@BruceET's answer has some nice information, and simulations is a good way to go for this sort of thing in practice. In this answer, I'll detail an exact approach, which demonstrates that the exact answer is actually $20.275$, **the same as in the sampling without replacement approach**. 

Note that this is unsurprising due to the linearity of expectation and the fact that, marginally, each $X_i$ has the same distribution in either case. Note that the variances will NOT be the same for each approach, as @BruceET points out in his answer. 


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In the sampling with replacement case, you can view each draw as an *independent* draw from the given discrete distribution $X_1, X_2, X_3 \stackrel{\text{iid}}{\sim} p(x)$. In the sampling *without replacement* case, you have to consider the *joint* distribution of the random variables ${\bf X} = (X_1, X_2, X_3) \sim p(x_1, x_2, x_3)$. Then the expected value of $Y = X_1 + X_2 + X_3$ is taken as the weighted average with weights given by $p(x_1, x_2, x_3)$. 


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More generally, let $S = \{s_1, s_2, \ldots s_n\}$. A general strategy for finding $E(X_1+X_2+X_3)$ starts by listing all of the possible outcomes,
$$\mathcal S = \{(x_1, x_2, x_3) | x_1 = s_i, x_2 = s_j, x_3 = s_k, i \neq j \neq k),$$
of which there are $n(n-1)(n-2)$. Since the sum doesn't care about the order in which each number is drawn, we can simplify things a bit by looking at the set of all combinations: 
$$\mathcal S = \{(x_1, x_2, x_3) | x_1 = s_i, x_2 = s_j, x_3 = s_k, i < j < k),$$
of which there are $\binom{n}{3}$ options. Then the expected value can be computed as
\begin{align*}
E(X_1+X_2+X_3) &= \sum_{k=3}^n\sum_{j=2}^k\sum_{i=1}^j(s_i + s_j + s_k)\times 6 \times p(s_i, s_j, s_k)\\
&= 6 p(s_i, s_j, s_k)\sum_{k=3}^n\sum_{j=2}^k\sum_{i=1}^j(s_i + s_j + s_k)
\end{align*}
where the $6$ comes from the various orderings of $s_i, s_j, s_k$. Since each element $s_i (i=1,\ldots n)$ has the same probability of being selected, we have $p(s_i, s_j, s_k) = \frac{1}{10\cdot 9\cdot 8}$ for each combination. 


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## Example R Code



    compute_expected_value <- function(S){
      n <- length(S)
      res <- 0
      for(i in 3:n){
        for(j in 2:(i-1)){
          for(k in 1:(j-1)){
            res <- res*6*(S[i] + S[j] + S[k])/10/9/8
          }
        }
      }
      return(as.numeric(res))
    }  

    S <- c(0,0,0,0, 22.95, 0,0,0, 19.125,25.5)
    compute_expected_value(S)

    > 20.2725