Computing the variance of two card pulls from numbered deck without replacement

Question

I have a deck of 10 cards numbered 1-10 and I draw two cards, A being drawn first, then B drawn, without replacement. How would I go about calculating the variance of A + B? And how would you extend this computation to a deck of cards numbered 1-n? I've solved the initial question using casework but this process was tedious and looking for a faster way to do this.

Answer 1

Here's how to generalise the problem even further:

Let there be a deck of $n$ cards, and label them from 1 to $n$ so that card $i$ has value $y_i$.

Suppose you draw $k$ cards from the deck. Define a random variable $\delta_i$ for each card so that it is equal to 1 if card $i$ was drawn, and 0 if it wasn't.

Then define $p_i$ as the probability that card $i$ was drawn, and $p_{ij}$ as the probability that cards $i$ and $j$ were both drawn. Notice that

$$p_{ij} = \begin{cases} \frac{k}{n} = p_i & \textrm{if } i = j \\ \frac{k(k-1)}{n(n-1)} & \textrm{if } i \neq j \end{cases}$$

and additionally you have $E(\delta_i) = p_i$ and $E(\delta_i \delta_j) = p_{ij}$.

Then let $Y = \sum_{i \in [n]} y_i \delta_i$ be the sum of the values of the drawn cards. You can calculate the expected value of $Y$ as

$$\begin{eqnarray} E(Y) & = & E(\sum_{i \in [n]} y_i \delta_i) \\ & = & \sum_{i \in [n]} y_i E(\delta_i) \\ & = & \sum_{i \in [n]} y_i p_i \\ & = & \frac{k}{n} \sum_{i \in [n]} y_i \end{eqnarray}$$

and you can find a similar expression for $Var(Y)$ that will look something like $\frac{n}{k}(1 - \frac{k}{n}) \sigma^2_Y$ where $\sigma^2_Y$ is the standard population variance of the $y_i$ values. But I will leave the specific derivation as an exercise.