Collecting all six cereal-box toys

Question

A cereal brand hides one toy in every box. There are 6 different toys, each equally likely, and each box is independent of the others.

On average, how many boxes must you buy before you own at least one of every toy?

Accepted Answer

Think about your progress in stages, counted by how many different toys you already have. Suppose you currently hold k different toys. The next box shows a toy you are missing whenever it is one of the 6-k types you lack, which happens with probability \frac{6-k}{6}. Waiting for an event with probability p takes on average 1/p tries, so the average number of boxes to jump from k types to k+1 types is \frac{6}{6-k}. To go all the way from 0 toys to 6 toys, add these stage times for k=0,1,2,3,4,5: E = \frac{6}{6} + \frac{6}{5} + \frac{6}{4} + \frac{6}{3} + \frac{6}{2} + \frac{6}{1} = 6\left(1+\…

Collecting all six cereal-box toys

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