Slitherine

we have three bgs of 8 lb (x,y and z) acing 4 bgs of 6 lb (1,2,3 and4). same frontage. bg 1 of the 4 is fragged, so is bg 4 of 4.

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how many dice are fired at x, y and z and why?


TIA hunter

BG 1 will fire all bases at BG X - 3 front rank bases (3 dice) + 3 rear rank bases ( 1 die) / 2 because fragged = 2 dice, with an extra rear rank base (fragged). BG 2 fires one front rank base at BG X (1 die) but the one rear rank base is combined with the extra rear rank base of BG 1 (1 die). There is no reduction for being fragged as there is only one fragged base contributing. So BG X takes 4 dice from BGs 1 and 2. BG Y takes 2 front rank dice and 1 rear rank dice each from BGs 2 and 3. for a total of 6 dice, and BG Z takes 4 dice from BGs 4 and 3 in a mirror image of BG X.

Sounds good to me.

thanks fellahs. thats how we played it.
hunter