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Go First Dice for Five Players and Beyond.

Robert Ford, James Grime, Eric Harshbarger and Brian Pollock

Abstract

Before a game begins, the players need to decide the order of play. This order of

play is determined by each player rolling a die. Does there exist a set of dice such

that draws are excluded and each order of play is equally likely? For four players the

solution involves four

12

-sided dice, sold commercially as Go First Dice. However,

the solution for ve players remained an open question. We present two solutions.

The rst solution has a particular mathematical structure known as binary dice,

and results in a set of ve 60-sided dice, where every place is equally likely. The

second solution is an inductive construction that results in one one

36

-sided die;

two

48

-sided dice; one

54

-sided die; and one

20

-sided die, where each permutation is

equally likely.

Imagine four friends are about to start a game and need to decide who goes rst, who

goes second, who goes third and who goes fourth.

So the friends decide that each player will roll a die, and order play from highest to

lowest. Unfortunately, if two players roll the same number, those players will have to roll

again. Potentially this could go on forever.

Is it possible to make a set of dice so that no ties occur, and each player is equally

likely to be placed rst, second, third and fourth?

In 2012 Robert Ford and Eric Harshbarger solved the problem with a set of four

12-sided dice, [Bel12] [Har18]:

A: [01, 08, 11, 14, 19, 22, 27, 30, 35, 38, 41, 48]

B: [02, 07, 10, 15, 18, 23, 26, 31, 34, 39, 42, 47]

C: [03, 06, 12, 13, 17, 24, 25, 32, 36, 37, 43, 46]

D: [04, 05, 09, 16, 20, 21, 28, 29, 33, 40, 44, 45]

©

2023 Robert Ford, James Grime, Eric Harshbarger, Brian Pollock. This is an open ac-

cess article licensed under the Creative Commons Attribution-NonCommercial-NoDerivs License

(https://creativecommons.org/licenses/by-nc-nd/4.0/).

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Go First Dice for Five Players and Beyond.

76

With these dice each player is equally likely to be placed rst, second third and fourth.

However, even more remarkably, this property remains true for any subset of dice. So,

any subset of three dice is equally likely to be placed rst, second and third; and any

subset of two dice is equally likely to be placed rst and second.

These dice became known as

Go First Dice

. However, a set for ve players was an

open problem. We present two sets of ve Go First Dice, with details of how each set

was constructed.

Our rst solution is a set of ve 60-sided dice found by James Grime and Brian Pollock

with a `binary construction' and computer search.

A: [002, 008, 012, 018, 024, 029, 032, 038, 044, 049, 053, 059, 063, 067, 073, 078, 083,

088, 092, 098, 103, 109, 113, 117, 122, 127, 133, 138, 143, 148, 153, 159, 164, 167,

173, 178, 183, 188, 194, 199, 202, 208, 214, 217, 224, 227, 233, 238, 243, 248, 253,

257, 263, 269, 272, 278, 284, 289, 292, 298]

B: [003, 007, 013, 019, 023, 028, 033, 037, 043, 048, 052, 058, 064, 068, 074, 079, 084,

087, 093, 097, 104, 108, 112, 118, 123, 128, 134, 137, 142, 149, 154, 158, 163, 168,

172, 177, 184, 187, 193, 198, 203, 207, 213, 218, 223, 228, 234, 239, 242, 249, 252,

258, 264, 268, 273, 279, 283, 288, 293, 297]

C: [004, 009, 011, 020, 022, 027, 031, 039, 042, 047, 051, 060, 065, 069, 075, 077, 082,

089, 091, 099, 102, 110, 114, 119, 124, 129, 132, 136, 144, 147, 152, 157, 162, 169,

171, 176, 182, 189, 192, 200, 204, 209, 215, 219, 222, 226, 235, 237, 244, 247, 254,

259, 262, 267, 271, 280, 282, 290, 294, 296]

