Losing Shooting Dice

[philqw78]

Shooting dice are totalled by target. What if one the the shooting groups is disrupted and the other not? e.g.


etc




Not disrupted Shooter
Disrupted Shooter
target in question
Other targets

Shooters are Bow. Half the shots at unit are from the disrupted unit. Half from a non-disrupted unit . They would lose 1 dice per 3 if all shooters were disrupted, however do they still get three dice as they have half the second rank and all the first, 1 an a half disrupted, not 3, and 1 and a half of a non-disrupted, total 3?

[bigdamnhero]

Im trying to get my head around your description. How im reading this is that ther is more than one BG firng at a target? If it is one BG shooting at a target, then it is all disrupted through either combat, shooting or terrain.

Also it is the SHOOTER than has dice reduced, not the target.

If what you are saying is that the front rank of disrupted bow and half the second rank can fire, this would mean a total of three dice initially, less one for being a disrupted BG. So a total of two dice are rolled.

Non-disprupted BG,s roll their allocated dice and are not affected by a disrupted BG in their shooting totals.

If this disruption is due to terrain disruption, total up the dice per rank/file. So the disrupted bases lose one dice, all others as per normal.

[philqw78]

One file from each of 2 BGs is firing at a single target. One file is disrupted, one file not.

Dice reduction when shooting is done by target.

If both files were disrupted and from different BG's they would get 2 dice, losing one per three out of the one and a half dice from each.

If shooting was by BG each file would only ever get one dice. One per 2 bases from the rear rank gives none because there is only one rear rank base per BG.

Also if shooting dice was by BG those of 6 bases would only ever get 4 dice if two ranks deep, however if they are shooting with another BG their rear ranks add together. So 2 BG's of six all shooting at the same target would add 3 dice for their rear rank, half of six. If done by BG they would only add one dice per BG, half of three rounded down plus half of three rounded down.

[bigdamnhero]

Hiya. This sounds awfully complicated. If a BG had only one base in the rear rank, then you are right, they wouldnt fire at all owing to the ratio of bases at the back. You will need a minimum of two bases to qualify for the one shot from the rear rank.

Ive never heard that you can combine two BGs for dice reduction, i thought it was per BG for dice allotment. Remember, it is not the files that are disrupted that count, the dice reduction counts on the whole Bg (unless by terrain disruption).

I maybe confused with this. But your diagram is TWO BG's? So simply reduce the dice accordingly as per the BG, not as a combined BG as i understand it.

Thanks!

[philqw78]

This sounds awfully complicated

_ Thats why I posted it. It is 2 BG's. But you get the same problem when firing with some files in disordering terrain from the same BG. (Although it is difficult to get a rear rank firing in terrain that would disorder MF or LF, I can only think of woods and Villages. (Please don't somebody else hijack the thread to tell me which terrain does anybody) )

_ Perhaps simpler to question on though.

_ I might delete my first post and go with this.

_A BG of 4 Cav, one file in rough going, therefore disordered, shoots. How many dice does it get? 3 or 2. The file in the rough going loses one dice per 3 but only has 1.5 dice, so doesn't lose any. The file in good going gets 1 and a half and doesn't lose any. So it can ignore the rough? This gives a bit of cheese with 2 BG's or a single BG shooting on a join between units as they won't lose dice for being disrupted or disordered as in my first post.

_ In fact a rear rank of 4 mounted in rough going with a front rank of 4 in good going would not lose any dice as they only get 2 dice for the rear rank so the BG would shoot at full effect with 6 dice?

[philqw78]

In fact a BG of 4 cav with Bw* in rough would shoot at the same effect as a BG of cav with Bw in rough. Both on 2 dice.

[JanChris]

_A BG of 4 Cav, one file in rough going, therefore disordered, shoots. How many dice does it get? 3 or 2. The file in the rough going loses one dice per 3 but only has 1.5 dice, so doesn't lose any. The file in good going gets 1 and a half and doesn't lose any.

We've played this as: Cv get full shots. As you said, you loose the third die, but there is no third die, so you get the third die. Plain and simple

If both files are disordered, you would loose a die. An alternative interpretation is that you get 1.5 dice for the Cv in GGoing and less than that for the Cv in Rough, so would get less than 3 dice. This seems logical but is not supported by the rules imvho.

_ In fact a rear rank of 4 mounted in rough going with a front rank of 4 in good going would not lose any dice as they only get 2 dice for the rear rank so the BG would shoot at full effect with 6 dice?

Each file is disordered so I think you loose a die.

[hammy]

Having had a quick look at the rules (but forgotten the page references) I believe that it is pretty clear that while the number of dice rolled when shooting is per target the dice lost for disruption are per battlegroup regardless of their target. In this situation I think that the disrupted BG loses one of the 3 shooting dice it will contribute (admittedly half a rear rank die per target) so that you would only get 2 dice per target rather than the 3 that this sutuation would normally generate.

I just found the page, look at P114. It clearly states that shooting dice are lost per BG, just the same as melee dice.

[SirGarnet]

That covers how to allocate lost dice against 2 different targets, but please confirm how we figure out the number of dice to roll in the first place when a file of cavalry is in rough and the other is out.

Is the following right?

First we count the number of shooting dice. That's 1 each for the two in the front rank, and 1 for the 2 in the rear rank.

