OK - so I don't know what's intuitively correct here. Assuming random dice, which of the following two troop layouts are easier to conquer?

--> 30 total troops

Layout A: 3 terts, 10 troops each

Layout B: 13 terts, with a assortment of 1, 2, and 3 troops each

To answer this question, I've tried comparing A and B at their extremes.

Layout A1: 1 tert, 30 troops on it

Layout B1: 30 terts, each holding 1 troop

Comparing A1 and B1, attacking B1 has the "advantage" that each attack will result in a maximal attack of 3 attackers vs 1 defender. So every attack has optimal odds to succeed. In A1, until and only if the tert has only 1 troop remaining, does it defend with anything other than 2 troops. So almost or sometimes all the attacks are of the less optimal 3v2 (assuming a big stack is attacking). As such, it would seem the B1 layout to be preferable for an attacker.

B1 however, requires the attacker to leave an "occupying" troop for each tert he conquers, so his stack gets reduced as he conquers. Put another way, if you did 32v30 for A1, there are great odds a conquer will occur, but if you did 32v(1x30) (so a stack of 32 attacking 30 terts each with one troop), I don't think anyone would count on a total victory. So in A1 vs B1, an equivalently-sized stack is more likely to defeat A1 compared to B1.

Does this give us any more information on which of A or B is a preferable target? I can't figure it out.

Thoughts?