Chess board probability

[Tanishq Nandan]

Homework Statement

2 squares are chosen at random from a chess board.What is the chance that these 2 squares will share exactly 1 corner?

Homework Equations

P=favourable possibilities/Total possibilities

The Attempt at a Solution

So,the total no of possibilities should be 64C2.
Now,for favourable...
For the 4 corners,only 1 square is possible,so 4 cases.
For one side(excluding the corners) there can be 2 squares for every square we choose.,i.e
6×2=12 cases in on side...48 cases for all sides.
Now,we are done with the borders and are just left with the middle ones.
For each middle one there can be 4 possible squares..i.e,
36×4=144 cases
Summing up,we get P=(4+48+144)/64C2
=196/64C2
Is this correct or have i missed (or added) something?

 

[Orodruin]

You are double counting.

 

[Tanishq Nandan]

Ooh yeah..I got it..so how do you suggest I find out the possible cases?
I see now that there are lots of repetitions,so it won't be possible to subtract them..i guess.
So,i think I need a new method.

 

[Orodruin]

Well, you can either count combinations (in which case you are double counting in the nominator) or you count ordered combinations (in which case you are under-counting the denominator by a factor of two).

 

[Tanishq Nandan]

I didn't get you..how exactly??
Say I take unordered pairs(I'll divide the entire thing by 2 in the end)
Now how do I find out how many pairs exist on a chessboard which have only one corner in common?

 

[Orodruin]

What do you mean? You already counted the ordered pairs in the first post. Your counting essentially was picking the first square of the ordered pair and checking the possibilities for placing the second.

 

[Tanishq Nandan]

Oh...so you are saying that just dividing the entire thing by 2 will give my answer?

 

[Orodruin]

Right.

 

[haruspex]

There is a slightly quicker way. Count the possible corners, then double.

 

[Tanishq Nandan]

There is a slightly quicker way. Count the possible corners, then double.

Umm...sorry??
Possible corners of which square?
The one which is itself at the corner,the one at the side or the one in the middle?

 

[Orodruin]

Umm...sorry??
Possible corners of which square?
The one which is itself at the corner,the one at the side or the one in the middle?

For every corner (i.e., where 4 squares meet) in the interior of the board, there are two possible (unordered) pairs, white-white and black-black. You are computing the total number of possible pairs of squares that are diagonally adjacent. Since there are 49 such crossings, the number of unordered diagonal pairs is 49*2 = 98. Divide with the total number of unordered pairs 64C2 and you get your probability.

 

[PeroK]

There is a slightly quicker way. Count the possible corners, then double.

There's also a slower way to do it. Work out the probability directly where the first square is a corner, edge or interior.

The good thing about the slow way is that you don't have to be so clever to get the answer.