Slitherine

Hi to all, in a recent game of v2, we had the following situation: a bg disrupted was left with 5 bases, but fighting in melee 2 enemy bgs...3 bases are engaged with the one and 2 with the other, the question is the dice lost is from the 3 bases fighting the first enemy bg as a result of the rule of losing 1 die per 3...or instead from the other 2 dice to the second group as a result of keeping more melee dice in proportion to the engaged bases!

dices are lost BEFORE their alocation between enemies so you have 5 dices that are reduced to 4 THAN you have to allocate giving at least 1 to each opponents , yoc can choose to give 2 and 2 or 3 and 1 (no 1 and 3 as the number of dices allocated to one units can't exceed the max you generate with bases facing it)
Ciao
Matteo

thx Matteo!

MatteoPasi wrote:dices are lost BEFORE their alocation between enemies so you have 5 dices that are reduced to 4 THAN you have to allocate giving at least 1 to each opponents , yoc can choose to give 2 and 2 or 3 and 1 (no 1 and 3 as the number of dices allocated to one units can't exceed the max you generate with bases facing it)
Ciao
Matteo

That's not strictly true though Matteo...

You allocate dice and then remove for disorder. Aggregating dice removal such that you always have at least one dice against each opponent.

Just to be clear here.

The dice are allocated and are then lost in proportion for the disruption.

So initially 3 dice versus 1 Bg and 2 dice versus the other.

Dice are then lost for disruption at a rate of 1 per 3.
So there are 5 dice in total so 1 must be lost.
Now look at how the dice have been split up. There is a group of 3 so 1 dice MUST be lost from this group.

This is a fairly simple dice allocation problem with only one possible result.
2 dice versus one BG and 2 dice versus the other.

Peter

What happens where (in this example of one BG vs two bgs) you start with only 4 dice? Allocate 2 dice per opponent and then remove one due to disorder. The owner of the bg gets to decide which dice is removed leaving 2 + 1?

zoltan wrote:What happens where (in this example of one BG vs two bgs) you start with only 4 dice? Allocate 2 dice per opponent and then remove one due to disorder. The owner of the bg gets to decide which dice is removed leaving 2 + 1?

Yep.

philqw78 wrote:

zoltan wrote:What happens where (in this example of one BG vs two bgs) you start with only 4 dice? Allocate 2 dice per opponent and then remove one due to disorder. The owner of the bg gets to decide which dice is removed leaving 2 + 1?

Yep.

Just can't bring yourself to say 'Zoltan is correct' can you....

Nope

A master of brevity .