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A company puts five different types of prizes into their cereal boxes, one in each box and in equal proportions. If a customer decides to collect all five prizes, what is the expected number of boxes of cereals that he or she should buy?

TRY

Let $X$ be the number of boxes customer buys. For $i=1,2,3,4,5$, write

$$ X_{ij} = \begin{cases} 1, & \text{ith prize is inside jth box} \\ 0, & \text{otherwise} \end{cases} $$

As I understand the problem, the number of boxes is not given so we may write

$$ X = \sum_{j=1}^{\infty} \sum_{i=1}^5 X_{ij} $$

So

$$ E(X) = \sum_{j \geq 1 } \sum_{i=1}^5 E(X_{ij}) $$

We know $E(X_{ij}) = P(X_{ij}=1)$ so we need to find proobability that ith prize is inside jth box. Here is the part where I get stuck. Am I appraoching this problem the correct way?

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James
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2 Answers2

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HINT: Consider instead: if I have found a given number of unique prizes, what is the probability that the next box I open has a prize I don't have yet? What is the expected time to get a new prize?

SPOILER: this is the Coupon Collector's Problem.

Dan Uznanski
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$\mathbb E[X]=1+\frac{1}{0.8}+\frac{1}{0.6}+\frac{1}{0.4}+\frac{1}{0.2}=\frac{137}{12}$

We know that the probability of selecting a different prize is $\frac{1}{5}$. But, when we purchase that first box of cereal, we have a 100% chance of getting a prize that we don't have. Applying a probability distribution, we are expected to purchase $\frac{1}{1}$ boxes of cereal initially. Next, we have a 80% chance of getting a prize that we don't have, so total we purchase $\frac{1}{1} + \frac{1}{0.9}$ so far. The probability of getting a toy that we don't have thus decreases with each successive purchase, down to a 20% chance of getting one that we don't have once we've gotten all the others.

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