# Average Sum and Standard Deviation When There Can Not Exist Repetition?

Imagine there are six cards with values from 1 to 6. Three cards are chosen at random, each card having an equal probability of being chosen, and their values are added together.

What will the average sum be? And what is the standard deviation?

I know how to calculate these quantities when it is possible to repeat the values, but how to calculate in this case when a value can only fall once? In the first choice of cards, all have the same probability of being chosen. However, what happens later when one of them is chosen and can no longer be chosen? Does this influence the mean sum and standard deviation account?

• I am not educated in calculating the variance. As for calculating the mean sum, see Linearity of Expectation, which includes a proof that the formula applies even when the events are not independent of each other. One of the consequences of this article is that it is irrelevant whether the numbers are sampled with or without replacement. Mar 16 at 15:26
• stats.stackexchange.com/questions/236299/… Mar 16 at 16:05

You are drawing a random subset of size 3 from the set $$\{1,2,3,4,5,6\}$$. There are $${6 \choose 3}=20$$ ways to do it; if $$X$$ is this triple then it can take values \begin{align} (1,2,3)\ & S=6\\ (1,2,4)\ & S=7\\ (1,2,5)\ & S=8\\ (1,2,6)\ & S=9\\ (1,3,4)\ & S=8\\ (1,3,5)\ & S=9\\ (1,3,6)\ & S=10\\ (1,4,5)\ & S=10\\ (1,4,6)\ & S=11\\ (1,5,6)\ & S=12\\ (2,3,4)\ & S=9\\ (2,3,5)\ & S=10\\ (2,3,6)\ & S=11\\ (2,4,5)\ & S=11\\ (2,4,6)\ & S=12\\ (2,5,6)\ & S=13\\ (3,4,5)\ & S=12\\ (3,4,6)\ & S=13\\ (3,5,6)\ & S=14\\ (4,5,6)\ & S=15 \end{align} So $$S=\begin{cases} 6,& \text{with prob. }\frac{1}{20}\\ 7,& \text{with prob. }\frac{1}{20}\\ 8,& \text{with prob. }\frac{2}{20}\\ 9,& \text{with prob. }\frac{3}{20}\\ 10,& \text{with prob. }\frac{3}{20}\\ 11,& \text{with prob. }\frac{3}{20}\\ 12,& \text{with prob. }\frac{2}{20}\\ 13,& \text{with prob. }\frac{2}{20}\\ 14,& \text{with prob. }\frac{1}{20}\\ 15,& \text{with prob. }\frac{1}{20} \end{cases}$$ from which you can get thar $$ES=10.5$$, $$E(S^2)=\frac{231}{2}$$.
Suppose that we sample in turn $$n$$ objects from a population at random $$S=\{x_{1}, x_{2}\dotsc, x_{N}\}$$ without replacement. Let $$X_1,\dotsc, X_{n}$$ be these values. Then $$P(X_{m} = j) = \frac{1}{N}\quad (j\in S,m\in[n])$$ by the counting argument $$P(X_{m} = j) =\frac{\binom{N-1}{n-1}(n-1)!}{\binom{N}{n}n!}$$ where the denominator counts the number of permutations of length $$n$$ from $$S$$ and the numerator counts those permutations where the $$m$$th position is fixed. It follows that $$EX_{m} = \frac{1}{N}\sum_{i=1}^Nx_{i}=\bar{x}$$
so that if we let $$S=\sum_{i=1}^nX_{i}$$, $$ES = n\bar{x}.\tag{0}$$ For the computation of the variance of $$S$$ we now assume without loss of generality that the $$X_{i}$$ have zero expected value (if not we can center the variables beforehand without changing the variance). Then because we are sampling with replacement the $$X_{i}$$ are not independent. Indeed, $$\text{Cov}(X_{i}, X_{j}) = EX_{i}X_{j} =\frac{1}{N(N-1)}\left( \sum_{k=1}^N x_{k}\sum_{l=1}^Nx_{l}-\sum_{i=1}^N x_{i}^2 \right) =\frac{-\sigma^2}{N-1}$$ for $$i\neq j$$ where $$\sigma^2 = EX_{1}^2= \text{Var}(X_{1}) =N^{-1}\sum_{i=1}^Nx_{i}^2$$. So \begin{align} \text{Var}(S) &= \sum_{i=1}^n\text{Var}(X_{i})+2\sum_{1\leq i In your specific case $$S=\{1,2,3,4,5,6\}$$. Using $$(0)$$, $$ES=3\times\frac{7}{2}=\frac{21}{2}$$ while $$\text{Var}(S) = \frac{3\times3}{5}\sum_{i=1}^6(i-3.5)^2=\frac{3\times3}{5}\frac{35}{2}=\frac{63}{2}$$