Assume a case where there are 30 doors. 29 have goats, and 1 has a car. You begin to chose doors one by one until there are only two doors left. All the doors you have chosen have been goats, leaving just one goat door, and one car door.

Now two cases:

- Assuming you have selected a "special" door from the start in your head, and have been picking every other door but that one, what is the probability that the "special" door will be a car.
- Assuming you have just been picking randomly up until now, what is the probability that that door will be a car?

Now if I understand correctly, I believe the answer to #1 will be 1/30, like the Monty Hall Problem, because when you first selected it the probability was 1/30. I think the answer to #2 is 50%, thus the Gambler's Fallacy because the probability of you picking one then is 50%, independent up to now.