@belkrax It's exactly because you aren't choosing from another set of doors. It's also exactly because only doors that definitely will not win are opened for you.
You should think of the doors as two sets of doors. There are
x doors. You may only choose one, which means the set of 1 door that you chose has a chance of 1/
x of being correct, which means the set of the other
x-1 doors has a 1 - 1/
x chance of having the correct door.
In the case of 3 doors, this means that the set you didn't choose has a 2/3 chance of
including the correct door. With 100 doors, the set you didn't initially choose has a 99/100 chance of
including the correct door.
When all the incorrect doors of the not-chosen set are opened (except one, if you happened to initially choose the correct door), what remains is a single door that is included in that set with the higher 1 - 1/
x chance.
To make it even easier: When presented with 100 doors, would you prefer to open 1 door, or 99 doors? With 3 doors, would you prefer 1 door, or 2 doors? With the first choice, you're essentially given the option to choose the one door that you
don't want to pick.
On a side note, Michael from Vsauce was right about it being fun to think of a way to explain the dilemma lol