PR 4 # 13
 

A castle and a bishop are placed at random on different squares of a chessboard. What is the probability that one piece threatens the other?

[spoiler]13/36[/spoiler] ?

Chris

[spoiler]259/263? No, it's too high...[/spoiler]

[spoiler]Zero, castles aren't chess pieces, rooks are.[/spoiler]:razz:

Castles/rooks
 

[QUOTE=Uncwilly][spoiler]Zero, castles aren't chess pieces, rooks are.[/spoiler]:razz:[/QUOTE]
:popcorn:
In India both terms are used. So on with the game!
Mally :coffee:
P.S. Killing time for the Brazil/Croatia match to kick off :grin:

[spoiler]91/256[/spoiler]

The probability that the rook threatens the bishop is 14/64. This is true no matter where you place the rook or the bishop.

The probability that the bishop threatens the rook depends on position. If the rook is in a corner, the bishop can only threaten from 7 positions. If the rook is on a center square, there are 13 possible positions for the bishop to threaten the rook. I'll leave it to others to figure out what the intervening numbers are and how to treat the set of probabilities.

I make it 24/64 probability that one piece threatens the other.

[QUOTE=Fusion_power]The probability that the rook threatens the bishop is 14/64. This is true no matter where you place the rook or the bishop.
[/QUOTE]
Are you sure? I make it 14/63

[QUOTE]Are you sure? I make it 14/63[/QUOTE]Oops, I forgot that the two pieces are not allowed to occupy the same square. I change my answer to [spoiler]91/252[/spoiler]

You are correct, I should have reduced to 63 since both pieces cannot occupy the same position. Revise the final answer to 24/63.

Some good tries. But I think that Chris got it right on the first shot.

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

"Official Solution"
 

Quoting from the "Answers" portion of PR 4:

[spoiler] 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. [/spoiler]

[spoiler] 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.[/spoiler]

[QUOTE=philmoore]Because the events "rook threatens bishop" and "bishop threatens rook" are independent…[/QUOTE]

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.

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