We derived the needed probabilities for this simulation as follows. Let N1 be the number of cases and N2 be the number of controls in the second population, and let p be the frequency of the A1 allele and q = 1-p be the frequency of A2. Then we note that for the 2 × 3 table of case status by genotype status, the marginals for the case and control counts are N1 and N2 respectively, and the marginals for the A1A1, A1A2, and A2A2 counts are p2(N1+N2), 2pq(N1+N2), and q2(N1+N2) respectively. Then the probabilities for the cells of the 2 × 3 table, recalling that the odds ratio is set to be 1 (corresponding to no effect in the second population), are the products of the marginal probabilities, so that Pr(case and A1A1)=N1N1+N2p2, Pr(ctrl and A1A1)=N2N1+N2p2, etc.
