Thank your lucky stars

Thank you for this solution Andrei (Andrei Lazanu, School 205 Bucharest) and for the link to the useful site:

For a 2x2 grid I need to make 2 moves.

There are 4 possible paths:

For a 3x3 grid I need to make 4 moves.

There are 16 possible paths.

Only 6 of these take me to the top left-hand corner of the grid, so the probability of getting to the opposite corner is: $$\frac{1}{2^4}\times 6=\frac{6}{16}=\frac{3}{8}$$

For a 4x4 grid I need to make 6 moves.

There are 64 possible paths.

Only 20 of these take me to the top left-hand corner of the grid, so the probability of getting to the opposite corner is: $$\frac{1}{2^6}\times 20=\frac{20}{64}=\frac{5}{16}$$

I found on the Internet, at the Math Forum, the formula together with the explanation.

The address is: http://mathforum.org/library/drmath/view/54218.html

The formula generating the number of ways to go from one corner to another is: $$\frac{[2(n-1)]!}{[(n-1)!{]}^2}$$

The formula generating the probability of landing in the opposite corner in a n x n grid is: $$\frac{1}{2^{2(n-1)}}\times \frac{[2(n-1)]!}{[(n-1)!{]}^2}.$$