You might be able to do something with: [[{(1d6-3),0d0}kh1]] You'd need to know how many dice you're going to roll up front, but here's an example where a dropbox lets you roll between 1 and 10 of them: [[ [[ { [[{(?{Number of Dice|1|2|3|4|5|6|7|8|9|10}),0}kh1]],1}kl1 ]]*{(1d6-3),0d0}kh1+ [[ { [[{(?{Number of Dice|1|2|3|4|5|6|7|8|9|10}-1),0}kh1]],1}kl1 ]]*{(1d6-3),0d0}kh1+ [[ { [[{(?{Number of Dice|1|2|3|4|5|6|7|8|9|10}-2),0}kh1]],1}kl1 ]]*{(1d6-3),0d0}kh1+ [[ { [[{(?{Number of Dice|1|2|3|4|5|6|7|8|9|10}-3),0}kh1]],1}kl1 ]]*{(1d6-3),0d0}kh1+ [[ { [[{(?{Number of Dice|1|2|3|4|5|6|7|8|9|10}-4),0}kh1]],1}kl1 ]]*{(1d6-3),0d0}kh1+ [[ { [[{(?{Number of Dice|1|2|3|4|5|6|7|8|9|10}-5),0}kh1]],1}kl1 ]]*{(1d6-3),0d0}kh1+ [[ { [[{(?{Number of Dice|1|2|3|4|5|6|7|8|9|10}-6),0}kh1]],1}kl1 ]]*{(1d6-3),0d0}kh1+ [[ { [[{(?{Number of Dice|1|2|3|4|5|6|7|8|9|10}-7),0}kh1]],1}kl1 ]]*{(1d6-3),0d0}kh1+ [[ { [[{(?{Number of Dice|1|2|3|4|5|6|7|8|9|10}-8),0}kh1]],1}kl1 ]]*{(1d6-3),0d0}kh1+ [[ { [[{(?{Number of Dice|1|2|3|4|5|6|7|8|9|10}-9),0}kh1]],1}kl1 ]]*{(1d6-3),0d0}kh1]]
You can see the basic idea is in this example: [[ [[ { [[{(?{v}-3),0}kh1]],1}kl1 ]]*{(1d6-3),0d0}kh1]]
If you put in a number greater than 3, the first term (before the *) will be 1, otherwise it will be 0, causing the term to drop out of the sum regardless of what was rolled. See in this example: The first 5 terms were used resulting in that 6 that was rolled 3rd turning into a 3. The 5 and two 6s rolled in the later 5 dice are ignored because of the 0 multipliers.