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?

The mean of the values in $S$ is $6.7576,$ as you say. So, with replacement, each of the three chosen values has an expected value of $6.7575$ and
thus the average of the three is again
$\bar X_3 = 6.7575.$
s = c(0,0,0,0, 22.95, 0,0,0, 19.125,25.5)
mean(s)
[1] 6.7575

In sampling without replacement, the mean $E(\bar X_3)= 6.7575$ and the variance $V(\bar X_3)$ of the sample mean will be
somewhat smaller because there are fewer possible samples
of size three that may be drawn. (For example, once value $25.5$ is chosen, it can't be chosen again.)
Without working through the combinatorial detains, it
is possible to get a good idea of the sampling
distribution of $\bar X_3$ according to each
sampling method by doing a simple simulation in R.
With 10 million iterations one can expect about three place accuracy.
set.seed(3033)
s = c(0,0,0,0,22.95,0,0,0,19.125,25.5)
a.wo = replicate(10^7, mean(sample(s, 4)))
mean(a.wo);  sd(a.wo)
[1] 6.760266 # aprx 6.7575
[1] 4.254535

summary(a.wo)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.000   4.781   6.375   6.760  10.519  16.894 

a.with = replicate(10^7, mean(sample(s, 3, rep=T)))
mean(a.with);  sd(a.with)
[1] 6.757315   # aprx 6.7575
[1] 6.017261

summary(a.with)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.000   0.000   7.650   6.757   8.500  25.500 

par(mfrow=c(1,2))
hist(a.wo, prob=T, xlim=c(0,26), col="skyblue2")
hist(a.with, prob=T, xlim=c(0,26), col="skyblue2")
par(mfrow=c(1,1))


