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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 enter image description here.

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Kiran
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  • For a question with such a lengthy history, its formulation lacks the expected clarity. How about defining what it means for two squares to "lie on a diagonal line." – hardmath Aug 25 '15 at 12:54
  • @hardmath. thx. editing the question for clarity. – Kiran Aug 25 '15 at 12:57
  • You have a notion of "the diagonal" of a single square. Does a square have two well-defined diagonals? Isn't the notion of lines being parallel and overlapping inconsistent? If you allow a line to be parallel to itself, surely this is at best a roundabout way of saying so. – hardmath Aug 25 '15 at 13:34
  • @hardmath, I wish i could upload a diagram here to make it clear. Sorry, not able to describe it better. – Kiran Aug 25 '15 at 13:43
  • @harmath, yes i created a diagram. how to post? – Kiran Aug 25 '15 at 14:00
  • @hardmath, not able to understand your question also fully. I hope you will understand it with the diagrams I created. – Kiran Aug 25 '15 at 14:07
  • @hardmath, yes, i uploaded image. thx – Kiran Aug 25 '15 at 17:23
  • I posted an answer, but you might have intended something more restrictive than what I counted with the phrase "diagonals are parallel and overlapping" (my emphasis). Specifically, I did not require that all the squares should "touch", and thus a $2\times 2$ square at each end of the main diagonal, plus three unit squares inside one or the other of those, would be considered five squares "on" the main diagonal in the way I counted them. – hardmath Aug 25 '15 at 20:00
  • @hardmath thx. in fact, while studying your answer, I infact learned more about my own question. I didn't even thought of some aspects you mentioned. My question was derived from another question as I mentioned. Your answer has provided a great opportunity for me to learn. Still spending time with your answer to understand it fully. thx again – Kiran Aug 25 '15 at 20:42

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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.

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