Posted: Sun Feb 21, 2010 8:26 pm
				
				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.