## If we are doing dice – try some different dice!

Posted by Karl Sharman on August 9, 2014 – 5:05 am

With a standard pair of six sided dice, there is one way of obtaining a 2, two ways of obtaining a 3, and so on, up to one way of obtaining a 12. Find all other pairs of six-sided dice such that:

a. The set of dots on each die is not the standard {1,2,3,4,5,6}.

b. Each face has at least one dot.

c. The number of ways of obtaining each sum is the same as for the standard dice.

August 9th, 2014 at 8:11 am

My first thought is that the only possibility is to multiply the standard face by a common factor k, then add a common offset c. Let s represent a sum using the standard dice, and n be the corresponding sum for the non-standard dice, then n = ks + 2c.

If we first added the offset and then did the multiple we’d have got n = ks +2kc. That’s only a subset of the previous result. So we can ignore it.

Repeating the process, using N for the next non-standard set, and using C and K for the offset and multiplier, we get N = Kn + 2C = K(ks + 2c) + 2C = Kks + 2(Kc +C). Because we can set k’ = Kc + C and c’ = Kc + C, we can rewrite that as n = k’s + 2c’, so nothing new there, so we needn’t bother with it either.

I can’t think of a neat way to show that every such set of die is linearly related to the standard set. However, I believe that there is no other scheme. At the moment that seems to me to be obvious. I always find it hard to prove the obvious.

August 9th, 2014 at 8:19 am

If the pair of dice don’t have to be identical, then, a priori, it might be that other solutions can be found. I haven’t, yet, bothered to see if that can easily be disproved (or not). Doing 36 sums is too much like hard labour for me.

August 9th, 2014 at 10:52 am

I’ve now got a general answer. It confirms my previous belief and extends it to the case where the dice aren’t identical. It also is readily extended (by induction) to any number of dice. I’ll keep it to myself, for now, so as not to totally hijack this page.

August 10th, 2014 at 4:19 am

I’m away for a few days (brother getting married), so I’ll post my scratchings now.

I’ll try for the most general solution. Let the dice have face values A1, A2,…, A6 and B1, B2,…, B6.

WLOG let A6 ≥ A5 ≥ A4 ≥ … ≥ A1 and B6 ≥ B5 ≥ B4 ≥ … ≥ B1.

There must be exactly one way to represent the highest sum. That’s A6 + B6. If either A5 = A6 or B5 = B6 then we’d have two or more ways of representing the highest sum. So A6 > A5 and B6 > B5. Similarly, the second highest sum must be represented in only two ways. That sum must be A6 +B5 = A5 + B6. If A5 = A4 or B5 = B4 we’d have more than two ways of representing the second highest sum. So A5 > A4 and B5 > B4. Proceeding in this fashion, we can conclude that

A6 > A5 > A4 > A3 > A2 > A1 and B6 > B5 > B4 > B3 >B2 > B1. Note that we have A6 – A5 = B6 – B5.

The third highest sum must (this is the first assertion that I’m not certain about – so all that follows may be wrong) be A6 + B4 = A5 + B5 = A4 + B6. Taking the first pair gives A6 – A5 = B5 – B4 = B6 – B5. The second pair => B6 – B5 = A5 – A4 = A6 – A5. The outer pair gives A6 – A4 = B6 – B4.

Continuing in this way we conclude that, if the smallest value on the A die is a, then the die must have faces: a, a+k, a+2k, a+3k, a+4k, a+5k and for the B die the faces must be b, b+k, b+2k, b+3k, b+4k, b+5k.

So each new die can be generated from the standard die, by multiplying the standard value by a factor k (the same k for both dice), and then adding any integer, to each. The additional integer need not be the same for both of the dice.

e.g. Let k = 5 and a =3 and b = 7. The two dice would be:

8,13,18,23,28,33 and 12,17,22,27,32,37

I’ll admit that I’ve been lazy, and have a copped out of providing every detail.

It seems clear that this can be extended to any number of dice. Each one can be multiplied by a common factor, k, and then each one can be offset independently by an arbitrary amount.

August 10th, 2014 at 11:09 am

I’m not sure if my last post is almost complete bilge.

August 11th, 2014 at 7:13 am

Chris…Post 4? I am looking for 2 dice which can have one way of obtaining a 2, two ways of obtaining a 3, and so on, up to one way of obtaining a 12. The dice won’t be the same, but are 6 sided with at least one dot on each face.

Anyway, have an enjoyable brother’s wedding, and if you are giving a speech, remember they are all looking at you, hanging on your every word – it had better be a GOOD speech!

August 12th, 2014 at 2:37 am

I give up. I can’t imagine how to do it, unless e.g. we use multiplication rather than addition (not that I can do it that way either).

August 13th, 2014 at 6:11 am

OK Chris, I went with a ‘generating function’ to represent each die.

Standard dice can be represented by the generating function f(x) = x^1 + x^2 + x^3 + x^4 + x^5 + x^6.

Here, the exponent represents the score on the die face; the coefficient is the number of ways each score can be obtained.

The sums that can be obtained by throwing two standard dice correspond to multiplying their generating functions:

(x^1 + x^2 + x^3 + x^4 + x^5 + x^6)^2 = x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 + 6x^7 + 5x^8 + 4x^9 + 3x^10 + 2x^11 + x^12.

Here again, each coefficient represents the number of ways each score (exponent) may be obtained. Any alternative dice (that have the same number of ways of obtaining each sum as the standard dice) must have generating functions whose product is the same as for the standard dice, above.

So the problem is reduced to finding an alternative factorization, into two factors, of (f(x))^2; a more amenable task.

f(x) = x(1 + x + x^2 + x^3 + x^4 + x^5)

= x(x^6 − 1)/(x − 1)

= x(x^3 + 1)(x^3 − 1)/(x − 1)

= x(x + 1)(x^2 − x + 1)(x^2 + x + 1)

Therefore (f(x))2 = x^2(x + 1)^2(x^2 − x + 1)2(x^2 + x + 1)^2

An alternative factorization must redistribute these terms into two factors, g(x) and h(x), which will be the generating functions for the alternative dice.

There are two further constraints:

•Each face has at least one dot. This corresponds to g(x) and h(x) each having a minimum exponent of 1. Therefore g(x) and h(x) must each get one copy of the factor x.

•Each die has six faces. This corresponds to a requirement that g(1) = h(1) = 6. Looking at the factorization of f(x), (x + 1) yields 2, (x^2 + x + 1) yields 3, while the remaining two factors yield 1. Therefore g(x) and h(x) must each get one copy of (x + 1) and (x^2 + x + 1).

This leaves only the two (x^2 − x + 1) factors to distribute among g(x) and h(x).

Clearly, if we give one copy to each function, we get g(x) = h(x) = f(x), yielding the standard dice.

The only alternative is to give both factors to one of the functions:

g(x) = x(x + 1)(x^2 + x + 1)

= x + 2x^2 + 2x^3 + x^4

h(x) = x(x + 1)(x^2 + x + 1)(x^2 − x + 1)2

= x + x^3 + x^4 + x^5 + x^6 + x^8

This yields unique alternative dice of {1,2,2,3,3,4} and {1,3,4,5,6,8}.

August 14th, 2014 at 1:01 am

Karl. That was amazing. I’m sorry that I didn’t rise better to the challenge.

Amongst many of my failures, I hadn’t thought to let two faces have the same value.

I’m sure that there is no way that I would ever have thought of using generating functions. I think the time has finally arrived for me to get to grips with them. This is at least the third time they have cropped up, although before now only as side-shows, not centre stage.