King and King

Probability Level 4

Dimitri places a Black King and a White King on an empty chessboard. If the probability that Dimitri places both Kings on the chessboard such that neither of the Kings is in check (that is, the White King is not adjacent to the Black King), can be expressed as $\dfrac{m}{n}$ , in which $m$ and $n$ are coprime positive integers, find $m+n$ .

As an explicit example, if a King is on $\text f2$ , the $\text f1$ , $\text g1$ , $\text e1$ , $\text f3$ , $\text g2$ , $\text e2$ , $\text e3$ and $\text g3$ squares are under attack and these are the adjacent squares to $\text f2$ . Note that if the other King is placed on any of the squares, the Kings are attacking each other, so both are in check, which is impossible.

The White King and the Black King cannot be placed in the same square.

This is the sixth problem of the set Look after the King!

The answer is 91.

2 solutions

Sam Bealing
Apr 17, 2016

This problems boils down to case consideration. WLOG we consider placing the white king first. There are three cases :

Case	Squares for white king	Squares for black king without check	Probability
Corners	$4$	$63-3=60$	$\frac{60 \times 4}{63 \times 64}$
Sides	$6 \times 4=24$	$63-5=58$	$\frac{58 \times 24}{63 \times 64}$
Middle	$6 \times 6=36$	$63-8=55$	$\frac{55 \times 36}{63 \times 64}$

The sum of these is:

$\frac{240}{4032}+\frac{1392}{4032}+\frac{1980}{4032}=\frac{3612}{4032}=\frac{43}{48}$

$n-m=48-43=5$

Moderator note:

Simple standard approach using the rule of sum on different cases.

We could simplify the cases further by counting the number of attacking pairs according to the direction. If the squares were side by side, then there are $7 \times 8 = 56$ possibilities for each of the 4 direction. If the squares were diagonally across, then there are $7 \times 7 = 49$ possibilities for each of the 4 directions.

Well done. I think this is the easiest problem of the set.

Mateo Matijasevick - 5 years, 1 month ago

K T
Nov 15, 2018

If the first king is at a corner: (4/64), the other king is okay in (60/63) of the cases

If the first king is at the edge, but not a corner: (24/64), the other king is okay in (58/63) of the cases

otherwise(36/64) the other king is okay in (55/63) of the cases

Total: (4×60+24×58+36×55)÷(63×64)=43÷48, 43+48=91

King and King

This is the sixth problem of the set Look after the King!

The answer is 91.

2 solutions

Moderator note:

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