Rooks Under Siege

First take a look at the chessboard below:

Alt text

As you can see, if you place a rook on any square, it will be able to attack $14$ other squares.

So place $\text{rook}_1$ on the chessboard. That leaves $63$ squares for you to place $\text{rook}_2$ . Out of those $63$ squares, $\text{rook}_1$ can attack $14$ squares. So, if $\text{rook}_2$ is on one of these $14$ squares, $\text{rook}_1$ can attack it.

So the probability that $\text{rook}_1$ can attack $\text{rook}_2$ is $\frac{14}{63}=\frac{2}{9}$ .

And $a+b=2+9=\boxed{11}$ .

No matter where the rook is on the board, it will be attacking 14 other squares out of the 63 squares. Since the no. of squares the rook attacks at any one square is the same, the probability that another rook will be in it's line of attack is: 14(squares the rook attacks)/63(no. of squares the second rook can be on) =2/9 Thus 2+9=11

Let the first rook be placed anywhere on the board. Regardless of where this rook is placed, there will be 7 places in the same row and 7 places in the same column to place the second rook. There are 63 total places to put the second rook. The probability is thus 14/63=2/9 and the answer is 11.

Consider the placement of the first rook. Wherever it lands, it attacks $7$ squares in the vertical direction (the whole column but the square it is on) and $7$ squares in the horizontal direction. The second rook can then land on $8^2 - 1 = 63$ squares, $14$ of which it will be attacked by the first rook. Therefore, our probability is $\frac{14}{63} = \frac{2}{9}$ and our desired sum is $2+9 = \boxed{11}$

$2$ rooks will attack each other when they are in same row or same column. There are ${8 \choose 2} = 28$ ways to place $2$ rooks a row or a column, there are $8$ rows and $8$ column, thus there are total of $16 \cdot {8 \choose 2} = 448$ ways to place $2$ rooks on a chess board such that they will attack each other. There are total of ${64 \choose 2} = 2016$ ways to place $2$ rooks on a chess board. Hence, the probability that $2$ rooks will be able to attack each other is $\frac{448}{2016} = \frac{2}{9}, a=2, b=9, a+b=11$ .

Let N be the total possible combination of moves of the 2 rooks and n be the total possible combination for an attack.

Since there are 64 squares where a rook can be placed and for each placement its opponent can be placed on the remaining 63 squares, then N is 64x63.

Since for every placement of a rook there are 7 squares on the column and 7 squares on the row where its opponent can be attacked, then n is 64x(7+7) or 64x14. Our working equation is n/N.

(64x14)/(64x63) = 2/9

Here the value of a is 2 and b 9. Therefore, a+b = 2+9 = 11. 11 is the answer to the challenge.

The total number of ways to put 2 rooks randomly on a chessboard is $\binom{64}{2}$ .

Now, 2 rooks can attack each other as long as they are the same row or column. Hence there are a total of $(8+8)\binom{8}{2}$ to do so.

Hence the desired probability is $\frac{(8+8)\binom{8}{2}}{\binom{64}{2}} = \frac{2}{9}$ .

Therefore $a + b = 2 + 9 = 11$

we have 64.63 = 4032 ways to place 2 rooks on 2 different squares of the chessboard and 64.14=896 ways to place 2 rooks on 2 different squares such that they are able to attack each other so P = 896/4032=2/9 2+9=11

A rook anywhere in the chessboard can be attacked by another rook in 14 ways. Consider that an union of a file and a rank requires 15 squares with the intersection of the file and rank is the location of the rook. Therefore, the other rook can be placed on the remaining 14 squares to attack the rook. Since the remaining squares unoccupied equals 63, the probability that the rook can attacked the other rook is 14/63 or 2/9. a=2 and b=9, hence a+b=11.

There are a total of 64C2=2016 ways to place the two rooks (either they are attacking each other or not). When the two rooks are attacking each other, they must be in the same row or the same column. For each row (or column), there are 8C2=28 ways to place two rooks. Since there are 8 rows and 8 columns, there are a total of 8+8=16 rows (or columns) to consider. There are 28x16=448 ways where two rooks will attack each other. Therefore, the probability is 448/2016=2/9. a=2, b=9. a+b=11

Each row and each column of the chessboard has 8 squares.

