Coupon Collecting With Unequal Odds

The classic coupon collector problem asks how many random draws (with replacement) it takes to collect all $n$ distinct coupon types, assuming each type is equally likely on every draw — the answer is $n H_n$, where $H_n$ is the harmonic sum $1 + \tfrac{1}{2} + \dots + \tfrac{1}{n}$. In reality, many collecting problems — cereal-box prizes, or waiting for every stock in an index to trade at least once in a given window — have unequal probabilities: some items are common, others rare.

When probabilities $p_1, \dots, p_n$ differ, there's no simple closed form like $n H_n$ anymore. Instead, the expected number of draws to collect all $n$ types is computed using inclusion-exclusion:

$$\mathbb{E}[T] = \int_0^\infty \left( 1 - \prod_{i=1}^n \left(1 - e^{-p_i t}\right) \right) dt = \sum_{\emptyset \neq S \subseteq \{1,\dots,n\}} \frac{(-1)^{|S|+1}}{\sum_{i \in S} p_i} .$$

In plain terms: because rare items dominate the waiting time, the expected total draws is driven heavily by the smallest $p_i$ — the rarest item essentially sets a floor on how long you'll wait, much more than the common items do.

For just two coupon types with probabilities $p$ and $1-p$, this simplifies to $\mathbb{E}[T] = \frac{1}{p} + \frac{1}{1-p} - 1$. With $p = 0.9$ and $1-p = 0.1$: $\mathbb{E}[T] = \frac{1}{0.9} + \frac{1}{0.1} - 1 \approx 1.11 + 10 - 1 = 10.11$ draws — dominated almost entirely by waiting for the rare 10%-probability item.