This problem has been analyzed many times, and the result is this: the probabilities do change.
The chance of initially picking the correct door with all doors closed is
1/n
where n is the number of doors.
The chance of picking the correct door by switching is
(n-1)/(n*(n-k-1))
where k is the number of doors opened by the host.
(n-1)/n is the complimentary chance of 1/n, or in words the chance the player picked incorrectly on the first pick. The two chances of picking correctly and incorrectly sum to unity.
(n-k-1) is the number of doors remaining closed, except for the first one picked. This is also the degrees of freedom in this problem.
To satisfy intuition, consider the case with one trillion doors. Pick one. Then Monty opens up all the doors except the one picked, and one other door. Are the odds 50/50? Of course not.
It is a quick proof to show that the second fraction is larger than the first whenever k=>1.
The solution is all over the web, I am surprised it wasn't posted first and finished this thread quickly.