probability that 5 square lie along a diagonal line (modified) This is a modified version of the question here and asked based on the clarifications obtained here. 
If 5 squares are chosen at random from a chess board, what is the probability that they lie on a diagonal line (squares can overlap and all 204 possible squares shall be considered)
Note: Assume squares lie on a single diagonal line if their diagonals are parallel and overlapping. for example, squares such as the ones given in the image are in the same diagonal
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 A: The images and (to a lesser extent) the language suggest "diagonal line" means a line with slope $1$ through vertices of the $8\times 8$ square grid.  For the sake of completeness I will address the possibility that squares on the chess board might alternatively share diagonals with slope $-1$.  Note Brian M. Scott's answer to the original Question where he remarks that there are "15 diagonals in each direction" (emphasis mine).
As we will see, considering both slopes $\pm 1$ will double the chance of five randomly chosen squares sharing a diagonal, but there is a little check that has to be done to conclude that.
First we compute the size of the sample space. Since there are a pyramidal number $204$ of squares (of varying sizes) on the board:
$$ 204 = 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 $$
there are $\binom{204}{5}$ subsets of five out of these squares.
Now we count how many of these subsets will share a diagonal with slope $1$.  Since a square has exactly one such diagonal, counting may be stratified by taking the individual diagonals of slope $1$ and determining the subsets of five squares that are "on" each of these lines.
The diagonals of slope $1$ may cross from one to eight unit squares.  In order to have five distinct squares "on" such a diagonal, it must cross at least three unit squares (because a shorter diagonal has less than five squares of all sizes "on" it).  Simple counting shows that we have one diagonal (of slope $1$) crossing eight unit squares, and a parallel pair of diagonals crossing each of $3,4,5,6,7$ unit squares.
Now if a diagonal crosses exactly $k$ unit squares, the number of squares of varying sizes "on" that diagonal is the triangular number $k(k+1)/2 = \binom{k+1}{2}$.  For $k \ge 3$ the number of 5-subsets of each is:
$$ \binom{\binom{k+1}{2}}{5} $$
We have only to add these figures up for $k=3,\ldots,7$, double that, and add the count for $k=8$, to get the number of $5$-subsets of squares which share a diagonal of slope $1$.
Of course by symmetry there are an equal number of $5$-subsets of squares which share a diagonal of slope $-1$.  However we need to exclude the possibility of some $5$-subsets sharing diagonals of both slopes $\pm 1$.
Fortunately this is not possible on the $8\times 8$ chess board.  If squares share both diagonals, they have a common center.  Since the smallest would be a unit square, to get five nested squares forces the largest to have side at least one plus $2\cdot 4 = 8$, and the chess board holds no squares that large.
