[RPG] How to use AnyDice with Classic World of Darkness / 20th Anniversary

anydicedicetoolsworld-of-darkness

How can I take advantage of AnyDice with 20th Anniversary rules? I'm new to AnyDice but very interested in using it.

I would like a wrapper function that accepts the following params.

the number of dice to roll
the action's difficulty
optional: a willpower being spent for a success.
optional: a character having a relevant specialty, allowing for second success on 10's.

If the dice pool is higher than difficulty I'd like and indication for automatic success.

Dice mechanics

Roll Xd10 against target difficulty, dice equal to or greater than the difficulty are successes.
Rolling 10's
- are always successes.
- count as 2 successes, if the character has a relevant specialty.
Rolling 1's
- each 1 negates a success down to failure.
- if there were no successes rolled, the action is botched.
optional: before rolling you may spend 1 willpower for an additional success.
- you can only spend 1 willpower per turn.
optional: if Xd10 is greater than the target difficulty you can take the action as an automatic success.
- not usable in stressful situations.
- only counts for 1 success.
- you can roll, or take the automatic success, but not both.

Success results

5 Phenomenal
4 Exceptional
3 Complete
2 Moderate
1 Marginal
0 Failure
-1: Botch (only possible when no successes are present)

Psuedocode (this is a loose sketch of what I have in mind)

function roll(int POOL, int DIFFICULTY, bool WILLPOWER, bool SPECIALTY, bool AUTO_SUCCESS = true)
{
  int  _roll = 0;
  int  SUCCESS = 0;
  int  FAILURE = 0;
  bool BOTCH = false;

  if (WILLPOWER) SUCCESS++;

  for (int i = 0; i < POOL; i++)
  {
    _roll = 1d10;
    if (_roll >= DIFFICULTY) SUCCESS++;
    if (SPECIALTY && _roll == 10) SUCCESS++;
    if (_roll == 1) FAILURE++;
  }
  if (SUCCESS == 0 && FAILURE > 0) BOTCH == true;

  if (BOTCH)
    output "Botch!\t" + tally(SUCCESS, FAILURE);
  else if (SUCCESS > FAILURE)
    output "Pass.\t" + tally(SUCCESS, FAILURE);
  else
    output "Fail.\t" + tally(SUCCESS, FAILURE);
}

function string tally(int S, int F)
{
  return (SUCCESS - FAILURE) + "\t(" + SUCCESS + " - " + FAILURE + ")";
}

Best Answer

I ended up getting a tweet from the creator of AnyDice with an optimized solution. The function names are clear and it works quickly.

@catlikecoding Hey Jasper, could you help me with an #AnyDice question? I'm still learning the ropes with the system. http://t.co/H0qMhIaIGD

— Taylor Wright (@redlamp) June 6, 2014

@redlamp Don't use a mega function, break it up. http://t.co/anUMab5ByK

— Jasper Flick (@catlikecoding) June 6, 2014

The code sample:

function: roll ROLL:s {
 if 1@ROLL < 1 & (#ROLL)@ROLL = -1 {
  result: -1
 }
 result: [highest of 0 and ROLL]
}

function: roll ROLL:n with willpower {
 result: [highest of 0 and ROLL + 1]
}

function: specialty N:n {
 if N = 10 { result: 2 }
 if N = 1 { result: -1 }
 result: N >= DIFFICULTY
}

DIFFICULTY: 7

NORMAL: d{-1:1, 0:(DIFFICULTY - 2), 1:(11 - DIFFICULTY)}

SPECIALTY: [specialty d10]

X: 4

output [roll X d NORMAL] named "normal"
output [roll X d SPECIALTY] named "speciality"
output [roll X d NORMAL with willpower] named "normal willpower"
output [roll X d SPECIALTY with willpower] named "speciality willpower"
\ automatic success is automatic, nothing to show \

Related Solutions

[RPG] How to model Burning Wheel probabilities in AnyDice

AnyDice handles exploding dice in a peculiar fashion that's poorly-suited for roll-and-count dice pools. I have cobbled together a somewhat inelegant solution, but it appears to work correctly.

Here's the link to the program.

Instead of using dice in the usual manner, I've created several functions that, given parameters of a die, return its output, i.e. the number of successes it has produced.

function: roll ROLL:n threshold T {
  if ROLL >=T {result: 1}
  result: 0
}

function: rollexploding ROLL:n threshold T{
  if ROLL =6 {result: 1+[rollexploding 1d6 threshold T]}
  result: [roll ROLL threshold T]
}

function: rerollfailures ROLL:n threshold T{
  if ROLL < T {result: [roll 1d6 threshold T]}
  result: [roll ROLL threshold T]
}

function: rerollthenexplode ROLL: n threshold T {
  if ROLL < T {result: [rollexploding 1d6 threshold T]}
  result: [rollexploding ROLL threshold T]
}

Then I made a wrapper function that would figure out which of these functions to call, and handle requests for multiple dice being rolled:

function: wrapper DICE:n threshold T explode E reroll R{
  RES:0
  loop N over {1..DICE} {
    if E & R {RES: RES+[rerollthenexplode 1d6 threshold T]}
    else {if E {RES:RES+[rollexploding 1d6 threshold T]}
      else {if R {RES:RES+[rerollfailures 1d6 threshold T]}
        else {RES:RES+[roll 1d6 threshold T]}
      }
    }
  }
  result:RES
}

Finally, I made a bunch of functions that simplify input by providing pre-determined combinations of parameters. I'll provide only 3 of them here:

function: b DICE:n{
  result:[wrapper DICE threshold 4 explode 0 reroll 0]
}

function: be DICE:n{
  result:[wrapper DICE threshold 4 explode 1 reroll 0]
}

function: gr DICE:n{
  result:[wrapper DICE threshold 3 explode 0 reroll 1]
}

To use it, type things like output [b 4]

[RPG] Help Writing an AnyDice Function for a Weird Dice Mechanic

output 3d{-1,1,1,1,1,2} named "fiveSixths"

Your mechanic returns 1 success exactly 5/6ths of the time. But it's a lumpy bi-modal distribution on 1 and 3.5 total successes. Make sure to provide free aspirin to players, as calculating results will be entailed.

Best Answer

Related Solutions

[RPG] How to model Burning Wheel probabilities in AnyDice

[RPG] Help Writing an AnyDice Function for a Weird Dice Mechanic

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