It’s possible for two D6s to be correlated in such a way that their sum is 1 + a D11. You’ll see how later. For convenience, the rest of this article will make a Dn be randomly from 0 to n-1 rather than 1 to n.
The first thing you might try after seeing an example might be trying D2 + D2 = D3. This one is simple, and also unique: (0,0) with probability 1/3, (0,1) and (1,0) with probability 1/6 each, and (1,1) with probability 1/3. The next may be D2 + D3 = D4, which is also unique, and in fact any D2 + Dn = D(n+1) is unique: (0,x) has probability (n-x)/(n^2+n), and (1,x) has probability (x+1)/(n^2+n).
It’s also possible for three D2s to be correlated so that any two of them make a D3 and their entire sum is a D4. A similar thing can be done for four D2s where any two make a D3, any three make a D4, and all four make a D5, and this can also be done for any number of D2s: the chance for any specific combination with m of n D2s landing on 1 is 1/(n*(n choose m)). This means it can also be represented by rolling a D(n+1) and picking a random combination with that many 1s, which provides a neat way to physically build this distribution.
This new probability distribution gives us an answer to the question at the start: How can we make D6 + D6 equal D11? The most elegant answer: use the construction above with ten D2s, and consider the first five as one D6 and the other five as the other D6.
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