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Note: the formula you provided will add @init to the number of successes. Example: Input: /roll 5d6>5+10 Lets assume that 5d6 rolled a 15 Result: 11 successes This is because 5d6>5 is one success (automatically) and then it adds 10 for 10 more successes. If you are trying to compare it to a target number which is (5+10) then you will need to rework the formula. Here is the rework: /roll {@{init}d6-@{init}}>5 - Gauss
He never provided the reason for the formula. I just fixed what he presented to get it to work in the manner he presented it. Never thought about the actually aspect of the formula and the results wanted. Good thing you did. :D
I think the purpose of this formula is: Roll Init number of d6, count 5 and 6 as a success. Then afterwards add Init again to the number of successes. I think Shadowrun uses a system like that. And I believe the formula should indeed look like what Metroknight said.