No, if the spectator happens to open an empty door, the probability collapses to 50/50. That's the point I'm making for how counterintuitive this is. Here's the full set of cases. Let's say that you select door 1 and spectator opens door 2 (all other cases are permutations of this):

Door1 Door2 Door3

Car Empty Empty

Empty Car Empty

Empty Empty Car

Let's suppose your strategy is to stick with your original choice. In the first case above then you get the car. In the second case the audience member stumbles upon the car and you lose. In the third case you lose because you stick with an empty door. All three cases are equally likely and since the second one ends, and you know that your game didn't end, you know that you're either in case 1 or case 2. Your chance is thus 50/50.

The issue is that in the classic MH problem, cases 2 and 3 are collapsed into one outcome (MH opens an empty door), but that's not true here.

More mathematically, you should ask yourself p(Door1 | Game Did Not End).

Using Bayes we see

p(Door1 | GDNE) = p(GDNE | Door1) * p(Door1) / p(GDNE).

p(GDNE | Door1) = 1 p(Door1) = 1/3 p(GDNE) = p(GDNE|Door1)p(Door1) + p(GDNE|Not Door1)p(Not Door1) = 11/3 + 1/22/3 = 2/3.

This, p(Door1 | GDNE) = 1 * (1/3) / (2/3) = 1/2.