Exact conditional Pearson test for a binary contingency table ($P(X=x\mid K)=\frac{{\prod }_i\left(\genfrac{}{}{0ex}{}{n_i}{x_i}\right)}{\left(\genfrac{}{}{0ex}{}{N}{K}\right)}$)

Consider independent binomial counts $X_i\sim \mathrm{Bin}(n_i,p)$ under the common-probability null hypothesis, with fixed positive row sizes $n_i$. Let $N={\sum }_i{n}_i$, $K={\sum }_i{X}_i$ and assume $0<K<N$. Conditional on $K$, the probability mass function is the displayed multivariate hypergeometric distribution: in the joint likelihood, the factor $p^K(1-p{)}^{N-K}$ is common to all admissible tables and cancels. The Vandermonde identity gives the denominator, so the conditional distribution is free of the unknown nuisance parameter $p$.

For a table $x$, its Pearson chi-squared statistic isAn exact conditional p-value sums the displayed probabilities over all tables with the same margins and $Q(x)\ge Q(x_{\mathrm{obs}})$. The ordering is by the Pearson chi-squared statistic, rather than by the probability of an individual table; different exact-test orderings can give different answers. Enumeration is practical when $K$ and the number of rows are small. Monte Carlo tests can approximate the same tail by sampling tables with fixed margins.^[ref]

An exact arithmetic comparison is easy too. Let $M$ be the least common multiple of the row sizes. Compare the integersince every table has the same positive denominator $MK(N-K)$. This includes boundary ties without a floating-point tolerance. Conditional p-values are discrete and can be conservative; “exact” concerns their null calibration, not equality of rejection probability with every chosen significance level.