# A Bishop And A Rook Puzzle

A white rook and a black bishop of a standard chess set are randomly placed on a chessboard

**What is the probability that one is attacking the other?**

** ANSWER** – 0.3611

__EXPLANATION-__

The probability of either the Rook or the Bishop attacking the other is 0.3611

A Rook and a Bishop on a standard chess board can be arranged in:

= 64P2 (64 Permutation 2)

= 64*63

= 4032 ways

Now,

There are 2 possible cases – __Rook attacking Bishop__ and __Bishop attacking Rook__.

As we know that the Rook and the Bishop never attack each other simultaneously. Let’s consider both the cases one by one.

Rook can be placed in any of the given 64 positions and it always attacks 14 positions.

Case I – Rook attacking Bishop

Hence, total possible ways of the Rook attacking the Bishop

= 64*14

=

*896 ways*

__ Case II – Bishop attacking Rook__View the chess-board as a 4 co-centric hollow squares with the outermost square with side 8 units and the innermost square with side 2 units.

If the bishop is in one of the outer 28 squares, then it can attack 7 positions.

If the bishop is in one of the 20 squares at next inner-level, then it can attack 9 positions.

If the bishop is in one of the 12 squares at next inner-level, then it can attack 11 positions.

If the bishop is in one of the 4 squares at next inner-level (the innermost level), then it can attack 13 positions.

Hence, total possible ways of the Bishop attacking the Rook

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

= 560 ways

Thus, the required probability is

= (896 + 560) / 4032

= 13/36

= **0.3611**

### Your Answer

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