Proportional Dice Loss

[iversonjm]

My memory of discussions on this forum is that when dice are lost in proportion, you first remove the dice in the majority until numbers are equal, and then can alternate. For example, if a unit with 6 dice (4 needing 4s and 2 needing 5s) is disrupted, it must roll 2 dice for 4s and 2 dice for 5s. If the same unit we fragmented, it would roll 2 dice for 4s and 1 die for 5.

Is my recollection correct?

[grahambriggs]

No, I don't think so. I believe the rules say you need to lose them in proportion.

[Jilu]

Disrupted : lose 1 out of 3 dice

ok put the dice in front of you

4 reds and 2 blues
first you lose 1 of the reds and none of the blues as you can not lose what you do not have
then you loose 1 red more as you have 5 dice left in total and you can devide 3 by 3 and again you do not lose a blue

That leaves you with 2 reds and 2 blues = 2/3 of your original dice

For fragmented it easier
as your total of dice must be half of the ones you started ;
4 reds and 2 blues
first you lose 1 of the reds and 1 of the blues
then you loose 1 red more as you have 4 dice left in total and you can devide 4 by 2 and you do not lose a blue as you canot devide 1 by 2

That leaves you with 3 dice so half the dice you started with so that is correct

[iversonjm]

No, I don't think so. I believe the rules say you need to lose them in proportion.

Please explain further. I understand that the rules say dice must be lost in proportion, but I am trying to determine how that general statement should be applied to specific circumstances.

[grahambriggs]

No, I don't think so. I believe the rules say you need to lose them in proportion.

Please explain further. I understand that the rules say dice must be lost in proportion, but I am trying to determine how that general statement should be applied to specific circumstances.

Well, for example. Let us say that you have 12 dice and that eight of them happen to be at an even POA and 4 are at a - POA - a proportion of 2 to 1. so say I use red dice for the guys on evens and blue for the - guys I have 8 red dice and 4 blue dice.

If you become fragmented, then you lose 1 dice in every 2, so you will have 6 dice. You need to lose dice so that the proportion remaining is as close to 2 to 1 as possible. You can do this exactly, so you will get 4 red dice at evens POA and 2 blue dice at a - poa. Note that because the proportion works out exactly you are able to remove half the red dice and half the blue dice.

If you then bolster your BG up to disrupted, you start off with 12 dice and lose 1 in 3, so will need to lose 4 dice. You need to do that so you lose as close to a 2:1 as possible (i.e. 66.6% red). The best way to do this when starting playing is:

take 6 of the red dice, remove 1 in 3. Put 4 reds to one side and discard 2 reds.
take 3 of the blues, remove 1 in 3. So, put 2 blues to one side and discard the other one.

You now have two reds and one blue that you haven't touched yet. You need to lose 1 in 3, so 1 dice. If you remove a red and discard it, that will leave you with 5 reds and 3 blues. the other alternative is to remove a blue and discard it. That would leave you with 6 reds and 2 blues.

To decide which is the right thing to do, you can work out the percentages. In fact 5 reds and 3 blues is closer, so that is the proportion you end up with.

I disagree with the way Jilu is doing it, by the way.

[hammy]

The way I play it and what I think the rules say although I don't have them to hand is as follows:

Consider Grahams 12 dice (8 red and 4 blue) and the BG is disrupted

There are four sets of three dice, two sets of 3 red, one set of 3 blue and one mixed set of 2 red and 1 blue.
From the two sets of three red you lose one dice each, from the set of 3 blue you lose 1 dice and that leaves the mixed set. There are 2 red and 1 blue, because there are more red than blue you lose a red meaning you have lost a total of 3 red and 1 blue so end up with 5 red and 3 blue as per Graham's calculation but a different (and I think easier way) to do it.

[iversonjm]

Thanks. That get us closer to an answer, but not quite all of the way there.

To go back to the original example (which is the one where dispute arises most often) you have six dice, 4 red and 2 blue (a 2:1 ratio), and are disrupted, losing two dice. Everyone agrees that the first die to go is red, leaving 3 red and 2 blue dice in a 3:2 ratio. Now comes the problem. You can either (a) lose a blue die (creating a 3:1 ratio) or (b) lose a lose another red die (creating a 1:1 ratio). Both solutions produce a ratio equidistant from the original 2:1 ratio. So which is it? My memory was that you take the next die from the majority, like in Jilu's solution. Granted, nothing in the rules says that; I just have that memory from forum discussion. Your solution is apparently to take the next die from the minority. Is this just because the guy rolling the dice should get to choose? This solution is no worse then mine, but I just want to get a definitive position on the rule as this particular circumstance comes up a lot and people's position on it invariably turns on whether the result benefits them or not.

