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A teacher has her class of 75 students correct their own homework. She collects the papers, shuffles them, and passes one to each student. What is the approximate probability that no student receives his own paper back?

My naive approach is the following:

All possibilities to hand back the papers: $75!$

For the first paper which she gives back, there are 74 possibilities, for the second one 73 and so on, thus I get that the approximate probability that no students gets his own paper back is $$ \frac{74!}{75!}=1/75\approx 0.013 $$

Am I correct or do I think too naively?

RobPratt
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Rhjg
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    See Derangements – lulu Apr 05 '22 at 13:15
  • The probability is very near to $\frac{1}{e}$. For exactly one correct assignment we have again a probability very near to $\frac{1}{e}$ – Peter Apr 05 '22 at 13:18
  • By the way, the expected number of correct assignments is $1$ no matter how many students there are. – Peter Apr 05 '22 at 13:21
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    "For the first paper there are 74 possibilities" Yes, that is correct. "...for the second one 73" No. It could have been $74$ possibilities again if the paper handed back to the first person was the second person's and $73$ otherwise. "...and so on" For the third person that depends on whether either of the first two persons got back the third person's test... It could have been $73$ or it could have been $72$ possibilities, and this gets more complicated to keep track of via a direct approach the further along we go. – JMoravitz Apr 05 '22 at 13:29
  • If I see the link right, then the probability I am searching for is $$ \sum_{i=1}^{75}\frac{(-1)^{i+1}}{i!}.$$ How to compute that? – Rhjg Apr 05 '22 at 13:59
  • You... add up all the numbers? There's no closed-form expression for that sum, if that's what you're asking. Also, note that your sum is different from the (correct) one given in @lulu's link. – Michael Seifert Apr 05 '22 at 20:29
  • (To clarify, you want the equation for $!n$ in that link; the one you provided in your comment is the number of permutations that leave at least one object fixed.) – Michael Seifert Apr 05 '22 at 20:36

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