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There are 100 numbered balls in an urn. We make 100 random draws with replacement. Of course, we can not expect to draw every number exactly once, there will be multiples. What is the expected value of multiply drawn numbers?

Browsing the questions I found the related problem Expected number of unique items when drawing with replacement, but I think that the solution given there counts the number of items picked at least once, not necessarly exactly once.

Added May $24^{th}$: There must be something wrong with the answers so far: Think of just 2 balls and 2 draws: Then the probability of multiples is 50%. But the answers given so far suggest a probability of 25%, which is for a fixed number to be drawn twice.

Community
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scenario
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The probability the ball numbered $1$ is picked exactly once is ${100 \choose 1}\frac{99^{99}}{100^{100}} = 0.99^{99} \approx 0.3697296 \approx e^{-1}$

So by linearity of expectation, the expected number of numbered balls chosen exactly once is $100$ times this, i.e. about $36.97296$

Henry
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  • OK, thanks, so the expected number of multiply picked balls would be the number of balls picked $100(1-1/e)$ minus the number of balls picked exactly once $100/e$, i.e. $100(1-2/e) \approx 26$? – scenario May 16 '16 at 16:27
  • Approximately yes – Henry May 16 '16 at 16:36