Please see the attached simple finite game

http://www.oddschecker.com/index.html?archive/oddsgame.php

The competition has closed but the game is still there for fun. Can
anyone provide me with the solution which appears trivial. I have won
just once but cannot repeat my success and it is driving me mad. I am
sure the game theorists amongst you can solve this in seconds.

Here is one way to win:

Write the number of horses in each row in binary.  The starting
configuration (7 8 9 10 11) looks like this:

0111
1000
1001
1010
1011

What you have to do is keep the total number of ones in each column
even.  Right now the numbers of ones are 4, 1, 3, 3.  You can make
them all even by taking all seven horses in the top row:

0000
1000
1001
1010
1011

Now whatever the computer does will not leave all columns with even
number of ones, and you can always make all the columns even again,
and so you win.  In order to save time writing things in binary, as
soon as you can make the four remaining rows into two matching pairs,
it's easy to win.  For example, if you can leave 6, 8, 8, 6 then
whatever the compter does, just copy it, to leave two matching pairs
again.

Web pages which explain it:

http://www.handsonlearning.org/games/games_info.html
http://www.geocities.com/chessdp/biospheres2.htm
http://www.thoughtbubbles.org/archives/000157.shtml


Another version of the game is here:
http://www.transience.com.au/pearl.html