D: [005, 006, 014, 017, 021, 030, 034, 040, 041, 050, 054, 057, 062, 066, 072, 076, 085,

086, 094, 100, 101, 107, 111, 116, 125, 130, 135, 139, 145, 146, 155, 156, 165, 166,

174, 179, 185, 190, 191, 197, 201, 210, 212, 216, 225, 229, 232, 236, 241, 250, 251,

260, 261, 270, 274, 277, 281, 287, 295, 299]

E: [001, 010, 015, 016, 025, 026, 035, 036, 045, 046, 055, 056, 061, 070, 071, 080, 081,

090, 095, 096, 105, 106, 115, 120, 121, 126, 131, 140, 141, 150, 151, 160, 161, 170,

175, 180, 181, 186, 195, 196, 205, 206, 211, 220, 221, 230, 231, 240, 245, 246, 255,

256, 265, 266, 275, 276, 285, 286, 291, 300]

This set was designed so that the dice are all the same size, and that each place (rst,

second, third, etc) is equally likely, including all subsets of dice.

The second solution was found by Eric Harshbarger using an inductive method.

A: [002, 015, 016, 018, 019, 032, 038, 051, 052, 054, 055, 068, 070, 083, 084, 086, 087,

100, 107, 120, 121, 123, 124, 137, 139, 152, 153, 155, 156, 169, 175, 188, 189, 191,

192, 205]

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77

B: [005, 006, 010, 014, 022, 023, 025, 031, 041, 042, 046, 050, 058, 059, 061, 067, 073,

074, 078, 082, 090, 091, 093, 099, 110, 111, 115, 119, 127, 128, 130, 136, 142, 143,

147, 151, 159, 160, 162, 168, 178, 179, 183, 187, 195, 196, 198, 204]

C: [003, 009, 011, 012, 020, 024, 028, 029, 039, 045, 047, 048, 056, 060, 064, 065, 071,

077, 079, 080, 088, 092, 096, 097, 108, 114, 116, 117, 125, 129, 133, 134, 140, 146,

148, 149, 157, 161, 165, 166, 176, 182, 184, 185, 193, 197, 201, 202]

D: [004, 007, 008, 013, 017, 021, 026, 027, 030, 040, 043, 044, 049, 053, 057, 062, 063,

066, 072, 075, 076, 081, 085, 089, 094, 095, 098, 109, 112, 113, 118, 122, 126, 131,

132, 135, 141, 144, 145, 150, 154, 158, 163, 164, 167, 177, 180, 181, 186, 190, 194,

199, 200, 203]

E: [001, 033, 034, 035, 036, 037, 069, 101, 102, 103, 104, 105, 106, 138, 170, 171, 172,

173, 174, 206]

This set uses dice of dierent sizes including one 36-sided die; two 48-sided dice; one

54-sided die; and one 20-sided die; and is designed with the stronger property that each

permutation is equally likely, including subsets.

We will now proceed to describe how each set was found or constructed.

1 Binary construction with search

Initially, a computer search seemed like a good option for nding a set of ve Go First

Dice. Although it was immediately clear that this was going to be a large task.

In particular, we knew the number of sides was going to be some multiple of 30.

That's because, if we start with ve

m

-sided dice, we will have

m5

equally likely out-

comes. If each place is to be equally likely, this number must be divisible by 5 and, since

5 is prime,

m

is also divisible by 5. Similarly, because we want all subsets of dice to be

fair,

m

must also be divisible by 2 and 3.

So we decided to restrict our search to sets of dice where the wins and losses were

particularly neat.

Let's start by assuming we have

n

dice with

m

sides. If we write our dice in separate

rows, then the sides will form

m

columns. Fill the dice with the values 1 to

nm

, column

by column, so that the last die contains the highest or lowest value of each column. The

following is an example of a valid set of Go First Dice made in this style:

A: [02, 05, 07, 12, 14, 17]

B: [03, 04, 08, 11, 13, 18]

C: [01, 06, 09, 10, 15, 16]

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Go First Dice for Five Players and Beyond.

