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 given $i − 1$ coupons is $p_{i} = \frac{n − (i − 1)}{n}$. Therefore, $t_{i}$ has geometric distribution with expectation $\frac{1}{p_{i}}$. By the linearity of expectations we have:
That's an except from wikipedia page. I understand $p_{i} = \frac{n-(i-1)}{n}$. But I don't quite understand why $t_{i}$ is $\frac{1}{p_{i}}$. I appreciate much if anyone can explain it.
Thanks
Community wiki answer to allow this question to be closed as answered:
As André has pointed out, this is a well-known fact; see e.g. Wikipedia.