Apportioning lost dice for disrupted Pike BG

[daleivan]

Several times in the past couple of battles Mark Sieber and I have played one of his pike BGs loses a base and goes disrupted while fighting one of my Roman legionaire BGs. The pike BG is 8 bases and in four ranks. Example below.


P
PP
PP
PP
LLL
LLL

Losing one base drops the POA one of the files, while of course being disrupted causes the BG to lose 1 dice per 3 bases. With four bases in total fighting, he would lose one dice, but how do you determine which POA gets two dice and which gets only 1 in this circumstance? The proportion of bases fighting is equal (unless you take into account which file is deeper--however each file is still fighting with two ranks). I've been giving the benefit of the doubt to Mark (who, honorable as always would prefer a definite ruling and letting his better POA fight with two dice.

Dale

[hammy]

In this situation each file should get 2 dice, you have to lose one per three. Neither file has three on it's own and they both have the same number so the owner of the pike gets to choose which dice to lose.

If the pike was a BG of 11 on a frontage of 3 bases and disrupted then there would be 4 dice at full POA and 2 with one POA less. This time you have more than 3 dice at full POA so you must lose 1 from them leaving 1 dice at full and 2 at the lower POA so you must lose one of the two at the lower POA.

You only get to choose when you have equal numbers of dice on each POA.

I find that the best way to work it is to lay out the dice behind the BG fighting with different colours for each POA. Then lump them together and if you have three of one colour lose one and put the other aside to roll, look at the remaining dice, see if you have three more the same, if so repeat, if you don't have three the same look at what is left, if there are three or more dice in total then lose one of the dice that you have the most of and put two more aside to roll etc. Try it next time you have a complex disruptec combat, it is actually quite simple and you can do it in your head very quickly 9 times out of 10.

[daleivan]

In this situation each file should get 2 dice, you have to lose one per three. Neither file has three on it's own and they both have the same number so the owner of the pike gets to choose which dice to lose.

If the pike was a BG of 11 on a frontage of 3 bases and disrupted then there would be 4 dice at full POA and 2 with one POA less. This time you have more than 3 dice at full POA so you must lose 1 from them leaving 1 dice at full and 2 at the lower POA so you must lose one of the two at the lower POA.

You only get to choose when you have equal numbers of dice on each POA.

I find that the best way to work it is to lay out the dice behind the BG fighting with different colours for each POA. Then lump them together and if you have three of one colour lose one and put the other aside to roll, look at the remaining dice, see if you have three more the same, if so repeat, if you don't have three the same look at what is left, if there are three or more dice in total then lose one of the dice that you have the most of and put two more aside to roll etc. Try it next time you have a complex disruptec combat, it is actually quite simple and you can do it in your head very quickly 9 times out of 10.

Got it. Thanks!

[philqw78]

What happens then if, fighting 3 bases wide, you have a 4 - 2 split on the dice and the 4 dice and 2 dice are at different POAs? If they become disrupted it says lose dice proportionately. Which can't be done. Obviously the 4 dice must lose 1, but who loses the other dice? Doing it your way Hammy this would become a 2-2 split, which is less proportional than a 3-1.

[rbodleyscott]

What happens then if, fighting 3 bases wide, you have a 4 - 2 split on the dice and the 4 dice and 2 dice are at different POAs? If they become disrupted it says lose dice proportionately. Which can't be done. Obviously the 4 dice must lose 1, but who loses the other dice? Doing it your way Hammy this would become a 2-2 split, which is less proportional than a 3-1.

4-2

Take away 3 dice from the 4 (and lose one of them), leaving

1-2

2 is larger than 1 so lose the 2nd dice from the 2

So the actual dice fighting will be

3-1

[philqw78]

thanks, thought so spike

[philqw78]

thanks, thought so spike

[shall]

philqw78

wrote:
What happens then if, fighting 3 bases wide, you have a 4 - 2 split on the dice and the 4 dice and 2 dice are at different POAs? If they become disrupted it says lose dice proportionately. Which can't be done. Obviously the 4 dice must lose 1, but who loses the other dice? Doing it your way Hammy this would become a 2-2 split, which is less proportional than a 3-1.



4-2

Take away 3 dice from the 4 (and lose one of them), leaving

1-2

2 is larger than 1 so lose the 2nd dice from the 2

So the actual dice fighting will be

3-1

Indeed, which means there is chocie much less often than it may appear ... e.g. only if say 2-2 and losing 1 per 3, or 1-1 and losing 1 per 2. Ussually it is defined.

I found at Rampage that many people were thinking they had choice when they did not.

Si