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With this construction, a player will automatically win if they roll the higher side, or if

they roll the same side and have a higher value.

To emphasise the high and low values, we can rewrite the dice so that each successive

die contains only two values, one that is higher and one that is lower than all previous

values, while maintaining the order of each column.

A: [2, 2, 2, 2, 2, 2]

B: [3, 1, 3, 1, 1, 3]

C: [0, 4, 4, 0, 4, 0]

Notice that A is essentially a dummy die, containing only one value, while B is con-

structed so that half the values are higher than A and half lower than A. This will ensure

A and B are equally likely to be placed rst and second in a two-player game.

Next, C is written with one value higher and one value lower than all previous values,

in this case we have used the values 0 and 4. If half the values are high values then each

place is equally likely when C plays two-player games against A or B.

A three-player game is trickier, as the position of the high values matter. To nd out

where to place our high values, let's consider which of the three dice will win the roll.

In this construction, a die automatically wins if it rolls a higher side. For example,

when

m= 6

, each die automatically wins 55 outcomes and loses 55 outcomes, leaving 51

outcomes in contention. These are the outcomes when two or more dice roll the same

side, and who wins depends on the placement of the high and low values. If the dice are

to be fair, C must win 1/3 of the contended outcomes.

If we number the sides from 1 to 6, then a high value on side

s

of die C will increase

its wins by

2s−1

. These are the outcomes where either A or B roll the same side as C

and the other die rolls a lower side, or where A, B and C all roll the same side. In our

example, the high values of C are found on sides 2, 3, and 5, which increases the number

of wins by 17, one-third of the contended outcomes as required.

In general, for

m

-sided dice, a high value on side

s

of the third die contributes

2s−1

wins, and these wins must add up to

m(3m−1)/6

.

However, since half of the values on C must be high values, we can tidy-up the pre-

vious condition to say that the sides with high value must add up to

m(3m+ 2)/12

.

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Robert Ford, James Grime, Eric Harshbarger, Brian Pollock

79

So when

m= 6

, we need 3 high values on dice B and C, and for C we need sides with

high values to add up to 10, which is true in our example.

We must stress that these are only necessary conditions, it is still possible to satisfy

the summation conditions and still not be a valid set of Go First Dice. However, we may

quickly eliminate any sets that do not satisfy these conditions.

1.1 For more than three players

We may now generalise the ideas found for three players to sets of 4, 5 and potentially

n

dice.

As before, the number of outcomes in contention are when two or more dice roll the

same side. And because the last die in our set only uses values that are higher or lower

than all other dice, counting the wins is relatively easy.

In general, if we have

k

dice, then a high value on side

s

of the

k

th

die contributes

sk−1−(s−1)k−1

wins. For the dice to be fair, these wins must add up to

1/k

of the

contended outcomes. The number of contended outcomes is

mk−k

m−1

P

i=0

ik−1

. And the

k

th

die must satisfy all conditions for sets of

k

dice and fewer.

Dening each die by a list of its high sides,

si

, results in the following conditions:

1. First die is an

m

-sided die;

2. Second die: All of the above and

Ps0

i=m/2

;

3. Third die: All of the above and

Psi=m(3m+ 2)/12

;

4. Fourth die: All of the above and

Ps2

i= (m2(m+ 1))/6

;

5. Fifth die: All of the above and

Ps3

i=m(5m2(3m+ 4) −4)/120

.

These four conditions are remarkably compact and only require us to perform various

sums with the sides that have high values.

We can also derive some column-wise conditions. For any subset of three dice, the

second die must win half of the contended outcomes not won by the third die. So let's

consider columns where the second die has a high value, and the third die has a low

value. If both dice satisfy condition 3, high-low values will occur in 1/3 of the columns.

