← The Last Card Problem · Lesson 2 of 3
The collector curve
Add up the waits. For six faces: 6/6 for the first, 6/5 for the second, then 6/4, 6/3, 6/2 and 6/1. That is 6 × (1 + ½ + ⅓ + ¼ + ⅕ + ⅙), about 14.7 rolls to see every face at least once.
The same sum works for any set of n equally likely items: n × (1 + ½ + … + 1/n). Mathematicians call it the coupon-collector problem, after cereal-box coupons. It grows a little faster than the size of the set, because the sum in the brackets keeps creeping up.
A set has 100 different cards, all equally likely, one random card per pack. About how many packs does a whole set take on average?
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