Slitherine

Not sure on this one. Is there rounding when determining whether a battlegroup took 1 HP2B/3B?

For instance, If a BG with 4 bases is to meet the 1 HP3B criteria, does it take 2 hits or one, as it doesn't have a full 3 bases...

Thanks
Cole

There is no rounding in FoG

If you have to loose 1 dice for every 3 then with 5 dice you loose 1 and with 6 you loose 2

If you need to take 1 hit for every 3 bases then with a BG of 7 bases 2 hits is not enough but 3 is.

Hi Hammy

In the glossary on pg134 of rules it does say in ref to '1 Dice per X bases' & '1 Dice per X' that there is rounding down and rounding up respectively.

I know that 1HP2B/1HP3B is slightly different on pg35 in that no rounding mentioned but I do confess I am a bit confused at times as to how many dice to subtract in different situations on occassions

Cheers
Gary

Although we say there is no rounding up or down, in general the following normally applies:

a) When calculating how many dice you use before losses for disorder etc you round down:
i.e. you 'gain' 1 dice for each full 1, 2 or 3 bases that you have.
So if you have 8 dice counting 1 per 3 you get 2 dice [round down!]

b) When calculating how many dice you keep when disrupted etc you round up.
i.e. you 'lose' 1 dice for each (full) 2 or 3 bases that you have.
So if you have 8 dice losing 1 per 3 you get 6 dice (losing 2) [round up]

If you have both - you first calculate how many dice you should have in (a).
Then, using that total, you work out how many dice you lose in (b)


e.g.
You have a disrupted BG of 6 bowmen shooting within effective range:
You get 1 dice for each front rank (3)
You get 1 dice for each 2 in the back rank (1)
Total: 4
You lose 1 dice per 3 (1)
Leaving you with 3 shooting dice.

It says "lose 1 dice per x" = lose 1 dice per full x dice, i.e. round up.

It does not say round up. It says lose one dice for every two dice you have or in other words round up.

I repeat there is NO rounding in FoG, if you need to do so many hits per so many bases then if there is an extra base over so many you have to do another lot of hits. There is effective rounding but the rules do not at any point say that you round. They do say that in other words you round.

Personally I think the best way to remember everything on this topic is not to think about rounding at all as sometimes the effect is to round up and other times to round down.

Hi Hammy/Terry

Ok I think I have it now there is NO rounding in FOG except in other words when you round up or down

Terrys examples give the clarity I sought

Maths was never my strong point and I hate having to remove my socks every time I count above 10 !

Cheers
Gary

I think that part of the reason for not having rounding is to make the maths simpler.

If you consider Terry's example of the archers and work it by counting out the dice based on the bases shooting and put them out to roll then remove one form each set of three you will get the right result.

I am not convinced the question has been answered - lots has been said about 1 dice per x bases, and loose 1 dice per x.

The glossary defines 1HP2B & 1HP3B in terms of "1 hit per x bases", and defines how many bases you have.

If I have 4 bases what is 1 hit per 3 bases?

The nearest definition in the rules is 1 dice per 3 bases which gives the value of 1.

The definition of 1 dice per 3 bases is effectively "1 dice for every 3 full bases."

I have played that 1 hit per three bases is "1 hit for every 3 full base plus one hit if there are any left."

I'm glad this thread was posted, I was just going to ask myself.

My uncertainty was the modifier on the cohesion test >= 1HP3B from close combat.

2 base elephant unit (counts as 4) did you need 1 or 2 hits to get this modifier. You have confirmed that you do indeed need 2 hits on the battlegroup to count this modifier.

I don't care whether the term "rounding" is used or not, as long as players understand when the number of bases is not an exact multiple of the appropriate factor, how they should count it.

I think it would be a good idea to take whichever answer to this thread is deemed the least confusing and add it to the FAQ.

Andy
PS Where is the QRF i'm impatient!!!

sagji wrote:I am not convinced the question has been answered - lots has been said about 1 dice per x bases, and loose 1 dice per x.

The glossary defines 1HP2B & 1HP3B in terms of "1 hit per x bases", and defines how many bases you have.

If I have 4 bases what is 1 hit per 3 bases?

The nearest definition in the rules is 1 dice per 3 bases which gives the value of 1.

The definition of 1 dice per 3 bases is effectively "1 dice for every 3 full bases."

I would say that your interpretation above is correct, and so...
1HP3B for 4 bases of Cv = 1 hit
1HP3B for 3 bases of Cv = 1 hit
1HP3B for 2 bases of Cv = 0 hits, etc.

Exceptions as listed in the Glossary, of course.

Cheers,
Scott

Scott,

I think you have given the opposite interpretation to the one that Hammy gave at the beginning of the thread.

My understanding is that Hammy's is correct?

Or am I confused?

Andy

miffedofreading wrote:Scott,

I think you have given the opposite interpretation to the one that Hammy gave at the beginning of the thread.

My understanding is that Hammy's is correct?

Or am I confused?

Andy

Hammy appears to be talking about something completely different, i.e. dice per x bases, rather than hits per x bases. Hence the confusion.

Cheers,
Scott

OK,

This is my understanding and the way I have always played.

Hits per X bases are at least 1 hit per x bases so to get 1HP3B you need:

1 hit for 1-3 bases
2 hits for 4-6 bases
3 hits for 7-9 bases
etc.

Dice per x bases are dice for every full x bases so if you get 1 dice per 2 bases then:

1 base = no dice
2-3 bases = 1 dice
4-5 bases = 2 dice

Losing x dice per y means that for every x dice you lose y so if you have to lose 1 per 3 if you have:

1 or 2 dice you loose nothing
3-5 dice you lose 1
6-8 dice you lose 2

I have never thought of these any other way and having looked again in detail at the glossary I can see there could be other interpretations but they have never occured to me.

hammy wrote:OK,

This is my understanding and the way I have always played.

