Expected boxes to get both toy figures

Question

A cereal box contains one of two collectible toys, each equally likely and independent. You buy boxes until you have both toys. What is the expected number of boxes you must buy?

Accepted Answer

The first box always gives a new toy, so stage j = 1 starts after 1 box. From holding 1 of 2 toys, a new box matches the missing toy with probability 	frac12. Let e1 be the expected boxes to get it: e1 = 1 + 	frac12 e1 \;\Longrightarrow\; e1 = 2. Total E = 1 + e1 = 1 + 2 = 3. (Same as n(1 + 	frac12) = 2 \cdot 	frac32 = 3.) Takeaway. Two equally likely types average 3 boxes: one free plus a 1/p = 2 wait for the second.

Expected boxes to get both toy figures

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