A friend and I have a disagreement about the number of unique ways to label the faces of a six-sided die with integers from one to six so that each pair of opposing faces sums to seven.

I think the answer is 8 since swapping each pair of opposing faces results in (I think) a different handedness for the two hemi cubes centered on the two faces being swapped, without affecting the other hemi cubes. So I think the problem is represented by the possible independent states of the three opposing pairs, giving $2^3=8$ permutations.

My friend thinks the answer is 2, as he argues that a single swap changes the handedness of the entire die, and that further independent swaps can’t further change the die’s handedness, and thus don’t generate unique states.

Who is right? (Or is neither of us right?)

  • 1
    $\begingroup$ How is handedness defined for a hemicube? $\endgroup$ – coffeemath Jan 1 '20 at 20:19
  • $\begingroup$ If all else fails, try making them with, I don't know, plasticine or something. $\endgroup$ – Shaun Jan 1 '20 at 20:21
  • $\begingroup$ @coffeemath I mean the order the sides of the hemicube occur, CW or CCW. Not sure I’m right though. $\endgroup$ – bob Jan 1 '20 at 20:47

Your successive face-swapping doesn't produce as many unique labelings as you think. Once you swap faces twice, you are back in the original orientation, as this figure shows:

Some symmetries of a cube.  The cube is labeled with the numbers 1 through 6.  The numbers on opposite sides are swapped twice, and the cube is rotated 90 degrees twice.  We see they produce the same configuration.

The large numbers are on the faces, and the small ones are on the hidden faces that they are nearest. Swapping the top/bottom and left/right pairs produces the same labeling as simply rotating 180° around the front-back axis.

So here is another way to go about counting the labelings: Fix a single corner of the cube and consider the three faces incident on that corner:

  • There are six possibilities for the first face: any number from 1 to 6;
  • Once the first face is chosen, there are four possibilities for the second face, because neither the number on the first face, nor its complement, which has to be placed opposite the first face, are available.
  • Once the first two faces are chosen, there are only two possibilities for the last face.

So there are $6 \times 4 \times 2$ ways to label the cube if it's in a fixed position. Of course, we are allowed to rotate the cube, and labelings of the cube should be considered equivalent if one can be rotated into the other. There are $4!$ rotations of the cube. So the number of distinct labelings modulo rotation is $$ \frac{6 \times 4 \times 2}{4 \times 3 \times 2 \times 1} = 2 $$


Note that the faces with labels 1, 2, and 3 must appear in a 3-cycle (share a corner), and labeling these three faces uniquely determines the labels of the other three faces. There are two ways to do that.


Your friend is correct.

We can orient the die so that 1 is on the bottom, which forces 6 to be on the top, and so that 2 is facing us, which forces 5 to be in the back. That leaves us two choices. We can either place 3 on the left and 4 on the right or 3 on the right and 4 on the left.


48 ! :-)
(special case)

A friend and I have a disagreement about the number of unique ways to label the faces of a six-sided die with integers from one to six so that each pair of opposing faces sums to seven.

It is clear from both your initial disagreement and comments and answers here that the question is somewhat ambiguous - the meaning of "unique ways" needs refining. eg does 'handedness' matter, or .... .

So, based on what's been said since, while it's clearly not what you originally intended, here is an interpretation that seems valid (to me :-) ) that yields the result 48!
This is based on the assumption that the unnumbered die is NOT rotation agnostic but itself has uniquely identifiable faces apart from the numbering. While that's not how die are usually made, it's a valid (I think) variation on the original question.

Imagine a cube with faces inlaid with 6 different semi-precious stones. (It could equally well have 6 different colours but Semi precious stones seems more interesting.) If you have multiple copies of a standard die with colours assigned in the alignment specified below, and then assign valid numbers to faces and produce a set of varying dice, how many unique dice can you create?

Assume a standard SPS die, with say topaz up, chalcedony down, beryl left, jasper right, carnelian front and amethyst back. If we decide this is going to bend the brain we can resort to calling the sides Up, Down, Left, Right, Front, Back - or even U D L R F B.

Start with Up / Topaz.
We have 6 numbers to choose from.
Assigning any number also sets the number of Down/Chalcedony to make the opposing faces sum to 7.

Now assign a number to say Left / Beryl
We have 4 choices - and this also sets the value of Right / Jasper.

That leaves 2 choices for Front / Carmelian
and sets Back / Amethyst accordingly.

Total unique die are 6 x 4 x 2 = 48.

But wait, there's more!

Here's an arguably* new question arising which I may propose as such.
But, it's worth outlining here. *"Arguably" as this too seems to be a legitimate interpretation of the original question. Just as I randomly assigned numbers to face-pairs, I could assign stone types to faces. But as there is no sum-to-7 constraint there are 6! assignable patterns.
However, many of these are non unique if the unnumbered cube is rotated.
How many unique numbered cubes can we make if the cubes can be assigned stones to faces randomly before numbering?


The special collectors edition

uses a Sapphire, Emerald, Sardonyx, Chrysolite, Chrysoprase & Jacinth die.

Bonus question - where did I get my two stones lists from?

  • $\begingroup$ Wow! -2 and no comments why. Why this is "not useful" is inapparent. Not what was intended, but fits the original spec nonetheless. $\endgroup$ – Russell McMahon Jan 3 '20 at 3:34
  • $\begingroup$ Would anyone care to comment on this? I have two unexplained (as per usual) downvotes and no upvotes. I don't care about the upvotes in their own right but am surprised that SOMEONE has not said ~= "Well. Fancy that. Yes - if the cube is unique in its own right then this appears correct.Fancy that (again)". | Whatever. $\endgroup$ – Russell McMahon Jan 3 '20 at 8:32

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