Anydice: Counting Successes on dice and rerolling any (+ bonus dice) below the target of 6 on d6's

Question

I'm trying to model the roll and reroll probabilities of skill checks in Coriolis RPG.

Here the skill rolls use a pool of d6, and you count how many 6 you rolled. You can optionally reroll any dice that were not 6s the first time you rolled. Sometimes you can add bonus d6 dice to the reroll pool.

I've been trying to model this in AnyDice, but I'm struggling to make it work! Especially around the reroll that need to be the number of dice less as many 6's from the initial roll.

My starting point was this:

N:5
output [count {6} in N d6]

then for the reroll element i tried this

N:5
output [count {6} in N d6] + [count {6} in (N - [count {6} in N d6]) d6]

But for sure that's not working, as the output from this alone a count up to 1

[count {6} in (N - [count {6} in N d6]) d6]

Any help greatly appreciated!

Answer 1

I'm going to give you two different approaches, because they highlight two different and very powerful techniques in AnyDice.

Treating it like a dice pool

In order to work with dice pools, especially when caring about more than one property of the rolled dice (as a collective) or caring about it more than once, we need to cast it to a sequence. To do so, we make a function which takes our roll as a sequence. We can apply logic as if working with a single roll, and at runtime AnyDice will iterate over every possible roll and spit out the distribution.

function: count T:n in A:s with reroll {
   result: [count T in A] + [count T in (#A-[count T in A])d6]
}
output [count 6 in Nd6 with reroll]

Treating it as independent dice

Now, in mathematical terms the dice aren't really a pool. Each dice is rerolled independently of the results on other dice. Dependance would be if you could roll a die if a different die rolled a 5, or some other such rule.

We can then construct a die which encapsulates a d6, but rerolling non 6s (once). In concept, we want a die which has one side which has a 6, and 5 which are a d6 roll. Now due to how AnyDice converts dice to sequences and sequences back into dice this is slightly trickier (in short, it doesn't handle fractional faces). The construction actually looks like:

d{1d6:5, 6:6}

Which is in essence a 36 sided dice which has the same net distribution as d6 rerolled once on non-6. Conceptually, if you split its faces in 6 sets, one is all sixes, and five of them has the 1-6 series of a normal d6 roll.

You can then just ask AnyDice to count the number of 6s in a roll of N such dice.

output [count 6 in Nd{1d6:5, 6:6}]

These two methods give the same results. The former might be easier to generalize, especially if you have more complex pool/rerolling rules, but the latter will perform better (which can matter for large pools).

Answer 2

Paradox: Math is simpler than Anydice!

This application is another, beautiful use case of the Binomial distribution: you actually do not need anydice.

The probability of getting \$t\$ successes in a Nd6 roll is given by $$ P(t,N)= \sum_{i=0}^t \mathcal{B}(N,i,1/6)\cdot\mathcal{B}(N-i,t-i,1/6) $$

where \$\mathcal{B}(N,t,s)\$ is the Binomial distribution of parameters \$N,s\$. The mathematical explanation is given further in the text.

To infinity and beyond!

Someone_Evil's anydice program works fine, but the problem with anydice programs is that it can not run them for a large value for \$N\$: in this case just for a 7d6 roll the site stop working. Instead, using direct formulas you can simulate rolls with an high amount of dice, in the case you are interested into the distribution analysis, or in the case you can add bonus dice to the reroll. The figure below refers to a roll 100d6.

---

Mathematical explanation

Given \$N\$ independent trials with the same change of success \$s\$, the probability to get \$t\$ success is given by $$ \mathcal{B}(N,t,s) = \begin{pmatrix}N\\t\end{pmatrix}s^t(1-s)^{N-t} $$ where \$\begin{pmatrix}N\\t\end{pmatrix}\$ is the binomial coefficient.

Let's compute the probability \$P(2,6)\$ of getting exactly two 6s on a 5d6 roll under the proposed rules: this is given by $$ P(2,5)= P(0,5)\cdot P(2,5) + P(1,5)\cdot P(1,4) + P(2,5)\cdot P(0,3) $$ where

• the first term computes the probability of getting zero 6s on the first roll and two 6s on re-rolling all the dice,
• the second term computes the probability of getting just one 6 on the first roll and another one in the re-rolling of the remaining 5 dice,
• the last term accounts for getting two 6s on the first roll and none on re-rolling the remaining 4 dice.

Each probability above can be computed by the Binomial distribution: hence $$ P(2,5)= \mathcal{B}(5,0,1/6)\cdot\mathcal{B}(5,2,1/6) + \mathcal{B}(4,1,1/6)\cdot(3,1,1/6)+\mathcal{B}(5,2,1/6)\cdot\mathcal{B}(3,0,1/6) $$ or, more compactely, $$ P(2,5)= \sum_{i=0}^2 \mathcal{B}(5,i,1/6)\cdot\mathcal{B}(5-i,2-i,1/6) $$ and \$P(2,5)=0.35531228235766055 \sim 35.53\%\$, accordingly to SomeoneEvil's answer.

It can be easily generalized to the case of getting \$t\$ success in a Nd6 roll: $$ P(t,N)= \sum_{i=0}^t \mathcal{B}(N,i,1/6)\cdot\mathcal{B}(N-i,t-i,1/6) $$ The following python code produces the results for \$N=5, t\in\{0,1,2,3,4,5\}\$:

from scipy.stats import binom
# Number for dice
n = 5
# Probability of success
p = 1/6

print("Succ\tProbability")
for i in range(n + 1):
  P = 0;
  for j in range(i+1):
    P = P + binom.pmf(j, n, p)*binom.pmf(i-j, n-j, p)
  print(str(i) + "\t" + str(P))