Is this a modified Monty Hall problem (numbered doors)? On a job interview, I got this question:

Monty placed a car and two goats behind three identical doors (and the things do not move during the game). You receive the prize which is behind the door you picked in the second round.
You choose door number 1. Then, Monty opens door number 3 and you see a goat there. If you want to win a car, should you change your guess from door no 1 to door no 2?

I answered: YES and was told I was wrong. The interviewer explained me, that the difference between classical MH problem and this problem is that this problem clearly states, that Monty opens door number 3, which reduces the state space and after that my chance is 50/50.
I still believe chance of picking right door at first attempt is 1/3 and this probability is not changed by the information that Monty opens door number 3.
Am I missing something or was the interviewer wrong?
The interviewer is wrong in stating that Monty "clearly" picked door 3 and that therefore the chances are 50-50.  Monty picked door number 3  in this particular case but he didn't say why.

The interviewer is correct that if we do not know why Monty showed door number 3 (does he always show door number 3 no matter what is behind it or what you pick-- does he always pick a door at random that you did not pick no matter what is behind it-- does he always show a door with a goat you did not pick?) then the question can not really be answered. (But he is wrong in assuming that it is 50-50).  We have to make an assumption but... what assumptions are valid and which aren't?  

Here are several possible rules Monty could be played by:

Classic:  Monty always shows you a goat you do not pick.  Strategy: switch. 2 out of 3 in your favor.

Random:  Monty picks a door you did not pick and this time it just randomly happened to be a goat.  Strategy: doesn't matter. 2 out of 4 whether you switch or stay.

Devious: If you pick the car Monty will show you a goat in the hopes that you will assume a classic game.  If you pick a goat he won't give you a choice.  Strategy: stay.  100% in your favor.

Tough luck: Monty will always show you the car if he can.  He'll only show you a goat if you pick the car.  Strategy: stay. 100% in your favor.

Warped: There is one goat with a spotted tail.  Monty will always show you a door you did not pick that does not have the spotted tail goat.  Strategy: stay.  If you switch it is 2 in 3 that you will get the goat with the spotted tail.  So it is 2 in 3 if you don't switch you get the car.

Hierarchical: If you pick the goat with spotted tail, Monty will show you the goat without the spotted tail.  If you pick the goat without the spotted tail Monty will show you the car.  If you pick the car Monty will show you the goat without the spotted tail. Strategy: 50-50.

etc.

Which is the more likely one he is playing?  We can't tell.  And obviously these are not the only strategies.

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Actually what would be fair is if it were worded like this:

You are an a game show and where you have a chance to pick a car or two goats.  The hosts goal is to give you a goat and keep you from picking a car.  You pick door 1 and he shows you door 3 has a goat and offers you a chance to switch to door 2.  Should you?

Answer: it doesn't matter.  Whichever door you pick he will put the goat behind it  after you pick it.
This answer is based on statistics and variable change if the person changes his decision and choses door number two instead of door number one the probability changes from 66.6% to 33.33% ...