On the probability that a random Latin square is all-even

In a Latin square, every row can be interpreted as a permutation, and therefore has a parity (even or odd). We prove that in a uniformly random $n\times n$ Latin square, the probability that all rows are even is of the form $2^{-n+o(n)}$. This fills a gap left by recent work of Kwan, Petrova and Sawhney: they proved that the row parities behave like a sequence of independent coin flips in a variety of different senses, but they were not able to lower-bound the all-even probability, as their approach had a small number of “exceptional rows” whose parities they could not control. In fact, we prove a more general theorem on the probability of any possible prescription of row parities, column parities and symbol parities. Our approach extends the ideas of Kwan, Petrova and Sawhney, introducing a new way to “bootstrap” intercalate switchings to control more complex switching operations.