PLINK 1.0 used the SNP-HWE algorithm in a paper by Wigginton et al. [13] to execute this test. SNP-HWE exploits the fact that, while the absolute likelihood of a contingency table involves large factorials which are fairly expensive to evaluate, the ratios between its likelihood and that of adjacent tables are simple since the factorials almost entirely cancel out [14]. More precisely, given n diploid samples containing a total of n1 copies of allele A1 and n2 copies of allele A2 (so n1+n2=2n), there are \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document} $\frac {(2n)!}{n_{1}!n_{2}!}$ \end{document}(2n)!n1!n2! distinct ways for the alleles to be distributed among the samples, and \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document} $\frac {(2^{n_{12}})(n!)}{((n_{1}-n_{12})/2)!n_{12}!((n_{2}-n_{12})/2)!}$ \end{document}(2n12)(n!)((n1−n12)/2)!n12!((n2−n12)/2)! of those ways correspond to exactly n12 heterozygotes when n12 has the same parity as n1 and n2. Under Hardy-Weinberg equilibrium, each of these ways is equally likely. Thus, the ratio between the likelihoods of observing exactly n12=k+2 heterozygotes and exactly n12=k heterozygotes, under Hardy-Weinberg equilibrium and fixed n1 and n2, is \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb}
