Alternate d4 labelling
When I was a kid, I used to play a little Dungeons and Dragons. If you’ve played, you are probably pretty familiar with polyhedral dice. These dice are used to generate random numbers in various ranges. A basic set of dice comprises a traditional six-sided die in the form of a cube (typically referred to as a d6), along with a four-sided die d4, an eight-sided die d8, a ten-sided die d10, a twelve-sided die d12, and a twenty-sided die d20.

All but the d10 are platonic solids (tetrahedron d4, cube d6, octahedron d8, dodecahedron d12, icosahedron d20). The d10, whilst a fair die, typically takes the form of a truncated pentagonal trapezohedron. Whilst some might consider that this makes the d10 the ‘odd one out’ of the bunch, to me, it was always the d4 that seemed weird.

d4 Weirdness
The d4 is weird because you do not read the result in a conventional way. All the other dice land with one face flat, horizontal, and parallel to the landing surface. To read the ‘roll’ you simply note the label on the topmost surface. Because the d4 is tetrahedral in shape, it lands with one face flat, and an apex pointing upwards. There is no way to place a visible label on the point of the apex (nor a way to read the number on the face flat on the table).
To remedy this, the dice is read in an alternate way. There are two common styles (they might be given formal names but I am calling them Base Face Read, and Apex Read). In the first variant, the number ‘rolled’ is the number that is shown, in common, along the edges of the base face. In the second variant, you read the result from the number show at the apex. In both variants the number is repeated on all visible sides.

Alternate Numbering
The other day I was wondering if it would be possible to number a d4 such that, to ‘read’ the roll, you simply total numbers that are printed on the three visible faces. It turns out that it is possible. (I’m not, for any moment, suggesting you do this! It’s overly complicated, and just a recreational puzzle). The only benefit I can see to this method is that it requires only one number per face!
An advantage of this system is that it only needs one number per face.
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Math
The math really isn’t hard at all. It’s just a collection of simultaneous equations. If we define the values of the faces to be: a, b, c, d we just need to solve these equations:
a + b + c = 1
a + b + d = 2
a + c + d = 3
b + c + d = 4
Rearranging the first gives:
a = 1 - b - c
Which we can substitute in the second and third to give:
d = c + 1
d = b + 2
And equating these gives:
c = 1 + b
Substituing into the fourth gives:
b + c + d = 4
b + (1 + b) + (2 + b) = 4
3b = 1
b = 1/3
From here it all follows out that:
a = -2/3 , b = 1/3 , c = 4/3 , d = 7/3
Table
Here is the data in tabular form. A check shows that this face is visible.
-2/3 | 1/3 | 4/3 | 7/3 | ||
---|---|---|---|---|---|
Score: 1 | ✓ | ✓ | ✓ | ✗ | |
Score: 2 | ✓ | ✓ | ✗ | ✓ | |
Score: 3 | ✓ | ✗ | ✓ | ✓ | |
Score: 4 | ✗ | ✓ | ✓ | ✓ |
d6 version
How about a regular six-sided die? What numbers would you need to place on each side so that adding the five visible faces would results in an answer in the range [1-6]?
Here's the answer:
-9/5 , -4/5 , 1/5 , 6/5 , 11/5 , 16/5
-9/5 | -4/5 | 1/5 | 6/5 | 11/5 | 16/5 | ||
---|---|---|---|---|---|---|---|
Score: 1 | ✓ | ✓ | ✓ | ✓ | ✓ | ✗ | |
Score: 2 | ✓ | ✓ | ✓ | ✓ | ✗ | ✓ | |
Score: 3 | ✓ | ✓ | ✓ | ✗ | ✓ | ✓ | |
Score: 4 | ✓ | ✓ | ✗ | ✓ | ✓ | ✓ | |
Score: 5 | ✓ | ✗ | ✓ | ✓ | ✓ | ✓ | |
Score: 6 | ✗ | ✓ | ✓ | ✓ | ✓ | ✓ |