Puzzle! Okay, let's make this fun: I'll buy a Boba tea for the first person who either gives me a correct solution to the following problem, or sends me a demonstration that no solution exists.
Background
Undergrad statistics classes spend a lot of time thinking about dice. In particular, suppose you have two six-sided dice, one red and one black. When you roll them and add their faces together, you'll get something between 2 (when both dice read 1) and 12 (when both dice read 6).
For convenience, represent a roll as an ordered pair (a,b). For example, suppose you roll, and the black die reads 5 and the red die reads 4. Then you would represent that roll as (5,4). The "value" of the roll is the sum of the two entries--9 in this case.
Now, you're more likely to get some values than others. For example, (1,1) is the only way to roll a 2, but there are six ways to roll a 7--(1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). Note that order matters: if the black die comes up 4 and the red die comes up 3, that's a different result than if the black die comes up 3 and the red die comes up 4, even though the sum is the same.
A distribution tells you how many ways you can roll each possible value. Here's the distribution for two regular dice:
2: 1 way3: 2 ways4: 3 ways5: 4 ways6: 5 ways7: 6 ways8: 5 ways9: 4 ways10: 3 ways11: 2 ways12: 1 way
Just to make things explicit, here are all the ways to roll the possible values. You can check for yourself that (a) every roll is accounted for, and (b) the number of ways you can get any particular value match the numbers given in the previous chart.
2: (1,1)3: (1,2), (2,1)4: (1,3), (2,2), (3,1)5: (1,4), (2,3), (3,2), (4,1)6: (1,5), (2,4), (3,3), (4,2), (5,1)7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)8: (2,6), (3,5), (4,4), (5,3), (6,2)9: (3,6), (4,5), (5,4), (6,3)10: (4,6), (5,5), (6,4)11: (5,6), (6,5)12: (6,6)
The problem
Does there exist another way to number two six-sided dice with positive integers that gives an identical distribution? That is, by putting different numbers on the dice, is it possible to still have exactly one way of rolling 2, exactly two ways of rolling 3, exactly three ways of rolling 4, etc.?
Here are the restrictions:
(1) The numbers must be positive. No negative numbers allowed. Zero is also not allowed; otherwise a cheap solution is {0,1,2,3,4,5} and {2,3,4,5,6,7}, which is too easy.
(2) The numbers must be integers. No fractions or irrationals allowed. Adding imaginary components won't help you at all, so don't even try.
(3) The distribution must match that of the regular six-sided dice. If there are more than six ways to roll a 7 or fewer than three ways to roll a 10, for instance, it won't work.
The solution
Since rolls higher than 12 are taboo, no number on either of the two dice can be larger than 12. This puts a very low upper bound on the number of possibilities, which makes a solution (or demonstration of a lack of a solution) well within the realm of trial and error.
If it turns out that there's no way to do this, I leave the structure of the counterargument to your discretion. If you find a solution, however, I don't need the distribution or the roll table; just the numbers on the two dice, in the form {*, *, *, *, *, *} and {*, *, *, *, *, *} where the stars are replaced by the numbers you've put on the dice.
Hints
Since this problem is fun and not very hard in the grand scheme of things, I'll post hints every few days until someone comes up with a solution. Just to get you started, here's
Hint 1
Note that 1 is the least number you can put on either die. Suppose that one of the dice doesn't have a 1. Then your pair will look like
{1, *, *, *, *, *}{a, *, *, *, *, *}
where a > 1 is the least element on the second die. But then the least possible roll of these dice is 1+a > 2, which means that there's no way to roll a 2. This doesn't fit the distribution of the regular dice, since there's exactly one way to roll a 2 with regular dice. Therefore, any solution (if one exists) will take the form
{1, *, *, *, *, *}{1, *, *, *, *, *}
where the stars are between 1 and 11 (inclusive).
Suppose further that one of the dice has more than one 1:
{1, 1, *, *, *, *}{1, *, *, *, *, *}
With this configuration, there are at least two ways to roll a 2; but this is too many, since a normal pair of dice has only one way to roll a 2. Thus any solution (if one exists) will look like
{1, *, *, *, *, *}{1, *, *, *, *, *}
where the stars are between 2 and 11 (inclusive).
Now let's work down from the top. Suppose that one of the dice has an 11:
{1, *, *, *, *, 11}{1, *, *, *, *, *}
Since the second die doesn't have any more 1s, every starred value will be 2 or greater; but if one of these is rolled together with the 11 from die 1, the sum will be greater than 12. This isn't acceptable, since there's no way to roll something that high with regular dice. Therefore any solution (if one exists) will look like
{1, *, *, *, *, *}{1, *, *, *, *, *}
where the stars are between 2 and 10 inclusive--and maybe 10 is too high? I leave that to you.

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