This is actually an interesting variation of the problem. If a random door opens but by chance happens to not have the prize, then the result is different.
There's a 1/3 chance you picked right, a 1/3 chance you picked wrong and a wrong door was revealed, and a 1/3 chance you picked wrong and the right door was revealed. Throw the last third out of your sample space and you get that the odds are 1/2.
This is what that forced Marilyn to admit the odds can be 1/2 on some readings of the problem, which was ambiguously phrased the first time it was printed. (But I think most people read the problem for the 1/3 solution and still came up with 1/2.)
I could be wrong, but I think of it this way: You have a choice between picking one door, or picking two doors. If you choose to pick two doors, the host will happen to open one of them to reveal the non-prize. Irrespective of the differences between the wording of my problem and the MH problem, how are the actions different between the two?
Here's the flaw in your logic:
There's a 1/3 chance you picked right, and a 2/3 chance you picked wrong. In the 2/3 chance you picked wrong, there is a 1/2 chance of the correct door being revealed. No matter if the correct door is shown or not (in this case, it isn't shown), it doesn't change your original odds.
If the host is choosing randomly, no information is passed.
With regards to your last paragraph, the odds of the host picking wrong are either 1/2 or 0 depending on the intentions of the host. I'm a tiny bit annoyed that you think you're pointing out a "flaw" in my logic when you missed something so simple, but whatever.
If you don't like my explanation, direct your attention to the OP's link to Wikipedia, where there are many more variations and explanations that you might like.