However, the rook is already on one square. It could then attack 14 squares because it could attack 7 squares on its same row and 7 squares on its same column.

A rook can move only forward or to its sides and it occupies one square. Hence the other rook can be placed in 14 other possible positions out of the
remaining 63 squares.

Probability to attack = \frac {14} {63} = \frac {2} {9}

then a = 2 and b = 9 and a + b = 11

For second rook to attack the first it has to be in same row or column of the first one.

Thus 7 (row places) +7 (column places) = 14 places are suitable for other rook to be placed so that the two rooks attack each other.

Total places at which second rook can be placed is 63.

Hence P(second rook attacks first) = (14/63) = (2/9)

ANS: 2+9=11

Note that a rook always attack $14$ squares, no matter where it's placed. (excluding the square it stands on.)

Let the first rook be placed anywhere on the board. Then the second rook have $64 - 1 = 63$ choices of squares, with $14$ of them being attacked by the first rook.

Thus, the desired probability is $\frac{2}{9}$ , so the answer is $2 + 9 = \boxed{11}$ .

We'll take on an easy generalisation to an $n$ by $n$ chessboard.

WLOG randomly place one of the rooks first. Then it is clearly covering one square and can attack the $n-1$ other squares in its row and the other $n-1$ other squares in its column, so it can attack $(n-1)+(n-1)=2(n-1)$ squares.

Therefore, to be attacked, the other rook has $\frac{2(n-1)}{n^2-1}=\frac{2(n-1)}{(n+1)(n-2)}=\frac{2}{n+1}$ choices of squares, and this is the desired probability.

Plugging in $n=8$ : $\frac{2}{n+1}=\frac{2}{9}$ $\therefore a+b=2+9=\boxed{11}$

It does not matter where the first rook is placed on the board. No matter where it is, it will always be able to attack $7$ other squares in its row and $7$ squares in its column, for a total of $14$ .

The second rook can be placed on $63$ spaces on the board because it has to be placed on a square other than that of the first rook. The probability that it can be attacked is $\frac{14}{63}=\frac{2}{9}$ . $A=2$ and $B=9$ , so $A+B=\boxed{11}$

One liner: Let the first rook be in any location, there are 14 squares that result in rooks attacking each other and 63 tiles in total. 14/63=2/9, 2+9=11

To find the solution, you find the amount of possibilities there are to place the two rooks. That is 64 choose 2, or (64 times 63)/(2 times 1) or 2016. Then, find the amount of possibilities satisfying the condition for each row, which is 8 (per row) choose 2 which is 28. There are 8 rows and 8 columns, so you multiply the 28 by 16 and get 448. After simplifying 448/2016, you get 2/9, which means a+b is 11.

If we place one of the rooks randomly on any of the squares, then there are $14$ places available, out of the remaining $63$ for the next rook to be placed in so that the two are along the same line, such that they are able to attack each other. Required probability: $64\cdot(\frac{1}{64}\cdot\frac{14}{63})=\frac{2}{9}$ Hence, $a+b=2+9=11$ .

Note that it doesn't matter where we put the first rook, because it always covers precisely one row and one column. Therefore the probability is the same as that of placing just the second rook while assuming the first rook has already been placed somewhere, let's say the top left corner.

In this case, there are $7$ free places in the top row and $7$ free places in left column giving a total of $14$ where the rooks are in mutual attack. And there are $63$ places where to put the second rook altogether (one less than $64$ since the corner is already taken by the first rook). So the probability is ${14 \over 63} = {2 \over 9}$ , giving an answer of $\bf 11$ .

We shall find the probability that the Rooks cannot attack, then subtract that value from 1.

Note that no matter where the first Rook is placed, there is always $49$ squares where the second Rook cannot attack it. From the remaining $63$ squares, the second Rook has a $\dfrac{49}{63}=\dfrac{7}{9}$ chance of landing on one of those squares. Therefore, the probability that the second Rook lands on a square in which the first rook can attack is $1-\dfrac{7}{9}=\dfrac{2}{9}$ and our answer is $\boxed{11}$ .

