Dmitry Kamenetsky asked the following two questions:

Making two fair dice

Making three fair dice

My question is a generalization of Dmitry‘s questions.

You are given N unlabelled standard 6-sided dice. For which values of N>3, if any, can you write every number from 1 to 6N on the dice, such that in any given throw each die has the same chance of being the highest?


2 Answers 2


You can't do it for

$n\geq 7$, since whichever die has the largest number wins at least $1/6$ of the time, which is too much.

Or for

$n=5$, since the state space has $6^5$ elements, which is not divisible by $5$.

It is possible for

$n=4$, using the dice $(24,17,13,10,7,4)$, $(23,18,14,12,5,1)$, $(22,19,15,11,8,2)$ and $(21,20,16,9,6,3)$.

I found these by greedily adding the largest numbers one by one to the die with the lowest probability so far, until some dice were sufficiently close to $1/4$ that I could determine what their other sides needed to look like.

Finally, it's not possible for

$n=6$. This is because all dice have to have a total of $7776$ winning combinations. Whichever die has number $36$ gets that many from that face alone, so if that die has exactly $7776$ winning combinations then none of the other faces on that die can possibly win. But then every face on every other die would have to be the winning score in a number of combinations that is a multiple of $5$ (possibly $0$ combinations, of course). This is because changing the outcome of the first die between the five small numbers makes no difference to which number wins. Thus if the first die has exactly $7776$ winning outcomes, none of the other dice can do so.

So the answer is

only $n=4$ is possible.


This is pairwise possible for all Values of N>1

Number the dice from 0 to N-1 Then assign numbers for each dice X as

  1. X + 1
  2. 2N - X
  3. 2N + X + 1
  4. 4N - X
  5. 4N + X + 1
  6. 6N - X

Every dice will then have an average of (3N + 0.5) - and (18N+3) total pips

Now if we take any 2 dice at random - we can see that numbers on sides 1,3,5 will be larger on one and sides 2,4,6 on the other. With every random pair being fair - we can see that the entire set should be fair (Still trying to figure a proof of that)


  • 2 dice {1,4,5,8,9,12} and {2,3,6,7,10,11}
  • 3 dice {1,6,7,12,13,18}, {2,5,8,11,14,17} and {3,4,9,10,15,16}
  • 4 dice {1,8,9,16,17,24}, {2,7,10,15,18,23}, {3,6,11,14,19,22} and {4,5,12,13,20,21}

When we have more than 6 dice whichever dice has the largest number (6N) has at least a 1 in 6 chance of winning. So N>6 cannot work

  • $\begingroup$ The three dice {1,2,3,16,17,18}, {4,5,6,13,14,15} and {7,8,9,10,11,12} are "pairwise fair", but when throwing all three dice, the first one has winning probability 1/2. Hence, your final claim needs some further arguments. $\endgroup$
    – gerw
    Commented Jun 6, 2023 at 9:39
  • $\begingroup$ @gerw Yeah my method gives pairwise fair dice regardless of N - but there are some obvious errors in my thinking/reasoning $\endgroup$
    – Collett89
    Commented Jun 6, 2023 at 10:30

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