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That's right! This is useful for me, by the way, for tightening up the language in the rules. Thank you.

I have a other question for you. The game play loop has been so engaging now after you helped me with my confusion. Now the next question I have is about the "Steel" action. 

In your rules example you have the wanderer's resolve split into 2 separate pools. 

Question #1a. If I wanted to play the "Skeleton Key" conquest side up would I still have 2 resolve left? Since it is two separate pools of resolve? 

Question #1b. In that same situation what if I used the "Shady Accomplice" instead? [My guess is still 2 Resolve because one of the pools doesn't make change] 

Question #2 continuing with the Wanderer, after Steeling once my resolve pools are 2 + 2. Steel 2nd time 2+1+1, because I slide the mini boss/hero to cover the 2 pool resolve. But if I Steel a 3rd time is it 1+1+1+1? Is that what the upside down 1 columns are for? 


I hope these questions are coming off as a sign of appreciation and love for the game loop because I really have been having a blast with the game!! Take care Alfred. 

1a. If you'd already steeled an appropriate hero's resolve once, and have 2/2, then yes. You'll spend one 2 and have another left.

1b. Correct, you'd be left with 2.

2. Yes, steeling 3 times without spending anything will get you to 1/1/1/1

Let me know if you have any more questions!

Wait, hold on, so to clarify:

Steeling your resolve ends your turn? 

I've been steeling resolve and playing a conquest in the same turn (which is to say, before drawing cards) this whole time

Ah, yes you should be drawing a card.

Well. This makes a lot of sense.

Ha, glad I could clarify!

i have one last question which is, you can only play a peril card if you can also steel resolve, right? 

Yes, steeling and playing a peril go hand in hand.