Assume n people who are initially assigned seats from 1 t0 n but they don't know their seat number, now they randomly pick up a seat on their own and what would be the probability that no one is seating to its assigned seat?
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Please let me know if this question is duplicate. – Xiaonan Jun 18 '18 at 19:23
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It is, but the difficulty is in selecting which question to set it as a duplicate of. This is just a rephrasing of the question of counting derangements (and then taking the ratio to get the probability) and has an easy answer of $\dfrac{!n}{n!}$ – JMoravitz Jun 18 '18 at 19:34
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Thank you! I think the wiki page helps me a lot and I still needs to figure out how final result !n is calculated from $!n = (n-1)(!n-1 + !n-2)$. – Xiaonan Jun 18 '18 at 23:38
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These permutations, where no element is mapped to itself, are known as Derangements. You get the probability by dividing by the number of all permutations. So it is
$$\frac{\left\lfloor\frac{n!}{e}+\frac{1}{2}\right\rfloor}{n!}.$$
(This most certainly is a duplicate but I couldn't find any post that had just this basic derangment question).
ploosu2
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Prettier if you remove the $\frac12$ and then round to the nearest integer rather than rounding down – Henry Jun 18 '18 at 22:05
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@ago is the bracket (used on Wikipedia) a standard notation for the nearest integer function. For me (in my mind), it doesn't immediately register as nearest integer but could be the same thing as floor function, or actually the integer part, (which for negative values is the ceiling). – ploosu2 Jun 19 '18 at 06:31