You use a trick with failure group roll ( on the wiki here ). You can count successes, and count failures. So if you treat rolls of 6 as successes, and count rolls of 1-3 are failures, then add the number of dice you get the right answer. Like so: {{result=[[ {[[@{valeur_competence}+@{itemBonus}+@{modificateur}]]d6}>6<3 + [[@{valeur_competence}+@{itemBonus}+@{modificateur}]] ]]}} or {{result=[[ {(@{valeur_competence}+@{itemBonus}+@{modificateur})d6}>6f<3 + (@{valeur_competence}+@{itemBonus}+@{modificateur})]]}} Both methods work, it just depends how you want the output to look. Imagine you roll 6 dice, and get 1,2,3,4,5,6: Under the rolling-at-table method, you get 3 rolls of 4 or higher, and 1 at 6, giving you 3 +1 = 4 successes. Under the failure-calculation method above, you get 3 failures, 1 success, and 6 extra successes, which comes to -3 +1 +6 = 4 successes. Same result, by a slightly more convoluted method.