The problem is such that the event has already occured, ie the door with the prize has already been decided.Probablity is associated with the event space and this space does not change at any point when we are making the choice of selecting the door the first time or the second time.The switch is a illusion giving us option of increasing the probablity when the event space has not changed at all.

For eg. if you are rolling a dice, each roll of a die can lead to event space of six possibilities.When you are about to roll a die you make a choice, and when the dice is in flight you change your choice.This switch does not increase your probablity of getting the correct answer and probability remains constant.

I know what you mean, actually you are right, but your conclusion is wrong.

====> after we choose our door,
there is prob 2/3 that the prize is behind the other 2 doors.
As you point out, this remains true when the game master opens one of those 2 doors.
But then that 2/3 probability collapse on the remaining door.
So at this point we better change door!
(isn't it obvious when looked at this way)

We do increase our chances by switching because we now have new information, given by the game master opening a door with no prize.
This doesn't change anything about our door, but it change everything about the other door left.

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Let me include a statement taken from the website
https://montyhallproblem.com/

"Imagine that there were a million doors. Also, after you have chosen your door; Monty opens all but one of the remaining doors, showing you that they are “losers.” It's obvious that your first choice is wildly unlikely to have been right. And isn’t it obvious that of the other 999,999 doors that you didn’t choose, the one that he didn’t open is wildly likely to be the one with the prize?"
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This is easily seen by listing all probabilities.

We have 3 doors A,B,C.
Only one has the prize.
(In what follows, let UPPER = has prize, lower = has no prize.)
So we have 3 possibilities:{Abc,aBc,abC}
So each door has probability 1/3 to have the prize.

Let say we choose door_A:
we have still 3 possibilities:{Abc,aBc,abC}.
we win if {Abc}, loose if {aBc,abC}.
Prob(winning)=1/3, Prob(loosing)=2/3.

With prob 1/3, we have Case 1:
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Case 1a: door_A has prize, game master open door_B.
Case 1b: door_A has prize, game master open door_C.
in both case, we win if we stay, we loose if we change.
Prob(win if stay)=1
Prob(win if change)=0

With prob 2/3, we have Case 2:
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Case 2a: door_A no prize, game master open door_B.
so door_C has the prize,
Case 2b: door_A no prize, game master open door_C.
so door_B has the prize,
in both case, we loose if we stay, we win if we change.
Prob(win if stay)=0
Prob(win if change)=1

But Case 2 is more probable! this is the important point.

So,
Prob(win if stay) =(1/3)*1+(2/3)*0=1/3
Prob(win if change)=(1/3)*0+(2/3)*1=2/3

That's it. (cool!...)