I am trying to figure out how to write an equation to solve the following problem:
- S = # of sides on a die
- R = # of dice being rolled
- T = Minimum result to count as a success event on each die (Aka "Threshold")
- M = Minimum # of success events desired
For example, if I want to know the probability (P) of getting at least two "5's" or higher when rolling three 6-sided dice, then: \begin{align} S &= 6\\ R &= 3\\ T &= 5\\ M &= 2 \end{align}
Is there a formula I could use to determine P?
Thanks!
Let $A$ be the number of acceptable numbers, that is numbers at or above the threshold. Then $A=S-T+1$. In your example, $A=2$, since $5$ and $6$ are the only acceptable numbers. The probability that a particular die has an acceptable number is $A/S = 1 -(T-1)/S$ and the probability that it is not acceptable is, of course, $(T-1)/S.$
For a given value $k,$ the probability that exactly $k$ rolls out of $R$ are acceptable is $${R\choose k}\left(1-{T-1\over S}\right)^k\left(T-1\over S\right)^{R-k}$$ since there are ${R\choose k}$ ways to choose the $k$ acceptable dice. For a success, we must have $k\ge M$ so the formula you seek is$$P=\sum_{k=M}^R{R\choose k}\left(1-{T-1\over S}\right)^k\left(T-1\over S\right)^{R-k}$$
P.S.
If $M$ is less than half of $R,$ it will be more convenient to compute the probability of success as one minus the probability of failure, since there will be fewer term is the sum. That is, $$P=1-\sum_{k=0}^{M-1}{R\choose k}\left(1-{T-1\over S}\right)^k\left(T-1\over S\right)^{R-k}$$