I have fun trying to explain this problem. Let me see if I can give an explanation that will help you.
So lets say you have just picked a door in the beginning. You know you have a 1/3 chance of being right.
If I then tell you, "I will give you two options... you can either bet you are right, or bet that you are wrong"
You would obviously choose to bet you are wrong, correct? Because you know you only have a 1/3 chance of being right with your guess, which means you have a 2/3 chance of being wrong. The smart bet is that your original guess was wrong.
This is actually what is happening in the game if you think about it. You pick a door and it has 1/3 chance of being the right one; since we know Monty is only going to ever reveal a goat and never the prize, we don't even NEED Monty to reveal the door at this point - we know he is going to reveal a goat, no matter what. We don't even have to wait to see which door he reveals, since that isn't going to give us more information (it is going to be a goat, no matter what). So when he asks you if you want to switch doors, he isn't asking you to switch to ONE of the other two doors, he is asking if you want to switch to having BOTH other doors as your choice. Whether he reveals the goat before or after you choose to switch doesn't matter, because you know it will always be a goat.
If that is still not clear, lets just write out all the options:
There are three doors, A B C. One has a prize, the other two have goats. Let see what happens with your two options (switch or dont switch).
In our first example, you pick door A and you are going to switch.
1/3 of the time the prize is behind door A. If the prize is behind door A, and you switch, you lose. This is 1/3 of the time, and you lose for switching.
1/3 of the time the prize is behind door B. You picked door A, so Monty reveals door C. You switch to the remaining door (B) and you win.
1/3 of the time the prize is behind door C. You picked door A, so Monty reveals door B. You switch to the remaining door (C) and you win.
Add up all those choices, and 2 out of the 3 times you win.
Now lets imagine that we DON'T switch.
1/3 of the time the prize is behind door A. Monty reveals one of the other doors, but you don't switch. You win.
1/3 of the time the prize is behind door B. Monty reveals door C, but you don't switch from A. You lose.
1/3 of the time the prize is behind door C. Monty reveals door B, but you don't switch. You lose.
So in this not switching world, you win 1/3 of the time.
In summary, switching wins 2/3rds, not switching wins 1/3.
That does help - if nothing else, adding up the outcomes manually helps to demonstrate it in a way that's not really easy to grasp on first glance. It still is kind of a spooky result which makes no sense, but at least I feel a little better about the correctness.
So lets say you have just picked a door in the beginning. You know you have a 1/3 chance of being right.
If I then tell you, "I will give you two options... you can either bet you are right, or bet that you are wrong"
You would obviously choose to bet you are wrong, correct? Because you know you only have a 1/3 chance of being right with your guess, which means you have a 2/3 chance of being wrong. The smart bet is that your original guess was wrong.
This is actually what is happening in the game if you think about it. You pick a door and it has 1/3 chance of being the right one; since we know Monty is only going to ever reveal a goat and never the prize, we don't even NEED Monty to reveal the door at this point - we know he is going to reveal a goat, no matter what. We don't even have to wait to see which door he reveals, since that isn't going to give us more information (it is going to be a goat, no matter what). So when he asks you if you want to switch doors, he isn't asking you to switch to ONE of the other two doors, he is asking if you want to switch to having BOTH other doors as your choice. Whether he reveals the goat before or after you choose to switch doesn't matter, because you know it will always be a goat.
If that is still not clear, lets just write out all the options:
There are three doors, A B C. One has a prize, the other two have goats. Let see what happens with your two options (switch or dont switch).
In our first example, you pick door A and you are going to switch.
1/3 of the time the prize is behind door A. If the prize is behind door A, and you switch, you lose. This is 1/3 of the time, and you lose for switching.
1/3 of the time the prize is behind door B. You picked door A, so Monty reveals door C. You switch to the remaining door (B) and you win.
1/3 of the time the prize is behind door C. You picked door A, so Monty reveals door B. You switch to the remaining door (C) and you win.
Add up all those choices, and 2 out of the 3 times you win.
Now lets imagine that we DON'T switch.
1/3 of the time the prize is behind door A. Monty reveals one of the other doors, but you don't switch. You win.
1/3 of the time the prize is behind door B. Monty reveals door C, but you don't switch from A. You lose.
1/3 of the time the prize is behind door C. Monty reveals door B, but you don't switch. You lose.
So in this not switching world, you win 1/3 of the time.
In summary, switching wins 2/3rds, not switching wins 1/3.
Does that help at all?