How many flavors after five picks?

Question

A gum machine sells one piece at a time, each an independent uniform pick from 4 flavors. You buy 5 pieces. On average, how many distinct flavors will your 5 pieces cover?

Accepted Answer

Counting distinct flavors looks tangled, but linearity of expectation makes it easy: the expected number of distinct flavors is the sum, over the 4 flavors, of the probability that each one appears at least once. A specific flavor is missed by a single piece with probability \frac{3}{4}, so it is absent from all 5 pieces with probability \left(\frac{3}{4}ight)^5. Hence it appears at least once with probability 1 - \left(\frac{3}{4}ight)^5 = 1 - \frac{243}{1024} = \frac{781}{1024}. All 4 flavors have the same chance, so E[	ext{distinct}] = 4\left(1 - \left(	frac34ight)^5ight) = 4 	i…

How many flavors after five picks?

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