How do you calculate the expected value of a discrete distribution without replacement? Say I have a set of 10 values I want to draw 3 values from, uniformly, without replacement. For instance:
$$S = \{0,0,0,0,22.95,0,0,0,19.125,25.5\}$$
With replacement, this seems simple, you just add all the values up and divide by $10$, giving $6.7575$, and multiplying by $3$ you get $20.2725$. However, if you don't replace values, I imagine it could effect the expected value, but uncertain how to figure out the specifics.
I guess thinking about it as discrete distributions means that the distribution changes every time a value is drawn, and maybe isn't the best way to think about this problem?

When sampling without replacement from the set $S=\{s_1,s_2,\dots,s_n\}$, the sampled values $X_1,X_2,\dots,X_m$ have an exchangeable distribution.  Hence,
for any $k$, $E(X_k)=\frac1n\sum s_i=\bar s$ and $\operatorname{Var}X_k=\frac1n\sum_{i=1}^n s_i^2-\bar{s}^2=\bar{s^2}-{\bar s}^2$.  Furthermore, for any pair $k\neq l$,
\begin{align}
E(X_kX_l)
  &=\sum_{i \neq j} s_is_j\frac{1}{n(n-1)}
\\&=\frac1{n(n-1)}\sum_{i\neq j} s_is_j
\\&=\frac1{n(n-1)}\left(\sum_{i,j} s_is_j-\sum_i s_i^2\right)
\\&=\frac{n}{n-1}\sum_{i,j} s_is_j\frac1{n^2}-\frac{1}{n-1}\sum_i s_i^2\frac1n
\\&=\frac{n}{n-1}\bar{s}^2-\frac{1}{n-1}\bar{s^2}. \tag{1}
\end{align}
Hence,
\begin{align}
\operatorname{Cov}(X_k,X_l)
  &=E(X_kX_l)-(EX_k)(EX_l)
\\&=\frac{n}{n-1}\bar{s}^2-\frac{1}{n-1}\bar{s^2}-\bar{s}^2
\\&=\frac{1}{n-1}\bar{s}^2-\frac{1}{n-1}\bar{s^2}
\\&=-\frac{1}{n-1}\operatorname{Var}X_k, \tag{2}
\end{align}
and
\begin{align}
\operatorname{Var}(X_1+X_2+\dots+X_m)
  &=m\operatorname{Var}(X_k)+m(m-1)\operatorname{Cov}(X_k,X_l)
\\&=m\operatorname{Var}(X_k)\frac{n-m}{n-1}. \tag{3}
\end{align}
Thus, relative to sampling with replacement, the effect of sampling without replacement is to reduce the variance of $\sum_{k=1}^m X_k$ (and $\bar X$) by a factor of $\frac{n-m}{n-1}$.
It can also be noted that (2) can be derived more directly by setting $m=n$ in (3) such that the variance of the sum on the left hand side is zero and solving for the covariance.
Unsurprisingly, the variances of the hypergeometric and binomial distributions differ by the exact same factor.
