, 2023

Solution:

We can deduce information about $a$ from the grid $b$ by looking at the largest element of it, say $m$. If $m$ fills an entire row, then the value of $a$ corresponding to this row must be equal to $m$. Otherwise, $m$ must fill an entire column, and the value of $a$ corresponding to this column must be equal to $m$. We can then ignore this row/column and continue this reasoning recursively on the remaining part of the grid.

Near the end, there are two cases. We could have a $1 \times 1$ remaining grid, where there are 2 permutations $a$ consistent with $b$. We could also have a case where one of the dimensions of the remaining grid is 1, the other dimension is at least 2 (say $k$), and the number $k+1$ fills the entire remaining grid. In that case, there are $k!$ ways to arrange the other elements $1, \ldots, k$.

It follows that there are exactly 2 permutations $a$ consistent with the grid if and only if one of $1$ and $2$ is assigned to a row and the other is assigned to a column, or they are both assigned to the same type and $3$ is assigned to the opposite type. The probability that this does not occur is the probability that $1, 2, 3$ are all assigned to the same type, which happens with probability $\frac{19}{39} \cdot \frac{18}{38} = \frac{18}{2 \cdot 39} = \frac{3}{13}$, so the answer is $1 - \frac{3}{13} = \frac{10}{13}$.