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Probability that numbers 1…6 show up at least once when rolling 10 dice. I know that it is a duplicate question here. The problem is I do not understand the answer. Can someone explain it more intuitively? Specially case 2 and 3 from the link. I understand, btw the inclusion-exclusion principle in set theory, but not here among probabilities. I know I am asking a simple question. Sorry for that!

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    $|A\cup B|=|A|+|B|-|A\cap B|$ is how the two-set version is written in the set theoretic sense. The only thing that changes when moving to probabilities is instead of $|~~|$ we use $Pr(~~)$, so the two-event version of inclusion-exclusion is $Pr(A\cup B)=Pr(A)+Pr(B)-Pr(A\cap B)$. With this knowledge and induction we can come up with a general statement for arbitrary number of sets or events: $Pr(\bigcup A_i) = \sum Pr(A_i)-\sum Pr(A_i\cap A_j) + \sum Pr(A_i\cap A_j\cap A_k)-\sum Pr(A_i\cap A_j\cap A_k\cap A_l)\pm\dots$ – 2017-02-27
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    I can see that. But in this case where are the combinatorics coming from? and the principle looses all its intuition when applied to probabilities. That's in fact my concern... – 2017-02-27
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    How does it lose any intuition, he is in fact only using the set theoretic version in the first place, he is counting how many ways you can roll dice. Letting $A_1,A_2,A_3,\dots$ be the events that the number $1,2,3,\dots$ dont appear respectively, he calculates $|A_1|$ to be $5^{10}$ as each die has $5$ possible outcomes which avoids $1$, he calculates $|A_1\cap A_2|=4^{10}$ as each die has $4$ possible outcomes which avoid both $1$ and $2$, $|A_1\cap A_2\cap A_3|=3^{10}$, $|A_1\cap\dots\cap A_4|=2^{10}$, $|A_1\cap\dots\cap A_5|=1^{10}$ and $|A_1\cap\dots\cap A_6|=0^{10}$ – 2017-02-27
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    Similarly, any intersection of three of these events will be of cardinality $3^{10}$ just as any intersection of five of these events will be of cardinality $1^{10}$. He is counting how many ways of rolling dice have each face appearing at least once. This is the opposite of finding how many have at least one of the numbers not appearing, i.e. $|A_1\cup\dots \cup A_6|$. – 2017-02-27

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