A lopsided three-way collection

Question

Each pack contains one of 3 cards, drawn with probabilities 0.6, 0.3, and 0.1 respectively, independently per pack. On average, how many packs until you have collected all three cards?

Accepted Answer

Unequal odds call for the weighted-collection formula (from inclusion-exclusion on "a type is still missing after t packs"): E[T] = \sumi \frac{1}{pi} - \sum{i<j}\frac{1}{pi+pj} + \frac{1}{p1+p2+p3}. Plug in p = (0.6,\,0.3,\,0.1): - Singles: \frac{1}{0.6}+\frac{1}{0.3}+\frac{1}{0.1} = \frac{5}{3}+\frac{10}{3}+10 = 15. - Pairs: sums 0.9,\ 0.7,\ 0.4 give \frac{10}{9}+\frac{10}{7}+\frac{5}{2} = \frac{635}{126} \approx 5.0397. - Triple: \frac{1}{1} = 1. E[T] = 15 - \frac{635}{126} + 1 = 16 - \frac{635}{126} = \frac{1381}{126} \approx 10.96 	ext{ packs.} The rare 0.1 card, needing on average 10 p…

A lopsided three-way collection

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