Slitherine

Ok here is a "simple" case for people to talk through:-

3 units of disrupted Skirmishers shooting at two units of HF in column




Unit 1 splits it's dice two base to Unit A, and one to Unit B

So does unit A take two dice and Unit B take 1 dice or do both take 1 dice?

then does unit B get 2 dice each from Unit 2 and 3, or does this add only another 3 dice from the two disrupted Units (2 & 3)


Thanks in advance

The LF shoot with 2 dice at A.

Against B there would be 5 dice. All shooters are disrupted so you lose 1 per 3, ie 1 and shoot with 4 dice.

kevinj wrote:The LF shoot with 2 dice at A.

Against B there would be 5 dice. All shooters are disrupted so you lose 1 per 3, ie 1 and shoot with 4 dice.

So if 3 units in column where to advance on to a single disrupted unit, so that each base provide 1 dice, there would be no lose of dice even thou the shooting unit is disrupted. this seem a bit strange, but OK if that's the rules

The LF unit A if not disrupted would have 2 dice on A and 1 on B.
However, as it is disrupted it loses a dice. It must lose one of the dice on A therefore it has one dice on A.

Paul

elysiumsolutions@fsmail.n wrote:The LF unit A if not disrupted would have 2 dice on A and 1 on B.
However, as it is disrupted it loses a dice. It must lose one of the dice on A therefore it has one dice on A.

Paul

I think this is right, but reading the whole paragraph when a superior and poor unit shot at a unit it seems to suggest that you do not re-roll the superior dice and you need re-roll 6 as these dice are now poor, and if you have two groups one at 0POA and one group at ++POA, then all shoot at -POA, is that right.


so 2 Superior Dice with ++POA and one poor 0POA would become 2 poor at 0POA

Leslie

elysiumsolutions@fsmail.n wrote:The LF unit A if not disrupted would have 2 dice on A and 1 on B.
However, as it is disrupted it loses a dice. It must lose one of the dice on A therefore it has one dice on A.

Paul

Now would that mean the LF only get 2 dice then? since they are disrupted they lose 1 per 3

LeslieMitchell wrote:

kevinj wrote:The LF shoot with 2 dice at A.

Against B there would be 5 dice. All shooters are disrupted so you lose 1 per 3, ie 1 and shoot with 4 dice.

So if 3 units in column where to advance on to a single disrupted unit, so that each base provide 1 dice, there would be no lose of dice even thou the shooting unit is disrupted. this seem a bit strange, but OK if that's the rules

No bullet point 4 on Page 94 handles this as you must take the 1 per dice before allocating dice to units these dice, and then no longer subject to 1 per reduction. so from the picture unit A takes one dice, and unit B now take 1 from unit 1 and 3 from unit 2 and 3.

Leslie

I disagree. My understanding is that, for shooting, you allocate the dice by target and then apply any reductions for cohesion or disorder. So in this instance the LF do not lose a dice shooting at A.

kevinj wrote:I disagree. My understanding is that, for shooting, you allocate the dice by target and then apply any reductions for cohesion or disorder. So in this instance the LF do not lose a dice shooting at A.

Bullet point four on page 94 says "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, and 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 battle group, leaving at least 1 dice (if possible) against each enemy battle group."

so in this case Unit A get 1 dice from the light foot and unit B gets the other

OK, let's consider this with a slightly simpler case. Ignore BG LH3 and assume just LF1 and LH2 are shooting at A and B. Here are the options:

1) LF1 assigns 2 dice against A and 1 against B. LH2 assigns 2 against B. 1 of the 3 dice against B is lost due to Disorder.

2) LF1 loses one of its 3 dice due to disorder and so assigns 1 against A and 1 against B. LH2 adds its 2 dice. Since there are now 3 disordered dice against B one is lost.

3) LF1 loses one of its 3 dice due to disorder and so assigns 1 against A and 1 against B. LH2 adds its 2 dice. However, since LF1 has already lost a dice due to disorder, only 2 of those shooting at B count for this so no further dice is lost and B receives 3 dice.

Which of these do you think makes most sense?

Option 3 is correct, because once the dice from battle group 1 have determined i.e., 1 to each battle group those dice are consider to be no longer dice disordered so you only get 2 dice from unit 2 which are then disorder, so no further lose of dice.

