[pontus]

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.

[darekkay]

Thanks for the explanation! As with the original MH problem, your "100 doors" example helped me better understand your variation of the problem.

[basch]

The thing is, 1/3rd of games ends with the audience member picking the car. So, probability collapses to 50% in 2/3rds of cases.

It's a completely different problem than the Monty Hall problem, which can never end in the car being discarded.

[pontus]

The point I was making is that the problem (MH) is a lot more subtle than people give it credit for. Many of the arguments people make for how the MH problem is "obvious" seem to work for this modified game too unless you're careful with them.