"Welcome to Blackburg" uses Cultist of Baphomet as the only Henchmen. It also has the location Middle of Nowhere (which has an empty 'location card list').
By RAW, during setup I am supposed to shuffle 1 Villain/Henchman in the otherwise empty MoN. However, what made me suspicious is that, in 6-play, this leaves no Cultists in the box. This led to a few questions:
Am I really supposed to put 1 Villain/Henchman in Middle of Nowhere?
If "Yes", this leaves the possibility that a Crazed Cultists barrier is flipped open, which requires you to summon the Cultist - but then, there's no Cultist in the box to summon. By RAW, we should ignore the "summon Cultist" instruction, as there are not enough cards and it's therefore "impossible instruction".
However, Crazed Cultists also says:
"Each character that does not evade or defeat the henchman banishes an ally. To defeat this barrier, all of the summoned henchmen must be defeated."
- obviously, none of the players defeats or evade the Henchman (as none are summoned). Are we really supposed to banish allies, or do we treat this as part of the "impossible instruction'?
- also, is the barrier supposed to go undefeated? Technically, no Henchmen are summoned, so "all of the summoned henchmen" CANNOT be defeated.
Don't spoil your game :-)
In our case, we just considered that the Scenario was in fact saying :
Henchmen : Demoniac of Baphomet, Cultist of Baphomet
.. and no problemo.
Now for and official FAQ you would need to replace the Demoniac of Baphomet by a AD0, 1 or 5 henchman by the Afghanistan principle. But our own games do not need to follow it :-)
1. Yes, you put a villain or henchman into Middle of Nowhere. See also this post by Vic which covers a question that naturally follows from that ("what if I encounter the villain at Middle of Nowhere and lose, while all other locations are either permanently or temporarily closed?")
Each character at an open location summons and encounters the henchman Cultist of Baphomet. Each character that does not evade or defeat the henchman banishes an ally. To defeat this barrier, all of the summoned henchman must be defeated.
After you act, banish this barrier.
I'll just be looking at the first power, instruction by instruction, to show how this would be handled when you are unable to summon any Cultists of Baphomet:
A. "Each character at an open location summons and encounters the henchman Cultist of Baphomet." -- There aren't any in the box, so we ignore this instruction.
B. "Each character that does not evade or defeat the henchman banishes an ally." -- This sentence has two possible interpretations: either it only applies to people that actually encountered the Cultist of Baphomet, or it applies to each character. RAW, it's everyone, so if you were at a closed location you'd be banishing an ally even if there were cultists to summon. Similarly, you'd banish an ally if you were unable to summon any cultists. RAI, this seems farcical, because it would mean that bane is having you banish allies without any ability for you to prevent that should you happen to be at a closed location (I'd personally expect every bane that has you banish something would be preventable in some way based on just the bane's powers; as in, "examining the card beforehand and knowing it is coming up so you can reposition yourself accordingly" doesn't count as a preventative measure). As such, I believe the intended interpretation is that this instruction only applies to characters that actually encountered a Cultist of Baphomet. As no characters encountered one in this example, this instruction has no effect and therefore no cards are banished.
C. "To defeat this barrier, all of the summoned henchman must be defeated." -- There were 0 summoned henchmen, so therefore 0 of the summoned henchmen must be defeated. This statement is always true, so the barrier is always defeated.
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C. "To defeat this barrier, all of the summoned henchman must be defeated." -- There were 0 summoned henchmen, so therefore 0 of the summoned henchmen must be defeated. This statement is always true, so the barrier is always defeated.
That's how I see it too. Congratulations, you caught a (rare) lucky break!
C. "To defeat this barrier, all of the summoned henchman must be defeated." -- There were 0 summoned henchmen, so therefore 0 of the summoned henchmen must be defeated. This statement is always true, so the barrier is always defeated.
I agree with all of your other statements, and they match the consensus reached on our table.
However, this last part (while probably being closer to intent in such extreme edge-case) does contradict classical logic. We can look at this sentence as:
"For A (barrier is defeated)) to be True, all of B (X number of Henchmen) must be C (are Defeated)"
Now, even if X=0 , i.e. no Henchmen are summoned, then still none of these 0 Henchmen can acquire a Defeated status.
Again, I'm not so much arguing on intent (in the rare -if not impossible, in WotR- circumstance where it would matter, I'm sure most people would house-rule the barrier is defeated), as I'm wondering if there's some official reference that would patch this 'logic gap'.
C. "To defeat this barrier, all of the summoned henchman must be defeated." -- There were 0 summoned henchmen, so therefore 0 of the summoned henchmen must be defeated. This statement is always true, so the barrier is always defeated.I agree with all of your other statements, and they match the consensus reached on our table.
