Consider a 10-cards deck: two cards have face value 1, two cards have face value 2,..., two cards have face value 5. You pick 4 cards without replacement from this deck. What is the expected number of pairs in the hand ?
My Solution:
Let $x$ denote the expectation under consideration. I proceed with a sum of indicator variables.$$x = E[\sum_{i=1}^5 1_{\text{both cards of value }i \text{ are in the hand} }] = 5P(\text{both 1's are in the hand)}$$To compute this last probability, I relabel the cards so they are all distinguishable and the $1's$ are now cards number $1$ and $2$. The random hand is now modeled with a random $4$-subset of $\{1,\ldots,10\}$, hence$$x = 5\frac{\binom{10-2}{2}}{\binom{10}{4}}=\frac 23.$$
Question:
Is my approach correct ? Is there a quicker way ?
