Finishing a partly complete collection

Question

A blind-box line has 10 distinct figures, each equally likely and independent per box. You have already collected 4 different figures.

Starting from here, how many more boxes do you expect to open before you own all ten?

Accepted Answer

Only the missing types matter now. Break the remaining journey into stages by how many distinct figures you hold, starting at 4 and ending at 10. When you already hold k different figures, a new box helps with probability \frac{10-k}{10}, so on average it takes \frac{10}{10-k} boxes to go from k to k+1 distinct figures. Add the stage averages for k = 4,5,6,7,8,9: E = \frac{10}{6} + \frac{10}{5} + \frac{10}{4} + \frac{10}{3} + \frac{10}{2} + \frac{10}{1} = 10\left(	frac16+	frac15+	frac14+	frac13+	frac12+1ight). The bracket equals 2.45, so E = 10 	imes 2.45 = 24.5 	ext{ more boxes.} No…

Finishing a partly complete collection

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