For every pair of individuals, solves the observed IBS counts against their expectations under Purcell et al.'s (2007) Table 1 to get the IBD sharing probabilities \(Z_0, Z_1, Z_2\), and reports \(\hat\pi = Z_1/2 + Z_2\), the proportion of the genome shared IBD. Expected values for a full sib pair are \((0.25, 0.5, 0.25)\) with \(\hat\pi = 0.5\); for a parent-offspring pair \((0, 1, 0)\), also \(\hat\pi = 0.5\); for unrelated individuals \((1, 0, 0)\) with \(\hat\pi = 0\).
