Open nstephenh opened 10 months ago
I still don't understand what "@" does but I did figure out what I was trying to do with * and then, instead of setting the number of dice on each roll, using "repeat" on the whole expression.
The @
operator comes from anydice, and it gets the nth element in a set. For example 2@3d6
would get the middle die of the 3d6 (dice sets are sorted highest to lowest). I don't think it applies to what you're trying to do.
For 10 hits, it seems like ((10)d6cs>3)d6cs>3
would be the correct formula. I'll see if I can figure out why its causing an error. repeat(d6cs>3, 10d6cs>3)
would also work, but I get the same error there too.
Some errors appeared to be coming from the parser, so I circumvented it by creating the parsed output object manually. The values I had returned didn't seem to be accurate, but perhaps I messed up something else with my logic. (repeat(d6,2d6cs>=4))cs>=4
What does the "@" operator do exactly? I'm looking to calculate warhammer saves, which would be (number of wounds)d6cs>3 for example, where number of wounds is the cloud from (number of hits)d6cs>3, and number of hits is another cloud. I think this would be
output (d6cs>3)d6cs>3
, but that gives me Cannot read properties of undefined (reading 'values') "@" does seem to do something close to what I want for one die,output (1d6cs>3)@(1d6cs>3)
gives a 75,25 split, but it doesn't appear to work with multiple dice, and I don't quite understand what it does.