Command to roll 8d6 where 1 and 2 count as 3's

I'm playing a character where each die rolled if below my modifier (which is 3) can be counter as the modifier, 3. So I want to roll a fireball i.e. 8d6 where the 1s and 2s count as a 3 and give me the total damage. Kindly help with the command. Thanks! 

There's no streamlined way to do this, however, this general formula should work for you; simply repeat as many times as you have dice to roll: [[{1d6,3d1}kh1]] So, for your 8d6, it would look like this: [[{1d6,3d1}kh1+{1d6,3d1}kh1+{1d6,3d1}kh1+{1d6,3d1}kh1+{1d6,3d1}kh1+{1d6,3d1}kh1+{1d6,3d1}kh1+{1d6,3d1}kh1]] If you are putting this in a field on a character sheet (like the 5e by Roll20 sheet), you can most likely remove the inline roll brackets ([[ and ]]), although keeping them will probably make the output look cleaner. EDIT: could even make it dynamically handle changes to your modifier like so (I'm assuming you don't want it to be possible for a very high number to allow for a higher "die" roll than would be possible with that type of die): [[{1d6,[[{6,@{someModifier}}kl1]]d1}kh1]]