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Numbering the Faces of Dice

```
Date: 02/27/2001 at 06:35:12
From: Alison Cotterill
Subject: Platonic solids as dice

If we were to make dice out of solids such as the tetrahedron,
octahedron, etc., would it be possible to number them so that all
opposite face pairs would add up to the same number? If so, what would
the opposing face sums be for each solid type?

What if we wanted the sums of the faces at each vertex to be equal? Is
such a die possible?

Are there other ways we can number the dice as well?
```

```
Date: 02/27/2001 at 12:15:43
From: Doctor TWE
Subject: Re: Platonic solids as dice

Hi Alison - thanks for writing to Dr. Math.

As you probably know, there are exactly five Platonic solids: the
tetrahedron, cube, octahedron, dodecahedron, and icosahedron. If
you're not familiar with them or their characteristics, check out our
Regular Polyhedra FAQ at:

http://mathforum.org/dr.math/faq/formulas/faq.polyhedron.html

Each has an even number of faces (4, 6, 8, 12, and 20, respectively)
and this makes the answer to your first question 'yes'. To number the
opposite faces on an n-hedron, simply pair up the numbers that add up
to n+1, as we do on a 'normal' cubic die. On a normal 6-sided die,
opposite faces add up to 6 + 1 = 7: 1 and 6, 2 and 5, 3 and 4.
Similarly, on a tetrahedral die, the opposite faces would add up to
4+1 = 5: 1 and 4, 2 and 3. On an octahedral die, the opposite faces
would add up to 8+1 = 9: 1 and 8, 2 and 7, 3 and 6, 4 and 5. And so
on.

(Having said this, note that in a tetrahedron, no side is 'opposite'
any other side.  However, we could still pair up the sides, making two
of them one color, and the other two another color, and the rest of
the discussion would still make sense.)

The answer to your second question, however, is 'no' - at least, not
if we retain the requirement that the faces be numbered with the
values from 1 to n, inclusive. As a simple counterexample, consider
the tetrahedron. Each vertex touches all but one of the four faces.
Thus, the vertex sum for any vertex is equal to (1+2+3+4) - x, where
x is the value of the face opposite the given vertex. In order for all
vertices to have the same sum, the value of x must be the same for all
vertices, and thus all faces must have the same value. This
contradicts the requirement that the faces be numbered with the values
from 1 to n.

A similar proof can be constructed for the cube and the dodecahedron.
In these shapes, each vertex connects exactly three faces. Thus, any
two neighboring vertices share two faces but have different third
faces. In order for the vertex sums to be equal, the values of the
third faces must be equal, thus requiring two different faces (the two
'third faces') to have the same value.

Because vertices on the octahedron connect four faces and vertices on
the icosahedron connect five faces, this "disproof" doesn't work for
them - perhaps it is possible to construct such a die for these
shapes. I'll think some more about that and write back if I can either
prove or disprove it.

As to your final question, when making a die, you can number the faces
in any manner you choose, and you don't have to use a cube. There are
many games that use dice with different numbering schemes and/or
different numbers of faces. I have a pair of (six-sided) dice where
one is numbered:

1, 2, 2, 3, 3, 4

and the other is numbered:

1, 3, 4, 5, 6, 8

Although the numbers don't match a pair of traditional dice, they
produce the exact same probabilities when rolled as a pair!

I hope this helps. If you have any more questions, write back.

- Doctor TWE, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 03/07/2001 at 04:51:00
From: Alison Cotterill
Subject: Re: Dice and the formula behind them

Given that the faces can be numbered in any way to make an infinite
number of dice, is there a formula to show how this can be done, or an
attempt to calculate how many possibilities of the different dice
there are?

Thanks.
```

```
Date: 03/07/2001 at 12:06:03
From: Doctor TWE
Subject: Re: Dice and the formula behind them

Hi Alison - thanks for writing back.

We could put any number on any face. So we could have, for example,
faces numbered {1,1,1,1,1,1} or {1,2,3,4,5,6} or {3,7,0,8,5,9,12}
or... As you said, there are infinitely many possibilities, and we
can't make a list of them all.

But suppose we restrict it to a cube with six specific and distinct
values. I'll use {1,2,3,4,5,6} for simplicity. Let's place the values
on the faces of the cube one at a time and count the number of ways we
can do it.

