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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.

RobPratt
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Marko
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    Why not just compute the probability that you get at least three $6's$? – lulu Jul 27 '23 at 02:26
  • Can yuu give us an example? I do not understand the rules to count. – callculus42 Jul 27 '23 at 02:26
  • $\frac{5^{2}\operatorname{nCr}\left(5,3\right)}{6^{5}}+\frac{5\cdot\operatorname{nCr}\left(5,4\right)}{6^{5}}+\frac{1}{6^{5}}$

    @callculus42 This is by cases (3 6s, 4 6s, 5 6s), I guess this is fine. I just want to know how the PIE works in a case like this.

     – Marko Jul 27 '23 at 02:28
  • I mean your simplified exercise. If you roll 5,2,4 what would be the outcome? – callculus42 Jul 27 '23 at 02:32
  • 5,2,4 you choose the largest one, and that is 5. But 5 is not 6 so it would not count as a successful case. Basically, we look for cases with at least one 6. Does that explain it? – Marko Jul 27 '23 at 02:43
  • @Marko Yes, it does. – callculus42 Jul 27 '23 at 02:52
  • @Marko The application of the PIE does not make sense here, IMHO. In the second case you should calculate $P(X\geq 1) =P(X= 1)+P(X= 2)+P(X= 3) $, where $X$ is the number of 6's on the upfaces of the three dice. – callculus42 Jul 27 '23 at 03:04
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    The principle of inclusion exclusion gives the probability of at least one event happening, but you want the probability of rolling at least three sixes. Therefore, you need the generalized PIE. – Mike Earnest Jul 27 '23 at 14:45

1 Answers1

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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 $3.5\%$.

TShiong
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Marko
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