Then we look at disorder/disruption loss. It's lose 1 per 3 for disorder to the affected bases, which are one front rank base and one in the rear rank. Since one of the two in the back is disordered, the disorder applies to the die it shares with the non disordered base. So two dice are affected, but not 3, and no dice are lost? Same result if just both front rank bases or both rear rank bases were disordered, right?

I think this must be right since The whole "Lose 1 per" idea dispenses with the practice of calculating fractional damage rates per base, such as saying a base does 0.5 dice. or 0.333.

?

[hammy]

I am not 100% sure what is right but the way I would play it is that the disrupted BG would normally contribute 3 shooting dice from 2 front rank and 2 rear rank bases. Because it is disrupted it must lose one of those dice so effectively it looses it's rear rank dice. This means that if there are two disrupted and two steady bases shooting at one BG you get 2 front rank dice and only 1 second rank dice per target.

I may be wrong but if I was called as an umpire I am pretty sure that is how I would rule it.

[SirGarnet]

But disorder is per base, not the whole BG, and I think following the sequence of the player grouping for dice, then applying reductions works. EXAMPLES

So 4 bases with a file disordered: Total 3 dice
One front and one back disordered + one non-disordered paired for a die: = 2 dice
One front non-disordered: 1 die

Consider the same 2x2 with a severely disordered file: Total 2 dice (like being all disordered)
One front and one back disordered + one non-disordered paired for a die: = 2-1 = 1 die
One front non-disordered: 1 die

Taking it to the 6 base level (normally 4 dice), if one file is disordered I think: Total 4 dice
One front and one back disordered = 1 die
Two front and two back non-disordered = 3 dice

and if 2 files are disordered: 3 dice
Two front and two back disordered = 3-1 = 2 dice
One front and one back no disordered = 1 dice

And 8 bases (two BGs, say), which normally roll 6 dice in all:

If one file is disordered: Total 6 dice
One front and one back disordered+one back not disordered but added together for a dice: 2 dice
Three front and two back not disordered: 4 dice

If two files disordered: Total 5 dice
Two front and two back disordered: 3-1 =2 dice
Two front and two back non-disordered: 3 dice

If three files disordered: Total 5 dice
Three front and three back disordered + 1 disordered to pair up: 5-1=4
One front non-disordered: 1 die

If all files disordered: Total 4 dice
Four front and four back disordered: 6-2=4 dice

Mike

[philqw78]

And what if those adding their dice are of different quality. I thought it might get complicated, but I obviously underestimated you all.

[hammy]

If shooting dice are the result of combining shooting bases of different quality or POA you count the worst of all worlds.

[philqw78]

So half a dice from a superior shooter and half a dice from a poor shooter wouldn't make an average shooter then

[SirGarnet]

So half a dice from a superior shooter and half a dice from a poor shooter wouldn't make an average shooter then

No, the poor shooter shoots the superior shooter in the foot, throwing off his aim.

[pyruse]

I'd be interested in a definitive answer to the question of how many dice you lose when some files are disordered and some not, because it's not obvious how you do it.
Does the loss of dice apply only to those bases which are disordered?

Consider a BG of 6 bases, 2 deep, 1 file of which is disordered.
If the BG were not disordered, it would shoot with 4.5 dice = 4 dice.
If the whole BG were disordered or disrupted, it would lose one dice, so 3 dice
How many dice does it shoot with if 1 file is disordered?
If only the disordered bases lose any dice, then the answer is 4.

[hammy]

I'd be interested in a definitive answer to the question of how many dice you lose when some files are disordered and some not, because it's not obvious how you do it.
Does the loss of dice apply only to those bases which are disordered?

Consider a BG of 6 bases, 2 deep, 1 file of which is disordered.
If the BG were not disordered, it would shoot with 4.5 dice = 4 dice.
If the whole BG were disordered or disrupted, it would lose one dice, so 3 dice
How many dice does it shoot with if 1 file is disordered?
If only the disordered bases lose any dice, then the answer is 4.

With one file dissordered the dissordered file is only providing 1 and a bit dice so there aren't 3 dice from dissordered bases so you lose nothing.

Two dissordered files would contribute 3 shooting dice so you lose 1.

[SirGarnet]

Does the loss of dice apply only to those bases which are disordered?

No. The loss of dice applies only to DICE from bases which are disordered.

If a single die is from a mix of bases which are disordered and bases which are not disordered, then it is affected by the worst disordered of those bases.

If there are two ways to group "1 dice per 2 base" troops, such as a case with 3 bases shooting from a rear rank, the phasing player should think a moment before deciding how to group them.

This question has helped clarify that there is no such thing as half dice - just groupings of bases for shooting dice.

Mike

[bigdamnhero]

Blimey Phil, what HAVE you started?!!!!

I agree with the last comment. Its quite simple. In the example of having a BG of 6 bases two deep, with two bases in rough going (so disordered), the amount of dice to roll in missile fire is simple.

5 dice would be rolled overall (if the whole BG was NOT disordered), However as one file (two deep) is disordered you would lose a dice only if there were 3 bases in disorder. So 5 dice still apply dont they? You only take away dice if the 1 per 3 ratio applies!

I think.......hahahaha

[pyruse]

Where do you get 5 shooting dice from in a 6 base, 2 deep BG?
3 front rank, and 1 rear rank dice makes 4.