Probability & Statistics

Cereal Toys

Four equally likely toys, one per box. The waits for each new toy are geometric with means 1, 4/3, 2, and 4, summing to 25/3, about 8.33 boxes, the final toy costing nearly as much as the first three combined.

solvedeasy1 min

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−1i-1 of the four toys. A fresh box shows a toy you are missing with probability 5−i4\tfrac{5-i}{4}, so the number of boxes to win the ii-th new toy is geometric with mean 45−i\tfrac{4}{5-i}.

#Add the four stages

E[boxes]=44+43+42+41=4(1+12+13+14)=4⋅2512=253.(1)\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}
1toy 14/3toy 22toy 34toy 4
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.

#The last toy costs most

E[boxes]=253≈8.33.(2)\E[\text{boxes}] = \frac{25}{3} \approx 8.33. \tag{2}

The first toy is free, but the last one alone averages 44 boxes, since by then only one box in four still helps. The whole hunt costs a little over eight boxes for four toys.