20d6 w/ adv.?

I need to roll 20d6 with each dice roll having advantage. Keeping the higher of the two for each roll

I'm afraid your out of luck wthout a Pro subscription. Roll20 macros cant do that without manually entering each die, like this: /roll {2d6}kh1 + {2d6}kh1 + {2d6}kh1 + {2d6}kh1 + {2d6}kh1 + (and so on) If every die has advantage, then you need to break out every die into a check as above.

If you don't mind not seeing the rolls for both dice, you can make a Rollable table which has the correct statistical distribution and just roll against that. Grab the probability from AnyDice:&nbsp; <a href="https://anydice.com/program/a91" rel="nofollow">https://anydice.com/program/a91</a> Multiply by 100 and put that in for the weight: Roll against the table: [[20t[6kh1]]]

That's a great solution, but your screenshot is confusing. Wouldn't you only need a single rollable table, with the probabilities for a single 2d6kh1? Edit: never mind, I see what you've done :)

Yup, I wanted to show each of the Rollable Table Items with their weights. =D

The weights can be simplified as 1, 3, 5, 7, 9, 11 and that pattern continues for two of any size dice kh1

Oh, good to know!&nbsp; Do you know the math behind that progression?

Works best visually The table colours show the different ways to get the values 1 to 6. As the result value gets larger it represents the edge of a larger and larger square; with the final count of the edge being the area of the square less the area square of the smaller square. Eg for six it's 6^2 - 5^2 or for four it's 4^2 - 3^2. Generally this means n^2 - (n-1)^2 Which expands to n^2 - (n^2 - 2n +1) And simplifies to 2n - 1 and thus ends up being the odd numbers