BTW, it seems many want to try to solve this problem by programs or trial sampling - the enumeration is really very simple, and no one's done it yet, I think, so here goes:

Let G = Goat, C = Car, S = Selected, and N = Not-selected. We then have the following 9 possibles scenarios:

SNN NSN NNS
GGC GGC GGC
GCG GCG GCG
CGG CGG CGG

Now throw out any N-Goat from all 9 configurations. 
We now have (reordered so as to make the results clear):

SN  SN  SN
GC  GC  CG
GC  CG  GC
CG  GC  GC

Hence, the 6/9 = 2/3 chance of winning with a switch.