I saw a post about "How to count successes for a Warhammer type system", and the answers there seemed a bit off to me. Since the subject is closed for being too old, I'm opening one for reference, hoping it might be useful to some : The Warhammer system (be it the the Fantasy or the 40K one) use the difference between the Skill score and the Dice score to count DoS (Degree of Success) and DoF (Degree of Failure). It is like this : if you have a Skill score of 54, every result of the Dice from 45 to 54 count as 0 DoS. From 44 (10 below your Skill score) and below, it is 1 DoS. Then from 34 (20 below Skill) and below, it is 2 DoS, etc... On the opposite side, if your Dice roll from 55 to 64, you have 0 DoF, from 65 to 74 it is 1 DoF, etc... Well, you get the gist of it : You succeed as long as your Dice score under you Skill value, and fail as long as your Dice score over your Skill value (with 1 being automatic success and 100 automatic failure). And when you need to count your Degrees, you count the tens from the difference between your Dice score and your Skill score. As for a formula which do that... Let's say you stored your Skill score in an Attribute named @{Skill}, and you're using this line as an Attribute (you should be able to adapt the code from here) : /roll (@(Skill} - 1d100)/10 This line shall suffice, you just have to mind only the units and ignore everything after the dot. Well, you will tell me "Why not use the abs function or the floor one ?". For the simple reason that if you have a Skill score of 54 and your Dice roll 56, it will result in : abs((54 - 56)/10) = 0. Which wouldn't let you know that you have failed. But if you just do (54-56)/10 = -0.2, you will know that you have failed because it is Negative, and that you have 0 Degree of Failure. Hence, just mind the sign (Positive for Success, and Negative for Failure) and the units, and it will do the trick.