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Pawns are placed randomly on each square on a $n$ by $n$ chessboard, removed and placed randomly again. What's the probability of their new square being adjacent to their previous one for $n$ even? Adjacent squares are such squares that share an edge.

I know that for $n$ odd the probability is zero. This is because each of the pawns needs to be placed on a square of opposite colour to their previous one, and so there needs to be an equal number of white and black squares on the board, which is not true for $n$ odd.

I asked this question on Quora (one answer there states a formula that might be correct) but maybe Stack Exchange is more appropriate for it.

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    " needs to be placed on a square of opposite colour to their previous one" is this one of the requirements in the statement of the problem? – 2017-02-11
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    @Test123 It's implied. Adjoining squares are always of opposite colour. – 2017-02-11
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    What do you mean by adjoining square then? You need to clarify if you don't consider the diagonal ones. Also, how is the probability 0 in the $n=odd$ case? – 2017-02-11
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    @Test123 If you take a look at a standard chess board you will see that every pair of adjoining squares (that share an edge) are of opposite colour. I didn't mean to consider diagonal ones, those are always of the same colour. – 2017-02-11
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    Its non-zero. I guess that is trivial. – 2017-02-11
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    Doesn't seem trivial to me though. And non-zero is how much, exactly? – 2017-02-11
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    Since there are an even number of squares in each row you can swap adjacent pawns $p1$ and $p2$, $p3$ and $p4$.... where $pN$ is the pawn in a given row in the Nth column. That should give you one possible way to get an arrangement you are looking for. – 2017-02-11
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    Yeah, I see it now. I was also thinking about "circling" the pawns by one in each "layer" of the board. So the probability is indeed non-zero. Still looking for the exact value though. – 2017-02-11
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    Is there a reason why you are sure that there is a closed formula for the probability? – 2017-02-11
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    There doesn't have to be but the possible number of events is finite so I expect an exact value to exist. – 2017-02-11
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53458/discussion-between-the-cryptic-cat-and-daphne). – 2017-02-11
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    There are of course $(n^2)!$ ways to place the pawns. To count how many ways have all pawns adjacent to their old position, note that for such a placement, if you choose a starting position and then move in the direction that its original pawn moved, then from the new square do the same thing, you get a path, which must eventually go back to the original square, forming a cycle. So the count of such placements is the count of ways of partitioning the entire board into closed cycles of length at least 2. – 2017-02-11

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