Dice Probabilities - How often does 3, 4, 5 show up in 5 dice on one roll?

Question

I want to find out the probability for rolling 5 dice and the result containing at least 3, 4, and 5 without caring what the other 2 numbers are. I'm not very familiar with this kind of probability so I've written a computer program to go through each combination and give me the result. I'd like to know how to do this without doing that.

Answer 1

We first determine the probability that at least one of those numbers does not show up, and then subtract from one to get the probability that all three do show up.

What is the probability that out of five dice, there is no $3$ OR no $4$ OR no $5$? We might naively begin by assuming that we add up the probability that there's no $3$, the probability that there's no $4$, and the probability that there's no $5$.

However, in doing so, we've duplicated some rolls, in which there is no $3$ AND no $4$. Similarly for rolls where there is no $3$ AND no $5$, and where there is no $4$ AND no $5$. So we need to subtract those out.

But now we've subtracted, once too many, the rolls where there is no $3$ AND no $4$ AND no $5$, so we need to add that back in.

$$ \begin{align} P(\mbox{no 3 OR no 4 OR no 5}) & = P(\mbox{no 3})+P(\mbox{no 4})+P(\mbox{no 5}) \\ & - P(\mbox{no 3 AND no 4})-P(\mbox{no 3 AND no 5})-P(\mbox{no 4 AND no 5}) \\ & + P(\mbox{no 3 AND no 4 AND no 5}) \\ & = 3(5/6)^5 - 3(2/3)^5 + (1/2)^5 = 1091/1296 \end{align} $$

so

$$ P(\mbox{3 AND 4 AND 5}) = 205/1296 $$