My brain still wants to short circuit to two doors, one prize, random selection, 50% chance.
This seems to be a wide-spread belief/bias ... if there are two choices, two options, two outcomes, then it's 50:50.
OK, I buy a lottery ticket ... either I win or I lose, so it's 50:50, right?
Of course not. If you can start to see, by default, that two options are not usually equally likely, then you start to lose this misconception.
But really, you always have this situation: the prize is behind the door you chose, or it's behind one of the other two. That always remains true, so your probability of having the prize is always 1/3, even if you get to see behind one of the other doors. You're still sticking with your original choice, or switching to both the other doors (although one of them is open, so you ignore it).
But suppose Monty only opens the other doors if you picked the prize, then obviously you shouldn't switch.
In Monty Hall, Monty always opens a door, and you're left to choose between your picked door and one other door.
If you think that this description means that the host always opens a dud after you open the first, then you are wrong. It doesn't say that anywhere, and there is no reason to believe that this game show works that way. The only reason you think it does is that you connect it to the original Monty Hall problem which worked that way, which causes you to make the mistake thinking the same rules apply here.
If the host always opens a dud then its a monty hall problem. but if the host sometimes opens a door, the host still can't do it every time you land on the price because then you wouldn't switch. so the host would have to sometimes open the door when you're on the price and sometimes not. But then when he opens a door it would indeed become a 50/50 so an evil host's best option is to never open a door in the first place.