How many dice?
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You are mixing apples and oranges. The bullet on page 93 applies to multiple shooters at a single BG, in which case all bases shooting at it are combined and the dice which are subject to reduction are then reduced. This is exactly what occurs with the bases shooting BG "Y". It is important to recognize which bullet applies in each case.
The bullet on 94 refers to BGs engaging more than one enemy BG.
To fully represent the entire process of reduction, let us look at the full process.
BG A has a total of 4 bases shooting. 2 at BG X, 2 at Y. BG B has a total of 4 bases shooting at BG Y.
Each BG starts with 4 dice. Because they are light horse this is reduced to 2 per BG. BG a splits its 2 dice against its two targets. BG B shoots its 2 dice at BG Y. BG X is shot at by 1 die. Not enough to reduce. BG Y is shot at by 3 dice, all subject to the same penalty, (lose 1 per 3), so only ends up with 2 dice against it.
So you see, when multiple BGs shoot at a target, the reductions are based on the original total of all bases shooting. It is not figured per BG. This is in contrast to what normally happens in close combat or when splitting shooting between BGs. What if both BGs were bow armed cav instead? BG A shoots 1 front rank and 1 rear rank die at BG X, and 1 front rank die and 1 rear rank die at BG Y, BG B shoots 2 front rank dice and two rear rank dice at BG Y.
Because there are not 2 rear rank bases shooting at BG X it receives 1 die of shooting without further reductions due to the disruption of BG A. BG Y it shot at by 3 front rank bases and three rear rank bases. This produces a total of four dice, reduced to 3 because the shooters were all disrupted. BG A contributes 1 die, BG B contributes the other 2.
The bullet on 94 refers to BGs engaging more than one enemy BG.
To fully represent the entire process of reduction, let us look at the full process.
BG A has a total of 4 bases shooting. 2 at BG X, 2 at Y. BG B has a total of 4 bases shooting at BG Y.
Each BG starts with 4 dice. Because they are light horse this is reduced to 2 per BG. BG a splits its 2 dice against its two targets. BG B shoots its 2 dice at BG Y. BG X is shot at by 1 die. Not enough to reduce. BG Y is shot at by 3 dice, all subject to the same penalty, (lose 1 per 3), so only ends up with 2 dice against it.
So you see, when multiple BGs shoot at a target, the reductions are based on the original total of all bases shooting. It is not figured per BG. This is in contrast to what normally happens in close combat or when splitting shooting between BGs. What if both BGs were bow armed cav instead? BG A shoots 1 front rank and 1 rear rank die at BG X, and 1 front rank die and 1 rear rank die at BG Y, BG B shoots 2 front rank dice and two rear rank dice at BG Y.
Because there are not 2 rear rank bases shooting at BG X it receives 1 die of shooting without further reductions due to the disruption of BG A. BG Y it shot at by 3 front rank bases and three rear rank bases. This produces a total of four dice, reduced to 3 because the shooters were all disrupted. BG A contributes 1 die, BG B contributes the other 2.
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- Corporal - Strongpoint
- Posts: 58
- Joined: Wed Apr 02, 2008 4:38 pm
Gozerius
Does that mean that if Battlegroup A was not disrupted and B was then:
Since battlegroup's A and B both start with only 2 die each as per page 90 and they get 1 die per 2 bases. There is no reduction for light horse and the bulletpoint on page 93 makes no reference to reductions and only shared dice take the worst results and where as the bulletpoint on page 94 is explicit:
"If a battle group which is losing shooting or close combat dice (due to DISRUPTION, FRAGMENTATION, DISORDER, SEVERE DISORDER or being light foot or light horse in close combat) is fighting against more than one enemy battle group, first determine the total number of dice the battle group should lose. Then apportion the lost dice, if possible, in proportion to the number of bases fighting each enemy battlegroup, leaving at least 1 die (if possible) against each enemy battle group."
That BG X gets one die and BG Y would get 3 dice as only 2 dice were disrupted the third being fine.
Does that mean that if Battlegroup A was not disrupted and B was then:
Since battlegroup's A and B both start with only 2 die each as per page 90 and they get 1 die per 2 bases. There is no reduction for light horse and the bulletpoint on page 93 makes no reference to reductions and only shared dice take the worst results and where as the bulletpoint on page 94 is explicit:
"If a battle group which is losing shooting or close combat dice (due to DISRUPTION, FRAGMENTATION, DISORDER, SEVERE DISORDER or being light foot or light horse in close combat) is fighting against more than one enemy battle group, first determine the total number of dice the battle group should lose. Then apportion the lost dice, if possible, in proportion to the number of bases fighting each enemy battlegroup, leaving at least 1 die (if possible) against each enemy battle group."
That BG X gets one die and BG Y would get 3 dice as only 2 dice were disrupted the third being fine.
My God!!!! If I was a newbie looking at this I'd just run to get my copy of WAB / DBA and throw FoG in the bin!!! I thought these rules were supposed to be clear and illustrated, you almost need a degree in maths to follow this!
No disrespect intended to the posters here, they are just trying to answer the OPs sensible question and I've been grateful enough myself for assistance on previous occasions but wow, I just can't follow this.
It seems to me that half the problem is that you have to look at how many BGs are shooting at each target, rather than what each one of your shooting units is doing. Looking at enemy BGs rather than your own which seems a little "contrived" and "unnatural" somehow.
If I read it right you need to;
1. Pick a target.
2. Add up the dice for all your undisrupted and unfragmented BGs shooting at it.
3. Add up dice for all your disrupted / fragmented BGs firing at that target, deduct 1 per 3 or 1 per 2 as appropriate and add the result to what you got in Stage 2.
4. Throw that number of dice.
No disrespect intended to the posters here, they are just trying to answer the OPs sensible question and I've been grateful enough myself for assistance on previous occasions but wow, I just can't follow this.
It seems to me that half the problem is that you have to look at how many BGs are shooting at each target, rather than what each one of your shooting units is doing. Looking at enemy BGs rather than your own which seems a little "contrived" and "unnatural" somehow.
If I read it right you need to;
1. Pick a target.
2. Add up the dice for all your undisrupted and unfragmented BGs shooting at it.
3. Add up dice for all your disrupted / fragmented BGs firing at that target, deduct 1 per 3 or 1 per 2 as appropriate and add the result to what you got in Stage 2.
4. Throw that number of dice.
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- Corporal - Strongpoint
- Posts: 58
- Joined: Wed Apr 02, 2008 4:38 pm
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- Corporal - Strongpoint
- Posts: 58
- Joined: Wed Apr 02, 2008 4:38 pm
Yes I know that but the way it was being played each LH battlegroup shot individually, therefore lots of 2 dice rolls not being reduced by disruption as 2 dice is less than 3 and after my manoeuvre phase it is 2 rounds of shooting before I can do anything. So yes it is over 2 turns but it feels like 1 turn and if your KN battlegroup drops a level from the first round of shooting it is like watching wolves gathering round a wounded animal in your opponent's manoevre phase.
P.S. Welcome to the Holy Land.
P.S. Welcome to the Holy Land.