Are magic squares inevitable? Consider: If we are given a reasonably well-behaved statistical population of real numbers, then samples that are not small will have mean approximately equal to that of the population, right? Then: let n be a large positive integer, and choose n positive integers at random, without replacement, from among the integers from 1 to n squared. Do this a total of n times. (The n squared numbers are depleted at the end of this process, no two selections containing common members.) Let these be the rows of a matrix. The all the rows, columns, and the two diagonals constitute a random sample of size n, and will have mean approximately that of the mean of the integers from 1 to n squared. Therefore this matrix, generated so easily, is already nearly a magic square. It is plausible that all it needs is some tweaking (swapping a few entries here and there) to be exactly a perfect square. Thus, magic squares are, contrary to one’s naïve initial reaction, inevitable, and, indeed, abundant, rather than unlikely or rare. The only surprise is that non-trivial magic squares of very low order exit.
So, here is my first question: In all that I have ever seen about magic squares, this statistical perspective has never been given. But perhaps there is mention that I simply haven’t noticed. That is why I made one of the tags for this question “reference-request”. If I’ve overlooked it, please tell me.
My main question is this: Is it indeed the case that, contrary to first blush, one can plausibly assert, in light of the statistical argument given above, without the need of exhibiting any magic square at all, that magic squares are inevitable and exist in abundance?
 A: There are 362,880 ways to put the 9 digits into a $3\times3$ array. Of these, exactly 8 are magic squares. They are about as abundant as needles in haystacks. It gets worse for larger sizes. 
A: This kind of reasoning leads to false results. For instance, to construct a projective plane of a given order $N$, take $N^2+N+1$ points and $N^2+N+1$ lines, randomly choose $(N+1)(N^2+N+1)$ pairs $(l,p)$ of lines and points and say that $p$ lies on $l$ iff $(l,p)$ is one of the chosen pairs. Then each line will on average have $N+1$ points on it, and each point will on average lie on $N+1$ lines, so it is plausible that all it needs is some tweaking (moving a few points to different lines here and there) to get a projective plane. In fact, however, it is known that there are orders for which there are no projective planes. The fact that the constraints are fulfilled in the mean says little about whether it's possible to fulfill them all exactly.
A: One problem with your argument is that rows are not independent of columns or diagonals.
Another is that the usual definition of magic square requires all elements distinct.
