Coupon collector's problem

Let T be the time to collect all n coupons, and let t_i be the time to collect the i-th coupon after i − 1 coupons have been collected. Think of T and t_i as random variables. Observe that the probability of collecting a new coupon is ${\displaystyle p_{i}={\frac {n-i+1}{n}}}$. Therefore, ${\displaystyle t_{i}}$ has geometric distribution with expectation ${\displaystyle {\frac {1}{p_{i}}}}$. By the linearity of expectations we have:

${\displaystyle {\begin{aligned}\operatorname {E} (T)&=\operatorname {E} (t_{1})+\operatorname {E} (t_{2})+\cdots +\operatorname {E} (t_{n})={\frac {1}{p_{1}}}+{\frac {1}{p_{2}}}+\cdots +{\frac {1}{p_{n}}}\\&={\frac {n}{n}}+{\frac {n}{n-1}}+\cdots +{\frac {n}{1}}\\&=n\cdot \left({\frac {1}{1}}+{\frac {1}{2}}+\cdots +{\frac {1}{n}}\right)\\&=n\cdot H_{n}.\end{aligned}}}$

Here H_n is the n-th harmonic number. Using the asymptotics of the harmonic numbers, we obtain:

${\displaystyle \operatorname {E} (T)=n\cdot H_{n}=n\log n+\gamma n+{\frac {1}{2}}+O(1/n),}$

where ${\displaystyle \gamma \approx 0.5772156649}$ is the Euler–Mascheroni constant.

Now one can use the Markov inequality to bound the desired probability:

${\displaystyle \operatorname {P} (T\geq cnH_{n})\leq {\frac {1}{c}}.}$

Using the independence of random variables t_i, we obtain:

${\displaystyle {\begin{aligned}\operatorname {Var} (T)&=\operatorname {Var} (t_{1})+\operatorname {Var} (t_{2})+\cdots +\operatorname {Var} (t_{n})\\&={\frac {1-p_{1}}{p_{1}^{2}}}+{\frac {1-p_{2}}{p_{2}^{2}}}+\cdots +{\frac {1-p_{n}}{p_{n}^{2}}}\\&<\left({\frac {n^{2}}{n^{2}}}+{\frac {n^{2}}{(n-1)^{2}}}+\cdots +{\frac {n^{2}}{1^{2}}}\right)\\&=n^{2}\cdot \left({\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\cdots +{\frac {1}{n^{2}}}\right)\\&<{\frac {\pi ^{2}}{6}}n^{2}\end{aligned}}}$

since ${\displaystyle {\frac {\pi ^{2}}{6}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\cdots +{\frac {1}{n^{2}}}+\cdots }$ (see Basel problem).

Now one can use the Chebyshev inequality to bound the desired probability:

${\displaystyle \operatorname {P} \left(|T-nH_{n}|\geq cn\right)\leq {\frac {\pi ^{2}}{6c^{2}}}.}$

A different upper bound can be derived from the following observation. Let ${\displaystyle {Z}_{i}^{r}}$ denote the event that the ${\displaystyle i}$-th coupon was not picked in the first ${\displaystyle r}$ trials. Then:

${\displaystyle {\begin{aligned}P\left[{Z}_{i}^{r}\right]\leq \left(1-{\frac {1}{n}}\right)^{r}\leq e^{-r/n}\end{aligned}}}$

Thus, for ${\displaystyle r=\beta n\log n}$, we have ${\displaystyle P\left[{Z}_{i}^{r}\right]\leq e^{(-\beta n\log n)/n}=n^{-\beta }}$.

${\displaystyle {\begin{aligned}P\left[T>\beta n\log n\right]=P\left[\bigcup _{i}{Z}_{i}^{\beta n\log n}\right]\leq n\cdot P[{Z}_{1}^{\beta n\log n}]\leq n^{-\beta +1}\end{aligned}}}$