Two-thirds of the way through the badges

Question

An app grants one random badge per task, uniformly from 15 badge types, independently. You already have 10 different badges.

On average, how many badges must you earn before one of them is a type you are missing?

Accepted Answer

Only the next new badge matters. A single badge is new when it is one of the 15 - 10 = 5 missing types, with probability p = \frac{5}{15} = \frac{1}{3}. The geometric wait averages \frac{1}{p} = 3 badges. Takeaway. With k of n types collected, the next new type averages \frac{n}{n-k} draws; here \frac{15}{5} = 3.

Two-thirds of the way through the badges

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