Squares of chess

On a chess-board if two squares are chosen, what is the probability that they have side in common.


The answer is 0.055.

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3 solutions

Nishant Rai
May 21, 2015

Sample space will be total ways of selecting 2 squares (order doesn't matter) out of available 64 squares. So sample space = 64 C 2 ^{64} C_2

Now there are 7 unique adjacent square sets in each row and each column. i.e. favorable cases will be 7×(8 rows + 8 columns) = 112. Hence Required Probability= F a v o u r a b l e c a s e s S a m p l e S p a c e = 112 64 C 2 = 1 18 \huge \frac{Favourable~ cases}{Sample ~Space} = \frac{112}{^{64} C_2} = \frac{1}{18}


Also there could be other solution too if we consider ALL possible squares and not only the smallest squares. (i.e. if we consider that total squares are more than 64).

In that case let us first calculate the sample space i.e. total number of squares on a chess board.

1, 8x8 square

4, 7x7 squares

9, 6x6 squares

16, 5x5 squares

25, 4x4 squares

36, 3x3 squares

49, 2x2 squares

64, 1x1 squares

Therefore, there are actually: 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 8 2 + 7 2 + 6 2 + . . . 1 2 = 204 64+49+36+25+16+9+4+1= 8^2+7^2+6^2+...1^2=204 squares on a chessboard!

So sample space = 204 C 2 ^{204} C_2

Now if we assume that 2 squares can have a side common only if they are of the same dimensions:

Let the length of the smallest square be 1 unit.

So the possibility of selecting 2 squares with a common side (As calculated in the first method above):

having the side length greater than 4 unit length = 0

having the side length of 4 unit = 1×(2 rows + 2 columns) = 4

having the side length of 3 unit = 3×(6 rows + 6 columns) = 36

having the side length of 2 unit = 5×(7 rows + 7 columns) = 70

having the side length of 1 unit = 7×(8 rows + 8 columns) = 112

So total number of favorable cases: = 4 + 36 + 70 + 112 = 222 = 4+36+70+112=222

Hence Required Probability= F a v o u r a b l e c a s e s S a m p l e S p a c e = 222 204 C 2 \huge \frac{Favourable~ cases}{Sample~ Space} = \frac{222}{^{204} C_2}

Replied 0.55 got a wrong answer

Dhiman Bhattacharyya - 6 years ago
Shashank Gupta
May 24, 2015

Consider all possible cases There are 64C2 ways to select any 2 squares... That is 2016 ways Now looking for favourable cases... Instead of looking for the blocks we count the edges... Every edge signifies the two adjacent blocks connected to it. There are 7 vertical and 7 horizontal lines... Each decided into 8 parts(edges). So we count and get (7x2)x8

So the probability is 112/2016 = 0.0555555.....

Tanishq Varshney
May 21, 2015

Any one with the solution

Adarsh Kumar @Raghav Vaidyanathan

Tanishq Varshney - 6 years ago

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Total number of ways to select two squares is 64 C 2 ^{64}C_2 .

Each side in the chessboard corresponds to a unique pair of squares which have that side common. So just count the number of sides to get the required number of favorable cases. This comes out to be 112 112 .

Raghav Vaidyanathan - 6 years ago

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Next question is: if three squares are chosen at random, what is the probability that there are two (pairwise) "shared" sides? Or in general, if n n squares are chosen, what is the probability that there are n 1 n - 1 (pairwise) shared sides?

And the NEXT question is: why do I have to keep on thinking of what "the next question" should be? :)

Brian Charlesworth - 6 years ago

Exactly. : ) :)

Keshav Tiwari - 6 years ago

@Raghav Vaidyanathan

You chose the squares of dimensions ( 1 1 ) (1*1) , what if someone chooses squares of ( 2 2 ) (2*2) , i mean they can also have a side in common .

Nishant Rai - 6 years ago

I got a generalization : ( n 1 ) × ( 2 n ) n 2 C 2 \dfrac{(n-1) \times (2n)}{^{n^2}C_{2}} , where n n denotes the no of rows/column of a n-sided chessboard.

Vishwak Srinivasan - 6 years ago

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