I'll try to give a simple answer.
As you note, there are three choices involved:
Where is the prize?
What will the contestant choose?
What door will monty open?
Let's call the doors A B and C, and let's get to write down the possible scenarios in a simbolic manner;
(A,B,C), for instance, would represent the scenario in which the prize is in A, the contestant chooses B and monty opens C (i kept the order of the questions).
(A,B,C) is a possible scenario, as opposed to, let's say, (A,B,A) (where monty would open the prize door), or (A,B,B) (where monty would open the contestant's choice).
So, the possibile scenarios are these:
(A,A,B) (A,A,C) (A,B,C) (A,C,B)
(B,B,C) (B,B,A) (B,C,A) (B,A,C)
(C,C,A) (C,C,B) (C,A,B) (C,B,A),
and only half of these would make the contestant win if he chooses to switch.
Now, the question is:
are these scenarios equally likely to happen?
Is (A,A,B) equally likely as, let's say, (A,B,C)?
Well, let's take a step back.
The blurry territory comes with monty's choice;
before it, we're sure that there are nine possibilities, and those are equally probable.
(A,A) (A,B) (A,C)
(B,B) (B,C) (B,A)
(C,C) (C,A) (C,B)
So... couldn't we see what happens next like this?
(A,A,(B or C)) (A,B,C) (A,C,B)
(B,B,(C or B)) (B,C,A) (B,A,C)
(C,C,(A or B)) (C,A,B) (C,B,A)
I mean, let's look at (A,A,B) and (A,A,C).
Aren't they just two faces of the same possibility, (A,A)?
What i'm suggesting is that at the end there aren't twelve equally probable possibilities.
There are nine, and three of those have two possibile manifestations.
Or, there are... 18, and 12 of those have an identical twin.
Like, maybe monty will open the door either slowly or fast at random if you don't choose the prize at first, and at normal speed if you choose the right door;
and, you wouldn't be able to see the difference between those speeds, because you just see him open the door once, without any other reference; still, what you see must be one of three possibilities: normal, slow or fast.
So you would have
(A,A,(B or C),n) (A,B,C,(s or f)) (A,C,B,(s or f))
(B,B,(C or B),n) (B,C,A,(s or f)) (B,A,C,(s or f))
(C,C,(A or B),n) (C,A,B,(s or f)) (C,B,A,(s or f)),
at least with the schematization i proposed.
With yours, we would have 18 possibilities instead of nine two faced, but by both schematizations, we would come to the conclusion that it's better to switch.
Ok this is getting confusing, I'll stop...