By drawing a 8 by 8 chessboard and choosing any point along the diagonal, one will observe that the total number of squares vertically and horizontally to that point is 14.

Since the total available number of squares to put the second rook is 64-1=63, the probability of the 2 rooks being able to attack each other is 14/63 which when expressed as coprime form is 2/9. Hence 2+9=11. The answer is 11.

1. Draw a $4 \times 4$ board, and place one rook anywhere.

2. Notice that regardless of where you place the rook, it can attack three squares in horizontal direction, and 3 squares in the vertical direction. These "attackable" squares are where the second rook should go to fulfill the conditions of the problem.

3. Extrapolate from this fact that given a board of size $N \times N$ , a rook with random placement can attack $(N - 1) \times 2$ squares. Also, since the first rook takes up one space, the number of remaining possible locations for the second rook is $(N \times N) - 1$

4. Set up a ratio: $\frac { ((N - 1) \times 2) }{(N \times N) - 1}$ , and simplify to get $a$ and $b$ .

5. Add $a$ and $b$ to get the answer.

Where ever the first rook is placed, in order to attack the other rook, the other rook must be in the same column or row... Without the place of the first rook, there are $(64-1)=63$ places where the second rook can be... And there are total $(7+7)=14$ places in the same column and row of the first rook... Hence the probability is...

$\dfrac{14}{63} = \dfrac{2}{9}$

$a+b=2+9=\fbox{11}$

Pick any square on the chessboard for the first rook to be placed on. For the two rooks to be able to attack each other, the second rook must be placed either in the same column or the same row as the first, and since they have to be placed on different squares, there are seven choices for each, for a total of $7+7=14$ choices that work for the second rook. The total number of choices for the second rook is equal to $64-1 = 63$ , since the two rooks cannot occupy the same square, so our probability is equal to $\frac{14}{63} = \frac{2}{9} \implies 2+9 = \boxed{11}$

Here's the thing.

First place the first rook anywhere on the board (not outside the board). There are 64 ways to do this.

Now place the second rock on the board which can be done in 63 ways.

But if you want to them to attach each other, place it in the same row or column in which first rook has been placed which can be done in 7 + 7 ways. 7 ways are available on the same row and 7 ways are available on the same column on which first rook has been placed. (You cannot place both the rooks on the same place. You actually can but you can't.)

All done. Just do the following thing now.

Total no of events = $64 \times 63$

Favorable Events = $64 \times 14$

Probability = $\frac{Fav \ events}{Tot\ events} = \frac{2}{9}$

9 + 2 = 11 ANSWER

My take on this requires only basic calculation skills and ability to imagine. The probability that the 2 rooks will be able to attack each other depends only on the probability where the second rook is placed.

Imagine an $8 \times 8$ chessboard, and place the first rook anywhere on one of the squares. Now, to place the second rook, we have 63 squares to choose from. In order for the second rook to be able to attack the first rook (and vice versa), the second rook must be placed on a square that is along the same row or column as the first rook. There are 14 other squares that can satisfy this condition; 7 squares along the same row as the first rook, 7 squares along the same column as the first rook. You can observe that this will be true no matter which square the first rook was placed.

Therefore, the probability for the two rooks to be able to attack each other is simply $\frac{14}{63}$ , which can be simplified as $\frac{2}{9}$ . 2 and 9 happen to be coprime numbers and thus the answer is simply 2+9=11

• After placing the first rook, there are 63 places for the second rook.
• Out of these, 7 are on the same row as the first rook, and 7 are on the same column.
• So the answer is $\frac{7 + 7}{63} = \boxed{\frac{2}{9}}$ .

There are 64*63/2 = 2016 positions where we can place the rooks. Notice that we can place the first rook wherever we want. Then there are 14 choices to place the second rook- 7 across the row, 7 across the column. We then divide by two since we overcounted.

So our fraction is 64*7/2016= 2/9. Answer is 11.

for a rook to attack another rook they must be in the same row or column

for the first rook at i,j square, there are 7 places lifted in the row i and 7 places in column j so we have 14 events.

the total number of events are 8*8-1=63

probability=14/63=2/9

so a+b=2+9=11