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Help with Avalon Hill Dice

I'm looking to run some old Avalon Hill board/ tabletop games in Roll20. However they have odd dice that do I not know how to make in roll 20. Its 3d6's, but they come up with odd results. The first D6 is White and has a 1 on 1 side, a 2 on two sides, and a 3 on 3 sides. The other two dice are black, one has 0-5 on its sides, the other has 0-4 with a second 0. The die results make a number 10-39, the white die corresponds to 10, 20, or 30; while the black dies are added together to reach a result of 0-9. The games are built around the idea that die results are not all equal chance. But I am not familiar enough with Roll20 to know how to implement this. Any help would be appreciated.
If you just want the job done you can use this macro [[ceil(1d6**0.6)*10 + 1d6-1 +{1d6-2,0d0}k1]] It's not pretty, Roll20 doesn't like it, but it works . An alternate way is to use Rollable Tables to create custom dice One table 'white' with items 1,2,3 with weightings 1,2,3 respectively One table 'black' with items 0,1,2,3,4 with weightings 2,1,1,1,1 respectively The other black die you can just use 1d6-1 for a final macro [[ 1t[white]*10 + 1d6-1 + 1t[black] ]]
RainbowEncoder said: If you just want the job done you can use this macro [[ceil(1d6**0.6)*10 + 1d6-1 +{1d6-2,0d0}k1]] It's not pretty, Roll20 doesn't like it, but it works . An alternate way is to use Rollable Tables to create custom dice One table 'white' with items 1,2,3 with weightings 1,2,3 respectively One table 'black' with items 0,1,2,3,4 with weightings 2,1,1,1,1 respectively The other black die you can just use 1d6-1 for a final macro [[ 1t[white]*10 + 1d6-1 + 1t[black] ]] That would still create a 4 when I roll a 6, but I could work with that by keeping the lower of the dice roll or 3. Thanks for the assist.