Here is the question
Suppose we roll 5 standard fair dice and sum the upfaces of the
largest 3 values showing. Find the probability that the sum is
18.
I tried an easier version of this question using Inclusion Exclusion and it worked
Suppose we roll 3 standard fair dice and sum the upface of the largest value showing. Find the probability that the sum is 6.
$\frac{1}{6}\cdot3-\frac{1}{6}^{2}\cdot3+\frac{1}{6}^{3}$
So first we count getting at least one six and multiply by 3 choose 1, then subtract getting two sixes and mutliply by 3 choose 2, and the finally add back getting 3. I think this only works because the PIE is revolving around only one die.
Here is me applying the same logic but to at least 3 out 5, instead of at least 1 out of 3.
$\frac{1}{6}^{3}\cdot\operatorname{nCr}\left(5,3\right)-\frac{1}{6}^{4}\cdot\operatorname{nCr}\left(5,4\right)+\frac{1}{6^{5}}$
I calculated the right answer a different way using cases, so I know this answer is wrong and is overcounting.
Best Answer
Ok I'm back a month later and I figured it out.
Basically rewording the first small question.
Rewording the more complex.
Mike Earnest linked the generalized PIE, which was a little hard to understand. So I will try to explain it here a little more gently.
--- Second questions revised and answered.
$$\frac{1}{6^{3}}\cdot\operatorname{nCr}\left(5,3\right)\ -\ \frac{1}{6^{4}}\cdot\operatorname{nCr}\left(5,4\right)+\frac{1}{6^{5}}$$
So why didn't this work with less than 3?
Because we counted wrong!
We first add the at least three $6$-face-up subsets, once for each one. But how many times did we add the at least $4$ $6$-face-up subsets?
Well for each $4$ there are $\operatorname{nCr}\left(4,3\right)$ subsets of $3$ within it, so we we did not add twice like my previous attempt tries. We added it $4$ times! So we need to subtract it $3$ times!
Well ok, how about the all $5$ subset.
Well we know we added it 10 times for the first case, and now we are going to subtract it $\operatorname{nCr}\left(5,4\right)\cdot3$ times. So we will get a total of $10-15 = -5$ counts of the $5$ six set, so we need to add it back $6$ times!
So what we wanted was $\frac{1}{6^{3}}\cdot\operatorname{nCr}\left(5,3\right)\ -\ \frac{1}{6^{4}}\cdot\operatorname{nCr}\left(5,4\right)\cdot3+\frac{1}{6^{5}}\cdot6$
Which is around $3.5\%$.