Make a 10 count as two successes, and the 1 take the 10 ath the same time in WoD

Hi, we are playing a game that uses a system that resembles the one in vampire, werewolf and WoD in general. But instead of exploting the 10 we count them as two successes , I have a cuestion ¿is it posible to tell the system that make the 10 count as two successes instead of one and, at the same time, make the 1 remove the 10? For now I know how to make the 10 conut as two successes like this: /r ?{Number of dice}d10>10f<?{Dif}sd+?{Number of dice} I also know how to revome succeses with the 1 like this: /r ?{Number of dice}}d10>?{Dif}f1sd But I have no clue how to combine these two principles, We want the 10 counts as two successes and 1 remove successes at the same time. Thanks in advance if anyone have the solution.

If you roll 2 dice and get a 10 and 1, how many successes do you have? I'm not sure if its 1 or 0.

It looks like there is also a target number here? So if the target number is 7: 1: 2 failures 2-6: 1 failure 7-9: 1 success 10: 2 successes? Is this correct? If so, the number of conditionals is too much for normal macros & dice operators, that will need the API. You can manage 3 outcomes with the >success and >fail operators, but not 4. The best you can do is automatically count 3 of the outcomes, like your first macro, and use the red critical fail coloring to count the 1's manually. You can do some number filtering by juggling some math on a single die roll, but it doesn't work on multiple rolls.

I don't know what an API is, but I will look for it, thanks. 

It's a Pro subscriber perk which allows you to run custom scripts - unfortunately not available to Free or Plus accounts. The only other option would be to find a sheet which already uses this roll system - maybe one of the WoD sheets available uses this? I'm not sure. Creating your own sheet also requires Pro.

I made this macro: &{template:wod} {{name=Generic Roll}} {{Result= [[?{How many dice?|0}d10>?{Dificulty|6}f1!]]}} this asks you the dices, the dificultu and every 1 you toll, takes away one success. it may help you, but dont make the 10 become 2 successes, it just rerolls the 10 again

JJ thanks a lot mate, I am going to prove your macro, I appreciate you take the time and efforts.

f you want to keep the same mechanic and count 10s as 2s, you can do it with a mix of manual work. Use JJ's formula, but remove the !, so it looks like this &{template:wod} {{name=Generic Roll}} {{Result= [[?{How many dice?|0}d10>?{Dificulty|6}f1]]}} Now when you roll, keep an eye out for a green or blue border around the roll. That means you have at least one 10 in the roll. When that happens, hover your mouse over the die result, to see something like this: You can immediately see all tens, because they are green. So just add one extra success per green die to the result, to get the correct number of successes. That would take the above roll to 5 successes, the correct amount when treating 10s as 2 successes. It would be nicer if this could be done automatically, but this is next best thing I think.

GiGs said: &{template:wod} {{name=Generic Roll}} {{Result= [[?{How many dice?|0}d10>?{Dificulty|6}f1]]}} Now when you roll, keep an eye out for a green or blue border around the roll. That means you have at least one 10 in the roll. When that happens, hover your mouse over the die result, to see something like this: This is probably how I would do it with a free account. The exploding dice method (rerolling 10's) is only going to work for 10% of the tens.

Thanks a lot!