No, IMO, when a door is opened, the problem is restated anew. You start with 3 doors, one of which is the correct door there is a 33% possiblity of any given door being the right one. You then open 1 of the doors showing that nothing is behind it. You now have 1 door with a 0% possibility of being right and 2 doors each with a 50% chance of being right. The fact that you picked in the first place is inconcequential if they don't open the door you picked.
Possibilities:
1 (no car when they open door 3)
2 (no car when they open door 3)
3 (find car when they open door 3)
If you always pick door 1 and they always open door 3, there is a 1/3 possibility that the car is behind that door. Since you said that that case never happens, the only real possibilities are that the car is behind door one or door two, thus your odds of either being correct is 50%
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