Hi @ all,

I spotted some difficulties for me in understanding the following phrase:

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.

Does this mean, e.g.

4 players game, none eliminated: player 1 has 2 successes , followed by player 2 with 3 successes. (accumulated 5 succeses <-> 4 players)

Do I now:

a) remove 1 doom token and reset successes to 0, then go on with players 3 and 4 or

b) add to those 5 successes the successes from players 3 and 4, then remove doom token(s) and reset successes to 0

Thanks in advance

Neither. Remove one doom token immediately and carry forward the extra success for 3 and 4. If 3 and 4 have successes not adding up to a full token after their attacks, 1 starts from that position assuming they survive the counter attack.

Essentially, you just need to get investigators*doom track hits in total. (So for 4 investigators and Ithaqua, that would be 44 hits, for example). The doom tokens provide a convenient way to track the large parts of this - I like to use the activity markers (hexagons numbered 1-3) to track the parts of doom tokens.

This is clarified in the FAQ - if you haven't found that yet it's definitely worth reading.

Thx - that sounds much better than the original rules (just downloaded the FAQ).

But now I wonder why the German reprint of AH from December 2008 and the English pdf-rules doesn't have this more than 2 years old change of rules implemented^^

All that matters is that you got a total of 5 successes. That means that one doom token will be removed, and one success will still remain until three more successes are dealt, regardless of who survives the AO's attack or not.