Monty Hall where host doesn't know where the 'gold' is I have been given this version of the 'Monty Hall' Problem and am having some issues:
On this version of the show, you are, as usual, shown 3 face down cards, one of which is the Q♥ and the other two of which are black spot cards. Your objective is to select
the Q♥, which wins $1,000.
As usual, you have to select a face-down card, and then the host turns over
one of the other two cards. The twist this time is that the host himself does not know where the Q♥ is,
and so may reveal the winning card. In this case you would win immediately!
The answer is that it doesn't matter you switch or not, but I am having trouble understanding why. If anybody could help me out using the law of total probability, that would be great. I understand 'regular' iterations of the Monty Hall but not this specific version where the host doesn't know where the card is.
 A: Due to symmetry, we can assume you always select the first card, and the host always select the second card.
If you don't switch, you win if the queen of hearts is either the first or the second card.
If you switch, you win if the queen of hearts is either the second or the third card.
Clearly in both cases, your probability of winning is $2/3$.
A: This is interesting. Initially I also thought that this variations does not change the answer and you should switch. Then I thought about it like this:
Imagine a variation of your description (that I will show it's equivalent to yours): You choose a door. The host (who does not know where the prize is) picks the door to remain unopened. You can then make your choice "switch or not". After you have chosen, we proceed with opening the door not chosen by the host. If the prize is there, you win. If not, we open your door (your final selection, switched or not). If the prize is there, you win. If we set it up like this, I think the answer is obvious: it does not matter whether you switch or not. The host has the same information as you in picking a door, so your selections have the same probability of winning. (Of course you have extra opportunities to win overall via the process of the host opening doors. Your overall probability of winning is $2/3$). 
This setting is equivalent to your one. It does not matter if the host chooses the door to open, or the door to remain closed. It also does not matter if you make your choice to switch or not before any door opening. If the door opened by the host have the prize, you win anyway. 
A bit more insight: I started thinking about the host choosing the unopened door instead of opening one door, because I thought of the paradox's variation, where we have 100 doors, you pick one, and the host opens 98 before you are called to make your choice of switching or not. This is a nice variation to teach the intuition behind the regular Monty hall paradox. But it serves us here as well. So the host keeps opening doors until one (beside yours) is left, and then you are called to choose "switch or not". In our case since the doors are opened at random, this is equivalent to the host choosing the last unopened door. And since the opened doors automatically win you the game, you can make your selection of "switch or not" beforehand. I think it becomes much clearer this way, that it does not matter whether you switch or not. Note of course that the overall probability of winning is $99/100$, the same as the original $100$-door Monty Hall paradox, but in our case it does not matter whether we switch or not.
