[Out_of_Characte]

How is it wrong in this case? your picked door had a 33% chance to have a price, since its one of three doors, now that a goblin is revealed the other door has a 66% chance of hiding the price. exactly like the monty hall problem

[Jensson]

No, you are wrong. The way the comic described it only describes a single instance of the host opening the door in this case. It could very well be that this host doesn't want you to win, and thus reveals another door and tries to goad you into switching only when you have picked the correct door. That is a very reasonable interpretation of a game like this, and in that case the correct answer is to not switch.

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.

[Out_of_Characte]

I thought about it some more, if the game is never played more than once then there's no strategy to develop for both the monty hall problem and the problem described in the comic, therefore that's a moot point. but if its played more than once then your evil host that sometimes opens a door doesn't make sense.

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.