Note that this is not necessary if one of the dice does not satisfy condition 3.

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Go First Dice for Five Players and Beyond.

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For example, here is a set of four 12-sided dice:

A: [3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3]

B: [4, 2, 2, 4, 2, 4, 2, 4, 2, 4, 4, 2]

C: [5, 1, 5, 1, 1, 5, 1, 5, 5, 1, 5, 1]

D: [6, 0, 0, 6, 6, 0, 6, 0, 0, 6, 6, 0]

This set satises all necessary summation conditions, with C and D satisfying the column-

wise condition. It is equivalent to the original set of Ford and Harshbarger.

1.2 Five players

We were now ready to nd a solution for ve players, armed with a few conditions

that allowed us to quickly reject any incorrect solutions.

We knew the numbers of sides was to be some multiple of 30. Unfortunately, when

m= 30

, there are no solutions to conditions

1−5

above, meaning a binary construction

of ve 30-sided dice was impossible. So we continued our search with

m= 60

.

Using a computer search, our method was to start with the potential fth dice that

satised all necessary summation conditions. We would then add a fourth dice that

satised both sum conditions and column-wise conditions, and continued to be build

backwards in this way. Sets that were not immediately eliminated could then be tested

to see if each place was equally likely.

Since this was still a very large task, we tried to speed up the search with a few

hunches. These hunches included making the third and fourth dice interchangeable, and

making the fth die a bit-inverted palindrome.

We found one set that satises all necessary conditions, and checks out as a valid set

of ve Go First Dice. It is the set of ve 60-sided dice given in the introduction.

This method was investigated by James Grime and Brian Pollock, and was the rst

set of ve Go First Dice found.

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Robert Ford, James Grime, Eric Harshbarger, Brian Pollock

81

It should be noted that while this set has the property that each

place

is equally likely,

not every

permutation

is equally likely. For example, while A and E are both equally

likely to be rst, the ordering ABCDE is not as likely as EDCBA. See [Enr19] for an

online tool that checks dice for place-fairness and permutation-fairness

Next, we describe a construction that is not only place-fair but also permutation-fair.

2 Inductive construction

Originally, Go First Dice were imagined with all dice having the same number of

sides. However, this does not need to be the case.

For example, the following is a set of three permutation-fair dice, consisting of one

2-sided die, one 4-sided die and one 6-sided die.

A: [03, 10]

B: [02, 04, 09, 11]

C: [01, 05, 06, 07, 08, 12]

These dice have 48 possible outcomes, with each ordering of A, B and C appearing equally

frequently.

We can denote which value appears on which dice with a simple two-row notation:

1 2 3 4 5 6 7 8 9 10 11 12

CBABCCCCBA B C

Or, more simply, in one-line notation as follows:

CBABCCCCBABC

We will now describe how to construct a set of

n+ 1

permutation-fair dice from a

known set of

n

permutation-fair dice. This method is believed to have been rst used

by Paul Vanetti [Kno12], although no details were given. The method was then reverse-

engineered by Robert Ford, and more fully investigated by Eric Harshbarger.

We'll start with the previous set of three permutation-fair dice. To create a fourth

die, D, we will need to create some gaps in the numbering of A, B and C. Let's repeat

the values on the dice, thus making each die larger, with suitably large gaps between the

copied values.

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Go First Dice for Five Players and Beyond.

82

For example, we could copy each die three times, adding 100 to the values of the rst

copy, 200 to the values of the second copy, and 300 to the values of the third copy:

A

: [103, 110, 203, 210, 303, 310]

B

: [102, 104, 109, 111, 202, 204, 209, 211, 302, 304, 309, 311]

C

: [101, 105, 106, 107, 108, 112, 201, 205, 206, 207, 208, 212, 301, 305, 306, 307, 308, 312]

Importantly, if the original set are permutation-fair then so is the expanded set. This

may feel intuitively true, but let's check an example. Since dice A, B, C are permutation-

fair, then

P(

A

>

B

>

C

)=1/6

. We will show that the same is true for

P(A>B>C)

.

