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Macro for Rolling when a specific number comes up

September 19 (4 years ago)

The roll i'm using is: 

/r [[(?{Occult rank?|1}+?{Bonus dice?|0})d6>4!]] 

It works well but I need to figure out how to have it roll another 1d6 when I get 6. Not just reroll it 0_o I've been having a hard time figuring it out let alone how to add it to my current roll. Just hit a wall really. Any help is great. Thanks ^^ 

September 19 (4 years ago)
GiGs
Pro
Sheet Author
API Scripter

You can use the exploding symbol !, like

/r [[(?{Occult rank?|1}+?{Bonus dice?|0})d6!>4!]] 

This will roll another die for every die that rolls 6.

September 19 (4 years ago)
GiGs
Pro
Sheet Author
API Scripter

Note that if you are using inline rolls, you dont need the /r at the start. You can do

[[(?{Occult rank?|1}+?{Bonus dice?|0})d6!>4!]] 

Or include in a sentence like

My Occult Rank Roll is [[(?{Occult rank?|1}+?{Bonus dice?|0})d6!>4!]] 
September 19 (4 years ago)

Doesn't seem to work for me, for some reason :/ Doesn't even give an error just nothing. ><; 

September 19 (4 years ago)
GiGs
Pro
Sheet Author
API Scripter

I didnt notice you already had an ! symbol in there. It should be this:

[[(?{Occult rank?|1}+?{Bonus dice?|0})d6!6>4]]

or if you want to see the individual dice

/roll (?{Occult rank?|1}+?{Bonus dice?|0})d6!6>4
September 19 (4 years ago)

Edited September 19 (4 years ago)

This one ending up working. Perfect. ^^ since i wanted it to count any numbers above 4. Thank you so much! You are truly Amazing : ) 

/roll (?{Occult rank?|1}+?{Bonus dice?|0})d6!6>4

Edit: They both work perfectly actually. ^^ thanks. I'll use the hidden dice one. it works amazing. 



September 19 (4 years ago)
GiGs
Pro
Sheet Author
API Scripter

Great :)