How many dice?

[kevinj]

This occurred in a game last night and we couldn't come to a definitive answer, so I'd be grateful for any thoughts.

A and B are enemy Average LH with Bow. Both are Disrupted.
X and Y are my BGs.
All are facing the direction indicated by the arrows.

In shooting File 1 of A shoots at X, no problem.

File 2 of A and 3+4 of B shoot at Y. Normally, that would be 3 dice but, as both are disrupted, do you:

A) Take the total and reduce by 1 per 3, i.e. 2 dice.

or

B) Calculate the dice for each BG, i.e. A shoots with 1 and B with 2.


Thanks,

Kevin

[JFAS]

good day

B is correct

john

[rogerg]

Page 93 at the bottom, the combined dice for shooting mentions the 'one dice per x bases' rule. In the table at the top of 93, this phrase is in the overhead shooting section. It is different to the 'losing 1 dice per x dice' for disruption and disorder.
Page 94 on losing dice for disruption mentions loss by BG.
It could be clearer, but B appears to be correct.

[hammy]

The way I have always played it and that I believe to be correct is that with shooting you apply the 1 base per x rules per target BG so option A.

I agree that the use of shooting in the paragraph on P94 could be considered to apply to shooting but it then goes on to use the word fighting which IMO implies close combat.

If a BG of 4 cavalry in a single rank is dirsupted and shooting with 2 bases at 2 enemy BGs then I would rule that it gets 2 dice against each target. If 6 bases of disrupted light horse from 2 BGs shoot at one BG then I would rule that they get 2 dice.

[christospap]

I don't have the rule book handy, but let me propose another reading of the rules, which I would have applied if the situation had happened in one of my games:

1. A is not correct, because I've always assumed that the "1 per X" rule applies to the BG as one unit. So, taking each BG separately, each one only has 2 dice to shoot with - meaning, no taking out a 3rd dice for being disrupted.

2. B is not correct either, because A has two dice to allocate: one dice against X and another against Y.

Consequently, I would consider that a third option might be correct:
A shoots 1 dice at X and 1 dice at Y.
B shoots 2 dice at Y.
X will be struck by 1 dice.
Y will be struck by 3 dice (1 from A and 2 from B - not affected by the "1 per X" rule, because the shots are coming from different BGs).

Bye all,
Christos

[hammy]

I don't have the rule book handy, but let me propose another reading of the rules, which I would have applied if the situation had happened in one of my games:

1. A is not correct, because I've always assumed that the "1 per X" rule applies to the BG as one unit. So, taking each BG separately, each one only has 2 dice to shoot with - meaning, no taking out a 3rd dice for being disrupted.

There are two possibly relevant passages.

On P93 it says "If more than one BG is shooting at the same target, add the total number of bases to which a '1 dice per x bases' rule applies before calculating the number of dice to roll"

On P94 it says "If a BG which is losing shooting or close combat dice (due to ...list...) is fighting against more than one enemy BG, first determins the total number of dice the BG should lose. Then apportion the lost dice, if possible, in proportion to the number of enemy bases fighting each enemy BG, leaving at least 1 dice (if possible) against each enemy BG."

These are tow different bullet points in the same list so IMO only one of them can be used at a time. That is the whole point of bullet points. In the example two BGs are shooting at one so the P93 bullet point applies. Now that I am reading the section more closely it seems to me that the second bullet point applies to the example I gave of a BG of 4 disrupted cavalry firing at two enemy BGs but not losing any dice. In this case only one BG is shooting so there must be a dice loss as per the second bullet point.

Thoughts?

[batesmotel]

I believe the bullet from P. 93 is the one that applies for shooting. The bullet from p. 94 is for impact and melee.

Chris

[hammy]

I believe the bullet from P. 93 is the one that applies for shooting. The bullet from p. 94 is for impact and melee.

Chris

That is what I thought but on closer reading the P94 one also mentions shooting. I am therefore assuming it applies whena disrupted BG shoots at more than one target.

[christospap]

On P93 it says "If more than one BG is shooting at the same target, add the total number of bases to which a '1 dice per x bases' rule applies before calculating the number of dice to roll"

Hmmm, yes, I can see your point Hammy and it would seem that you are right. Which means that, when shooting with multiple BGs against 1 target, what should be counted is the total number of Disrupted/Fragmented shooting bases in all BGs together, and not just decide the dice for each BG separately.

