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:

\frac{1}{2^{2}} \times 2 = \frac{2}{4} = \frac{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:

\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}} .

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