Another look at the Monty Hall Problem

The book I was referring to looks at the Monty Hall Problem the following way which sounds logical to me:

Suppose Door 1 and Door 2 has a goat behind it and Door 3, the Car. Possible outcome:

You choose Door 1, the host reveals Door 2.
You switch doors, you WIN

You choose Door 2, the host reveals Door 1.
You switch, you WIN

You choose Door 3.
The host reveals either Door 1 or 2
You switch, you LOSE

So if you switch, your chance is 2 in 3 or 66.7%.

In other words, if you don't switch, actually nothing has changed. The host is just revealing the doors to you one by one. So your chance is the same as the original, 1 in 3 and not 1 in 2.

Revealing a door when you have already made your choice does not improve your chance if you do nothing.

Hahaha, logical or not?