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I do not know if there is a correct solution to this puzzle, but I can't stop thinking about it, so here goes: is it possible to accurately simulate a single, fair, six-sided die using three custom six-sided dice?

Here are the more specific stipulations:

  1. The only possible roll totals between the three dice must be one through six.
  2. These six numbers must be rolled with equal probability.
  3. No negative numbers are allowed.
  4. The dice do not have to be identical.
  5. When a six is rolled, all three dice must contribute some number greater than zero to the total (to avoid the trivial solution).

I feel like I'm getting closer but I'm not there yet. I know that you can't simplify the dice down to two- or three-sided dice (as in, having each die have only two different numbers, equally represented), because then the number of combinations possible isn't divisible by six.

TheSoundDefense
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    Do you have to combine them by summing? – Deusovi Sep 04 '17 at 04:01
  • @Deusovi yes, by summing. Is there a solution for other methods of combination? – TheSoundDefense Sep 04 '17 at 04:04
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    If other methods of combination are allowed, you could just roll 3 standard dice and add them mod 6 (if you get a result above 6, subtract 6 until you're low enough). – Deusovi Sep 04 '17 at 04:10
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    Here is a harder version of this: Simulate two dice with three indistinguishable dice: https://www.youtube.com/watch?v=xHh0ui5mi_E and solutions: https://www.youtube.com/watch?v=hBBftD7gq7Y – Fabian Röling Sep 04 '17 at 07:36
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    @Fabian It was also asked on this site here https://puzzling.stackexchange.com/questions/31101/make-2-dice-out-of-3-dice – Kruga Sep 04 '17 at 07:54
  • What if you drop condition 3 but keep the others? It isn't clear that you would pick up any additional solutions. – John Coleman Sep 04 '17 at 10:37
  • @Deusovi: Mod 6 gives you scores between 0 and 5, so you'd need to add 1 to the sum. I was very surprised to see that every sum has the same probability (36/216), though. Nice one! – Eric Duminil Sep 04 '17 at 13:08
  • @EricDuminil - I didn't mean the binary modulo operator - instead I mean dividing the integers into $\mathbb{Z}_6$, where $[0]=[6]$. (I'm picking 6 as the representative of its equivalence class - but I didn't think the formalism was necessary.) And that trick always works as long as at least one die is standard - imagine rolling all the other dice first. Then the last die will shift you a random number of spaces around the mod-6 circle. – Deusovi Sep 04 '17 at 14:00
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    @JohnColeman: Then you'd pick up the solution 1/1/1/1/1/1, 1/1/1/1/1/1, -1/0/1/2/3/4, among others. – Deusovi Sep 04 '17 at 14:01
  • On a related issue, it is possible to emulate "two standard distinguishable dice" from "three standard indistinguishable dice" – Hagen von Eitzen Sep 04 '17 at 17:06

2 Answers2

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I believe this set of dice satisfies all your requirements:

Die 1: 111111
Die 2: 002244
Die 3: 010101

Deusovi
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@Deusovi's answer is totally correct, but I want to add here the general approach for solving such problems as well. No need to upvote, since I did not invent the technique, and you can see it described in this puzzle as well.

The idea is to use generating functions. Basically, we try to check if there is a factorization of $x^1 + x^2 + x^3 + x^4 + x^5 + x^6$ as a scaled product of three polynomials with integer non-negative coefficients.

\begin{align*} x^1+ x^2 + x^3 + x^4 + x^5 + x^6 &= x(1+x)(1 + x+ x^2) (1-x+x^2) \\ &= x^1(x^0+x^1)(x^0+x^2+x^4) \\ &= x^1(x^0+x^1+x^2)(x^0+x^3) \end{align*}

Therefore the two possible relabelings are:

1, 1, 1, 1, 1, 1

0, 0, 0, 1, 1, 1

0, 0, 2, 2, 4, 4

or

1, 1, 1, 1, 1, 1

0, 0, 1, 1, 2, 2

0, 0, 0, 3, 3, 3

Puzzle Prime
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  • Nice analysis, Artur K., related to a couple of those at Relabeling two 20-sided dice without changing their total – humn Sep 04 '17 at 15:48
  • Thanks, @humn, I have added link to that problem in the solution. The most classic example is about relabeling 2 dices. – Puzzle Prime Sep 04 '17 at 15:57
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    I'll have to look further into this as this still seems somewhat magical. Thanks! – TheSoundDefense Sep 04 '17 at 17:46
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    It might be easier to see what's going on if you write the polynomial for which the coefficient of $x^n$ is the probability of an outcome $n$, i.e., $$\frac{1}{6} x^1+ \frac{1}{6} x^2 + \frac{1}{6} x^3 + \frac{1}{6} x^4 + \frac{1}{6} x^5 + \frac{1}{6} x^6 = x^1\left(\frac{1}{2} x^0+ \frac{1}{2} x^1\right)\left(\frac{1}{3} x^0+\frac{1}{3}x^2+\frac{1}{3}x^4\right) = x^1\left(\frac{1}{3}x^0+\frac{1}{3}x^1+\frac{1}{3}x^2\right)\left(\frac{1}{2}x^0+\frac{1}{2}x^3\right).$$This makes it a bit more obvious that the original polynomial and its factors all must have coefficients that sum to 1, ... – Michael Seifert Sep 05 '17 at 15:09
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    ... and that multiplying any coefficient of one of the polynomials by 6 must yield an integer (if we're using 6-sided dice.) – Michael Seifert Sep 05 '17 at 15:12