Can a dice-roll experiment involving re-rolls and “negative successes” be modeled as a distribution?

I’ve been looking into statistical distributions lately, having little background in statistics myself, and the way it allows you to modelize complex experiments is absolutely fascinating to me. Lately, though, I’ve been getting into tabletop RPGs again and have found myself unable to put the World of Darkness success-based roll mechanic into something that could make it work as a distribution.

The mechanic is simple on paper : For any given roll you define two constants based on the involved characters' abilities and the situation at hand: a difficulty between 2 and 10 (more recent books have it always set to 8) and how many dices you’re allowed to roll. The game only uses 10-sided dices, and any dice with a value equal or greater to the difficulty value counts as a success. The number of successes you end up with then determines the outcome of the event based on success thresholds and such, but that’s beyond the scope of my question.

So far, this sounds like a simple Binomial distribution, right? $n$ is however many dices you can roll and $p=1-\frac{difficulty-1}{10}$. But the problem arises from 2 optional mechanics, which are the “ten-again” rule and the “botched roll” rule. Ten-again is a subcase of the usual success where any 10s can also be rolled again for additionnal successes for as long as the player keep rolling 10s, which sound somewhat like a Geometric distribution, but I don’t know how to integrate it with the initial binomial experiment. The other one, the botched roll, is another optional mechanic whereby any 1s are substracted from the success tally unless no successes were rolled, in which case this counts as some sort of “critical failure”.

Again, since these two mechanics make this falls outside of the conventional definition of a Bernouilli experiment, I don’t know how to put it together with the rest. I’ve been wondering if I’m maybe looking in the wrong direction and if there's actually a better suited distribution for this sort of case, or if I should use binomial coefficients instead or polynomial fractions. This question comes extremely close to what I'm looking for, but doesn't cover the possibility of a re-roll. What am I missing? Can this be adapted into a distribution at all?

• You have seem to have completely described the possible outcomes and how they are randomly generated, which is enough to define the probability distribution of the roll. So it sure can be modeled as a distribution, namely, the one you've just described. It's not a familiar named distribution, but obviously, only a few families of distributions have names. So what are you looking for? A way to compute the probability of getting each number of successes, perhaps? – Kodiologist Aug 29 '16 at 1:04
• See also rpg.stackexchange.com/questions/12776/…, anydice.com/articles/new-world-of-darkness, others, which do it numerically. But we should be able to calculate the PMF symbolically if you'd like. Question: if you get a 10 and then a 1, is that 0 successes or 1? Is a 10 then a 9 one success or two? Is (1, 1, 1, 9) distinct from (1, 1, 2, 2)? – Dougal Aug 29 '16 at 1:10
• @Dougal : Once you roll a 10, it counts as if you had rolled a regular success, but you get to roll it again. If both optional rules apply, that would equal 0, yes. Rolling a 10 then a 9 gives you 2. Rolling (1,1,1,9) with a difficulty of 8 gets you the same outcome as, say (8,9,1,1) because you end up with a net "No successes" (negatives don't count here). Rolling (1,1,2,2) would get you no successes but 2 ones, which would count as a critical failure. – Santo Guevarra Aug 29 '16 at 1:26
• @Kodiologist : I'm mainly looking for a way to estimate the probabilities of getting X amount of successes, yes. When someone first explained the mechanic to me, it sounded rather unfair and I would really like to have a quick way to determine the possible outcomes and their respective likelihood before blindly letting luck decide what happens. – Santo Guevarra Aug 29 '16 at 1:30

Here's the probability of getting a critical failure. Assuming the difficulty is 8, this happens when all n dice come up below 8, and at least one die comes up 1. Call the dice $D_1, D_2, … D_n$. Then:

\begin{align*}p(\text{critical failure}) &= p(∀n D_n < 8 \quad \text{and} \quad ∃n D_n = 1) \\ &= p(∀n D_n < 8) p(∃n D_n = 1 \;|\; ∀n D_n < 8) \\ &= p(∀n D_n < 8) (1 - p(∀n D_n > 1 \;|\; ∀n D_n < 8)) \\ &= (7/10)^n (1 - (6/7)^n) .\end{align*}

Now let's look more broadly. All possible outcomes that aren't critical failures can be represented as a number of successes, which is a nonnegative integer. Because of the 10-again rule, all numbers of successes are possible for any number of dice.

Let's first consider a single die. To get any positive number of successes $n$, you could roll:

• 8 or 9 on roll $n$ (counting the first roll as roll 1),
• or 2, 3, 4, 5, 6, or 7 on roll $n + 1$,
• or 1 on roll $n + 2$.

Getting to roll $m$ in the first place, for any positive $m$, requires rolling 10 $m - 1$ times in a row. So, the probability of exactly $n$ successes is:

$$\tfrac{2}{10} / 10^{n - 1} + \tfrac{6}{10} / 10^n + \tfrac{1}{10} / 10^{n + 1} = 261 \cdot 10^{-(n+2)} .$$

For $n = 0$ (and no critical failure) the probability is just $\frac{6}{10} + \frac{1}{10} \cdot \frac{1}{10} = \frac{61}{100}$.

Notice that the chance of critical failure is $\frac{1}{10}$, so the sum of the probabilities of all the distinct outcomes is

\begin{align*} \tfrac{1}{10} + \tfrac{61}{100} + \sum_{n=1}^{∞} 261 \cdot 10^{-(n+2)} &= \tfrac{71}{100} + \tfrac{261}{100} \sum_{n=1}^{∞} (1/10)^n \\ &= \tfrac{71}{100} + \tfrac{261}{100} \cdot \tfrac{1}{9}\\ &= 1, \end{align*}

which is good reassurance that all is well.

I leave to you the problem of extending this to multiple dice (which is analogous to the more standard problem of finding the distribution induced by summing $m$ independent dice each with $d$ sides), with the caveat that I don't think you'll find a closed-form solution for $m$ dice in terms of $m$, only a closed-form solution for each fixed $m$.

• Interesting, I didn't think of using boolean algebra this way. Though that still leaves the 10-again rule. – Santo Guevarra Sep 1 '16 at 4:45
• @SantoGuevarra I've expanded my answer. See above. – Kodiologist Sep 4 '16 at 23:36