Your sentence the particular way you worded it is not the correct mathematical model.
The player does not get to switch to BOTH OF THE OTHER 2 boxes as an alternative to just the 1st box. Therefore the 2/3rd probability doesn't apply.
Where the non-intuitive 2/3rds probability becomes the answer instead of 50/50 is the host's perfect knowledge of always choosing the door without the car.
It wasn't really my sentence I just quoted the article. Nonetheless I disagree with you. The probability that the other boxes have the key is 2/3 and that is all that really matters. Opening a door doesn't change anything.
The Monty Hall problem distills to simply "would you like one or two doors, (if the prize is behind any door in the set you pick you win.)"
When we pick the original box, we know that the probability that the keys will be in there is 1/3. The probability that the keys will not be in the box you originally chose is 1 - 1/3 = 2/3. Just from this knowledge alone, you could decide that you will always switch, since the probability that the other boxes have the keys is 2/3.