Four trading cards, one per pack

Question

A card game sells booster packs, each containing exactly one card drawn uniformly at random from 4 different cards. Packs are independent.

On average, how many packs must you open to end up with all four cards?

Accepted Answer

Track how many distinct cards you have. If you already hold k different cards, the next pack is useful (a new card) with probability \frac{4-k}{4}, so the average number of packs needed to gain one more distinct card is \frac{4}{4-k}. Summing over the four stages (k = 0,1,2,3): E = \frac{4}{4} + \frac{4}{3} + \frac{4}{2} + \frac{4}{1} = 4\left(1 + 	frac12 + 	frac13 + 	frac14ight). The bracket equals \frac{25}{12}, so E = 4 	imes \frac{25}{12} = \frac{100}{12} = \frac{25}{3} \approx 8.33 	ext{ packs.} The last missing card alone averages 4 packs, more than the first two cards combined. …

Four trading cards, one per pack

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