A Binomial Distribution from generalized dice

Question

Given the following inputs:

1. No. of Dice Rolled
2. No. of "Success" sides per die
3. Total sides (all dice have same # of sides)
4. Total "Success" rolls required

What would the formula be to calculate the probability that you will roll enough Successes?

Thanks!

Answer 1

Each die has $a$ Success sides out of $s$ sides, where $a$ and $s$ are integers with $0 \le a \le s.$ Suppose we roll $n$ such dice independently, and that $X$ is the number of Successes seen.

Then $X \sim \mathsf{Binom}(n, a/s);$ thus the formula for getting exactly $k$ Successes is $$P(X = k) = {n \choose k}\left(\frac as\right)^k\left(1 - \frac as\right)^{n-k},$$ for $k = 0, 1, \dots , n.$

If you want at least $k$ Successes then

$$P(X \ge k) = P(X = k) + P(X = k+1) + \cdots + P(X = n).$$

Per @Masacroso's Comment, please look at the discussion of the Binomial distribution in your text. There you will find essentially the first formula, known as the binomial PDF (or PMF), where $n$ is the number of independent 'trials' and $p = a/n$ is the probability of Success on each trial.

Some statistical calculators and software will compute such quantities easily. For the case in which you're rolling ten ordinary six-sided dice and considering even faces as successes, here is a computation of $P(X = 7)$ and $P(X \ge 7)$ using R statistical software. [In R, dbinom is the binomial PDF, and pbinom is the binomial CDF.]

n = 10;  s = 6;  a = 3;  p = a/s;  k = 7
dbinom(k, n, p)            # P(X = 7)
## 0.1171875
1 - pbinom(k-1, n, p)      # 1 - [P(X=0) + P(X=1) + ... + P(X=6)]
## 0.171875
need = 7:10;  sum(dbinom(need, n, p))
## 0.171875