First we'll place the 1. It doesn't matter which face we put it on
because of the symmetry of the cube. Here is a net of our die so far:

+---+
|   |
+---+---+---+
|   | 1 |   |
+---+---+---+
|   |
+---+
|   |
+---+

Now to place the 2, we have two non-symmetric possibilities; we can
place it on one of the 4 faces adjacent to the 1 face, or we can place
it on the face opposite the 1. Placing it on any of the adjacent faces
is symmetric with placing it on any other adjacent face. (We could
rotate the dice to make them "look the same.") Here are the nets now:

+---+            +---+            +---+            +---+
|   |            |   |            | 2 |            |   |
+---+---+---+    +---+---+---+    +---+---+---+    +---+---+---+
|   | 1 |   | or | 2 | 1 |   | or |   | 1 |   | or |   | 1 | 2 |
+---+---+---+    +---+---+---+    +---+---+---+    +---+---+---+
| 2 |            |   |            |   |            |   |
+---+            +---+            +---+            +---+
|   |            |   |            |   |            |   |
+---+            +---+            +---+            +---+

(The above are all symmetric)

+---+
|   |
+---+---+---+
|   | 1 |   |
+---+---+---+
|   |
+---+
| 2 |
+---+

For simplicity, let's look only at the first net of 1 and 2 adjacent
(since the others are just rotations of it) and the net of 1 and 2
opposite:

+---+                +---+
|   |                |   |
+---+---+---+        +---+---+---+
|   | 1 |   |   or   |   | 1 |   |
+---+---+---+        +---+---+---+
| 2 |                |   |
+---+                +---+
|   |                | 2 |
+---+                +---+

Now consider placing the 3. For the left net, each remaining place is
unique relative to the 1 and 2 (i.e. no rotation would make any two
choices look the same), so we have four possibilities. On the right
net, each of the four remaining places is symmetric to every other
one. So we now have:

+---+            +---+            +---+            +---+
|   |            | 3 |            |   |            |   |
+---+---+---+    +---+---+---+    +---+---+---+    +---+---+---+
| 3 | 1 |   | or |   | 1 |   | or |   | 1 | 3 | or |   | 1 |   |
+---+---+---+    +---+---+---+    +---+---+---+    +---+---+---+
| 2 |            | 2 |            | 2 |            | 2 |
+---+            +---+            +---+            +---+
|   |            |   |            |   |            | 3 |
+---+            +---+            +---+            +---+

or

+---+
|   |
+---+---+---+
|   | 1 |   |
+---+---+---+
| 3 |
+---+
| 2 |
+---+

Now each of the remaining three places is unique on each net, so for
each of the five nets, there are three possible faces to place the 4,
giving us:

5*3 = 15

possible combinations. Of course, the remaining two faces on each of
those fifteen nets will be unique (since they were previously unique),
so placing the 5 on each of those fifteen nets, we will have:

15*2 = 30

possible nets. Then, of course, each of those thirty nets will have
only one remaining face on which to put the 6, so there are thirty
unique ways to number a die with the numbers {1,2,3,4,5,6}.

I hope this helps. If you have any more questions or comments, write
back again.

- Doctor TWE, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 03/14/2001 at 09:55:57
From: Alison Cotterill
Subject: Re: Dice and the formula behind them

Can you tell me please how many different ways there are to make a
die, with the traditional restriction that opposite faces add up to 7?

Thank you.
```

```
Date: 03/14/2001 at 12:59:28
From: Doctor TWE
Subject: Re: Dice and the formula behind them

I'll assume that you're not counting a rotation as a 'different way'.

We can use a similar line of reasoning as in my last response; let's
start with a blank cube and put the numbers on the faces.

To place the number 1, we have six faces, but whichever face we put it
on, we can rotate the 1 to the top. So there is only one 'different'
way of putting the number 1 on the cube. (I'll draw the net again.)

+---+
|   |
+---+---+---+
|   | 1 |   |
+---+---+---+
|   |
+---+
|   |
+---+

Now instead of placing the 2, let's first place the 6. We know that it
has to be opposite the 1 by the restriction you stated (it must be
'traditional'). So there's only one place we can put it - opposite
the 1, like this:

+---+
|   |
+---+---+---+
|   | 1 |   |
+---+---+---+
|   |
+---+
| 6 |
+---+

Why did we place the 6 next instead of the 2? Because we had a
restriction on numbers adding to 7; if we had placed the 2 next, we
might have put it opposite the 1 (as we did in the last response), and
that would have violated our restriction. So there's only 1*1 = 1 way
to place the 1 and 6.

Now let's place the 2. There are four remaining blank faces, but each
is adjacent to both the 1 and the 6, so each is a rotation of the
others. Thus, it doesn't matter which of the faces we choose, we have
1*1*1 = 1 way to place the 2.

+---+
|   |
+---+---+---+
|   | 1 |   |
+---+---+---+
| 2 |
+---+
| 6 |
+---+

Again, let's place its "opposite" next. The 5 must go opposite the 2,
based on our 'traditional' restriction, so there's only one place for
it, and we still have only 1*1*1*1 = 1 way:

+---+
| 5 |
+---+---+---+
|   | 1 |   |
+---+---+---+
| 2 |
+---+
| 6 |
+---+

In other words, up to this point, no matter which faces you place the
1, 2, 5, and 6 on, so long as the 1 and 6 are on opposite faces and
the 2 and 5 are on opposite faces, any placement will be a rotation of
any other placement.

Now there are two blank faces left to place the 3. These two faces are
_not_ equivalent. Placing the 3 on the left face on my net will cause
the 1, 2, and 3 to appear to go clockwise when looking at their shared
corner. Placing the 3 in the right face on my net, however, will cause
the 1, 2, and 3 to appear to go counterclockwise when looking at their
shared corner. Thus there are two different ways to place the 3, and a
total of 1*1*1*1*2 = 2 different ways so far.

+---+                +---+
| 5 |                | 5 |
+---+---+---+        +---+---+---+
| 3 | 1 |   |   or   |   | 1 | 3 |
+---+---+---+        +---+---+---+
| 2 |                | 2 |
+---+                +---+
| 6 |                | 6 |
+---+                +---+

Finally, in either net, there's only one face left to place the 4, so
we have 1*1*1*1*2*1 = 2 different ways altogether. Here are the final
nets:

+---+                +---+
| 5 |                | 5 |
+---+---+---+        +---+---+---+
| 3 | 1 | 4 |   or   | 4 | 1 | 3 |
+---+---+---+        +---+---+---+
| 2 |                | 2 |
+---+                +---+
| 6 |                | 6 |
+---+                +---+

Incidentally, these are often called "clockwise" and
"counterclockwise" dice, or "left-handed" and "right-handed" dice.

I hope this helps. If you have any more questions or comments, write
back again.

- Doctor TWE, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Permutations and Combinations
High School Polyhedra
High School Symmetry/Tessellations

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