People have been worried that since we don't seem to have a built in point system like Netrunner that playing a round of each side could result in a tie.

I think we do already have a built in thoguh.

abs(((# obj in LS victory pile) * 3) - (Death Star dial value)) == Strength of Win

One could argue that since the DS dial advances on a slope the more Objectives are in thier victory pile that maybe the LS side of the equasion should be sloped as well. I think that would make the DS SoW numbers too high as I think the number of times you can lose on the LS with 2 Objectives will be pretty high. Perhaps it should be a reverse slope however, 6 for the first Obj, 4 for the second, and 2 for the last.

Interested in any thoughts.

I don't understand your equation. Is that for one game, or the two-game round against a single opponent?

So if I win as the light side, meaning I have 3 objectives, but the Death Star Dial is on 11, I get a negative score? 3 x 3 = 9; 9 - 11 = -2. Is that right? What am I missing?

Budgernaut said:

Abs == absolute value. And damnit I meant to make it *4 not *3. that gives 12 points for either side. And yes it's just one game. If you win both games it largely irrevalant, but if you split the wins the one with the greater SoW for their win will win the round. This could also be used for ranking between rounds as well, to definitevly rank the people who won both games for pairing function.

Got it.

|(# objectives in LS victory pile x 4) - death star dial|

I like it. It seems straightforward enough.

Budgernaut said:

It's funny, I didn't use pipes for abs because I thought it would blend in to easy, and would be less understood to begin with

So I take it a DS victory is: 12 - (# number of DS objectives destroyed x 4)

Eg. Light side destroy 2 DS obs. So 12 - 8 = DS +4 ?

Don't want to put a spanner in the works but what would happen to the player setting there deck around trench run, all good if the win but if they lose the potentially can lose by 12.

Same could be said if the DS player has out 'Heart of the Empire' from the start and the LS player focuses on it. Both of them are effectively all or nothing gambles.

True.

But my top statement is correct?

Roman_Sandal said:

Yes, that is a correct interputation.

Why not just play a best of 3 round? With decks switching in between each game?

ScottieATF said:

This is what I figured it would be like. However you have to realize unless one player wins both of the first rounds, there is going to be a 3rd round. Who plays which side on that round? I say there should be a random selection to see which side plays which. If you guys don't like random selection, allow the player who won the first game to decide which side they would like to play.

We are running a tourny tomorrow but we are running it like the old star wars ccg tournies. Using the above for and against system. Therefore no need for a tie breaker. I dnt think the net runner style works with this game. NR works because the scoring system is constant. There are too many varibles in SW.

sinnerfold said:

The problem with that is if the metagame ever reaches a state where there is a LS/DS imbalance. I'm sure it will happen. If, for instance, the DS currently has an advantage then that method might as well be "whoever played DS first gets the advantage," which might as well be random. I'm personally in favor of not switching within matches, but alternating which side you play between matches (i.e. each person would play the same number of matches as LS or DS, but each game would stand by itself, no best-out-of-X series).