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32 members have voted

Gialmere Joined: Nov 26, 2018
• Posts: 2049
February 18th, 2021 at 7:27:44 AM permalink 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?
Have you tried 22 tonight? I said 22.
Ace2 Joined: Oct 2, 2017
• Posts: 1046
February 18th, 2021 at 8:15:12 AM permalink
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?

Any chance of a problem for people who don't play chess?
It�s all about making that GTA
chevy Joined: Apr 15, 2011
• Posts: 86
Thanks for this post from: February 18th, 2021 at 8:33:05 AM permalink

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

ThatDonGuy Joined: Jun 22, 2011
• Posts: 4925
Thanks for this post from: February 18th, 2021 at 9:00:42 AM permalink
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?

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

charliepatrick Joined: Jun 17, 2011
• Posts: 2424
Thanks for this post from: February 18th, 2021 at 9:22:27 AM permalink
Assume 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 AA B B BA B C CA B C D`
Gialmere Joined: Nov 26, 2018
• Posts: 2049
February 18th, 2021 at 5:59:02 PM permalink
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: charliepatrick

Assume 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 AA B B BA B C CA B C D`

Correct!
------------------------------ ---------------------------------

Quote: Ace2

Any 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.
Have you tried 22 tonight? I said 22.
Gialmere Joined: Nov 26, 2018
• Posts: 2049
February 19th, 2021 at 7:27:04 AM permalink 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?
Have you tried 22 tonight? I said 22.
EdCollins Joined: Oct 21, 2011
• Posts: 1306
February 19th, 2021 at 9:59:49 AM permalink
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?

64.8 rolls
Gialmere Joined: Nov 26, 2018
• Posts: 2049
February 19th, 2021 at 10:44:40 AM permalink
Quote: EdCollins

64.8 rolls

Incorrect, but you're on the right track.
Have you tried 22 tonight? I said 22.
ThatDonGuy Joined: Jun 22, 2011
• Posts: 4925
February 19th, 2021 at 11:24:51 AM permalink
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?

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