Basic Swing Dice Theory

by James Ernest
May 1999

Here’s the basis of Swing Die theory, as I first wrote it up for a colleague in Berkeley, CA. I’ve made some adjustments to the discussion, but it’s the basic theory responsible for the Avis chart above. If you really hate math, you’ll really hate this.

This discussion pertains to the ideal choice of Swing Die versus a particular opponent. It assumes that you know not only who you are fighting, but also what their Swing Die is. In tournament play, both players choose their initial Swing in secret, knowing their opponent’s fighter. After the first round, only the loser is allowed to re-tune, which makes this chart the most useful.

It’s understood that the bigger of two characters will have to keep a die to win. More specifically, he will have to keep a certain number of -sides- to tie the game, which is equal to two thirds of the difference between the characters’ side totals, or (big-little)*2/3. Actually, the general case is (A-B)*2/3 where A isn’t necessarily the bigger character, and when the result is negative it means that B must keep that number of sides.

The exception to this rule is when the two characters are “roughly equivalent,” i.e., either character will win by keeping his smallest die. In this case, either character -must- keep one die to win, because there is no existing way for both players to take all of each others’ dice, and no existing way for both characters to have dice remaining at the end.

Here’s an example. S(Avis6)=36. S(Hammer4)=62. Hammer must keep (62-36)*2/3, or 17.3 sides to tie, more to win. This means either keeping a 12 and a 6 (18 sides, the closest possible thing above 17.3) or, more likely, a single 20.

(Notations above: Avis12 means Avis with a 12 as her Swing Die. S=Sum of all sides.)

Avis can go for more power in this case and not compromise her position. In fact, she should choose a 12, because Hammer must still keep either 12+6 or 20 sides to win: (62-42)*2/3=13.3, which is essentially equivalent to the 17.3 she got with her 6. Avis gets more attack power with no loss of scoring position by choosing a 12 over a 6. If, however, Avis chose a 20, Hammer could tie by keeping only 8 sides.

It’s a matter for further debate whether the number of initiative rolls Avis will lose makes up for her added attack power, but I don’t think it does. The primary liability of larger dice is that they are worth more points, and this is irrelevant in the range of 6 to 12 in this case.

So, when playing a small character, the ideal choice is to pick the largest possible die, while still forcing your opponent to keep the most possible points.

When playing a larger character, the goal is to choose the largest possible die without being forced to keep more sides. When Avis faces an even smaller character, Stark4, she must keep some sides to win. S(Stark4)=22. Avis is best off choosing either an 8, 10, or 12 in this case. Versus Stark4, these dice come up as -8, -9.3, and -10.7 for Avis. At Avis8 (-8), she can win by keeping any die larger than 8, and ties if she keeps her 8. At Avis10 (-9.3) she can win by keeping any die 10 or greater, which is three of her dice. At Avis12 (-10.7) she must keep one of her two 12’s to win. Of the three, I prefer Avis10, where she wins by keeping any one of her three largest dice.

When two characters are roughly equivalent, meaning that either must keep one die to win, I prefer to take the smallest possible die that won’t change that. Vs. Karl4, Avis can choose any die between 6 and 12, and the result is the same: Either character must keep one die to win. In this range, we’re assuming that Avis should try to go as fast as she can, and choose the 6. If she goes first more often, and the two characters are the same size, she’ll keep a die more often. That’s the theory, anyway.

All of this is mushy, and the Avis chart above is the result of squinting at the numbers, and the dice each character has, and taking the best shot. But it’s at least a starting point.