Another Shadows Question

By KristoffStark, in 2. AGoT Rules Discussion

I'm looking to clear up the interaction between the agenda City of Shadows, and the location Hidden Chambers.

City of Shadows: You may ignore the "House X only" deckbuilding restriction on any card with the Shadow crest. Whenever you bring any card with a "House X only" restriction that does not match your House card out of Shadows, pay 1 additional gold. If you do not have any cards in Shadows, you cannot claim power for unopposed challenges.

Hidden Chambers: If it is Winter, reduce the cost to bring your cards out of Shadows by 1. If it is Summer, return Hidden Chambers to your hand.

Let's just assume that I am playing House Stark, and it is Winter.

Let us also assume that I have Qyburn in Shadows. He is a Lannister Only character with a cost of s1

Now let's say that I have two Hidden Chambers in play, therefore lowering my cost to bring cards out of Shadows by 2.

1) Do City of Shadows and Hidden Chambers just add together: 1 (printed) + 1 (out of house) - 2 (two Hidden Chambers) = bringing him out of Shadows for free?

OR

2) Does the wording "pay 1 additional gold" on City of Shadows make it apply last, as follows: 1 (Printed) - 2 (Hidden Chambers, assumed minimum 0, as you can never pay less than 0 for something) + 1 (City of Shadows) = bringing him out of Shadows costs 1 Gold?

You would get to play him for free. All penalties and all reductions are all applied at the same time. So it you can do 1+1-1-1=0 or you can do 1-1-1+1=0. Either way you get to play him for free.

Note that both the gold penalty from City of Shadows and the reduction from Hidden Chambers apply only for cards brought out shadows with the game mechanic, not to cards brought out of shadows by their own triggered effects (there's a thread or three about that around here), even if those effects state the cost is to pay the rest (the number after the "s") of the card's cost.

Staton said:

You would get to play him for free. All penalties and all reductions are all applied at the same time. So it you can do 1+1-1-1=0 or you can do 1-1-1+1=0. Either way you get to play him for free.

So this isn't a 1+1-1-1=0 situation. It's a 1+1=2-1-1=0 situation. Of course, you get the same result in this case, though.

The distinction (that penalties are added in before the cost) is really only important where there are modifiers specific to the penalties. Say, for example, that you didn't have the Hidden Chambers, but instead had a card that said "lower all gold penalties by 2." If everything were added in together, you might be tempted to think that you'd pay 0 for Qyburn, but since it only applies to the gold penalty, and the gold penalty is modified from 1 to 0, which is then added into the cost (1), which you then have to pay.

Thanks, guys. I had assumed that version 1 was correct, but I have been tripped up by subtleties of wording before, so I wanted to double check that "1 additional gold" equated to "your cost to bring these cards out of Shadows is +1," which would be have been clearer, but taken up considerably more card space.