Hi there folks, I just wanted to see if I could get a quick clarification on the way damage is dealt in the final battle.
The rule in the rulebook is as follows:
Unlike a normal battle, the Ancient One cannot be defeated in a single attack. Instead, keep track of every success an investigator scores against the Ancient One. These successes are cumulative, and each successive player adds to them with his own attack. 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 reset their cumulative successes to zero.
The way I've interpreted this is that each player rolls all their attack dice, but can do a maximum of 1 damage per round, because the of the cumulative success wipe.
So, for example, a game with 4 players:
- player one rolls 8 dice, three successes. No damage is dealt.
- player two rolls 6 dice, five successes. The first success combines with the three existing, and one damage is dealt to the Ancient One.
- the other four successes are ignored because of the reset, and play moves on.
This seems to me to be what the rules intend, but this is a ruling by the game designer:
Defeating the Ancient One (07/09/05)
Basically, to defeat the Ancient One, the players must do a total # of successes equal to (# of players) x (# of doom tokens on 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.
This seems to ignore the reset mentioned in the rulebook. Does anyone have a firm example of how this is meant to work?
Also, just to be certain, cumulative successes carry over between player rounds, correct?