I was playing with Jenny Barnes against Yig. I had a shotgun (each 6 rolled gives me 2 success instead of only one).

I was playing alone (1 player game) and Yig awakened so i was cursed because of his "start of battle" ability.

My stats were: fight 4 and speed 4 (only stats that really matters against him). The shotgun gives me +4 combat dice for a total of 8 dice... against Yig's -3 modifier, i play 5 dice.

In the first round i don't roll to try to lose my curse because it's the following round i earned it (though i'll fail the check in all other rounds and i won't lose my curse for the entire combat).

Playing 5 dice i scored in the first round two 6s (4 successes).

I passed his first attack (speed check) so i don't lose my sanity and stamina in the first round (but i'll fail in all other rounds). I started combat with 4 sanity and 4 stamina.

In the next round i score another two 6s (more 4 successes) and fail his attack check (now i'm with 3/3 sanity/stamina).

In the third and last round of combat (though i'm almost sure i'm wrong) i score one 6 (more 2 success). So i scored 10 successes (1 for each of his doom tokens, Yig has only 10 doom tokens and it's the easiest of all ancient ones to battle). Did i win or i just take 3 doom tokens off? Each round i take one doom token and all other success of the round are discarded?

In the first round for example, i scored 4 success. One success is enough to take his first doom token away because i was playing alone, but all the other 3 are discarded because it resets to 0 each time a player has enough success to take one doom token off? If that's true, it'd be almost impossible to battle Yig alone (and impossible to fight any other ancient one). Because i'd need at least 10 entire rounds to take one of each doom token per round (assuming i'd have at least one success EVERY round O_o).

The rulebook is a little vague on this. The key point is that successes carry: the AO's hit points are (player number) times (doom track). It takes 10 hits to kill Yig in your case, no matter in what order or clustering you've scored them.

So yes, you won. There's a lot I can say about the spirit of your particular scenario but technically you did win.

Page 22 of the rulebook: "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) ."

Page 19 of the rulebook: "Each time they accumulate a number of success equal to the number of players or higher ( for example, 4+ successes in a 4 player game) , they remove one doom token from the ancient One's doom track and reset the succ ess tally to 0. "

Their cumulative successes to zero the rulebook says wouldn't be the other 3 i had in my first round? So i would just remove one doom token and ignore all the other 3 successes?

Let's suppose we are playing a 3 player game. 3 successes are needed on Yig (or any other ancient one to remove one doom token). I score 2 successes, the 2nd player scores more 2 successes. 4 successes! One doom token is removed, we reset the tally and the 3rd player (in the same round) try to score from the zero. So... the player that reaches the exact number of success to remove one doom token resets the tally to 0 (and all his/her cumulative successes are lost) and the next player (in the same round) starts from zero successes to try to remove the next doom token from the ancient one. That's what i understood after reading the rulebook a lot of times.

Yes, but like I said, the rulebook is vague. "Reset the cumulative successes to zero" should not have been in there. It means that, for a 3-investigator game, when you get to 3 successes, you remove a doom token, and keep on counting.

In your example, the investigators have scored four successes. Three of them go towards removing a doom token. You remove the doom token, then set the accumulated successes that removed the doom token to zero. That means you have one success leftover, which carries. It doesn't matter if the round has ended or a specific investigator's attack has ended; there is already one success "invested" in removing a doom token, so the next time two successes are scored, however or whomever, a doom token is removed.

In other words: successes are never lost and always count. Otherwise Cthulhu would be impossible to kill because he heals back one doom token with each of his attacks.

Hi, Nianphy, we had the same question in our first battle with the GOO and Tibs is right, the rulebook is to vague here. I think the FAQ clarifies it a bit:

So, yes, points that were not used to remove a doom token will carry over to the next round.

In my understanding " reset their cumulative successes to zero" is included to simplify the process of counting successes towards the next doom token removal. In case with 5-investigators game it goes like this:

Players rolls 7 successes, 5 points are used to remove one doom token, the successes reset and you get 2 more successes towards the next token removal. Now the next player makes his roll, he needs to roll 3 successes to remove the token... and so on.

Ok, thanks for the answers. So if you are playing alone and score five 6s with a shotgun you "kill" Yig (ancient ones aren't killed, just banished ^^). Wow! So i won my first game battling the ancient one in the end (the last 11 times i've played and won, i've won sealing the 6 gates with an elder sign). Epic fight! I think it's more awesome when you are against incredible odds and win. When you seal 4 portals in the game quickly, the game becomes much easier and it turns to be a piece of cake to seal the last two losing some part of the fun (though i was reading that's not the case when you are playing with large expansion boxes, i just have the core arkham, though my kingsport expansion arrives next monday: 12-16-2013). Unless if you are playing with one or two people and a very hard rumor appears. I imagine the game like a RPG. I was playing with Jenny Barnes - a DILETTANTE against a VERY ANCIENT being and win. Ok, they are called INVESTIGATORS for a reason. Though she's a dilettante, she collects clues along the game and we can imagine those clues giving her a way to defeat the supreme being. Totally girl power. I loved. LOL!

And.... congrats for that Indeed, sometimes the victory is the sweetest beverage.

Yes. Well 1-investigator with a shotgun is pretty much Yig's weakness.

Kevin Wilson once suggested the following house rule: during each round of final combat, each investigator may not spend more clues than the number of seals on the board. It's a great rule but it still can't account for 1-investigator games. Don't even play that way as it forces everything far out of whack.

Another question. Does Michael McGlen's (the gangster) special ability work against Ithaqua? He only takes 1 stamina damage against Ithaqua's attack if he fails it?

Yes. Strong Body works with losses, and Ithaqua's attack makes investigators lose Stamina.

Considering that Ithaqua uses checks Fight for both attack and defense, and McGlen starts with more than one weapon, he is pretty much made to crush Ithaqua in combat.

You definitely won your fight against Yig. The description of the Final Battle procedure in the Rulebook is ambiguous--it is actually just talking about resetting the accumulation of successes to zero in order to start on the next doom token, not about getting rid of any successes.

There is a much clearer description of the process in the FAQ from 5/12/2012, pp. 12-13 under "Battling the Ancient One", where it verifies that the total number of successes needed to defeat Yig is 10 times the number of players.

You are right that you would never defeat any Ancient One if you threw out successes in the way you describe. Try that method on Yog Sothoth or Quachil Uttaus to see how true that is!