However, this last part (while probably being closer to intent in such extreme edge-case) does contradict classical logic. We can look at this sentence as:
"For A (barrier is defeated)) to be True, all of B (X number of Henchmen) must be C (are Defeated)"
Now, even if X=0 , i.e. no Henchmen are summoned, then still none of these 0 Henchmen can acquire a Defeated status.
Again, I'm not so much arguing on intent (in the rare -if not impossible, in WotR- circumstance where it would matter, I'm sure most people would house-rule the barrier is defeated), as I'm wondering if there's some official reference that would patch this 'logic gap'.
Well, looking at set theory, I think you're still fine. You could say it as For A (barrier is defeated) to be true, Set B (the set of all henchmen encountered) must be equal to Set C (the set of all henchmen defeated or evaded). Since both Set B and Set C are empty sets, they are equal, and therefore, the requirement of A is met.
Well, looking at set theory, I think you're still fine. You could say it as For A (barrier is defeated) to be true, Set B (the set of all henchmen encountered) must be equal to Set C (the set of all henchmen defeated or evaded). Since both Set B and Set C are empty sets, they are equal, and therefore, the requirement of A is met.C. "To defeat this barrier, all of the summoned henchman must be defeated." -- There were 0 summoned henchmen, so therefore 0 of the summoned henchmen must be defeated. This statement is always true, so the barrier is always defeated.I agree with all of your other statements, and they match the consensus reached on our table.
However, this last part (while probably being closer to intent in such extreme edge-case) does contradict classical logic. We can look at this sentence as:
"For A (barrier is defeated)) to be True, all of B (X number of Henchmen) must be C (are Defeated)"
Now, even if X=0 , i.e. no Henchmen are summoned, then still none of these 0 Henchmen can acquire a Defeated status.
Again, I'm not so much arguing on intent (in the rare -if not impossible, in WotR- circumstance where it would matter, I'm sure most people would house-rule the barrier is defeated), as I'm wondering if there's some official reference that would patch this 'logic gap'.
Yes, I was approaching my logic for C based on a set theory argument. Let H be the set of henchmen summoned by this barrier, and D(b) be true if and only if b is defeated, where b is a bane. Therefore, C encodes the following proposition: (∀h∈H: D(h)) → D(Crazed Cultists)
Corollary: ∀x∈X: Q is always true for all propositions Q when X=∅.
Proof: Assume to the contrary that the statement is false. This means ∃x∈X: ¬Q. However, we know that X is empty, so there can be no such element x that satisfies this condition, a contradiction.
Since we have proven that ∀h∈H: D(h) is always true when H=∅, we can apply modus ponens to see that D(Crazed Cultists) is also always true, meaning that Crazed Cultists are always defeated should they not summon any henchmen.
Chief Technical Officer
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Since the last line on the card is "After you act, banish this barrier," it's going back to the box anyway, so the discussion of whether it's defeated or not is only relevant in the very rare circumstance* that something actually cares specifically about it being defeated or not.
*The rarity of that circumstance is further compounded by the fact that the situation in question can only come up in a 6-player game (or a 4-player game that isn't using the Add-On Deck) where the Crazed Cultists barrier has been randomly seeded into a location deck** and happens to be encountered before any of the Cultists of Baphomet have been banished.
**The rarity of that circumstance is compounded by the fact that the barrier has the Elite trait and Welcome to Blackburgh is the second scenario in Adventure Deck 5, meaning that the only copy of that barrier will have already been removed from the game if you happened to encounter it in the previous scenario.
But yes, if it didn't summon any henchmen, it's defeated.
I stand convinced :) Take that, Crazed Cultists!
EDIT: Aaand, since it seems Vic is on a roll to ninja all my posts tonight, I'm gonna show myself out. At least I'm glad I helped spot a problem with the card... whatever the mystery issue is :)
DOUBLE-EDIT: And now my EDIT has been ninja'd, and there's no more reference to the unknown issue that has been spotted with the Cultists!!
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Chief Technical Officer
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However, Crazed Cultists also says:
"Each character that does not evade or defeat the henchman banishes an ally."- obviously, none of the players defeats or evade the Henchman (as none are summoned). Are we really supposed to banish allies, or do we treat this as part of the "impossible instruction'?
I found a slightly different reason to address that second sentence in the FAQ, but the resolution clarifies this situation as well.
Since it's in the same vein:
In Ineluctable Prison (5-4), when you encounter the ally Waxberry, there may not be a free Stringy Fiend to summon. Does that mean I automatically lose my chance to acquire the ally?