Interesting one... Two squares are chosen at random on

Yep. This questionis unrealastic. I'm not a chest player, hence i wouldn't know how many squares the board has. Unless you're given this information, you could easily dispute this question to GMAC.

I got an answer of 7/128 (which is not in answer choices )
My logic is:
THere are three types of squares:
1. 4 Corners - These have exactly two other squares with a side common.
2. Along borders exluding corners - These have exactly three squares with a side common - 24 squares
3. All the other squares - These have exactly four squares with a side common - 36 squares.
Now, I tried to write the equation for all these cases. After simplifying, I get 7/128.
I know that answer is wrong. Can you guys explain why?

Here we go,

1) There total 64 squares on a chess board, arranged in 8*8 form
2) There are 4 corner squares where only 2 other squares share a side so total number of ways for such a square = (4*2)
3) There are 24 squares touching the boarder of the chess board (excluding 4 corner squares) where 3 other squares share a side.
=> (24*3)
4) There are remaining 36 squares where 4 other squares share a side => (36*4)

So, total number of ways in which a square can be selected with one side shared is = (4*2) + (24*3) + (36 * 4) = 224

& total number of ways in which a square can be selected = 64*63

So, probability of selecting one square such that one of it sides is in common = 224/(64*63) = 1/18

Thanks for the explanation Vivek. Atleast my approach is right.

Yeah :-D Your approach is correct! :good