Five sticker designs in a blind bag

Question

A vending machine dispenses sealed blind bags. Each bag contains one sticker chosen uniformly at random from 5 designs, independently of the others.

What is the average number of bags you must buy to collect all five designs?

Accepted Answer

Count progress by the number of different designs you own. When you already have k designs, a bag brings something new with probability \frac{5-k}{5}, so on average it takes \frac{5}{5-k} bags to move from k to k+1 distinct designs. Add the stage averages for k = 0,1,2,3,4: E = \frac{5}{5} + \frac{5}{4} + \frac{5}{3} + \frac{5}{2} + \frac{5}{1} = 5\left(1 + 	frac12 + 	frac13 + 	frac14 + 	frac15ight). The bracket is \frac{137}{60}, so E = 5 	imes \frac{137}{60} = \frac{137}{12} \approx 11.42 	ext{ bags.} Takeaway. The full-collection average is n Hn; for five equally likely designs tha…

Five sticker designs in a blind bag

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