Winterclaw,

  Let me go back to the two goats (called G1 and G2), a car (C) and three doors, because that's the specific case.  To simplify things, you always pick door 1, and Monte always opens door 3, unless it has the car, in which case, he has to open door 2. (The math works the same whether you simplify things or not.)

The prizes can be distributed 6 ways:

1   2     3
C  G1  G2
C  G2  G1
G1 C   G2
G2 C   G1
G1 G2 C
G2 G1 C

After Monte opens the door (Door 3, unless that's the car), here are the possibilities

1   2     3
C  G1 
C  G2 
G1 C  
G2 C  
G1       C
G2       C

In a third of the cases, Goat 1 is behind Door 1, Goat 2 is behind it in a third, and the car is behind  Door 1 in a third of the cases.

I didn't believe it, either, until I sat down and worked the problem for myself.  It's one of those counter-intuitive things that make some areas of math, like orbital mechanics, fun.