Two squares are chosen on a chessboard at random - Probability 
Two squares are chosen on a chessboard at random. What is the
probability that they have a side in common?

There are many explanations available in internet for this question and it is clear that answer is $\dfrac{1}{18}$. This is an easy question and I am very clear.
But, It seems the answer $\dfrac{1}{18}$ assumes that only 64 squares are there in chess board which is fine because the question may be considered in that way.
But what happens when we consider all the 204 squares of the chess board?
Most of the answers I saw did not consider this and I tried doing it myself. On further search, I found two answers  which I present it here to be clear.

When we consider all the 204 squares,
lofoya.com gives answer as
$\dfrac{222}{204C2}$
careerbless.com gives answer as $\dfrac{228}{204C2}$

I tried doing it with the help of a figure with small examples and see $\dfrac{228}{204C2}$ can be the right answer. Can anyone help in this regard?
 A: On an $n\times n$ chessboard, there are $(n+1-k)^2$ squares of size $k\times k$ that we have to count. So the number of total cases is 
$$\tag1 {1^2+2^2+\ldots + n^2\choose 2}={\frac{n(n+1)(2n+1)}{6}\choose 2}.$$
To have two horizontally adjacent $k\times k$ squares, there are $n+1-k$ choices for the vertical position and $n+1-2k$ choices for the horizontal position. As we have to consider the vertical case as well, we arrive at 
$$ 2\sum_{k=1}^{\lfloor\frac{n+1}{2}\rfloor}(n+1-k)(n+1-2k)$$
where the sum stops at $m:=\lfloor\frac{n+1}{2}\rfloor$ because only $k$ with $n+1-2k\ge 0$ make sense. Transforming we get
$$\tag2\begin{align}2\sum_{k=1}^m(n+1-k)(n+1-2k)&=
2\sum_{k=1}^m(n+1)^2-6(n+1)\sum_{k=1}^mk+4\sum_{k=1}^mk^2\\
&=2m(n+1)^2-3(n+1)m(m+1)+\frac23{m(m+1)(2m+1)}\\
\end{align} $$
favorable cases and a probability of $(2)/(1)$.
For $n=8$ (and $m=4$) this is
$$p_8=\frac{228}{204\choose 2}=\frac{38}{3451} $$
and for $n\gg 0$ we have
$$p_n\approx \frac{15}{2n^3} $$
A: Consider the problem on an $n\times n$ chessboard, and count the number of adjacent cells. There are $n(n-1)$ vertical pairs, which we see by first choosing one of $n$ columns and then choosing the topmost cell of a pair in $n-1$ ways. By symmetry there are $n(n-1)$ horizontal pairs, so the probability is
$$
\frac{2n(n-1)}{\binom{n^2}{2}} = \frac{2n(n-1)}{\frac{n^2(n^2 - 1)}{2}} = \frac{4}{n(n+1)}.
$$
A: The $\frac {228}{^{204}C_2}$ is correct.  They disagree on the number of pairs of $4 \times 4$ squares that share a border.  If the squares are above each other, there are five choices for the leftmost column.  Similarly if the squares are side by side you can choose the top border in $5$ ways, so there are $10$ pairs of $4 \times 4$ squares that share a border, which is part of the $228$ calculation.  The $222$ calculation only seems to have included the four pairs you get by taking quadrants of the board as the $4 \times 4$ squares.