Just to make sure, let me ask this clarification:
A 4-base (2X2) disrupted LH BG shoots with 2 dice (4 shooting bases).
A 6-base (3X3) disrupted MF BG shoots with 3 dice (5 shooting bases - - 1 per 1 for the front rank and 1 per 2 in the back rank = initially 4 dice, but disrupted, so loses 1 dice).
However, if those two disrupted BGs combined their shooting upon 1 target, they would not get 3+2=5 dice, but a total of 4, because, as per P39, the 2 BGs - had they been steady - would have thrown a total of 6 dice , meaning that, being all disrupted, they now have to lose 1 per 3, meaning 2 lost dice - so, 4 shots.

It would seem that it would be possible to reach different results, depending on what you count: total number of bases of all BGs, or count each BG separately and add the respective number of dice for each.

Is this correct?

And as far as the 4 disrupted Cavalry example is concerned, I think that yes, there should be a dice lost.

Thank you,
Christos[/quote]

[hammy]

Is this correct?

I believe so.

And as far as the 4 disrupted Cavalry example is concerned, I think that yes, there should be a dice lost.

I also agree on this one although I have played it differently in the past. It makes sense that a disrupted BG that is splitting its fire should be impacted at least slightly.

[philqw78]

That is what I thought but on closer reading the P94 one also mentions shooting. I am therefore assuming it applies whena disrupted BG shoots at more than one target.

Perhaps it mentions shooting because you can shoot in the impact.

[marioslaz]

I think hummy point is more in the spirit of rules. I don't want to read the exact wording, but since in shooting you can combine dice from various shooters against one target, it seems right to me you apply the deduction for disrupted also to this total, and not to each component.

[ShrubMiK]

That is my thinking too.

[gozerius]

I think that the bullets are covering two distinctly different situations. P 93 covers multiple BGs shooting and p94 covers split shooting at more than one BG. In the case presented above we have a situation where one BG is splitting fire against two BGs and one of those BGs is being shot at by two BGs. Let us apply the reductions accordingly. BG A splitting 2 (less 1 per 3) = 2 dice, so 1 die per target, BG B is combining 2 dice with 1 die from BG A (less 1 per 3) = 2 dice total against the enemy BG Y.

So roll 1 die vs X and 2 dice vs Y.

[kevinj]

I think Gozerius' post sums up exactly how I'd read it. But, just to add to the confusion, what if A were Fragmented instead?

A, in splitting its fire could opt to use its single die against Y, as it would be wasted against X.

You then have 2 dice from B and 1 from A, but now a different '1 per X' rule applies to both BGs and it would seem all 3 can be used against Y. It does not seem logical that you should actually improve through being Fragmented as opposed to Disrupted.

[marioslaz]

Not sure, but I think that rules say in some point when you combine different situation (troop types for example) you apply the worst to all. In this case you could apply the same (that is all shoot as they was frag). Of course, you can choose not to shoot with fragmented if in such way you would get a worst result.

[fgilson]

Not sure, but I think that rules say in some point when you combine different situation (troop types for example) you apply the worst to all. In this case you could apply the same (that is all shoot as they was frag). Of course, you can choose not to shoot with fragmented if in such way you would get a worst result.

Except if you have shooting dice to use that could cause a result you have to shoot them...right?

[marioslaz]

Not sure, but I think that rules say in some point when you combine different situation (troop types for example) you apply the worst to all. In this case you could apply the same (that is all shoot as they was frag). Of course, you can choose not to shoot with fragmented if in such way you would get a worst result.

Except if you have shooting dice to use that could cause a result you have to shoot them...right?

I'm pretty sure shooting is not mandatory, but optional, that is you are not obliged to shoot with your troops.

[TERRYFROMSPOKANE]

Page 37: "Both sides must shoot with any troops able to shoot...."

Terry G.

[donkiesrus2003]

Sorry to pour petrol onto the fire but. The bulletpoint in page 93 is for counting the no. dice from firers. The bulletpoint in page 94 is the duduction to a BG dice du to conditions.
This allows 2 6 base BG 3x3 of bow to shoot at 1 target with 9 dice (4 from each and combine the spare bases from each rear rank to give the extra die).
If 1 BG was disordered then then they would shoot with 8 dice (four from the okay BG and four from the other BG and combined bases as 5 dice less 1 per 3 is 4).

In the example stated the claim that one of the 3 dice is lost due to the shooting BG's being disordered is WRONG!
If you apply that logic then a 12 base 6 x 6 Warband fighting 3 four bases 2 x 2 legionaries where all the units are disordered.
Both the warband and the legionaries are on 8 dice each.

If you follow as written in the rules the then warband is on 8 dice (12 less 1 per 3) and each legionaries BG is on 3 dice (4 less 1 per 3) making a total of 9.
Good luck trying to tell the Roman player that despite what the rules say they should only have 8 dice.