Expected rolls to see five of six faces

Question

You roll a fair six-sided die repeatedly and stop the moment you have seen 5 distinct faces (any five of the six). What is the expected number of rolls?

Accepted Answer

Decompose the wait into stages by how many distinct faces you already hold: T = T1 + T2 + T3 + T4 + T5, where Tk is the number of rolls to see the k-th new face given k - 1 distinct faces already seen. While holding k - 1 faces, a roll is new with probability pk = 	frac{6 - (k-1)}{6}, so Tk is geometric with mean 	frac{6}{7 - k}. You stop at the fifth new face, so you never wait for the sixth: E[T] = \frac66 + \frac65 + \frac64 + \frac63 + \frac62 = 1 + 1.2 + 1.5 + 2 + 3 = 8.7 	ext{ rolls.} Compare with collecting all six faces, which costs 6H6 = 14.7 rolls. Dropping just the last coupon s…

Expected rolls to see five of six faces

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