Can you design two six-sided dice (different from the standard ones), where each face has a nonzero number of dots, so that the probability distribution of their total is the same as for two standard dice?

To be clear, you may only choose how the dice are labeled, you're not allowed to load the dice so certain faces are more likely.

This is one of the more famous classic math puzzles.

Edit/Clarifications: The goal is relabel the faces of two standard die so that

  • Each face is labeled with a positive integer.
  • The set of labels on each die is not the usual {1,2,3,4,5,6}.
  • For every number $n$, the probability of rolling an $n$ with your dice is the same as the probability of rolling an $n$ with standard dice. "Rolling an $n$" means rolling both dice and having the numbers on their top faces adding to $n$.
  • The labels on a die don't have to be all different.
  • You don't have to label both dice the same way.
  • $\begingroup$ If you mean by "their total" the sum of the top faces of each die, you could start with a pair of dice that are standard except for having their 1 and 6 faces swapped in each case. $\endgroup$
    – Lawrence
    Commented Jul 28, 2015 at 22:04
  • $\begingroup$ Do you mean with a different numbering such that they can't both have 1,2,3,4,5,6. but have to be weird numbers? $\endgroup$ Commented Jul 28, 2015 at 22:06
  • $\begingroup$ @Lawrence by "different from the standard ones", I mean "at least one die doesn't feature the numbers 1,2,3,4 5,6." Order doesn't matter, as this doesn't change any probabilities. $\endgroup$ Commented Jul 28, 2015 at 22:13

2 Answers 2


2nd Edit:
This is the answer you're probably looking for;

Die 1: 1, 2, 2, 3, 3, 4
Die 2: 1, 3, 4, 5, 6, 8


To maintain the distribution of sums while using positive non-zero whole numbers of dots, we would need to decrease some faces on one die and increase on the other. If we have a sequence from 1 to 6, then to decrease we will need to double up faces. We need exactly one 1-dot face on each die to maintain P(2) = 1/36, however we can start by placing two 2-dot faces on one of the die and work our way up from there.

One useful constraint when deriving the faces of the dice is that the highest value face on each die can only appear on a single face since P(12) = 1/36.

Previous answers:


Die 1: 0.5, 1.5, 2.5, 3.5, 4.5, 5.5
Die 2: 1.5, 2.5, 3.5, 4.5, 5.5, 6.5.

Edit: As requested, whole numbers!

Die 1: -6, -5, -4, -3, -2, -1
Die 2: 8, 9, 10, 11, 12, 13


I think the answer is


Reason is:

If you take a standard die and multiply each face by two, it should have the same probability distribution as the normal dice.


4: 1 ways (2,2) 6: 2 ways (2,4; 4,2) 8: 3 ways (2,6; 6,2; 4,4) 10: 4 ways (2,8; 4,6; 6,4; 8,2) 12: 5 ways (2,10; 4,8; 6,6; 8,4; 10,2) 14: 6 ways (2,12; 4,10; 6,8; 8,6; 10,4; 12,2) And then all the way back down to 24


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