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:

Only 2 of these take me to the top left-hand corner of the grid, so the probability of getting to the opposite corner is: $$ {{1 \over{2^2}}\times2} = {2 \over4} = {1 \over2} $$

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: $$ {{1 \over{2^4}}\times6} = {6 \over16} = {3 \over8} $$

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: $$ {{1 \over{2^6}}\times20} = {20 \over64} = {5 \over16} $$

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: $$ {[2(n-1)]!} \over{[(n-1)!]^2} $$

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

I verified my results and they worked for n = 2, 3 and 4.