When we roll a die from the expanded set, the face it lands on might be from the

rst copy, the second copy or the third copy. If we want

A>B>C

then the dice may

land on the following copies:

ABC = 333,332,331,322,321,311,222,221,211,111

In other words, these are the weakly decreasing sequences of the values 1, 2 and 3.

If three dice occupy the same copy, such as 333, 222 and 111 then they will automat-

ically inherit permutation-fairness. If two dice occupy the same copy, such as 332, 331,

322, 311, 221, 211 then the subset of two dice inherit permutation-fairness. If each die

land on a distinct copy, such as 321, then permutation-fairness isn't a concern since we

already know that

A>B>C

.

So, for each possibility on our list, we can calculate the probability that

A>B>C

.

Since we are equally likely to land on any of the three copies, the total probability can

be calculated as follows:

P(A>B>C)=1

33

1

3! +1

2!1! +1

2!1! +1

1!2! +1

1!1!1! +1

1!2! +1

3! +1

2!1! +1

1!2! +1

3!

=1

6

So our example has inherited permutation-fairness.

In the general case, imagine we roll

n

dice,

X1, X2,· · · , Xn

. Make a sequence of the

copies they land on, and let's say

ni

dice occupying the

i

th copy. If the sequence is weakly

decreasing then the probability of

X1> X2>· · · > Xn

is

Q

i

(1/ni!)

and 0 otherwise.

The number of sequences where

ni

dice occupying the

i

th copy is

n!/Q

i

(1/ni!)

, but

only one of these sequences is weakly decreasing. So, for each multiplicity, the probability

Recreational Mathematics Magazine, pp. 7587

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Robert Ford, James Grime, Eric Harshbarger, Brian Pollock

83

of

X1> X2>· · · > Xn

is

1/n!

.

Finally, summing over all multiplicities, the total probability of

X1> X2>· · · > Xn

is

1/n!

as required. Similarly, all other orderings of

X1, X2,· · · , Xn

will be equally likely,

as well as their subsets.

So, expanded sets do inherit permutation-fairness. We will now create a fourth die,

D, using values that t in the gaps of the expanded set, while maintaining permutation-

fairness.

2.1 Constructing a new die

Let's return to our previous example of three permutation-fair dice. We made three

copies of these dice to make the expanded set

A

,

B

and

C

. We now want to create a new

die, D, by insert values for D between the gaps of the expanded set. Our problem is to

nd how many values we need to insert while maintaining permutation-fairness.

For simplicity, let's refer to the expanded dice as A, B and C. Then, in one line

notation, we want to create something that looks like this

d0

CBABCCCCBABC

d1

CBABCCCCBABC

d2

CBABCCCCBABC

d3

where the

di

are placeholders for values of D that are to be inserted between the gaps.

Let

ki

denoted the number of values inserted in each position

di

. This means die D

will have

K=P

i

ki

faces. To nd the values of

ki

, imagine rolling the four dice together,

and calculating the probability of various permutations.

Let's start by calculating the probability of order ABCD. In that case, A, B and C

have all landed in copies with larger values than D. For example, if D lands on

d1

, then

the other dice have landed in the second and third copies. This occurs with probability

(2/3)3

, and the probability that A

>

B

>

C is

1/6

, due to the expanded set inheriting

permutation-fairness.

In full, if D lands on

d0

,

d1

,

d2

or

d3

, then the probability A, B, C have landed on

copies with larger values will be

1

,

(2/3)3

,

(1/3)3

and

0

, respectively, and in each case

the probability that A

>

B

>

C is

1/6

. We can now calculate the probability of order

ABCD to be:

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Go First Dice for Five Players and Beyond.

