WTA specialty Macro help

We are using the 20th anniversary rules from Werewolf the Apocalypse where if you use your specialty, a roll of 10 will result as a flat 2 success instead of 1. Does anyone know the macro formula for this?

Something like this should do the trick: [[?{How many dice?|6}d10>10f<[[?{Difficulty?|5}-1]]cf<[[?{Difficulty?}-1]] + ?{How many dice?}]] This is an inline roll, good for inserting into a template or an emote - the die rolls are on mouseover. If you want a bigger output with all the dice roll values, remove the leading [[ and final ]] and put /r at the start instead (then a space). Is it always a d10? You can throw another query in there to change the die type if you wish. You can also change the ?{How many dice?} Query into an Attribute call if it's tied to something on a character sheet (not familiar at all with the game, sorry). The basic formula was shamelessly stolen from GiGs... he posted it somewhere else, no idea if it was the same game or not.

Thanks!  And yes, only on d10 for game.

No problem. By the way, if you're not familiar with Queries, you can change the default values by changing the number after the | pipe. So the 6 for the dice and the 5 for the DC.

Btw, can you add something there that will deduct  any roll of 1 from the successes? The formula I was using was [[?{Amount of Dice|1}d10>?{Difficulty|6}f1/r ?{Amount of Dice|1}d10>?{Difficulty|6}]] and i thought it would be simple as adding f1 to your code but it wasn't working. This was the code that was not able to count 10 as 2 successes. Thanks again

That's getting a bit trickier and I think it would need the API, since the range is -1 up to +2 it's more than the success/fail operator can handle. The success and fail operators are already being used in the code I put above. I can't really think of any way to handle both the 10s and 1s without losing the ability to check the actual difficulty successes. If nothing else comes up, you can always use sd to sort the dice in descending order so you can manually count the 1s: /r ?{How many dice?|1} + ?{How many dice?}d10sd>10f<[[?{Difficulty?|6}-1]]cf1

Thanks for the help. I think we will just do that, it is more easier to add the 10's successes so we will use the code to have all the 1's auto deduct and just add 1 more success manually per 10's.