Ok. Here’s the way combat in Spartacus works.
You have from 1 to six speed, attack, and defence die (d6es).
The attacker rolls their attack die and arranges them in order.
The defender rolls their defence die and arranges them in order.
Line them up against each other, one to one, starting with the highest.
If the attack die beats its corresponding defence die, then that’s an hit.
If the defender has fewer defence die than the attacker has attack die, then the tail end is “undefended”, and the defence roll is treated as if it were a 2 (attacker needs 3 or better).
Each hit means you must sacrifice one of your dice. You must retain at least one attack, defence, and speed die. If you are dropped to 2 (or fewer) total dice, then you lose.
What’s your strategy?
Better stated, if you have Aa attack, Ad defence and As speed dice, and your opponent has Ba, Bd and Bs dice, and you have just taken N hits, which dice should you remove? I think it’s safe to say that removing from speed is always best, so the problem reduces down to As and Bs both == 1. That’s 216 cases assuming a maximum of 6 dice.
Why don’t we multiply it by a possible 6 hits? Because I assume that taking more than 1 hit reduces down to taking only one hit at a time, although I could be wrong about this – there may be local maxima. But I doubt it.
So we have two problems. Given Ba and Ad, what is the distribution of how many hits you are likely to take?
Given that distribution, what move gives you the best likelihood that you will win?
I don’t know. But Sunday is laundry day so I intend to find out.