Let's say we buy four scratch-off tickets. The clerk selling the tickets assures us that one of the four is a winner. You choose two tickets, and I take the remaining two.
I scratch one of my tickets, revealing that it is a dud. At this point, I suggest we should swap your two tickets for my two tickets. You start to object, but I point out that this is just like the Monty Hall problem.
When we made our initial choice, there was a 2/4 chance the winning ticket was in the set you picked, and a 2/4 chance that it was in the set that I picked. The fact that I revealed a ticket didn't change that. So you should be willing to trade, as both sets of tickets are equally likely to contain the winner.
You refuse, obviously. The difference is that whether there's two goats behind his doors or one, Monty will always choose a goat. He provides no information about the door you've already chosen, as the probability of him choosing a goat is unchanged based on whether you picked correct or not. I, on the other hand, could have revealed the winning ticket, but the odds of me doing so would be lower if you already had it in your possession. So, my dud updates the probability of all remaining tickets.
Yours is the first explanation I 'got', thanks!
Forget the opening of the door.
Start with picking a door at random, you have a 1/3 chance of having picked the car. On that I think we all agree.
Now let's say that the host offers to let you switch from the door you picked, to the other _two_ doors combined. He hasn't opened any doors, they are all closed, you're allowed to stick with your initial guess of one door, or switch to a combined guess of the other two doors.
If that's the case it should be fairly obvious that you have a 2/3 chance of getting the car by switching to the combined 2 doors.
Now that you've switched, would it really make a difference to your odds if the host opens one of your doors to reveal a goat? Would that lower your probably to 1/2 or would it remain 2/3?