I think what people are trying to say is that, if you have 100 doors and you open 98 and there is nothing, what are you supposed to do, follow the rule of 3 and switch since there is increased probability that way... or keep since what are the chances that you will get the prize by switching since the other 98 doors had nothing..

I mean its really hard to figure it out, since its alot more randomized chance then theory. What are your chances of getting the prize from those 2 doors?

The way I think it, it should be 50:50 % chance since the car can only be in one of the two so, which to choose? A or B

The way a statistician looks at it, is that he/she would count all 3 doors, in the 3 door problem, since they were there originally and even though one is taken away, it still effects your chances. This is because you chose a door from 3 doors and not two.

That should clear up any doubts.