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In the Coupon Collector problem with $n^2$ coupons, I want to compute the following probability corresponding to the iteration where the process ends:

$$Pr[I\leq i]=\sum _{k=n}^i \frac{n^2! \left(n^2\right)^{-i} {i\brace k}}{\left(n^2-k\right)!}$$

Given the CDF, I tried to use the complement to estimate the average iteration count:

$$E[I]=\sum _{i=0}^\infty \left(1-\sum _{k=n}^i \frac{n^2! \left(n^2\right)^{-i} {i\brace k}}{\left(n^2-k\right)!}\right)$$

However, though I looked at the post where the expected iterations for drawing the max amount of coupons is derived, I can´t find a way to simplify the above CDF due to the lower limit of the sum. Also, I tried with partial evaluations and $FindSequenceFunction[]$ in Wolfram and it doesn´t seem to help. So, is there any way to approach this sum to avoid such issue?

Cardstdani
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    Is your question specifically about evaluating this sum, or are you open to computing the expected value of number of iterations in a different way? – Wei Sep 07 '24 at 08:28
  • @wei Mainly I want to compute the sum as defined above. However, it can be rewritten in other forms such as the difference between the sum with $k\in[0,i]$ and $k\in[0,n]$ – Cardstdani Sep 07 '24 at 08:31

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