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PurpleSun
05-12-2010, 03:39
Anybody know what happens when you use a hex scroll against a unit of Pink Horrors?

Say the unit is 40 strong. Does it just turn the one Horror that is being used for LOS, etc. into a frog, or does the whole unit turn to frogs? Or does the whole unit turn into one frog (with 40 wounds because it doesn't affect wounds?)?

Seems like a good question for an FAQ.

Rakton
05-12-2010, 05:51
that would be funny of a unit of 40 frogs on the board (ribbit) but i'm not sure. i would play it as a unit of 40 frogs. (maybe i should make a regiment of 40 frogs just to replace horrors if that ever happens if i ever play demons)

TheRatsInTheWalls
05-12-2010, 05:59
I think the most literal interpretation of the rules would have the unit become an uberfrog. That would depend on the wording of the Horror rules, which I can't check at the moment. If it says the spell is cast from the horror who casts the spell, then he should be the only one removed. If the unit casts the spell, they should all ranatize. (Rana is the genus for the common frog)
Despite all that, I like the unit of frogs notion best. It's the most amusing, and therefore what I would probably go with in a friendly game.

DenWhalen
05-12-2010, 06:06
"A unit of Pink Horrors is considered to be a single wizard with its own magic level."

"Each time a spell is cast, nominate one Pink Horror in the unit as the caster for the purpose of line of sight, range, etc."

As cool as an uberfrog might be, I think this situation falls under the heading of "etc." If this happened to my Pink Horrors, however, I might not fight against them becoming an uberfrog--even if it was just to have a great story to tell.

TheRatsInTheWalls
05-12-2010, 07:58
Yeah, that was the type of wording I figured would be in there. Sadly, you'd only turn one horror into a frog then. The other ideas are both much more entertaining though, so it's up to you and your opponent.

Wakerofgods
05-12-2010, 08:00
However, the horrors don't have 40 wounds...they are 40 guys with one wound.

Surely the uber frog would have 1 wound? You don't stack all their str together, why all their wounds?

DenWhalen
05-12-2010, 18:03
If you count the unit as one wizard, it would be one wizard with 40 wounds. The hex scroll sets all other stats at 1, but leaves wounds unaffected. The "wizard" has 40 wounds. Therefore, if the unit were transformed into a single frog, it should still have all its wounds.

What we would get, however, is one W1 frog in a unit of 39 Horrors. It's not nearly as funny of a result, but it's more in line with the rules for a miscast by a group of horrors (which does d6 wounds to the unit with no saves as opposed to rolling on the miscast table).

Demon-cookie
05-12-2010, 20:43
What we would get, however, is one W1 frog in a unit of 39 Horrors.

As well as howls of laughter from the horrors themselves i would imagine.

The real question is what colour would the frog be?

RanaldLoec
05-12-2010, 21:51
Well there must be somebody who knows what colour you get when you mix pink and blue?

PurpleSun
05-12-2010, 22:26
I prefer the notion of one frog with 40 wounds, but I presume an FAQ would go with the one frog and 39 horrors.

An epic result would be a Waywatcher getting a lethal shot on that 40 wound frog. I don't think that I would stop laughing until they put me in my grave.

Zaonite
06-12-2010, 06:10
Well there must be somebody who knows what colour you get when you mix pink and blue?

I'm guessing octarine?

I wanna see a 40 wound frog... that I could then KB with one hit :D

solkan
06-12-2010, 08:16
It's good that someone else pointed out that the Hex scroll doesn't change the number of wounds that the "wizard" has. It's also worth pointing out the Hex Scroll doesn't change the model type, or the number of models, that the "wizard" is composed of.

So, how do you Killing Blow a frog composed of 40 individual 1 wound models? :shifty:

PeG
06-12-2010, 08:47
nominating one single horrors as the caster also helps when another big frog transfers a miscast so this would probably be the preffered way of playing.

ColShaw
06-12-2010, 13:26
I'm guessing octarine?

Well played, sir.

PurpleSun
06-12-2010, 16:04
It's good that someone else pointed out that the Hex scroll doesn't change the number of wounds that the "wizard" has. It's also worth pointing out the Hex Scroll doesn't change the model type, or the number of models, that the "wizard" is composed of.

So, how do you Killing Blow a frog composed of 40 individual 1 wound models? :shifty:

Drop a piano on them?

theunwantedbeing
06-12-2010, 18:15
A single chosen horror model suffers the effect, not every model in the entire unit.

TheRatsInTheWalls
06-12-2010, 18:24
A single chosen horror model suffers the effect, not every model in the entire unit.

Yeah, we've established that this is how the rule probably works. The other readings are just much more entertaining.

PurpleSun
07-12-2010, 12:44
A single chosen horror model suffers the effect, not every model in the entire unit.

Seriously dude, stop trolling. We conceded that interpretation already. We are just having fun.

mistrmoon
08-12-2010, 06:41
I'm guessing octarine?




Well played, sir.

seconded.

Also the real question is, since only 1 horror turns into a frog and a single horror has no magic level is the horror auto frogged?