The coupon collectors problem [closed]

probability coupon-collector

Wikipedia has a writeup of the coupon collector's problem. If there are $n$ coupons and you already have $k$ of them, the expected time to get the next is $\frac n{n-k}$ This leads to the time to get them all being $n$ times $H_n$, the $n^{\text{th}}$ Harmonic number. $H_n=\sum_{i=1}^n \frac 1i \approx \log(n)+\gamma$, where $\gamma \approx 0.577$ and you need $n$ times this, so $n \log(n)+n\gamma$

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