84

P(ABCD)

=X

di

P(D∈di)P(A, B, C > D)P(A > B > C)

=k0

K(1) 1

6+k1

K2

33

1

6+k2

K1

33

1

6+k3

K(0) 1

6

Calculating the probability of order ABDC is a little trickier, because D splits the

order into two parts. Namely, values that are higher than D and values that are lower than

D. Therefore, for each

di

, the probability of A and B landing on two higher copies and C

landing on a lower copy will be

(1)(0)

,

(2/3)2(1/3)

,

(1/3)2(2/3)

and

(0)(1)

, respectively.

And in each case, the probability of A

>

B is

1/2

. So the probability of order ABDC

will be:

P(ABDC)

=X

di

P(D∈di)P(A, B > D)P(D > C)P(A>B)

=k0

K(1)(0)1

2+k1

K2

32

1

31

2+k2

K1

32

2

31

2+k3

K(0)(1)1

2

Similarly we have:

P(ADBC)

=X

di

P(D∈di)P(A>D)P(D > B , C)P(B > C )

=k0

K(1)(0)1

2+k1

K2

31

32

1

2+k2

K1

32

32

1

2+k3

K(0)(1)1

2

P(DABC)

=X

di

P(D∈di)P(D > A, B, C)P(A > B > C)

=k0

K(0) 1

6+k1

K1

331

6+k2

K2

331

6+k3

K(1) 1

6

If each permutation is to be equally likely, we must set these probabilities to

1/24

.

Finally, solving for

ki

gives

k0= 1

,

k2= 3

,

k3= 3

and

k3= 1

with

K= 8

.

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Robert Ford, James Grime, Eric Harshbarger, Brian Pollock

85

We have now derived a fourth die, D. In one-line notation the set will look like this:

D CBABCCCCBABC DDD CBABCCCCBABC DDD CBABCCCCBABC D

Closing the gaps, so the dice use consecutive integers, results in this set:

A: [04, 11, 19, 26, 34, 41]

B: [03, 05, 10, 12, 18, 20, 25, 27, 33, 35, 40, 42]

C: [02, 06, 07, 08, 09, 13, 17, 21, 22, 23, 24, 28, 32, 36, 37, 38, 39, 43]

D: [01, 14, 15, 16, 29, 30, 31, 44]

This is a set of four permutation-fair dice. Establishing all orders of the full set are

equally likely is enough to ensure the orders of all subsets are also equally likely.

In general, if we start with a set of permutation-fair dice

X1,· · · , Xn

and an expanded

set made from

r

copies. Then, for all orders of

X1,· · · , Xn

and new die

Xn+1

to be equally

likely, we must nd solutions,

ki

, to the following set of equations:

1

(n+ 1)! =1

l!(n−l)!

r

X

i=0

ki

Ki

rlr−i

rn−l

,

for

l= 0,· · · , n

This can also be applied to all subsets of

X1,· · · , Xn

. In particular, when

n=l

we

get the following neat set of equations:

1

l+ 1 =

r

X

i=0

ki

Ki

rl

,

for

l= 0,· · · , n

2.2 Five or more players

Inductive construction can be used to construct permutation-fair dice from a known

set of permutation-fair dice, or build them from scratch. For example, the set of three

dice above was constructed from the trivial 1-sided die.

In general, if we start with

n

dice and make

r

copies, we get the following table of

results for the new die. Table entries are written in the form

(k0, k1,· · · , kr), K

:

