it's not. why did you add 2 different equations together? the last number is how likely 2 shots will kill the monster
Right in that case I misunderstood and got confused, never mind.
allmost every one in 2nd had hordes and they are still very prevalent
Yes, but generally speaking you didn't see armies with 4-5 hordes. So your horde-breakers had limited relevance as after using it a handfull of times you'd probably run out of viable targets.
You will however constantly see armies with that many heroes + behemoths & the occasional elite unit, ensuring this things remains relevant at every stage of the game.
and this become progresivly less efective the more you spend on defence. and starts off much worse
.... any attack becomes less effective the more defense you have, that's the whole point of defense.....
Regardless, that's not the point. The point is that hordebreakers are at maximum effectiveness the first time you hit that horde. The second time they'l do less damage cuz presumably you've killed part of that horde. Until eventually enough models have been killed and the hordebreaker attack does virtually nothing.
This on the other hand will always be equally effective against a 10 wound behemoth. Doesn't matter if he is at full health or only has 1 health left.
cool you tend to like it when that is the case. or at least you did the last 3 times you brought it up
O yes I approve of it having some sort of baseline. Was just putting it there to show how much it contrasts with horde-breakers as it has none of the drawbacks.
this is true of any high damage attack with low acracy
Yeah, and those inherently all have the exact same problems causing them to not be particularly fun for the victim. Especially in PvP focussed games as human players will eventually figure out if there's an efficient way to make the lack of accuracy essentially a non-issue.
