Coupon collector's problem extended

Question

I came recently upon this question. Here it is:

'Every box of cereal contains one toy from a group of 5 distinct toys, each of which is mutually independent from the others and is equally likely to be within a given box. How many distinct toys can you expect to collect if you bought 7 boxes?'

What do you think?

I tried using indicator variables. For $2≤i≤5$ let $X_{i}$ be a random variable if we get toy $i$. Then we want $$E(∑iX_{i})=∑iE(X_{i}),$$ by linearity of expectation. But $E(X_{i})$ is just the probability that we get toy $i$. This is simply $\frac{1}{5}$.

So the answer is $$1 + 6 \cdot \frac{1}{5} = \frac{11}{5} = 2.2$$

Is this correct?

Answer 1

Your current approach is mostly correct. I'll elaborate on lulu's hints in the comments.

If we let $X_i$ (for $1 \le i \le 5$) be an indicator r.v. for whether or not we got toy $i$ from any of the $7$ boxes, then the expected number of toys is $$E\left(\sum_{i = 1}^{5}X_i\right) = \sum_{i = 1}^{5}E(X_i),$$

and $E(X_i)$ is just the probability of getting toy $i$ in any of the $7$ boxes. So now, you just need to work out this probability.

The probability of not getting toy $i$ in box 1 is $\dfrac{4}{5}$. The probability of not getting toy $i$ in box 2 is also $\dfrac{4}{5}$. And similarly for all the other boxes. Can you figure out the probability of not getting toy $i$ in any of the $7$ boxes? The probability of getting toy $i$ in at least one box is the compliment of this. Finally, you can just add up these probabilities to get the expected number of distinct toys.

Answer 2

I would compute the probability distribution for the number of distinct toys at each number of draws. For one draw you have probability $1$ of having one toy. For two draws you have $\frac 15$ chance of having $1$ and $\frac 45$ chance of having $2$. Now for three draws you have $\frac 45 \cdot \frac 35$ chance of having three, $(\frac 15 )^2$ of having $1$ and the rest of having $2$. $7$ is not that far away. You could make a spreadsheet and copy down if the number of draws gets higher.