@ Dom365: my condolences in this sad and difficult moments.
@ Traintown
*is now being forced to try to figure out if there is a way to make 2 of the same dice, with any number of sides and any numbers on the sides, such that every number has an equal chance of being rolled*
For finite dice with real numbers, this is only possible if all numbers on the sides are equal (no matter how many sides there are).
Proof: suppose there are at least 2 sides with at least two different numbers. Then there must be a maximum, say "A". Define m as the number of sides that are "A". Then there are at least m*m possibilities to roll A+A and, since A is the maximum, no other possibilities exist to roll A+A, so there are exactly m*m. Similarly, there must be a minimum B with n sides that are "B". Following the same reasoning, there are exactly n*n possibilities to roll B+B. Thus, we must have m=n or the proposition can't be true.
Now we know that A+B, which is not equal to A+A, nor to B+B can be rolled at least 2*m*n times, which is larger than m*m=n*n=m*n.
Consequently, it is not possible for finite dice that have different numbers on the sides.
* For unfair dice, the same reasoning can be follewed as the "number of sides" could simply be replaced by the probability of rolling a certain number, which can always be defined.
* For non-real "numbers", this also holds when a complete order can be defined. When there is no complete order, this is possible. Counterexample: 2 fair sides "yes" & "no" where no+no=yes (double negative). P(sum=yes) = P(D1=yes)*P(D2=yes)+P(D1=no)*P(D2=no) = 0.5 = P(D1=yes)*P(D2=no)+P(D1=no)*P(D2=yes) = P(sum=no)