The main problem here is that bullet point 3 is read before bullet point 4, but you need to action bullet point 4 before bullet point 3

I think the full answer to the original query is

The LF unit A if not disrupted would have 2 dice on A and 1 on B.
However, as it is disrupted it loses a dice. It must lose one of the dice on A therefore it has one dice on A and one dice on B.
LH1 has 2 dice which while disrupted are not reduced as a BG as they have not got 3 which are assigned as 2 'disrupted' dice to B
LH2 has 2 dice which while disrupted are not reduced as a BG as they have not got 3 which are assigned as 2 'disrupted' dice to B.

That gives 1 dice on A, 1 dice on B and 4 'disrupted but not reduced within the BG' dice on B.
Bullet 3 on page 93 then says "if more than 1 BG is shooting at the same target add the total number of bases to which 1 per x reductions apply before calculating the number of dice to roll." So the 4 becomes a 3.

Final result 1 dice on A and 4 on B.

I suppose strictly as bullet 3 is before bullet 4 the sequence is 5 dice on B which is being shot by multiple BGs. Therefore, lose one dice which has to be off one of the LH. LF has 3 dice reduced to 2 due to disruption deducting the one firing at B leaves one on A. Which gives the same answer 1 dice on A and 4 on B.

Paul

elysiumsolutions@fsmail.n wrote:I suppose strictly as bullet 3 is before bullet 4 the sequence is 5 dice on B which is being shot by multiple BGs. Therefore, lose one dice which has to be off one of the LH. LF has 3 dice reduced to 2 due to disruption deducting the one firing at B leaves one on A. Which gives the same answer 1 dice on A and 4 on B.

apart from this line in Bullet point 4, which says "first determine the total number of dice the battle group should lose." therefore bullet point 4 comes before 3

So the shooting of a 4 base BG of skirmishers is never affected by dissorder or disruption ??

Are you sure about that chaps.....

So the shooting of a 4 base BG of skirmishers is never affected by dissorder or disruption ??

Are you sure about that chaps.....

Sorry Hammy thats a bit cryptic for me.

The not reducing 2 disrupted units of LH is a common error, see Phils comment on the other shooting thread which I think is throwing some people.

My final result of 1 dice on A and 4 on B is from a total of 7 disrupted dice so I assume you are refering to the other answers.

How many dice do you think they should get ?

Paul

elysiumsolutions@fsmail.n wrote:So the shooting of a 4 base BG of skirmishers is never affected by dissorder or disruption ??

Are you sure about that chaps.....

Sorry Hammy thats a bit cryptic for me.

The not reducing 2 disrupted units of LH is a common error, see Phils comment on the other shooting thread which I think is throwing some people.

You and Leslie seem to be saying that a 6 base BG suffers twice.

I always work on the basis that you look target by target as per Kevin's statement. I can see that there are situations where this seems odd but the logical extrapolation of the other way of looking at it i.e. you lose shooting dice by shooting BG is that 4 base skirmisher BGs never lose dice from disruption when shooting.

How would you treat three disrupted BG of LH shooting two target BGs such that each target had one full BG and half one other BG shooting at them?

How would you treat three disrupted BG of LH shooting two target BGs such that each target had one full BG and half one other BG shooting at them?

Bullet 3 on page 93 says "if more than 1 BG is shooting at the same target add the total number of bases to which 1 per x reductions apply before calculating the number of dice to roll."

So I would say that you have 2 dice on each BG.

I like Phil had always allowed 2 disrupted LH BGs 4 dice previously as I only reduced by BG but then I read the above rule and realised they should only get 3. This has been confirmed by RBS on the forum recently.

Paul

How would you treat three disrupted BG of LH shooting two target BGs such that each target had one full BG and half one other BG shooting at them?

In that case each target would normally have 3 dice shooting at them. However, since in both cases a 1 per 3 reduction applies to all of the dice, each receives only 2 dice of shooting (assuming 4 base LH BGs).

hammy wrote:You and Leslie seem to be saying that a 6 base BG suffers twice.

No once you have done a reduction those dice are no longer counted a disrupted, so if unit 3 was not there unit B would get three dice against it one from Unit 1 and two disrupted dice from unit 2

only taken the reduction once.

so when you add unit 3 back in you only get another 1 dice giving you 1 from unit 1 and 3 from unit 2 and 3

Leslie