Okay, let's go with your interpretation. Monty has now randomly chosen a door, and not revealed a car. If he had revealed a car you would have changed doors to

Okay, let's go with your interpretation. Monty has now randomly chosen a door, and not revealed a car. If he had revealed a car you would have changed doors to it and won. But he didn't.[0] Now you have a choice to make. His probability of not revealing a car in the case that both doors you didn't pick had goats behind them, is 100%. His probability of not revealing a car in the case that one door you didn't pick had a car behind it, is 50%. We can reason then, that of the three possible scenarios for the two non picked doors {g,c}, {c,g}, {g,g}, that it is equally likely now that since he didn't randomly reveal a car, {g,g} must be weighted twice as much as the other two. Therefore, we must be as likely to have a goat behind the remaining door as we didn't. In conclusion, Monty only changes the problem if he selects with information. [0] Notice that at this point, we have to throw away from the decision tree 1/3 of the histories, and in all of them switching was beneficial. In what remains then, switching is less beneficial than it was before this trimming.