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Topic: Rook & Bishop (Read 554 times) |
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ThudnBlunder
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Rook & Bishop
« on: Apr 15th, 2003, 6:17pm » |
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A rook and bishop are placed on two randomly chosen squares of a chessboard.
What is the probability that either one of them will attack the other?
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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cho
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I think, A rook attacks 14 squares. A bishop attacks an average of 8.75 squares. The attacking squares are always mutually exclusive, so for any rook position the odds of an attack are 22.75/64.
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Icarus
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Re: Rook & Bishop
« Reply #2 on: Apr 15th, 2003, 7:06pm » |
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By essentially brute force, I come up with 13/36.
Here is how:
Consider the first piece to be placed in the lower right quadrant. There are 7 positions in this quadrant adjacent to an edge. Each such position attacks/ or is attacked by the 7 other spaces on the same row, the 7 other spaces on the same column, and 7 diagonally adjacent spaces. Total is 21.
Of the squares one in from the edge, there are 5 such squares in the quadrant, the number of attack positions are 7 for rows, 7 for columns, and 9 for diagonals. Total is 23.
Two in from the edge gives 3 squares, and 7, 7, and 11 attack positions. Total is 25.
Three in from the edge has only 1 square, and 7, 7, and 13 attack positions. Total is 27.
Since there are 4 quadrants, the number of paired positions with one attacking the other is:
4 x(7x21 + 5x23 + 3x25 + 1x27) = 4x364=16x7x13
Since there are a total 64x63=64x9x7 pairs of positions, the probability of attacking positions is
16x7x13 / 64x9x7 = 13/4x9 = 13/36.
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redPEPPER
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Re: Rook & Bishop
« Reply #3 on: Apr 16th, 2003, 5:01am » |
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I came up with the same answer as Icarus, and with basically the same reasoning. I wouldn't call it brute force though. You enumerate the different group of occurences that have different properties and ponderate them correctly regarding how often they occur. I don't see how you could end up with the correct result without doing that.
Now that I look at it, Cho is doing the same, except that he hides all the enumeration and ponderation under the bishop's average. He is also correct. He doesn't find the same answer because he divides by 64 at the end. But for any rook position the bishop has only 63 other squares to go, not 64. Aside from that, 22.75/63 = 13/36 = 0.3611...
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ThudnBlunder
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Re: Rook & Bishop
« Reply #4 on: Apr 18th, 2003, 6:17am » |
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on Apr 16th, 2003, 5:01am, redPEPPER wrote:
Now that I look at it, Cho is doing the same, except that he hides all the enumeration and ponderation under the bishop's average. He is also correct. |
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And cho's answer is the more elegant. Using the fact that a rook and bishop can never simultaneously attack each other (and thus the same squares are never counted twice), he arrives at the correct answer in just two lines (although he did have to calculate the average for the bishop), except for his little slip at the end.
My solution was somewhere in-between. I reasoned that the required probability was the same as the probability of a queen attacking another square, and then I worked out the numbers for one quadrant.
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| « Last Edit: Apr 18th, 2003, 8:41am by ThudnBlunder » |
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