PR 4 # 13 - Page 2

Fusion_power · 2006-06-14, 21:19

Wacky,

Does your book have a solution to this? If so, would you please post it. I'm interested in seeing how it was derived.

Fusion

Wacky · 2006-06-14, 22:35

Quoting from the "Answers" portion of PR 4:

If the castle is on one of the 4 center squares, it threatens 14 squares and is threatened diagonally by 13 squares for a total of 27. This total is 25 for the squares bordering the center, 23 for the squares bordering these, and 21 for the outer border. The probability is therefore 4/64* 27/63 + 12/64 * 25/63 + 20/64 * 23/63 + 28/64 * 21/63 = 13/36.

philmoore · 2006-06-16, 22:44

Here is an alternative solution: No matter where the rook sits, it threatens exactly 14 squares. Since the bishop is on one of the other 63 squares, the probability is 14 out of 63, or 2 out of 9, that the rook threatens the bishop. On the other hand, the bishop threatens 7 other squares if it is in one of the 28 squares on the border of the chessboard, 9 other squares if it is in the next ring of 20 squares, 11 squares if it is in the next ring of 12 squares, and it threatens 13 squares if it is in one of the 4 central squares. So the probability that a randomly placed bishop threatens a randomly placed rook is: 28/64 * 7/63 + 20/64 * 9/63 + 12/64 * 11/63 + 4/64 * 13/63 = 5/36. Because the events "rook threatens bishop" and "bishop threatens rook" are independent, we can add the probabilities: 2/9 + 5/36 = 13/36. Moral: rooks are more powerful than bishops, but a bishop can partially compensate by placement near the center of the board.

Wacky · 2006-06-17, 09:51

Quote:

Originally Posted by philmoore

Because the events "rook threatens bishop" and "bishop threatens rook" are independent…

But, I do not agree that they are independent. In fact, they are mutually exclusive. If the rook threatens the bishop, then the bishop does not threaten the rook, and visa-versa.

philmoore · 2006-06-17, 21:17

Thanks, mutually exclusive is what I meant. So pr(A or B) = pr(A) + pr(B).

2006-06-14, 22:35	#13
Wacky Jun 2003 The Texas Hill Country 3²×11² Posts	"Official Solution" Quoting from the "Answers" portion of PR 4: If the castle is on one of the 4 center squares, it threatens 14 squares and is threatened diagonally by 13 squares for a total of 27. This total is 25 for the squares bordering the center, 23 for the squares bordering these, and 21 for the outer border. The probability is therefore 4/64* 27/63 + 12/64 * 25/63 + 20/64 * 23/63 + 28/64 * 21/63 = 13/36.

2006-06-14, 21:19	#12
Fusion_power Aug 2003 Snicker, AL 7×137 Posts	Wacky, Does your book have a solution to this? If so, would you please post it. I'm interested in seeing how it was derived. Fusion

2006-06-16, 22:44	#14
philmoore "Phil" Sep 2002 Tracktown, U.S.A. 3·373 Posts	Here is an alternative solution: No matter where the rook sits, it threatens exactly 14 squares. Since the bishop is on one of the other 63 squares, the probability is 14 out of 63, or 2 out of 9, that the rook threatens the bishop. On the other hand, the bishop threatens 7 other squares if it is in one of the 28 squares on the border of the chessboard, 9 other squares if it is in the next ring of 20 squares, 11 squares if it is in the next ring of 12 squares, and it threatens 13 squares if it is in one of the 4 central squares. So the probability that a randomly placed bishop threatens a randomly placed rook is: 28/64 * 7/63 + 20/64 * 9/63 + 12/64 * 11/63 + 4/64 * 13/63 = 5/36. Because the events "rook threatens bishop" and "bishop threatens rook" are independent, we can add the probabilities: 2/9 + 5/36 = 13/36. Moral: rooks are more powerful than bishops, but a bishop can partially compensate by placement near the center of the board.

2006-06-17, 21:17	#16
philmoore "Phil" Sep 2002 Tracktown, U.S.A. 3·373 Posts	Thanks, mutually exclusive is what I meant. So pr(A or B) = pr(A) + pr(B).