Suppose we draw 2 cards from a standard 52 card deck with value 1 - 13 (from Ace to King). How would expected values and variance compare with or without replacement?
$E[X] = 14$ for both cases (linearity of expectation)
$Var(X) = E[X^2] - E[X]^2$ however this seems like it would take incredibly long to compute for all the possible sums, is this the right approach? Would the variances even be different?
On
Let $X$ and $Y$ be the ranks of the two cards drawn.
$\small\begin{align}\mathsf {Var}(X+Y)&=\mathsf E(X^2)+2\mathsf E(XY)+\mathsf E(Y^2)-\mathsf E(X)^2-2\mathsf E(X)\mathsf E(Y)-\mathsf E(Y)^2\\&=2\mathsf E(X^2)+2\mathsf E(XY)-2\mathsf E(X)^2-2\mathsf E(X)\mathsf E(Y)\\&=2\mathsf E(X^2)+2\mathsf E(XY)-4\mathsf E(X)^2\end{align}$
These terms are not that hard to evaluate and only $\mathsf E(XY)$ changes if the cards are drawn with or without replacement.
You have already evaluated $\mathsf E(X)$ correctly, and $\mathsf E(X^2)$ will be evaluated similarly.
In the case of "with replacement" $\mathsf E(XY)$ will equal $\mathsf E(X)^2$ because $X,Y$ will be independent and identically distributed.
In the case of "without replacement" the joint probability function shall be $52/51$ times larger except in the selection of identical cards (which must be excluded†).
$$\begin{align}\mathsf E(XY) &= \begin{cases}\tfrac{16}{52^2}\sum_{k=1}^{13}\sum_{j=1}^{13}jk &:&\text{with replacement}\\ \tfrac{16}{52\cdot 51}\sum_{k=1}^{13}\sum_{j=1}^{13} jk-\tfrac{4}{52\cdot51}\sum_{k=1}^{13}k^2&:&\text{without replacement}^\dagger \end{cases}\\[1ex]&=\begin{cases}\mathsf E(X)^2&:& \text{with replacement}\\\tfrac{52}{51}\mathsf E(X)^2-\tfrac{13}{51}\mathsf E(X^2)&:&\text{without replacement}\end{cases}\end{align}$$
Even if enumerating all possible combinations (which for just two cards it would be difficult to significantly improve upon), there are only $\binom{52}{2} = 1326$ possibilities, which is trivial for a computer. The variances are 28 and about 27.45 with and without replacement respectively.