Rook & Bishop

If a rook and a bishop are placed on two randomly chosen squares of the board, what is the probability that either one of them will attack the other?

Originally posted by ThudanBlunder
If a rook and a bishop are placed on two randomly chosen squares of the board, what is the probability that either one of them will attack the other?

about 42.2% EDIT: never mind that doesn't work this one is tricky

Originally posted by ThudanBlunder
If a rook and a bishop are placed on two randomly chosen squares of the board, what is the probability that either one of them will attack the other?

Isn't it the probability that two queens attack each other? That can't be too hard to figure out...

...Edit: if you count placing them on the same square as "attacking each other" then it's

(64 + 64*14 + B)/(64*64)

where the B is the total number of squares a bishop can attack by placing it on all 64 squares of the chessboard. Starting from the middle four squares in the centre, and working outwards in concentric squares, we see

B = 4*13 + 12*11 + 20*9 +28*7 = 560

so the probability is 95/256. If you're not counting them being on the same square then take off the first 64 above to yield 91/256.

Originally posted by SPMars
Isn't it the probability that two queens attack each other? That can't be too hard to figure out...

...Edit: if you count placing them on the same square as "attacking each other" .

Right idea, but no need for two queens.

Originally posted by ThudanBlunder
Right idea, but no need for two queens.

Yeah, it's just the way my thought process was going at the time...the problem looked simpler that way.

Rook attacking bishop: 14/63 (on any square where the rook is)
Bishop attacking rook: (28 *7+20*9+12*11+4*13)/(64*63)

Both are mutually exclusive. Sum: 36.11% (my guess).

Originally posted by Mephisto2
Both are mutually exclusive. Sum: 36.11% (my guess).

🙂