Hits per X bases are at least 1 hit per x bases so to get 1HP3B you need:

1 hit for 1-3 bases
2 hits for 4-6 bases
3 hits for 7-9 bases
etc.

Yes, I agree 100% with Hammy. Please ignore my earlier post, which only muddies the waters!

Cheers,
Scott

I agree with everyone......

My head hurts, can i be excused?

hammy wrote: Hits per X bases are at least 1 hit per x bases so to get 1HP3B you need:

1 hit for 1-3 bases
2 hits for 4-6 bases
3 hits for 7-9 bases
etc.

This is correct.

As Hammy says 1 HP3B requires >= 1 hit per 3 bases.

For example, exactly 1 hit per 3 bases on a BG of 8 bases is 1/3*8 = 2.67 hits.

As there are no fractional hits, 3 hits are required to get 1 HP3B on a BG of 8 bases.

Woah! You are all overcomplicating it by trying to treat two different concepts as the same thing, which they are not.

They are just different and both very simple if you leave them that way.

When sorting out numbers of dice....

You have things that are per X bases or per x dice. You get or lose these for having a full set of x. So if you get 1 dice per 2 bases you get 1 dice for every set of 2 bases you have. This is not rounding, but is the same as rounding down for those who prefer to think of it that way. I simply say that 7 bases gives me 3 sets of 2 for 3 dice, and a spare that is wasted. End of story.

Similarly lose 1 dice per 3 is exactly what is says on the tin. For every 3 dice you have lose 1. Thus if you have only 2 dice lose none. If you have 8 dice lose 1 for the 1st 3 and 1 for the 2nd 3 and thats it. Again this isn't rounding its exactly what is says - take away very 3rd dice. or every 2nd dice.

When calculating casualty rates 1HP3 etc.

This say in the table the factors are for >= 1 hit per 3. Its the >= you are missing that makes the concepts different. There is no rounding, this is level to get equal to or above.

So if you have 4 bases 1.33 hits is 1 hit per 3. So 2 is >= 1HP3 and 1 is <1HP3>=) 1hit per 3 bases[/b]. So for a 6 this is 2 or more, for a 4 this is also 2 or more. But 1 per 2 is 3 or more for 6 and 2 or more for 4. So the 4s test and suffer a - every time, and thus are worse off.

Hope that helps. I have never found it confusion myself but heh! we all think we are clear when we read our own stuff!!

Shall, speaking as a newbie to the period and this only being the 3rd ancient ruleset I have played I found the concept the way its worded in the rules a little confusing. Not in what it said but perhaps in what it meant. An example such as your would have cleared it up in a hurry. I am seeing now that you need to take the rules literal. >=3 means 3 or higher sort of thing. When I first read it and played a game we used partials or rounding. But now I see the light : )

But once again thanks to everyone taking the time to explain things a little more as I know for me its done a world of good in helping me undestand the rules. And hopefully threads like these will help new people also looking for answers. If nothing else its feedback for you guys.

Brian

shall wrote:Woah! You are all overcomplicating it by trying to treat two different concepts as the same thing, which they are not.

When calculating casualty rates 1HP3 etc.

This say in the table the factors are for >= 1 hit per 3. Its the >= you are missing that makes the concepts different. There is no rounding, this is level to get equal to or above.

So if you have 4 bases 1.33 hits is 1 hit per 3. So 2 is >= 1HP3 and 1 is <1HP3>=) 1hit per 3 bases[/b]. So for a 6 this is 2 or more, for a 4 this is also 2 or more. But 1 per 2 is 3 or more for 6 and 2 or more for 4. So the 4s test and suffer a - every time, and thus are worse off.

Hope that helps. I have never found it confusion myself but heh! we all think we are clear when we read our own stuff!!

Where do you get 1.33 - there is no division in the rules. The rules say "1 hit per 3 bases" not 1/3rd of a hit per base. If you have 4 bases then 1 hit is >= 1 hit per three bases as you have only 1 lot of three bases.

I have always played it as you "round up" but I don't think the wording clearely supports - especially as the almost identical wording of "1 dice per 3 bases" is clearely stated as one for every full 3.

sagji wrote:

shall wrote:Woah! You are all overcomplicating it by trying to treat two different concepts as the same thing, which they are not.

When calculating casualty rates 1HP3 etc.

This say in the table the factors are for >= 1 hit per 3. Its the >= you are missing that makes the concepts different. There is no rounding, this is level to get equal to or above.

So if you have 4 bases 1.33 hits is 1 hit per 3. So 2 is >= 1HP3 and 1 is <1HP3>=) 1hit per 3 bases[/b]. So for a 6 this is 2 or more, for a 4 this is also 2 or more. But 1 per 2 is 3 or more for 6 and 2 or more for 4. So the 4s test and suffer a - every time, and thus are worse off.

Hope that helps. I have never found it confusion myself but heh! we all think we are clear when we read our own stuff!!

Where do you get 1.33 - there is no division in the rules. The rules say "1 hit per 3 bases" not 1/3rd of a hit per base. If you have 4 bases then 1 hit is >= 1 hit per three bases as you have only 1 lot of three bases.

I have always played it as you "round up" but I don't think the wording clearely supports - especially as the almost identical wording of "1 dice per 3 bases" is clearely stated as one for every full 3.

There is clearly room for confusion, so another one for the FAQ methinks as it is rather critical to the correct balance of the rules.