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There are travellers who have $n$ identical suitcases which they unwittingly did not label, and there suitcases are arriving on the baggage carousel. Each traveller selects a suitcase at random, being unable to tell it apart from all the others.

What is the probability that at least one suitcase ends up with its original traveller?

RobPratt
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MB1001
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  • Do complementary counting: Find the probability $p$ that no suitcase ends up with its original traveller, and the probability that at least one suitcase ends up with its original traveller is $1-p$. – KingLogic Apr 11 '22 at 05:35
  • So the way I'm thinking no one gets the right suitcase is when each of the 1,...n people have (n-1) suitcases to choose from, so the probability of 1 person not getting the right suitcase is (n-1)Cn = (n-1)/n. So probability no one gets the right suitcase is $(\frac{n-1}{n})^n$. Is this right? – MB1001 Apr 11 '22 at 06:09
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    @MB1001 Have a look at derangements. What you suggest is not correct because there is no independence. – drhab Apr 11 '22 at 06:37

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