I'm familiar with the problem and my intuition is still wrong. I wonder if you couldn't design a gambling machine that has a variant of the Monty Hall problem built-in favoring the house (naturally).
When you first pick a door, you have a 1/3 chance of it being the right door. There's a 2/3 chance of it being behind a door you didn't pick.
When the host then opens a door, there's still a 2/3 chance that it is behind one of the doors you didn't pick. However, there's now only one door in this set, so there's 2/3 chance that it's behind _that_ door.
To look at another way, imagine if the host didn't reveal the content of the door, but gave you the option to switch to BOTH of the other doors instead of your door. Your odds of winning clearly go up, as you now have two chances to win (and all are of equal probability). That's equivalent to what's happening here. By showing you the losing door of those two, he doesn't change anything - there's still twice the chance it was behind one of the doors you didn't pick compared to the one you did, and by switching you win if it was behind either of them.