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  1. Donation ketchuplover
    G.O.A.T.
    07 Feb '14 21:35
    CHESS 1,625,702,400 aka Random Pawn Placement
  2. Subscriber Ponderable
    chemist
    08 Feb '14 10:34
    Originally posted by ketchuplover
    CHESS 1,625,702,400 aka Random Pawn Placement
    would you elaborate? I for one can't do anything with the information provided.
  3. Subscriber dzirilli
    Duchampion
    08 Feb '14 14:42
    I'm assuming that's the number of possible starting positions if you place the pawns randomly
  4. Standard member ChessPraxis
    Cowboy From Hell
    08 Feb '14 15:52
    Originally posted by ketchuplover
    CHESS 1,625,702,400 aka Random Pawn Placement
    I saw this being discussed at another site.
  5. 08 Feb '14 16:07
    Originally posted by dzirilli
    I'm assuming that's the number of possible starting positions if you place the pawns randomly
    It does not seem to be. The number is equal to 8! squared, which can't possibly be the number of ways to place pawns on the board. It could be a miscalculated number of ways to place all the non-pawns on their respective starting ranks -- there the mistake would at least make sense.
  6. Standard member wolfgang59
    Infidel
    08 Feb '14 21:40 / 1 edit
    Originally posted by WanderingKing
    It does not seem to be. The number is equal to 8! squared, which can't possibly be the number of ways to place pawns on the board. It could be a miscalculated number of ways to place all the non-pawns on their respective starting ranks -- there the mistake would at least make sense.
    8! would be the number of ways of setting up your own pawns at the start
    (given that the pawns are distinguishable)
    Take into account the opponents permutations as well and you get (8!)x(8!)
  7. Donation ketchuplover
    G.O.A.T.
    08 Feb '14 23:58
    Originally posted by wolfgang59
    8! would be the number of ways of setting up your own pawns at the start
    (given that the pawns are distinguishable)
    Take into account the opponents permutations as well and you get (8!)x(8!)
    Does 8! = 1x2x3x4x5x6x7x8 ? Someone on another site thinks I am including missing pawns. Never thought of them.
  8. Standard member wolfgang59
    Infidel
    09 Feb '14 01:14
    Originally posted by ketchuplover
    Does 8! = 1x2x3x4x5x6x7x8 ?
    YES!
  9. Standard member DeepThought
    Losing the Thread
    09 Feb '14 06:04
    Originally posted by wolfgang59
    8! would be the number of ways of setting up your own pawns at the start
    (given that the pawns are distinguishable)
    Take into account the opponents permutations as well and you get (8!)x(8!)
    Except that pawns are indistinguishable so the correct number of ways of setting up the pawns at the start is 1.

    If you had the pawns go anywhere where there aren't pieces, and avoiding the initial position involving a check then there are C(46,8)*C(38,8) ways of placing the pawns (C(N,M) = binomial coefficient = N!/(N-M)!M!).
  10. Standard member wolfgang59
    Infidel
    09 Feb '14 06:48
    Originally posted by DeepThought
    Except that pawns are indistinguishable so the correct number of ways of setting up the pawns at the start is 1.

    Each pawn has its own character. I have one little fellow who is very aggressive ... always a tough call to start him on c2 or e2
  11. 09 Feb '14 08:08
    Originally posted by DeepThought
    Except that pawns are indistinguishable so the correct number of ways of setting up the pawns at the start is 1.
    Indeed. (I'm not checking the other half -- I have a hangover to cure.)
  12. Standard member wolfgang59
    Infidel
    09 Feb '14 08:40
    Originally posted by WanderingKing
    Indeed. (I'm not checking the other half -- I have a hangover to cure.)
    Greasy bacon sandwich washed down with
    a pint of fresh OJ & tonic water (50/50)

    trust me
  13. Subscriber sonhouse
    Fast and Curious
    09 Feb '14 13:09
    Originally posted by DeepThought
    Except that pawns are indistinguishable so the correct number of ways of setting up the pawns at the start is 1.

    If you had the pawns go anywhere where there aren't pieces, and avoiding the initial position involving a check then there are C(46,8)*C(38,8) ways of placing the pawns (C(N,M) = binomial coefficient = N!/(N-M)!M!).
    What does the 46,8 and 38,8 stand for? What does that number come out to?
    I assume N=46? or 38?
  14. 10 Feb '14 08:47 / 1 edit
    Originally posted by sonhouse
    What does the 46,8 and 38,8 stand for? What does that number come out to?
    I assume N=46? or 38?
    I think:

    C(46,8)*C(38,8) ways of placing the pawns (C(N,M) = binomial coefficient = N!/(N-M)!M!).

    translates into:

    46!/(46-8)!8! * 38!/(38-8)!8!

    So you apply the formula twice, once with N=46, M=8 and once with N=38, M=8. And then multiply those two.
    Whether the outcome is correct, is another story.
  15. Subscriber Ponderable
    chemist
    10 Feb '14 11:45
    Since I am quite dumb:

    You propose to put pawns on the second and seventh row rexpectively. Then you go and permuatet their position.

    What is the difference to the nomrla person with the transmutations?