Simple random sample

Remarks

Equivalently, one draws $n$ units one after the other, each uniformly at random among the units not yet drawn. Each unit is in $S$ with probability $n/N$. For values $y_1, \dots, y_N$ attached to the units, the sample mean $\bar y_S = \frac{1}{n}\sum_{i \in S} y_i$ is unbiased for the population mean $\bar Y$. Its variance is $\big(1 - \frac{n}{N}\big) \frac{S_y^2}{n}$, where $S_y^2 = \frac{1}{N-1} \sum_{i=1}^{N} (y_i - \bar Y)^2$. Sampling with replacement, where $n$ independent uniform draws are made, is a different design.

Example: $N = 4$ and $n = 2$, with $S$ uniform on the $\binom{4}{2} = 6$ pairs. Each pair has probability $1/6$, as required, and unit $1$ lies in $3$ of the pairs, so it is chosen with probability $1/2 = n/N$.

Non-example: with $N = 4$, choose $r \in \set{1, 2}$ uniformly and take $S = \set{r, r + 2}$. Every unit is chosen with probability $1/2$, but the pair $\set{1, 2}$ has probability $0$ instead of $1/6$. So equal inclusion probabilities are not enough for a simple random sample.