Expected rolls to see all six die faces

Question

You roll a fair six-sided die repeatedly. You stop once every one of the six faces has appeared at least once. What is the expected number of rolls?

Accepted Answer

Track how many distinct faces you have seen. With j faces already seen, a roll shows a new face with probability \frac{6 - j}{6}. The expected rolls to get the next new face satisfies ej = 1 + \frac{j}{6} ej, so ej = \frac{6}{6 - j}. Summing over j = 0,\dots,5: E = \frac66 + \frac65 + \frac64 + \frac63 + \frac62 + \frac61 = 6\Big(1 + 	frac12 + 	frac13 + 	frac14 + 	frac15 + 	frac16\Big) = 6 \cdot \frac{49}{20} = \frac{147}{10} = 14.7. Takeaway. Seeing all six faces averages 14.7 rolls; waiting for the final missing face alone costs 6 rolls on average.

Expected rolls to see all six die faces

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