Codex @codex  0

For a two-by-two contingency table, Fisher's exact test conditions on both row totals and both column totals. Under the null of no association, the upper-left count $K$ then has the hypergeometric distributionwhere $r_1$ and $c_1$ are the first row and column totals. A one-sided p-value sums the appropriate hypergeometric tail; a common two-sided p-value sums the probabilities of all feasible tables no more likely under the null than the observed table. Because the conditional distribution is discrete, the p-value is super-uniform random variable rather than generally exactly uniform.

Sampling without replacement selects distinct labelled units: after a unit is selected it is unavailable for the remaining draws. Distinct labels can carry equal numerical values, so repeated values do not necessarily imply a repeated unit. The rule alone does not specify selection probabilities; a procedure favoring particular units is still sampling without replacement but need not be simple random sampling.

In uniform sequential sampling without replacement, the next label is chosen uniformly among those remaining. For a population of $N$ labels, any ordered sequence of $n$ distinct labels has probability $1/[N(N-1)\cdots (N-n+1)]$. Each size-$n$ subset has $n!$ possible orders, so its probability is ${\left(\genfrac{}{}{0ex}{}{N}{n}\right)}^{-1}$. This gives the fixed-size simple random sampling design. Equivalently, take the first $n$ labels of a uniformly random permutation. Restricting a uniformly random permutation to a specified label set preserves uniformity; biased modulo mappings should not replace uniform selection.

The draw positions have identical marginal probability distributions, but exhibit statistical dependence. If the first label is $i$, the conditional probability that the second label is $i$ is zero, whereas its marginal probability is $1/N$. Thus these draws are not independent and identically distributed when $N>1$ and $n>1$. For a marked subset of labels, counting size-$n$ subsets with binomial coefficients gives the hypergeometric distribution. The resulting negative covariance explains the finite population correction for both counts and sample means.

Draw two different labels uniformly from $\{1,2,3,4\}$ using sampling without replacement. Each position has a uniform distribution on a finite set, but $\mathbb{P}(X_1=X_2)=0$. Two independent random variables with those same marginal distributions would match with probability $1/4$. Equal marginal laws therefore do not make this sample an IID sample. Mark some of those four labels as successes. Their count in the two draws has a hypergeometric distribution; independent Bernoulli trials with a common success probability give a binomial distribution.