Fayries |
When using a weapon, subtract 1 from each die rolled. […]
[…] If undefeated, examine the location deck until you find a monster; encounter it, subtracting 1 from each die rolled in your check. […]
[…] Subtract 1 from each die rolled in your check to defeat the Warlord.
Though these three cards instruct the same thing, if I were to encounter a Warlord at the Waterfront because of an Ambush (and fight it with a weapon), they would actually conflict, I would apply the Golden Rule and subtract 1 from each die rolled. Right?
(I'm really tempted to stack the penalties and subtract 3 from each die rolled, that's why I'm asking.)
Fayries |
What you would do is hopefully have an armor to chuck because making that check is going to be really, really hard as even Amiri or Val aren't going like 25%-30% of their dice being 0's. That's a brutal combo and is why having an actual (not Seelah) spellcaster in your group to send there.
Actually, it isn't that hard for Amiri, is it?
As a beginning character, she would have d12+2 for Melee and could bury a card from her hand for an additional d10, so she would be at 1d12 + 1d10 + 2. Certainly that wouldn't count as using a weapon, so there would be no penalties.
Vic Wertz Chief Technical Officer |
Firedale2002 |
The good news is that there is a minimum: No matter how many penalties are
applied to a die roll, the result can’t be reduced *below* 0.
Yay! Now we just need to find a Warlord that was misprinted to have a Check to Defeat: Combat 0!
LOL, I never did understand why those 'nothing below 0' rules existed. If you get 0 or -1, it's still under the check to win, so you lose.
Though, after typing that, I just realized that that can make the difference of 1 discarded card suffered for damage in PACG. If you had -5 and needed a 4, then you'd need to discard 9 cards, whereas with the 'cannot be below 0', you'd just take 4 damage.
So... Awesome rule early on! Later on, when you need a 15 and have to deal with a 0, though, it doesn't make much difference, your whole hand is probably going to go poof.
Mike Selinker Pathfinder Adventure Card Game Designer |
OberonViking |
Just to be extremely precise: The minimum result for the *roll* is 0, not the minimum for each die. So if you roll a 4, a 4, and a 1, and subtract 3 from each die, you get a result of 0, not 3.
Sorry, but since the cards say ”subtract one from each die rolled", doesn't your example result in 2?
4 - 3 = 1
4 - 3 = 1
1 - 3 results in a 0
Total = 1 + 1 + 0 = 2
Greyhawke115 |
Attempt the Roll. Roll the dice and add up their value, adding or subtracting any modifiers that apply to the check. If the result is greater than or equal to the difficulty of the check, then you succeed. If the result is lower than the difficulty, then you fail. No matter how many penalties are applied to a die roll, the result cannot be reduced below 0.
So that's the text from the rules, highlights mine. It seems clear that you add everything first, add/subtract modifiers next, then once you have the result, it cannot be less than zero.
Technically, the wording on the Waterfront might be better as "When using a weapon, subtract 1 for each die rolled..." or "When using a weapon, subtract 1 from the total result for each die rolled..."
Greyhawke115 |
So you have a 50% chance of being hindered by having Valeros there to add 1d4 in combat?
Seems like it, in this circumstance. Of course you are not required to have Valeros help. Considering there is a 50% chance it could hurt and only a 25% chance that it could add one point to the check, I would be inclined to ask Valeros to kindly not provide any "assistance" on this hopefully very rare occasion.
Vic Wertz Chief Technical Officer |
Casey Weston |
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The way I do these calculations is:
Add up the passive bonuses (the +1s, etc.)
Add up the result on the die
Subtract based on the number of die rolled.
In the given example if we assume the passive bonuses are 0, we add up the die (4+4+1) = 9.
Then we subtract the penalty which in this case is 3 times the number of dice, which is 3 (3x3 = 9).
So 9-9 = 0. If the result would be negative, we simply stop at 0 (no reason to go into the negatives).
Is that an accurate way to do the calculation?