TITLE: Do successive maximum permutations pick latin squares uniformly?
QUESTION [7 upvotes]: Suppose we start with a $n\times n$ matrix with entries sampled independently and uniformly at random from $[0,1]$. The weight of a set of entries will simply be the sum of those entries. A permutation refers to a set of $n$ entries, no two on the same row or column.
Pick a permutation whose corresponding entries have the largest weight, and place a label "1" in all those entries. Proceed to erase the values of those entries. At step k, pick a permutation from the remaining entries with largest weight and place a label "k" in all the selected entries.
This procedure ends with an $n\times n$ latin square. Do all latin squares appear with the same probability?

REPLY [3 votes]: Studying the results given in YBerman's answer gave me the following. Consider these two possibilities after drawing two symbols (in order) for $n=4$:
$$
\begin{align*}
A &= \begin{pmatrix}
1 & 2 & . & . \\
2 & 1 & . & . \\
. & . & 1 & 2 \\
. & . & 2 & 1
\end{pmatrix}, &
B &= \begin{pmatrix}
1 & 2 & . & . \\
. & 1 & 2 & . \\
. & . & 1 & 2 \\
2 & . & . & 1
\end{pmatrix} 
\end{align*}
$$
(Up to permutation of rows and columns, which doesn't change the probability distribution, any square with two symbols is equivalent to one of these.)
The square $A$ can be completed to $4$ different Latin squares, and $B$ can be completed to $2$ different Latin squares. So the probability of $A$ should be twice the probability of $B$. However, out of 10,000,000 trials, $A$ occurred 35,581 times and $B$ occurred 51,481 times.
So "almost surely" this is a counterexample, but I don't know if there's a simple proof for that.