Chasing the last two cards

Question

Each pack gives one card, uniformly from 20 card types, independently. You already own 18 different cards. On average, how many packs until you draw a card you are still missing?

Accepted Answer

Only the next new card counts. A draw is new when it hits one of the 20 - 18 = 2 missing types, with probability p = \frac{2}{20} = \frac{1}{10}. The geometric wait averages \frac{1}{p} = 10 packs. Near the finish line the wait balloons: with just 2 useful types out of 20, most packs are duplicates. Takeaway. The next-new-type wait \frac{n}{n-k} grows large as k approaches n; here it is already 10 packs with two types left.

Chasing the last two cards

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