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86

n\r 1 2 3 4 5 6

1 (1, 1),

K = 2 (0, 1, 0),

K = 1 (0, 1, 1, 0),

K = 2 (0, 0, 1, 0, 0),

K = 1 (0, 0, 1, 1, 0, 0),

K = 2 (0, 0, 0, 1, 0, 0, 0),

K = 1

2 - (1, 4, 1),

K = 6 (1, 3, 3, 1),

K = 8 (1, 0, 4, 0, 1),

K = 6 (0, 11, 1, 1, 11, 0),

K = 24 (1, 0, 0, 4, 0, 0, 1),

K = 6

3 - (1, 4, 1),

K = 6 (1, 3, 3, 1),

K = 8 (1, 0, 4, 0, 1),

K = 6 (0, 11, 1, 1, 11, 0),

K = 24 (1, 0, 0, 4, 0, 0, 1),

K = 6

4 - - - (7, 32, 12, 32,

7), K = 90 (19, 75, 50, 50,

75, 19), K = 288 (1, 5, 1, 6, 1, 5, 1),

K = 20

5 - - - (7, 32, 12, 32,

7), K = 90 (19, 75, 50, 50,

75, 19), K = 288 (1, 5, 1, 6, 1, 5, 1),

K = 20

6 - - - - - (41, 216, 27, 272,

27, 216, 41), K =

840

We could have made a smaller set of four dice in our previous example. Starting with

the set of three dice, and taking two copies instead of three, would have resulted in a

4-sided die A, an 8-sided die B, a 12-sided die C and a 6-sided die D.

If we then made six copies of this set, we will have a 24-sided dice A, a 48-sided die

B, a 72-sided die C, a 36-sided die D and a 20-sided die E. This gives a set with a total

of 200 sides, and is the smallest ve player set of permutation fair dice known to date,

when smallest means fewest number of faces in the set.

However, this set does include one large 72-sided die. Alternatively, if we want each

individual die to be smaller, we could start with the following four player set;

ACDBBDDCBCCDBAADAACDBBCBDDCCDBA

We may now continue as before, by taking six copies and creating a 20-sided die, E. This

results in one 36-sided die, two 48-sided dice, one 54-sided die and a 20-sided die. This

is 206 faces in total, but the individual dice are generally smaller and more usable. This

our preferred set, and is the set given in the introduction.

One important point to note about this set is that the initial set of four dice was not

found by the inductive construction. Instead, the set was discovered using a computer

Recreational Mathematics Magazine, pp. 7587

DOI 10.2478/rmm-2023-0004

Robert Ford, James Grime, Eric Harshbarger, Brian Pollock

87

program written by Landon Kryger - which has been extraordinarily helpful in searching

for permutation-fair sets, though it has yet to nd a ve player set on its own.

Finally, what if we wanted our dice to all be the same size? One set of permutation-fair

dice, with dice the same size, is based on the following set of four 18-sided dice:

DCBCDBBCDAAACAAADBBBDDAAACCCBDDCDCBB-

-BBCDCDDBCCCAAADDBBBDAAACAAADCBBDCBCD

By making ten copies of this set we may construct a fth die with 36 sides, giving

four 180-sided dice and one 36-sided die. However, making ve copies of the fth dice

will result in a set of ve 180-sided permutation-fair dice.

The two solutions given in the introduction should be compared and contrasted. Both

binary construction and induction were designed to make the search or construction of

Go First Dice easier, and both have nice features. Still, neither method is comprehensive

and other solutions, including smaller solutions, may still exist.

References

[Bel12] A. Bellos,

Puzzler develops game-changing Go First dice

, The Guardian, 2012.

http://www.guardian.co.uk/science/alexs-adventures-in-numberland/

2012/sep/18/puzzler-go-first-dice

[Enr19] B. Enright,

Dice checker

, 2019.

http://www.brandonenright.net/cgi-bin/dice_perm_util.pl

[Kno12] B. Knoll,

Five Player Go First Dice

, 2012.

http://byronknoll.blogspot.se/2012/10/five-player-go-first-dice.

html

[Har18] E. Harshbarger,

Go First Dice

, 2018.

http://www.ericharshbarger.org/dice/go_first_dice.html

[Har12] E. Harshbarger, Who Goes First,

Games Magazine

, Vol. 36, No. 10, Issue 286,

Dec 2012.

Recreational Mathematics Magazine, pp. 7587

DOI 10.2478/rmm-2023-0004