Just two flavors short

Question

A candy jar mixes 5 equally common flavors. You grab pieces one at a time (blindly, with replacement in effect since the jar is huge), and you already have 3 different flavors.

On average, how many more pieces must you take to have tasted all five flavors?

Accepted Answer

Only the 2 missing flavors matter. Advance by stages counting distinct flavors, from 3 to 5. Holding k flavors, a new flavor appears with probability \frac{5-k}{5}, so the average wait is \frac{5}{5-k} pieces. Sum over k = 3,4: E = \frac{5}{2} + \frac{5}{1} = 2.5 + 5 = 7.5 	ext{ more pieces.} Takeaway. Extra draws to finish is \sum{j=k}^{n-1}\frac{n}{n-j}; the final flavor, needing only 1 of 5, dominates at an average of 5 pieces.

Just two flavors short

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