid dice. When combined with our other 17 dice we get the sums
1921728616, 1921728617, 1921728618, . . . , 28825929240 where each sum appears
3 · 514 · 151921728599 times.
3.3
An Equal Labeling
We now show that for any m − 2x − γy ≤ λ ≤ m − x − γy, an equal albeling with
λ stupid dice is possible. Let λ = m − 2x − γy + t where 0 ≤ t ≤ x. Label the first
x − t dice as follows:
1, 1 + q i , 1 + 2q i , . . . , 1 + (q − 1)q i for 0 ≤ i ≤ x − t − 1
where each label is repeated p times. To find the sums this group of dice
produces, first subtract 1 from each of the labelings. The new labeling would be
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0, q i , 2q i , . . . , (q − 1)q i for 0 ≤ i ≤ x − t − 1.
These are exactly the numbers we need to write the numbers 0, 1, . . . , q x−t − 1 in
base q. Since each number can be written uniquely in base q, each of the sums
0, 1, 2, . . . , q x−t − 1
occur the same number of times. Note that this is a set of q x−t consecutive numbers.
Adding the 1’s back to the labeling adds x − t to each sum yeilding the equally
likely sums
x − t, x − t + 1, . . . , x − t + q x−t − 1.
Since there are p choices of which face to use on each die, each of these sums occurs
px−t times. Label the second x − t dice as follows:
1, 1 + pj q x−t , 1 + 2pj q x−t , . . . , 1 + (p − 1)pj q x−t for 0 ≤ j ≤ x − t − 1
where each label is repeated q times. To find the sums this group of dice
produces, first subtract 1 from each of the labels and divide each label by q x−t .
This new labeling would be
0, pi , 2pi , . . . , (p − 1)pi for 0 ≤ i ≤ x − t − 1.
These are exactly the numbers we need to write 0, 1, . . . , px−t − 1 in base p. Since
each number can be written uniquely in base p, then each of their sums,
0, 1, 2, . . . , px−t − 1,
occur the same number of times. Note that this is a set of px−t consecutive num­
bers. After multiplying each label by q x−t , we get the sums 0, q x−t , . . . , q x−t (px−t −1).
Then adding 1’s to each label, we get the sums
x − t, x − t + q x−t , . . . , x − t + nx−t − q x−t
which is px−t numbers each separated by a gap of q x−t . Thus, when we combine
the the q x−t consecutive sums from the first x − t dice with the sums from the
second x − t dice, the gaps from our second sums will be filled in giving us the nx−t
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consecutive sums
2(x − t), 2(x − t) + 1, . . . , 2(x − t) + nx−t − 1.
Since there are q choices on each die of which face to use, these sums occur
px−t q x−t = nx−t times. Label the next γy dice as follows:
1, 1 + pk nx−t , 1 + 2pk nx−t , . . . , 1 + (p − 1)pk nx−t for 0 ≤ k ≤ γy − 1
where each label is repeated q times. To find the sums this group of dice
produces, first subtract 1 from each of the labels and divide each label by nx−t .
This new labeling would be
0, pi , 2pi , . . . , (p − 1)pi for 0 ≤ i ≤ γy − 1.
These are exactly the numbers we need to write the numbers 0, 1, . . . , pγy − 1 in
base p. Since each number can be written uniquely in base p, then each of their
sums,
0, 1, 2, . . . , pγy − 1,
are attained the same number of times. Note that this is a set of pγy consecutive
numbers. After multiplying each label by nx−t , we get the sums
0, nx−t , 2nx−t , . . . , nx−t (pγy − 1). Then adding 1’s to each label, we get the sums
γy, γy + nx−t , . . . , γy + pγy nx−t − nx−t ,
which is pγy numbers each separated by a gap of nx−t . Thus, when we combine the
nx−t consecutive sums from the first 2(x − t) dice with the sums from the next γy
dice, the gaps will be filled in giving us the pγy nx−t consecutive sums
2(x − t) + γy, 2(x − t) + γy + 1, . . . , 2(x − t) + γy + pγy nx−t − 1.
Since there are q choices of which face to use on each die, the sums from this
group occur q γy times so the above sums occur nx−t q γy times. Label the last t dice
as follows:
1, 1 + nδ pγy nx−t , 1 + 2nδ pγy nx−t , . . . , 1 + (n − 1)nδ pγy nx−t for 0 ≤ δ ≤ t − 1
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where each label appears on exactly one face. To find the sums this group of
dice produces, first subtract 1 from each of the labelings and divide each labeling
by pγy nx−t . The new labelings would be
0, nδ , 2nδ , . . . , (n − 1)nδ for 0 ≤ δ ≤ t − 1.
These are exactly the numbers we need to write the numbers 0, 1, 2, . . . , nt − 1 in
base n. Since each number can be written uniquely in base n, then each of the
sums
0, 1, 2, . . . , nt − 1
occur the exactly once. Note that this is a set of nt consecutive numbers. After
multiplying each label by pγy nx−t we get the sums
0, pγy nx−t , 2pγy nx−t , . . . , pγy nx−t (nt − 1). Then adding the 1’s to each label, we get
the sums
t, t + pγy nx−t , . . . , t + pγy nx − pγy nx−t ,
which is nt numbers each separated by gaps of pγy nx−t . Thus, when we combine
the pγy nx−t consecutive sums from the first 2(x − t) + γy dice with the sums from
the next t dice, the gaps will be filled in giving us the pγy nx consecutive sums
2x − t + γy, 2x − t + γy + 1, . . . , 2x − t + γy + pγy nx − 1.
