Cereal Toys

A cereal company puts one free toy in every box, drawn equally likely from four different toys. How many boxes do you expect to buy to collect a complete set of all four?

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#Collect one new toy at a time

Suppose you already hold $i-1$ of the four toys. A fresh box shows a toy you are missing with probability $\tfrac{5-i}{4}$ , so the number of boxes to win the $i$ -th new toy is geometric with mean $\tfrac{4}{5-i}$ .

#Add the four stages

\E[\text{boxes}] = \frac{4}{4} + \frac{4}{3} + \frac{4}{2} + \frac{4}{1} = 4\left(1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4}\right) = 4\cdot\frac{25}{12} = \frac{25}{3}. \tag{1}

Winning the i-th new toy takes an expected 4/(5-i) boxes, climbing as the set fills. The four stages 1, 4/3, 2, and 4 sum to 25/3, with the last toy alone nearly half the total.

#Read it off

\E[\text{boxes}] = \frac{25}{3} \approx 8.33. \tag{2}

The first toy is free, but the last one alone averages $4$ boxes, since by then only one box in four still helps. The whole hunt costs a little over eight boxes for four toys.