AbstractGyárfás and Sárközy conjectured that every Latin square has a “cycle‐free” partial transversal of size . We confirm this conjecture in a strong sense for almost all Latin squares, by showing that as , all but a vanishing proportion of Latin squares have aHamiltontransversal, that is, a full transversal for which any proper subset is cycle‐free. In fact, we prove a counting result that in almost all Latin squares, the number of Hamilton transversals is essentially that of Taranenko's upper bound on the number of full transversals. This result strengthens a result of Kwan (which in turn implies that almost all Latin squares also satisfy the famous Ryser–Brualdi–Stein conjecture).