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List all samples in a small population
For a frame with four labelled units A, B, C and D, a size-two sample has six possible unordered subsets. A simple random design assigns probability 1/6 to AB, AC, AD, BC, BD and CD. The sample order is irrelevant for this definition.
Calculate one unit’s inclusion chance
A appears in AB, AC and AD, so its inclusion probability is 3/6 = 1/2. The same holds for every other unit. Equal unit probabilities follow from this design, but the reverse implication is not automatic.
Compare a different equal-chance design
A design that selects either AC or BD, each with probability 1/2, also includes every unit with probability 1/2. Four size-two subsets have probability zero, so it is not the same simple random design.
| Subset | Simple random | AC-or-BD design |
|---|---|---|
| AB | 1/6 | 0 |
| AC | 1/6 | 1/2 |
| AD | 1/6 | 0 |
| BC | 1/6 | 0 |
| BD | 1/6 | 1/2 |
| CD | 1/6 | 0 |
Carry the design into estimation
The distinction matters for the distribution and variance of an estimator. Equal marginal probabilities can make the same mean unbiased under both examples while producing different uncertainty. State the full design before using a simple-random variance formula.