If I choose Door 1 (1/3 chance) and the host opens Door 3, which is empty, my odds on Door 1 stay at 1/3 while my odds on Door 2 go up to 2/3????
Maybe the way to look at this is that you are, in effect, trading your single door for the other 2 doors. You would obviously trade your single door for the other 2 before the host reveals one of them as empty.
In other words, the winning door will in the doors you did not pick 2/3 of the time, and revealing one of them as a loser does not change this probability.
Don't feel bad if you missed this; when the puzzle was first presented back in the day, hundreds of PhD types argued against the answer.
More here: http://en.wikipedia.org/wiki/Monty_Hall_problem