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:

left, left
left, up
up, left
up, up


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
22
 ×2 = 2
4
 = 1
2

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
24
 ×6 = 6
16
 = 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:


1
26
 ×20 = 20
64
 = 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:


[2(n-1)]!
[(n-1)!]2

The formula generating the probability of landing in the opposite corner in a n x n grid is:


1
22(n-1)
 × [2(n-1)]!
[(n-1)!]2
 .
I verified my results and they worked for n = 2, 3 and 4.