Two down, four to go

Question

A promotion gives one random fridge magnet per purchase, chosen uniformly from 6 designs, independently. You already own 2 different designs.

On average, how many more magnets must you receive to own all six designs?

Accepted Answer

Only the 4 missing designs matter. Move through stages by how many distinct designs you hold, from 2 up to 6. Holding k designs, a new magnet helps with probability \frac{6-k}{6}, so the average wait to gain one more distinct design is \frac{6}{6-k}. Sum over k = 2,3,4,5: E = \frac{6}{4} + \frac{6}{3} + \frac{6}{2} + \frac{6}{1} = 6\left(	frac14+	frac13+	frac12+1ight) = 6 	imes \frac{25}{12} = \frac{25}{2} = 12.5. Takeaway. Extra draws to finish is \sum{j=k}^{n-1}\frac{n}{n-j}; from 2 of 6 that comes to 12.5.

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