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Tavern - I cant get the dice to act the way i want them to

2

I cant get the dice to act the way i want them to

Nara · 13. Jun 2015 08:54 by Nara · 3

The explode command is pretty close, but basicly here is what i want (its for a swedish role playing game, not sure if any other system uses it):

Any time a D6 comes up as 6, you take the original die and add a die, and then reroll both. If any of those become a 6, you take that die, add another and keep going.

So like

roll 2d6 = 6 + 6

take both dices, add 2 extra dices and reroll them all = 1 + 2 + 1 + 4 = 8.

So even tho you added extra dices, the outcome were lower. I only found commands to add an extra die to the already existing sum. I guess i have to do something like "every time a dice becomes 6, add +2 dice and subract -6" but im not sure how to make that keep it rerolling and im not sure what i should write in form of macro for it either :/

Thanks for all and any help!

1 · by udo · 04. Jun 2015 15:40

Hey, sorry I haven't responded sooner, been busy :)

I'm currently refactoring the dice roller code and when I'm done you should be able to use rx and rmax with explode, which would solve your problem.

Watch this space for updates!

3 · by udo · 05. Jun 2015 01:05 · rolls: 4 · changed: 05. Jun 2015 01:06
Rolls in this post: #1 reroll=max rx=2 4d6 = 23( (6: adding 1 die, 6→2 6→6 (6: adding 1 die, 6→6 (6: adding 1 die, 6→4 6→4) 6→1) ) = 11 +5 +5 +2)   #2 reroll=max rx=2 2d6 = 8(3 +5)   #3 reroll=max rx=2 2d6 = 5(4 +1)   #4 reroll=max rx=2 2d6 = 18( (6: adding 1 die, 6→5 6→6 (6: adding 1 die, 6→1 6→6 (6: adding 1 die, 6→5 6→3) ) ) = 14 +4)  

I've updated the dice roller, you should now be able to do what you wanted.

The code would be "reroll=max rx=2 2d6", meaning:

  • reroll=max > "reroll all dice having the max result" (reroll=6 would do the same thing)
  • rx=2 > "use two dice when rerolling" (rx default is 1)
  • without using the rmax code, this will keep happening until all sixes are gone

Let's try it out:

reroll=max rx=2 2d6 = 18( (6: adding 1 die, 6→5 6→6 (6: adding 1 die, 6→1 6→6 (6: adding 1 die, 6→5 6→3) ) ) = 14 +4)

1 · by Nara · 13. Jun 2015 08:54

Dude! Awesome!

This is going to be very useful when our group gets split up, i cant thank you enough! :D

8.6