Try this counting problem, its really interesting!
When two standard 6-sided dice are rolled, there are 36 possible outcomes for the sum of the two rolls: one sum of 2, two sums of 3, and so on, up to one sum of 12.
Find all possible ways of numbering two 6-sided dice with positive integers (not necessarily distinct), so that when they are rolled, the 36 possible outcomes for the sum of the two rolls are the same as the 36 possible outcomes for the sum of two standard 6-sided dice.
+214 I don't think that there is any way to do this other than the original way.
First of all, there must only be one way to make twelve. This can either be 5 and 7 or two 6s. If it is five and seven, then a possible outcome is 7 and 7, which sums to 14, so that cannot work. All the numbers are positive, so the lowest integer is one. Then, we prove that there are no double integer possibilities. Say the double number is 6. Then there are 4 ways to make 11 (\((6_1,5)(6_2,5)(5,6_1)(5,6_2)\)) as opposed to the original 2, 6 cannot be doubled. Similarly, 1 cannot be doubled. Similar proofs ensue for 2,3,4,5, no matter what number you remove, which I will not show here, I will leave you to do them yourself. So, there is only 1 way to do this, namely, the original way: 1,2,3,4,5,6.
