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Suppose we choose from n distinct items randomly with replacement for n times. What is the expected number of items that were never chosen?

I made a simulation and the answer seems to be 1/e * n. Also, the expected number of items that were chosen exactly once seems to be 1/e * n; number of items chosen twice is around 1/2e * n. However, I don't know how to approach this problem mathematically and derive the answer.

What about choosing for 2n times, n/2 times or any number of times?

user3701366
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    Linearity of expectation is incredibly powerful here. Focus on the first element. What is the chance it was not picked in the first picking? What is the chance it was not picked in any picking? Does this remind you of $e$? – JMoravitz Mar 02 '22 at 12:01
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    Oh, it relates to [(n-1)/n]^n – user3701366 Mar 02 '22 at 12:16

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