Losing shooting dice for disruption

[zoltan]

I thought I'd sussed this situation offline but a recent post by Dan and others (in the context of combat) has caused me to ask this online.

Target is a BG of 4 LH.

Shooters are:
a BG of 4 LH
a BG of 4 disrupted LH

Which is correct, option A or B?

Option A.
Each shooting BG gets 2 dice, total 4 dice.
However because one BG is disrupted, remove 1 dice per 3
Total of 3 shooting dice against the target BG

Option B.
Each shooting BG gets 2 dice, total 4 dice.
The disrupted BG loses 1 dice per 3
It only has 2 dice so none are lost
Total of 4 shooting dice against the target BG

[batesmotel]

Only 2 dice of the 4 are subject to disruption, so losing 1 per 3 has no effect, so you get 4 shooting dice.

If both BGs were disrupted and shooting at the same target, you would lose 1 per 3 out of 4 attacks, so would only get 3 shooting dice.

If each BG was shooting as a separate target by itself, each BG would get 2 shooting dice since losing 1 per 3 would cause no loses.

Chris

[lawrenceg]

Only 2 dice of the 4 are subject to disruption, so losing 1 per 3 has no effect, so you get 4 shooting dice.

If both BGs were disrupted and shooting at the same target, you would lose 1 per 3 out of 4 attacks, so would only get 3 shooting dice.

If each BG was shooting as a separate target by itself, each BG would get 2 shooting dice since losing 1 per 3 would cause no loses.

Chris

Or possibly as neither BG has more than 2 dice, neither loses any dice even if shooting at the same target. I note that the relevant bullet on p94 always talks about dice lost by a BG, not dice lost from a batch of dice.

[dave_r]

Only 2 dice of the 4 are subject to disruption, so losing 1 per 3 has no effect, so you get 4 shooting dice.

If both BGs were disrupted and shooting at the same target, you would lose 1 per 3 out of 4 attacks, so would only get 3 shooting dice.

If each BG was shooting as a separate target by itself, each BG would get 2 shooting dice since losing 1 per 3 would cause no loses.

Chris

Or possibly as neither BG has more than 2 dice, neither loses any dice even if shooting at the same target. I note that the relevant bullet on p94 always talks about dice lost by a BG, not dice lost from a batch of dice.

Batesmotel is right I think - shooting is done by the target. So in this instance, the target is taking four dice - if all four are disrupted then you lose one per three. If the two disrupted BG's were shooting at different BG's then they would each get two dice as the two targets are getting two disrupted dice each.

In the same way you can add up rear ranks from BG's shooting at the same target - so two BG's of six fire at one BG and would get nine dice (six front rank dice and three second rank dice). If they were to fire at two separate targets they would get four dice each.

Lawrence is wrong

[hazelbark]

Or possibly as neither BG has more than 2 dice, neither loses any dice even if shooting at the same target. I note that the relevant bullet on p94 always talks about dice lost by a BG, not dice lost from a batch of dice.

Did you read page 93 last bullet?

[petedalby]

Option B.
Each shooting BG gets 2 dice, total 4 dice.
The disrupted BG loses 1 dice per 3
It only has 2 dice so none are lost
Total of 4 shooting dice against the target BG

I believe this to be correct.

If both LH BGs were disrupted - then you would lose 1 dice. But in this case only 2 bases are disrupted - so no dice loss.

[zoltan]

Did you read page 93 last bullet?

Thanks for the replies. Most of you have agreed with the consensus view I obtained "offline" prior to posting. However, you'll note that Dan has reiterated the contra view that he expressed in another post (to do with combat) and that led me to post here.

The contra view is that the last bullet point on page 93 means you pool all shooting dice (at one target) and then adjust the pool for disruption. In my example, one disrupted BG of shooters would in effect taint the pool. 4 dice are reduced to 3 because one of the shooters is disrupted.

[zoltan]

Did you read page 93 last bullet?

Thanks for the replies. Most of you have agreed with the consensus view I obtained "offline" prior to posting. However, you'll note that Dan has reiterated the contra view that he expressed in another post (to do with combat) and that led me to post here.
The contra view is that the last bullet point on page 93 means you pool all shooting dice (at one target) and then adjust the pool for disruption. In my example, one disrupted BG of shooters would in effect taint the pool. 4 dice are reduced to 3 because one of the shooters is disrupted.

I've just caught up on the other (combat) post and Lawrence seems to have the contra to the contra argument. Page 94 first bullet, "If a battle group which is losing shooting.....dice (due to DISRUPTION....)....first determine the number of dice the battle group should lose."

Lawrence argues that this rule explicitly requires consideration of lost shooting dice to be made by shooting BG (and not pooled by target BG).

Dan will counter that this rule (bullet point) only applies where the shooter is shooting at more than one enemy BG. i.e. if the shooting BG is shooting at only one enemy BG then this bullet is irrelevant.

[philqw78]

Try applying some common sense to the rules. (Although this doesn't always work)

If a disrupted BG of 4 LF is shooting at the same target as a BG of 8 MF longbow the whole amount of dice do not count disrupted.

And if 2 disrupted BG of 4 LF shoot at the same target they should lose a dice

You're the umpire.

PS Try not to listen to Hammy or Dave, or Graham, which leaves, erm, no, not Shaun, or Mark, so just you then.