You may be wondering: What is my chance of rolling all the values on “n” dice equal to or higher than some number “y”?
This problems is very similar to our problem in part 2, but this time “p” is not simply equal to 1/s, but instead it is equal to 1/s multiplied by all the possibilities that fulfil the initial condition (in this case, being equal or higher than some number “y”).
For example, let’s imagine that we have a 6-sided die and y = 3. This means that we wish to roll anything above or equal to 3, i.e. a 6, 5, 4, or 3 with a single 6-sided die.
The probability p is then 4 possible outcomes that satisfy my condition p(c) multiplied by 1/s, i.e. p = p(c) * 1/s = 4 * 1/6 = 2/3 = 66.67%.
If we had several dice instead of just a single one, we must use the same rule as in part 2, and multiply p by itself by the number of dice n. Therefore the formula becomes: P = (p(c) * 1/s)ⁿ
With this you can, for example, calculate the probability of rolling only numbers equal or above 4 with three 10-sided dice:
P = (p(c) * 1/s)ⁿ = (7 * 1/10)³ = (7/10)³ = 34.3%