I came across with the problem of calculating the expected sum of two fair dice where if the two numbers are the same, the sum is zero. This can be done by simply summing all possible cases and then divided by 36. However, I got trapped by the case with three dice.
So the “cancel” rule works like this: if you roll (1,1,2) then your sum is 2, if you roll (1,1,1) then your sum is 1, since in the second case you only need to cancel two of the three 1’s.
What I have tried so far is I try to minus all the over-counted sum from sum of all possible sums, and then divided by 216. Not sure if there is simpler way, because I was expected to answer in a short time.
And the case can be further generalised to rolling 100 dice and cancel each pair of the same number…