# Distribution of complex results for pool of ten-sided dice, specific to custom tabletop RPG system

I'm no statistician, but I do find the area of mathematics fun and fascinating. I have a friend who has designed a "roll-mid" dice system for his custom tabletop RPG and as part of the game balancing, I'm interesting in finding out certain result possibilities.

An individual die has the following results:

1        : critical failure, total of -1 "success"
2-3, 8-9 : no "successes"
4-7      : +1 "success"
0        : critical success, total of +2 "successes"


A player will roll from one to ten individual ten-sided dice (d10s). I haven't any idea how to approach an issue like this with the additional layer of complexity. The concept, however, is just really cool to me.

My question: What is the distribution of "success" totals for any number of dice 1-10?

You might like a procedure to answer similar questions so you can play with your options. A good one uses the probability generating function, pgf. For dice, this is just a polynomial (or rational function) in a single variable (say, $t$). The powers of $t$ correspond to numbers of successes and their coefficients give the associated probabilities.

In your example, the pgf is

$$p(t) = \frac{1}{10} t^{-1} + \frac{4}{10} t^0 + \frac{4}{10} t^1 + \frac{1}{10} t^2 = \frac{1 + 4t + 4t^2 + t^3}{10 t}.$$

The pgf for the sum of successes with $k$ independent rolls of the dice ($k = 0, 1, 2,$ etc) is just the $k^\text{th}$ power of the pgf. Any computer algebra system will make short work of computing such powers. You can read off the desired probabilities from the results.

For example, with two dice $k=2$, so the pgf for the numbers of successes is

$$p(t)^2 = \frac{1 + 8t + 24t^2 + 34t^3 + 24t^4 + 8t^5 + t^6}{100 t^2}.$$

Noticing the factor of $100 t^2$ in the denominator, we get:

\eqalign{ \Pr[-2] &= 1/100 \\ \Pr[-1] &= 8/100 \\ \Pr[0] &= 24/100 \\ \cdots \\ \Pr[4] &= 1/100. }

I have written "$\Pr[i]$" for the chance of $i$ successes, which (with two dice) can range from $-2$ up through $4$.

Once $k$ gets larger than $5$ or so, you can approximate these probabilities with a Normal distribution. The expectation of the original die is

$$\mu = \frac{1}{10} \times (-1) + \frac{4}{10} \times 0 + \frac{4}{10} \times 1 + \frac{1}{10}\times 2 = 1/2$$

and therefore its variance is

$$\sigma^2 = \frac{1}{10} \times (-1 - \frac{1}{2})^2 + \frac{4}{10} \times (0 - \frac{1}{2})^2 + \frac{4}{10} \times (1 - \frac{1}{2})^2 + \frac{1}{10} \times (2 - \frac{1}{2})^2 = 13/20.$$

Consequently, the expectation for the sum of $k$ independent dice is $k\mu$ and the variance is $k\sigma^2$. Use these parameters to compute probabilities.

For example, with $k=10$ (counting successes on $10$ dice), the expectation is $5$ and the variance is $13/2$. What are the chances, say, of $10$ or more successes? We apply a "continuity correction" by equating $10$ or more with $9.5$ or more ($9.5$ splits $9$ and $10$, of course), and compute the upper tail area of a normal distribution of mean $5$ and variance $13/2$. This expression uses R, a free statistical calculator:

k <- 10 # An integer between -10 and 20
1 - pnorm(k-1/2, mean=(1/2)*k, sd=sqrt(13/20*k)) # Chance of k or more successes


It returns $0.03877808$; the correct value is $\frac{23617383}{625000000} \approx 0.0377878$. The normal approximation is only $1.03$ times greater than the correct value in this case: it would take a few million rolls to detect the difference.

• Thanks @whuber, this is exactly the breakdown I didn't have the statistics knowledge to compose. I will need to review this several times and give that statistical calculator a look (link I assume?) Commented Mar 9, 2012 at 19:24
• @whuber thank you so much for this. I've wondered idly about similar questions in the past. Commented Mar 9, 2012 at 19:31
• pan, that link is correct, but hardly necessary: see r!
– whuber
Commented Mar 9, 2012 at 20:08