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Rook versus bishop - Puzzle of the week

2022-07-31T13:58:35.654Z

A bishop and a rook are randomly placed on a chessboard. What is the probability that one character can beat the other?



Photo:

THE MIRROR

The rook can only move vertically and horizontally, the bishop only diagonally.

These two figures are the focus of the new puzzle.

The two pieces are randomly placed on a chessboard that is eight by eight squares.

Each character lands on any square.

The only thing that is excluded is that they stand together on one field.

What is the probability that one of the two figures can beat the other?

The probability is

1456/4032

.

That's around

36 percent

.

There are a total of 64 x 63 = 4032 different ways to place the two pieces on the board.

Let's start with the simple case: when can the rook capture the bishop?

When we place the rook on the board, it threatens seven squares vertically and seven squares across.

Altogether 14 fields.

The runner is not allowed to stand there.

A total of 14 x 64 = 896 different positions are possible in which the bishop is within striking distance of the rook.

A little more complicated is the question of when the bishop can capture the rook.

Here we have to distinguish four cases.

  • The bishop stands on one of the 28 squares on the edge of the board (top left in the drawing).

    Then he always threatens seven squares.

    That is 28 x 7= 196 positions.

  • The runner stands further in on the ring (top right).

    There are 20 fields there.

    He threatens nine squares - that's 180 positions.

  • The bishop stands on the ring with side length four (bottom left).

    This consists of 12 fields.

    With 11 threatened squares, that's 11 x 12 = 132 positions.

  • The bishop is on one of the middle four squares.

    In this way, 13 fields are threatened at a time.

    We get 4 x 13= 52 positions.

  • There are a total of 196 + 180 + 132 + 52 = 560 positions where the bishop threatens the rook.

    Rook and bishop cannot threaten each other at the same time.

    Therefore we can add the two cases, rook threatening bishop and bishop threatening rook.

    So there are 896 + 560 = 1456 positions where one piece can beat the other.

    The probability sought is therefore 1456/4032.

    I discovered this pretty chess puzzle on the math portal Mathigon.

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    Source: spiegel

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