## Poll

16 votes (50%) | |||

12 votes (37.5%) | |||

5 votes (15.62%) | |||

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3 votes (9.37%) | |||

6 votes (18.75%) | |||

5 votes (15.62%) | |||

10 votes (31.25%) | |||

7 votes (21.87%) |

**32 members have voted**

A white rook and a black bishop are randomly placed on an empty chessboard.

What is the probability that one of the pieces is attacking the other?

Any chance of a problem for people who don't play chess?Quote:Gialmere

A white rook and a black bishop are randomly placed on an empty chessboard.

What is the probability that one of the pieces is attacking the other?

I get 1456/4032 = 36.111%

for each possible position of rook, it can attack 7 vertical + 7 horizontal = 14

for each possible position of rook, it can be attacked by bishop.....depending on location of rook

-outer "ring" of squares (8x8 border only = 28 squares) attacked by bishop from 7 positions

-next "ring" of squares (6x6 border only = 20 squares) attacked by bishop from 9 positions

-next "ring" of squares (4x4 border only = 12 squares) attacked by bishop from 11 positions

-inside block of squares (2x2 = 4 squares) attacked by bishop from 13 positions

[ (64*14)+(28*7+20*9+12*11+4*13) ] / [64*63] = 1456/4032

Quote:GialmereA white rook and a black bishop are randomly placed on an empty chessboard.

What is the probability that one of the pieces is attacking the other?

Select the bishop's space first

Each of the 28 spaces on the board edge has 7 spaces that can attack the rook

Each of the 20 spaces one space away from the board edge spaces has 9 spaces that can attack the rook

Each of the 12 spaces two spaces away has 11 spaces

The 4 spaces in the center each has 13 spaces

No matter where the bishop is placed, there are 14 spaces from where the rook can attack it, none of which are on spaces where the bishop can attack the rook.

There are 64 squares on which the bishop can be placed; for each one, there are 63 on which the rook can be placed.

The probability is (28 x (7 + 14) + 20 x (9 + 14) + 12 x (11 + 14) + 4 x (13 + 14)) / (64 x 63) = 13/36

Wherever the rook is, it can take the bishop on 7 squares on the same vertical column and 7 on the same row.

If the rook is in on the edge, i.e. A, then there are 7 places (on the diagonal) the bishop can be where it could take the rook.

B means 9 places, C means 11, D means 13.

There are 7 A's, 5 B's, 3 C's and 1 D. Thus the average number of squares where the bishop could be taken is

(7*21+5*23+3*21+1*23)/16 = 22.75.

So the chances of being taken is 22.75/63 = 91/252 (x4) = 13/36 (/7).

A A A A

A B B B

A B C C

A B C D

Quote:chevy

I get 1456/4032 = 36.111%

for each possible position of rook, it can attack 7 vertical + 7 horizontal = 14

for each possible position of rook, it can be attacked by bishop.....depending on location of rook

-outer "ring" of squares (8x8 border only = 28 squares) attacked by bishop from 7 positions

-next "ring" of squares (6x6 border only = 20 squares) attacked by bishop from 9 positions

-next "ring" of squares (4x4 border only = 12 squares) attacked by bishop from 11 positions

-inside block of squares (2x2 = 4 squares) attacked by bishop from 13 positions

[ (64*14)+(28*7+20*9+12*11+4*13) ] / [64*63] = 1456/4032

Quote:ThatDonGuy

Select the bishop's space first

Each of the 28 spaces on the board edge has 7 spaces that can attack the rook

Each of the 20 spaces one space away from the board edge spaces has 9 spaces that can attack the rook

Each of the 12 spaces two spaces away has 11 spaces

The 4 spaces in the center each has 13 spaces

No matter where the bishop is placed, there are 14 spaces from where the rook can attack it, none of which are on spaces where the bishop can attack the rook.

There are 64 squares on which the bishop can be placed; for each one, there are 63 on which the rook can be placed.

The probability is (28 x (7 + 14) + 20 x (9 + 14) + 12 x (11 + 14) + 4 x (13 + 14)) / (64 x 63) = 13/36

Quote:charliepatrickAssume for simplicity the Rook is placed in one of the quarters, say NW. Then the other quarters would create identical chances.

Wherever the rook is, it can take the bishop on 7 squares on the same vertical column and 7 on the same row.

If the rook is in on the edge, i.e. A, then there are 7 places (on the diagonal) the bishop can be where it could take the rook.

B means 9 places, C means 11, D means 13.

There are 7 A's, 5 B's, 3 C's and 1 D. Thus the average number of squares where the bishop could be taken is

(7*21+5*23+3*21+1*23)/16 = 22.75.

So the chances of being taken is 22.75/63 = 91/252 (x4) = 13/36 (/7).A A A A

A B B B

A B C C

A B C D

Correct!

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Quote:Ace2Any chance of a problem for people who don't play chess?

Yes. I have been chess heavy lately but, I'm pretty much out of puzzles. I have a dice one on tap for tomorrow though.

A single die is rolled until a run of six different faces appears. For example, one might roll the sequence

535463261536435344151612534 with only the last six rolls all distinct.

What is the expected number of rolls?

My answer is...Quote:Gialmere

A single die is rolled until a run of six different faces appears. For example, one might roll the sequence

535463261536435344151612534 with only the last six rolls all distinct.

What is the expected number of rolls?

Quote:EdCollinsMy answer is...

64.8 rolls

Incorrect, but you're on the right track.

Quote:GialmereA single die is rolled until a run of six different faces appears. For example, one might roll the sequence

535463261536435344151612534 with only the last six rolls all distinct.

What is the expected number of rolls?

I'm not entirely sure, as I am having a little problem simulating it to confirm the answer...

Let E(N) be the expected number of rolls needed if the N most recent are all different

Suppose the last 5 rolls are 1, 2, 3, 4, 5, in order.

If the next roll is 1, then you have 2, 3, 4, 5, 1; if it is 2, you have 3, 4, 5, 2; if it is 3, you have 4, 5, 3; if it is 4, you have 5, 4; if it is 5, you have 5; if it is 6, you are done.

E(5) = 1 + 1/6 E(1) + 1/6 E(2)+ 1/6 E(3) + 1/6 E(4) + 1/6 E(5) + 1/6 x 0

Similarly, if the last 4 are 1, 2, 3, 4, then a 5 or 6 results in a "run" of 5:

E(4) = 1 + 1/6 E(1) + 1/6 E(2)+ 1/6 E(3) + 1/6 E(4) + 1/3 E(5)

E(3) = 1 + 1/6 E(1) + 1/6 E(2)+ 1/6 E(3) + 1/2 E(4)

E(2) = 1 + 1/6 E(1) + 1/6 E(2)+ 2/3 E(3)

E(1) = 1 + 1/6 E(1) + 5/6 E(2)

E(0) = 1 + E(1)

The first five can be written as five equations in five unknowns:

1/6 E(1) + 1/6 E(2) + 1/6 E(3) + 1/6 E(4) - 5/6 E(5) = -1

1/6 E(1) + 1/6 E(2) + 1/6 E(3) - 5/6 E(4) + 1/3 E(5) = -1

1/6 E(1) + 1/6 E(2) - 5/6 E(3) + 1/2 E(4) = -1

1/6 E(1) - 5/6 E(2) + 2/3 E(3) = -1

-5/6 E(1) + 5/6 E(2) = -1

Solving results in E(1) = 52.8, so the expected number = E(0) = E(1) + 1 = 53.8