Use Inclusion Exclusion On Dice

combinatoricsinclusion-exclusionprobability

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.

What is the probability we get at least one six, if we roll $3$ dice?

Rewording the more complex.

What is the probability we get at least three sixes if we roll $5$ dice?

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 $3.5\%$.

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