> The Multiple Opponent (MO) penalty only occurs on multiple defensive die
> rolls. If you only need to make one die roll, then it doesn't matter how
> many beings are contributing to the attack against you, there is no MO
> penalty.
>
> > For example: Lunar 1 attacks player 1. Lunar 1 gets augmented by
> > Lunar 2 and Lunar 3, and Player 1 gets -6 malus for the extra guys?
>
> No. The defender is only rolling one time, so there is no MO penalty (the
> benefits of having extra guys are reflected in the bonuses to the main
> attacker, rather than a penalty to the defender). Basically Lunars 2 & 3 are
> acting as followers to Lunar 1. Note that followers do not *cause* MO
> penalties, but they do *reduce* them.
The whole multiple opponents thing is a bit confusing because it really concerns rolls, not numbers of opponents, in some ways.
>
> I think what you're asking is three attacks, and so three defensve die
> rolls:
> Player 1 defends aganist Lunar 1 with no penalty.
> Player 1 defends against Lunar 2 with a -3 penalty
> Player 1 defends against Lunar 3 with a -6 penalty
>
Exactly. So that's the "straight up numbers" way to do it. The other is where Lunar 1 has been augmented by Lunars 2 and 3. As mentioned above, Lunar 1 gets bonuses from the augmentation, and Player 1 rolls as normal. It is effectively a one on one battle, with Lunars 2 and 3 tied to Lunar 1.
What is interesting to note is that if there were 3 players against the three Lunars, you could still have Lunars 2 and 3 augmenting Lunar 1. And in this case, even though Lunar 1 would be defending 3 times, he would get no MO penalties, because Lunars 2 and 3 would cancel the numerical advantage of the Players.
(At least as I understand it all)
LC
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