[philqw78]

I agree Ivor. They just jumped to 12 dice 'cos its easier to do. In practice I have seen it done both ways, but IMO it should end up 3:1. For no ther reason than I said so, as all this maths isn't backed up too much in the rules. 1 and a third from 1 pile and two thirds from the other. Makes 1 from each

[hammy]

For the six dice question you have one set of three and one mixed set

As you have three dice the same you have to lose one of them. As the mixed set has two of one and one of the other you have to lose one of the pair meaning you end up with 3 and 1.

[iversonjm]

For the six dice question you have one set of three and one mixed set

As you have three dice the same you have to lose one of them. As the mixed set has two of one and one of the other you have to lose one of the pair meaning you end up with 3 and 1.

OK, I see the logic in that, but just for the sake of argument, why wouldn't you divide the 4 red and 2 blue into two sets of 2 red and 1 blue? That would produce a final result of 2 red and 2 blue, and it seems equally logical. Assuming that you are charging into a unit with rear-rank bow (which is where this usually seems to come up), this approach would seem to make even more sense.

BTW, text of the rules (I'm looking at FOG-R here but I think FOG-AM is the same) says: "Similarly, if fighting at different POAs against parts of the same enemy battlegroup, apportion lost dice, if possible, in proportion to the number of bases fighting each part of the enemy battle group, leaving at least 1 dice (if possible) against each part of the enemy battle group."

This doesn't seem to address Hammy's dice division strategy. It actually (and unhelpfully) doesn't say anything at all about what to do when apportioning lost dice in a proportional fashion isn't possible, which seems to be what is producing the problem here.


This may actually be a FAQ thing.

[expendablecinc]

so to take an example:

10 pike (A)
vs
4 steady spear (B) and 4 steady spear (C)

Everyone is protected

BBCC
BBCC
AAA
AAA
AAA
A

If A goes disrupted thier 4:2 (6 in total) dice changes to 4 in total.
Rules: Dice must be lost in proportion
1:1 is just as close to 4:2 as 3:1 is.
In lieu of a written rule or FAQ I would allow the owner of A to decide the split.

NB: If they chose to roll 3 dice against B and 1 against C two dice would be at a lower factor

[lawrenceg]

For the six dice question you have one set of three and one mixed set

As you have three dice the same you have to lose one of them. As the mixed set has two of one and one of the other you have to lose one of the pair meaning you end up with 3 and 1.

OK, I see the logic in that, but just for the sake of argument, why wouldn't you divide the 4 red and 2 blue into two sets of 2 red and 1 blue? That would produce a final result of 2 red and 2 blue, and it seems equally logical. Assuming that you are charging into a unit with rear-rank bow (which is where this usually seems to come up), this approach would seem to make even more sense.
.

Suppose you had 6 red and 3 blue.

With your method, you divide into three groups of 2 red, 1 blue.
You remove 1 red from each group.
You end up with three red and three blue.

This is obviously wrong as you have not removed dice in proportion to the original number.

With Hammy's method you always get the exact proportion if this is possible. If not, you always get as close as possible to the exact proportion.

In effect Hammy's method divides each batch of dice by 3 and then rounds 2/3 up to 1 and 1/3 down to zero to give the number of dice removed from each batch, and takes another dice off if your rounding error adds up to a full dice.

This question comes up frequently and Hammy's method could usefully be included in the rules version 2.

[iversonjm]

Okay, that makes sense, but its a toughy to explain to your opponent in the middle of the game. Addition in 2.0 or a FAQ would be a good thing.

[imanfasil]

I understand the method when dice are uneven. What about a case where the dice are in equal proportion. A unit getting 2 'good' dice and 2 'bad' dice has to to lose 1 in 3? Is that 2 good and 1 bad or should it be 1 good and 2 bad?

[lawrenceg]

I understand the method when dice are uneven. What about a case where the dice are in equal proportion. A unit getting 2 'good' dice and 2 'bad' dice has to to lose 1 in 3? Is that 2 good and 1 bad or should it be 1 good and 2 bad?

As both possibilities are equally close to "in proportion" , you would have a choice.