Each of these sums appears nx−t q γy times. Finally, label the remaining
m − 2x − γy + t dice with 1’s on every face. This yields the sum m − 2x − γy + t.
Since there are n choice of which face to use on each of the stupid dice, the sums
m, m + 1, . . . , m + pγy nx − 1
all appear nx−t q γy nm−2x−γy+t = nm−x−γy q γy = pm−x−γy q m−x times.
We need to show that this labeling results in the standard sums. It is clear that
the lowest sum we can get is m and that each successive sum is increased by 1. This
means all we need to show is that the largest sum, m + pγy nx − 1, is in fact equal
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to mn. Since we know that mn − m + 1 = nx pγy we have that mn = m + nx pγy − 1
making the sums created by our dice the standard sums.
We will now show that the standard sums are equally likely by showing that
there is only one way to achieve each sum. Let δ be one of the standard sums minus
m − 2x − γy + t (i.e. without the stupid dices’ contribution). From our above sums
we know that the first x − t dice always result in a sum of the form x − t + r where
0 ≤ r ≤ q x−t − 1, the next x − t result in a sum of the form x − t + sq x where
0 ≤ s ≤ px−t − 1, the next γy dice result in a sum of the form γy + wnx−t where
0 ≤ w ≤ pγy − 1, and the final t result in a sum of the form t + vpγy nx−t where
0 ≤ v ≤ nt − 1. So δ can be written as
δ = (x − t + r) + (x − t + sq x−t ) + (γy + wnx−t ) + (t + vpγy nx−t ).
To show that this sum is unique assume there exists two sums that yield δ, i.e.
(x − t + r1 ) + (x − t + s1 q x−t ) + (γy + w1 nx−t ) + (t + v1 pγy nx−t )
= (x − t + r2 ) + (x − t + s2 q x−t ) + (γy + w2 nx−t ) + (t + v2 pγy nx−t ).
By subtracting constants that appear on both sides, we obtain
r1 + s1 q x−t + w1 nx−t + v1 pγy nx−t = r2 + s2 q x−t + w2 nx−t + v2 pγy nx−t .
Then we can consider these sums mod q x−t . Since q x−t |nx−t we get that r1 ≡ r2
(mod q x−t ). Then, since 0 ≤ r1 , r2 ≤ q x−t − 1 we have that r1 = r2 .
Next we consider our sums mod nx−t . Since p = 1 we have that nx−t |q x−t
making r1 + s1 q x−t ≡ r2 + s2 q x−t (mod nx−t ). Using r1 = r2 we get that
s1 q x−t ≡ s2 q x−t (mod nx−t ). Finally, since 0 ≤ s1 , s2 ≤ px−t we have that
0 ≤ s1 q x−t , s2 q x−t < nx−t − 1 making s1 q x−t = s2 q x−t which shows s1 = s2 .
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We will now consider our sums mod pγy nx−t . Since p = 1 we have that
pγy nx−t |nx−t making r1 + s1 q x−t + w1 nx−t ≡ r2 + s2 q x−t + w2 nx−t (mod pγy nx−t ).
Using r1 = r2 and s1 = s2 we get that w1 nx−t ≡ w2 nx−t (mod pγy nx−t ). Finally,
since 0 ≤ w1 , w2 ≤ pγy − 1 we have that 0 ≤ w1 nx−t , w2 nx−t < pγ y nx−t making
21 nx−t = w2 nx−t resulting in w1 = w2 .
Looking back at our original equation
r1 + s1 q x−t + w1 nx−t + v1 pγy nx−t = r2 + s2 q x−t + w2 nx−t + v2 pγy nx−t
and using the fact that the r’s, s’s, and w’s are equal we get that v1 pγy nx−t =
v2 pγy nx−t making v1 = v2 . Therefore δ is a unique sum from these groups of dice.
Thus the minimum number of stupid dice is m − 2x − γy, the maximum number
of stupid dice is m − x − γy, and it is possible to find an equal relabeling that uses
any number of stupid dice between m − 2x − γy and m − x − γy.
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Chapter 4
Further Questions
To continue this work we have left a few open ends that are worth investigating.
Can Conjecture 1 be proven? How many relabelings are there for a given m? The
labeling in 3.3 can be modified if we let our exponents be pm−x q m−x−γy instead of
pm−x−γy q m−x , is it a substantial difference? Are there other substantially different
ways to make general relabelings? What about the case when n is not a product
of 2 distinct primes?
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Bibliography
[1] F. Bermudez, A. Medina, A. Rosin, E. Scott. Are Stupid Dice Necessary?, The
College Mathematics Journal, Vol. 44, No. 4, September 2013.
[2] J. Gallian, Contemporary Abstract Algebra, Houghton Mifflin Company, 2008.
[3] W. Gasarch, C. Kruskal, When Can One Load a Set of Dice so That the Sum
Is Uniformly Distributed?, Mathematics Magazine 72, 1999.
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