[darekkay]

Correct me if I'm wrong, but in your particular example (spectator opens an empty door and I am asked if I want to switch), nothing changes in regards to the original Monty Hall problem. If a spectator opens a random remaining door, one of two things can happen:

- a car is revealed, I lose immediately (there is no option to switch anymore)

- no car is revealed, which means I again have 2/3 chances when switching, not a 50% chance as you've stated

In your example, the spectator opens an empty door, so there is no difference to the host opening an empty door in regards to the probability. Again, if the spectator opens a car, I just lose.

[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.

[n4r9]

I think you're correct. Whether or not the spectator had a non-zero chance of opening the non-empty door is irrelevant after the fact.

[pontus]

No. The fact that they happened to not open the door with the car tells you something. Take it to the extreme of 100 doors. If the spectator randomly opens 98 doors and doesn't randomly stumble upon the car you should take that as evidence that you might have the car yourself. This extra evidence in favor of staying with your door exactly cancels the original MH advantage of switching.

[n4r9]

Thank you for the explanation; I think I see it now.

Here's another way of seeing it which I think may be effectively the same. Suppose that there are 100 doors in a circle. You pick one and then all the doors are opened except for the one you picked and the next door along. If the car hasn't been revealed yet, it seems intuitively right that you have a 50-50 chance of winning whether you stick or switch at that point.