Expected Distinct Cards from Packs
James buys 6 packs of football cards. Each pack contains exactly one card, chosen uniformly at random from a set of 10 distinct cards (with replacement across packs). What is the expected number of distinct cards James ends up with?
Hints
- Think about each card type separately rather than trying to count distinct cards directly.
- For a specific card, the probability it is missed in all 6 packs is $\left(\frac{9}{10}\right)^6$. Use this with linearity of expectation.
- Define $X_i = 1$ if card $i$ appears at least once, then $\mathbb{E}[X] = \sum_{i=1}^{10} P(X_i = 1)$.
Worked Solution
How to Think About It: This is a classic coupon-collector-style question, but we are not collecting all coupons -- we just want the expected number of distinct types in a fixed number of draws. The right tool is indicator random variables. For each of the 10 card types, ask: what is the probability that card $i$ shows up at least once in 6 draws? Then use linearity of expectation to sum those probabilities. The complement (never seeing a given card) is much easier to compute.
Quick Estimate: Each specific card has a $\frac{1}{10}$ chance of appearing in any given pack, so it has a $\frac{9}{10}$ chance of missing. With 6 independent packs, the probability of never seeing card $i$ is $(0.9)^6 \approx 0.531$. So each card appears with probability about $1 - 0.531 = 0.469$. With 10 cards, the expected distinct count is roughly $10 \times 0.469 \approx 4.69$.
Approach: Formalize via indicator variables and linearity of expectation.
Formal Solution:
Define indicator variables $X_i$ for $i = 1, 2, \ldots, 10$:
$$X_i = \begin{cases} 1, & \text{if card } i \text{ appears in at least one pack,} \\ 0, & \text{otherwise.} \end{cases}$$
The total number of distinct cards is $X = X_1 + X_2 + \cdots + X_{10}$. By linearity of expectation:
$$\mathbb{E}[X] = \sum_{i=1}^{10} \mathbb{E}[X_i] = \sum_{i=1}^{10} P(\text{card } i \text{ appears at least once}).$$
The probability that card $i$ never appears in any of the 6 packs is $\left(\frac{9}{10}\right)^6$, so:
$$\mathbb{E}[X_i] = 1 - \left(\frac{9}{10}\right)^6 = 1 - 0.531441 = 0.468559.$$
Therefore:
$$\mathbb{E}[X] = 10 \times \left(1 - \left(\frac{9}{10}\right)^6\right) = 10 \times 0.468559 \approx 4.686.$$
Answer: $\mathbb{E}[X] = 10\left(1 - \left(\frac{9}{10}\right)^6\right) \approx 4.69$ distinct cards.
Intuition
The indicator variable trick -- decomposing a complex count into a sum of 0/1 variables -- is one of the most powerful tools in probability. It works because linearity of expectation holds regardless of dependence between the indicators. Here the $X_i$ are clearly not independent (seeing lots of card 1 reduces the slots for other cards), but we do not need independence to sum their expectations.
This problem is a finite-sample version of the coupon collector problem. The full coupon collector asks: how many packs until you have all 10 cards? That answer is $10 \times H_{10} \approx 29.3$ packs (where $H_{10}$ is the 10th harmonic number). Here we flip the question: given 6 packs, how many distinct types do you expect? The same indicator decomposition applies to both. In trading, analogous calculations appear when estimating how many distinct counterparties or instruments you will encounter in a fixed number of trades sampled from a large universe.