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3 Squares in a Chessboard

Posted by trickofmind on May 26, 2010 – 12:26 pm

What is the probability that if you randomly chose any three squares on a chess board, they would form a continuous diagonal? No dudes, you don’t hold a chess board in your hands to solve this one. Just think simply.

This post is under “Tom” and has 25 respond so far.
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25 Responds so far- Add one»

  1. 1. Karys Said:

    I assume a “continuous diagonal” means that one of the three squares shares a vertex (is it the word ?) with each of the two other squares.

    Then the diagonal can possibly be in one direction or the other, for example :
    o o o x o o o o
    o o x o o o o o
    o x o o o o o o
    o o o o o o o o
    o o o o o o o o
    o o o o o o o o
    o o o o o o o o
    o o o o o o o o

    With coordinates (x,y) :
    Upper left to bottom right line : One (x,y), with some conditions which we are to find, can be the upper left point of such a diagonal, it is {(x,y),(x+1,y-1),(x+2,y-2)} That means x and y cannot be above 6 on a 8×8 square.

    That gives 6^2 UL-BR diagonals.

    Same thing with UR/BL : 6^2
    We now know there are exactly 72 possible “continuous diagonals” from three points, divided by (8×8)!/5!/3! total possible sets (there’s a name for it (3 among 64 ?), don’t know the English word, if someone could help me).

    Simple division and P the probability that~
    =something very small, at least 60 zeroes following the point.

  2. 2. Karl Sharman Said:

    1 in 62?

  3. 3. Ernaya Said:

    3/how many squares there are overall

  4. 4. Dual Aspect Said:

    My guess: P = 0.000288

  5. 5. Chris Said:

    There are 72 possible diagonals. There are 6 ways that each diagonal could have been selected. There are 64*63*62 possible ways of choosing 3 (different) squares.

    So the probability is 6*72/(64*63*62) ≈ 0.17281%

  6. 6. Chris Said:

    Karys. The words you were looking for are combinations and permutations. nCr = n!/((n-r!)r!) and nPr = n!/(n-r)!

    For combinations, the order of selecting doesn’t matter; so 123 = 213 etc. For permutations, the order matters; so 123 is different to 213.

    I effectively used permutations when I posted my solution.

  7. 7. Chris Said:

    I made a typo above: nCr = n!/((n-r)!r!) is what I meant.

  8. 8. ananomos Said:

    whats up

  9. 9. Joseph Said:

    no chance. it couldn’t be continuous!

  10. 10. Nathan Said:


    If you place the first piece in the interior 6×6 grid, there are 36 possible places to put it. Then if you place the next piece, there are four possible places to make a diagonal. Then the last piece has two possible places to make a diagonal. There are 6 possible ways to order these pieces. 36*4*2*6=1728, 1/1728 = 0.05787%

  11. 11. Nathan Said:


    The interior grid I was thinking is a 4×4 grid.

  12. 12. Chris Said:

    I’m sticking with my answer. A modification is: if the 3 chosen squares could include choosing the same square (i.e. true blindfolded selecting), then the probability would be 6*72/64³ ≈ 0.16479%

  13. 13. Euclid's Brother Said:

    I’m going with Joseph’s answer. . 0%. No diagnal would be continuous, because the board has an 8×8 limit. The diagonal would stop at the edge of the board.

  14. 14. Chris Said:

    EB. I think you’ve misunderstood – you only need 3 squares in a row. Karys (post 1) has shown one of them.

  15. 15. jess Said:

    i hate chess. A dog came chewed up all the piece :( . too bad :) !!!!!!!!!!!!

  16. 16. Jason Said:

    Hey, Chris I’m not understanding where the six came from in your equation. I have a similar equation but no 6.

    first square chosen has 64 possibilities
    second square chosen has 63 possibilities
    third square chosen has 62 possibilities
    there are 72 possible placements on the board to achieve three consecutive diagonals

    therefore; 72/(64*63*62)=0.000288018433179724

  17. 17. Jason Said:

    Better odds than the Lottery.

  18. 18. Chris Said:

    Hi Jason. For each of the 72 diagonals, the first square chosen has 3 possibilities, the second has 2 possibilities and the third has only 1 possibility => 6 possible ways of choosing a given diagonal.

  19. 19. Nikhil Said:


  20. 20. bhavesh trivedi Said:


  21. 21. Ben Said:

    thinking Logically you can find out that there is

    12/64 of the time you have a 1/3906 chance(corner 3)
    16/64 of the time you have a 1/1853 chance(edges)
    4/64 of the time you have a 1/854.4375 chance (b2,b7,g2,g7)
    16/64 of the time you have a 1/732.75 chance (1 in from edges)
    16/64 of the time you have a 1/488.25 chance (middle 4×4)

    and with a bit of maths you turn that into…
    which is around
    1/931 or 0.11% chance

  22. 22. Nishant Said:

    for an 8*8 chess
    we have 64 small squares so total ways to select 3 squares is 64c3 = 4644 and the 3 squars lie in the diagnol ways = 2*8c3+ 4*[7c3+6c3+5c3+4c3+3c3] = 392 so probability = 392/4644 = 7/744…

  23. 23. Chris Said:

    Hi Nishant. I re-examined my answer (post 5) and realise it’s wrong, and too low. So maybe you’re right. I’ll have another think about it. Sadly, I can’t follow what you’ve done as you don’t give an explanation. (I understand your notation though).

    Thank you for re-awakening this problem.

    The question is a bit vague – I’d assumed that you pick three different squares. Perhaps two answers are required.

  24. 24. Chris Said:

    Brain fart over. I’m going back to my original answer.

    Hi Nishant. You’ve answered the wrong question correctly. You have found the probability that three randomly chosen squares lie on a diagonal. The question requires that the three squares should be touching.

  25. 25. Chris Said:

    I’ll explain Nishant’s post. That solution is for three squares anywhere on a diagonal. As it works out OK, I’ll use the NcR notation.

    On a chessboard, for the orientation bottom left to top right there are two
    diagonals of three squares, two of four squares, … , two of seven squares
    and one of eight squares. The same applies for the bottom right to top left orientation.

    In a diagonal of N squares, there are Nc3 ways of choosing 3 squares on it.
    So, altogether, we have 2(2(3c3 + 4c3 + 5c3 + 6c3 + 7c3) + 8c3) diagonals.
    3c3 = 1, 4c3 = 4, 5c3 = 10, 6c3 = 20, 7c3 = 35 and 8c3 = 56. That gives a total of 392 diagonals. There are 64c3 = 41664 ways of choosing 3 squares.
    So we get that the probability is 392/41664 = 7/744

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