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- From: shoham@ll.mit.edu (Daniel Shoham)
- Subject: Re: Chess Problem
- Message-ID: <1992Sep12.105358.16193@ll.mit.edu>
- Sender: news@ll.mit.edu
- Organization: MIT Lincoln Laboratory
- References: <BuFpLp.9nI@ecf.toronto.edu> <BuGA8t.DL8@cmptrc.lonestar.org>
- Date: Sat, 12 Sep 92 10:53:58 GMT
- Lines: 32
-
- In article <BuGA8t.DL8@cmptrc.lonestar.org> carter@cmptrc.lonestar.org (Carter Bennett) writes:
- >rairan@ecf.toronto.edu (RAI Ranjan) writes:
- >>Chess Problem:
- >> Eight rooks are placed at random on a chessboard.
- >> What is the probability that no two rooks can attack one another?
-
- After n rooks that cannot attack each other are placed, a total of n rows and
- n coulmns are now defended (i.e. a further rook may not be placed there). n
- rows + n coulmns represent 16*n-n^2 squares (remembering not to double-count
- the squares that are defended twice). This leaves (64-16*n+n^2=[8-n]^2) squares
- available for the next rook to be placed in without being attacked out of
- (64-n) empty squares. The probability that the next rook, when placed randomly
- would in fact be placed in one of those available squares is thus:
- P(n) = (8-n)^2/(64-n)
-
- So,
- P(8 rooks) = P(0)*P(1)*...*P(7)
-
- 8^2 7^2 6^2 1^2
- = --- * --- * --- ... ---
- 64 63 62 57
-
- (8!)^2
- = ------
- 64!/56!
-
- =~ 9.1E-6
-
- Or about 1 in a hundreed thousands.
-
- Dan Shoham
- shoham@ll.mit.edu
-