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Subject: Fighting Ancients - Mechanics question rss

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Jon Solo
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So, just a rules clarification question about engaging in combat with AO's.

The rule book states "keep a tally of successes, once you've reached the number of players, remove a doom token and reset the tally to zero".

[using 4 players, 11 doomtokens for example math]

To my thinking, this can be interpreted one of two ways:
A) He has "HP" equal to doomtokens*players(11*4 = 44), and takes that many successes to kill. So in other words, one player could stack clue tokens and just roll 44 successes for an instant kill in one round (Improbable, but legal).

B) Successes are cumulative (eg, player 1 does 2 damage, then player 2 only has to do 2 more to take off the doom token); however, exceeding one doom token on a single roll is impossible (tally gets reset to zero, and you remove one doom token). Therefor: 2 successes from player 1 followed by 5 successes from player 2 results in 1 doomtoken being lost, and player 3 starts at 0/4 needed for the next doomtoken. This results in the AO taking at minimum a number of attack rolls equal to his number of doom tokens.


We've been playing by (A), but after reading more carefully, I feel as if (B) is more akin to intent. We dropped ithaqua in 2 rounds, with noone taking any damage at all, because we had a couple decent weapons and enough clue tokens.... I feel like that shouldn't be possible.

Anyone else able to shed some light on the subject? I feel like Cthulu would be really hard, even using (A), but we've been kicking the AO's asses far too easily I think.
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Jon Solo
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Summary question: When do you reset the tally to zero? Mid-roll (eg: 2 before, 11 new successes = 2+2, +4, +4, +1) or after the roll: (2before, 11 new, therefor 2+11>=4, remove doom token, set tally = 0)
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Erik Berry
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Scenario A in your first post is correct. The Ancient Ones effectively have hit points = number of players * number of spots on the doom track. In your summary, it happens mid-roll (I find that language confusing, but if it helps you, that works!).

Happy gaming!
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Jon Solo
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The tally-reset terminology is what was in the rule book, and what caused my confusion as well! =]

The way it sounded was that 'after a roll, if you've got more than the # of players accumulated, then remove a doom token and start over', but I'm willing to keep playing with the "HP" interpretation, just wondered if it was the consensus.
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Erik Berry
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Yeah, we see that question a lot. The rules are really badly phrased. I think Innsmouth attempts to clean it up, but that doesn't help people just getting into it.

For more backing than just my word, here's another thread where it came up. There are many more too - that's just the first one I found.
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Chris Lawson
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Jalefor wrote:
I'm willing to keep playing with the "HP" interpretation, just wondered if it was the consensus.

Not only is it the consensus, it's in the official FAQ

Quote:
Q: In combat with an Ancient One, how do cumulative
successes in the "Investigators Attack" step work?

A: To defeat the Ancient One, the players must do a total
number of successes equal to the number of players multiplied
by the number of doom tokens on the Ancient One.
So, if 4 players are facing Yig (doom track of 10), they
need 40 successes to win. For every 4 successes they do,
they remove 1 doom token to track their progress. If the
investigators get 9 successes in the first round of combat,
they would remove 2 doom tokens, and 1 success would
carry over to the next round of combat.

Which can be found on the FFG Arkham Horror Support page > FAQ
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Brian Mc Cabe
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There are those who look at things the way they are and ask why . . . I dream of things that never were and ask why not
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A good way to remember that hit points are cumalative is to look at Cthulhu's sheet. His special ability is that after his attack, one doom token is added to his track, if it isn't already full. It wouldn't be much of a special ability if it applied to all of the AOs.

As an aside, I've always misinterpreted this. I always read it as the excess hits being taken off, and if you got lucky and took off exactly enough to remove the token nothing happened.

Boy, still learning. Not that I bother at a -6, anyway.

Brian
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Jack M
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Jalefor wrote:
So, just a rules clarification question about engaging in combat with AO's.

The rule book states "keep a tally of successes, once you've reached the number of players, remove a doom token and reset the tally to zero".

[using 4 players, 11 doomtokens for example math]

To my thinking, this can be interpreted one of two ways:
A) He has "HP" equal to doomtokens*players(11*4 = 44), and takes that many successes to kill. So in other words, one player could stack clue tokens and just roll 44 successes for an instant kill in one round (Improbable, but legal).


That's actually not too crazy to do in a single player game. Lets say you need 11 successes to win. If you have the fight skill, each clue token you spend gives two dice. If you're blessed with a shotgun, there's a decent chance to take him out in one round if you have something like 10 clues.
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Steve Tudor
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OK I'm thoroughly confused now. The rulebook and the FAQ seem to contradict each other. From the rulebook:
Quote:
When the players accumulate a total number of successes equal to the number of players (including any players that were eliminated from the game), remove one doom token from the Ancient One’s doom track and eset their cumulative successes to zero.

Which leads me to believe that scenario B is correct, which is how I've been playing it.

But from the FAQ
Quote:
So, if 4 players are facing Yig (doom track of 10), they need 40 successes to win. For every 4 successes they do, they remove 1 doom token to track their progress. If the investigators get 9 successes in the first round of combat, they would remove 2 doom tokens, and 1 success would carry over to the next round of combat.

Which supports scenario A

So the really question is; how and when is the tally reset to zero?
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brian
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Wahoffelmadenga wrote:
OK I'm thoroughly confused now. The rulebook and the FAQ seem to contradict each other. From the rulebook:
Quote:
When the players accumulate a total number of successes equal to the number of players (including any players that were eliminated from the game), remove one doom token from the Ancient One’s doom track and eset their cumulative successes to zero.

Which leads me to believe that scenario B is correct, which is how I've been playing it.

But from the FAQ
Quote:
So, if 4 players are facing Yig (doom track of 10), they need 40 successes to win. For every 4 successes they do, they remove 1 doom token to track their progress. If the investigators get 9 successes in the first round of combat, they would remove 2 doom tokens, and 1 success would carry over to the next round of combat.

Which supports scenario A

So the really question is; how and when is the tally reset to zero?

The problem is the way they wrote the rule book, they were explaining something else in mind. If you get to the Innsmouth FAQ, they go into more detail so you can finally understand what they were getting at.

Simply put: you need 1 success for each investigator x the doom track of the Ancient One. No more, no less (with the obvious exceptions like Cthulhu who regenerates and anyone else I might be missing). That means if an Ancient one has a track of 12 and there are 5 players, you need 60 successes. It doesn't matter how many are collected on each player's turn. One investigator can take out 2 or 3 doom tokens if he rolls enough. Successes are carried over.

....

What FFG was trying to explain in the contradictory rules was that you needed X successes to remove a doom token. X is the number of players. We all get that part. They then said that it resets. What we interpret is that any left over success are also reset and/or that each investigator can only get 1 token removed. What FFG was trying to say is that you need X for each token. Not that once you reach X, then it is a one for one removal. What they were afraid of is with the example above that people would think that once they reached 5 successes, then each additional success would take out a token. In other words, the first token "cost" 5 successes but the next 11 tokens cost 1 success each, i.e., for the 12-doom GOO and 5 investigators, only 16 successes were needed instead of 60. This is what they meant by "resetting" the count. Once you hit 5, take off a token, then start counting form zero again.

This is what they were trying to say all along. But it was a very convoluted way of saying that and no one understood that approach until Innsmouth.

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Steve Tudor
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Thanks Brian, I think the penny dropped for me as I was writing my question. It makes me wonder what other rules I'm playing wrong.
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