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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.

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    $\begingroup$ Do you have to combine them by summing? $\endgroup$
    – Deusovi ♦
    Sep 4, 2017 at 4:01
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    $\begingroup$ 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). $\endgroup$
    – Deusovi ♦
    Sep 4, 2017 at 4:10
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    $\begingroup$ Here is a harder version of this: Simulate two dice with three indistinguishable dice: youtube.com/watch?v=xHh0ui5mi_E and solutions: youtube.com/watch?v=hBBftD7gq7Y $\endgroup$
    – Fabian Röling
    Sep 4, 2017 at 7:36
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    $\begingroup$ @Fabian It was also asked on this site here puzzling.stackexchange.com/questions/31101/… $\endgroup$
    – Kruga
    Sep 4, 2017 at 7:54
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    $\begingroup$ @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. $\endgroup$
    – Deusovi ♦
    Sep 4, 2017 at 14:01

2 Answers 2

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I believe this set of dice satisfies all your requirements:

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

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  • $\begingroup$ Well I'll be, it does! Impressive work. Thanks! $\endgroup$
    – TheSoundDefense
    Sep 4, 2017 at 4:25
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    $\begingroup$ Nice use of what amounts to mixed bases $\endgroup$
    – humn
    Sep 4, 2017 at 4:26
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    $\begingroup$ This is very nice, but isn't die 1 incidental? $\endgroup$
    – frodoskywalker
    Sep 4, 2017 at 8:25
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    $\begingroup$ @frodoskywalker I agree, but I don't think that can be avoided while still resulting in only 6 equally probable outcomes. The number of possible outcomes on each die will have to divide 6, so it's either 1,1,6 or 1,2,3. $\endgroup$
    – Jaap Scherphuis
    Sep 4, 2017 at 8:43
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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

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  • $\begingroup$ Nice analysis, Artur K., related to a couple of those at Relabeling two 20-sided dice without changing their total $\endgroup$
    – humn
    Sep 4, 2017 at 15:48
  • $\begingroup$ Thanks, @humn, I have added link to that problem in the solution. The most classic example is about relabeling 2 dices. $\endgroup$
    – Puzzle Prime
    Sep 4, 2017 at 15:57
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    $\begingroup$ I'll have to look further into this as this still seems somewhat magical. Thanks! $\endgroup$
    – TheSoundDefense
    Sep 4, 2017 at 17:46
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    $\begingroup$ 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, ... $\endgroup$
    – Michael Seifert
    Sep 5, 2017 at 15:09
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    $\begingroup$ ... and that multiplying any coefficient of one of the polynomials by 6 must yield an integer (if we're using 6-sided dice.) $\endgroup$
    – Michael Seifert
    Sep 5, 2017 at 15:12

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