For example, in this basic, algebraic summary of swing die theory, a formula is given. I'm copying it verbatim -- except for adding the bullet points to make the definitions easier to read, adding one clarification, and adding some emphasis on a phrase that I think is mistaken:
Ax<=Bx-Di+3/2*(An)-1/2 where:The formula is certainly a correct, if somewhat ungainly restatement of the familiar two-thirds rule. While the author excludes poison dice, it's not that hard to add them into this (treat them as worth (-1/2) their size, as if captured, for the purpose of computing Di). And "An" could be the sum of two dice just as easily as a single die.
- Ax: Player A's swing die size,
- Bx: Player B's swing die size
- Di: Player A's initial hand size minus Player B's initial hand size (greater than -1 by definition), [clarification: the intention in this formulation is that this does not include the swing dice of either player]
- An: The size of one of Player A's dice which must be kept for him to win. This value can vary from Player A's smallest to largest die, including his swing die. Optimally, Player A wants An to equal his smallest die, while Player B wants An to equal Player A's largest die. However, in practice one must vary An to find the optimal swing die that fits in one's swing range
Everyone playing buttonmen should know the basic algebraic relationship between the total number of sides each player starts with and the victory conditions on the game, though you don't necessarily have to phrase it this formally. But the real unanswered question is:
WHAT SHOULD YOU CHOOSE "An" TO BE?
All things being equal (and maybe later I'll discuss the many exceptions to this), you'd like to be able to make Ax (your swing die) as large as possible, so long as you satisfy the inequality. Picking an unrealistically low value for An will needlessly restrict your value for Ax.
Example: if I'm playing slamkrypare (t1 10 10 z12 Y -- I've never actually had the pleasure of playing this button, but it illustrates my point)
vs. say, Kitty Cat Seven (4 6 8 10 X if you prefer, you can think of this as Ngozi)
and let's KCS has set her X swing to x=8 and soundly thrashed you, so you get to reset your
swing die.
if you believe "Optimally, Player A wants An to equal his smallest die", then you'll set An to equal 1 and the above formula (with Bx=8, Di=5) yields Ax <= 8 - 5 + 3/2 - 1/2 = 4. And it IS true that if you set s.'s swing die to X=4 in this matchup, you'll win the game if you capture all of KCS's dice and have at least 1 die uncaptured, whereas, if you picked anything higher for your swing, you would still lose if you captured all of KCS's dice, but had only your 1 sided die left at the end. BUT HOW LIKELY IS THAT TO ACTUALLY OCCUR? In order to win the game with ONLY your 1 sided die left uncaptured, on the penultimate move, your opponent would have had to capture your second-to-last die and rolled a 1 [actually, since your 1 sided die is a trip die here, there are some other possibilities, I suppose -- but the basic point is still valid == the idea that you'd lose all your mid-to-large dice and yet manage to strike the death blow with your littlest die is somewhat unlikely. You are much MORE likely to have one large die left uncaptured after the first few moves that you try to protect and avoid rerolling to keep it out of
range of your opponent. So, if you do the calculation with An = 10 [your next largest die, aside
from the swing die itself], you come up with Ax <= 8 - 5 + 30/2 - 1/2 = 17.5 and setting the swing equal to 17 gives a much more powerful button -- and you still win if you manage to keep any one of your dice (except the 1-sided die). If you are lucky, the 17 will be out of reach of KCS in the beginning; and even if not, she'll have to take it out early and expose herself to your speed die or at least give you a chance to get one of your 10 or 12 sided die out of reach. This example is muddied a bit since the t1 doesn't count for initiative... and setting your Y to 4 will increase your chances of getting initiative from "unlikely" to "somewhat unlikely" [though if someone does a computation or simulation to get hard numbers here, I'll stand corrected]. Anyhow, the same basic principle holds for other buttons; while it's certainly true that I'd like to be able to win with as few dice-sides as possible, I ought to stop to consider what a realistic endgame between these buttons is likely to be. If I'm playing Yeti (10 20 30 30 X) vs. Crab (8 10 12 f20 f20) , the formula above (modified just to account for the fact that Crab has no swing die), tells me I want my X to be <= -20 +(3/2)*An - 1/2 but what is realistic? An = 10 doesn't work, X would have to be negative. An=20 yields X>=9.5, i.e. X=9 is best. An=30 would permit Xn<=24.5, i.e. X=20 would be work. So there are really two options here: set X=9 and try to protect any one of the three big dice 20 30 30; or set X=20 and try to either protect the two big dice 30 30, or keep ANY two dice at the end. It isn't obvious to me which is the better game, though I personally feel more comfortable with the second option.
Example: if I'm playing slamkrypare (t1 10 10 z12 Y -- I've never actually had the pleasure of playing this button, but it illustrates my point)
vs. say, Kitty Cat Seven (4 6 8 10 X if you prefer, you can think of this as Ngozi)
and let's KCS has set her X swing to x=8 and soundly thrashed you, so you get to reset your
swing die.
if you believe "Optimally, Player A wants An to equal his smallest die", then you'll set An to equal 1 and the above formula (with Bx=8, Di=5) yields Ax <= 8 - 5 + 3/2 - 1/2 = 4. And it IS true that if you set s.'s swing die to X=4 in this matchup, you'll win the game if you capture all of KCS's dice and have at least 1 die uncaptured, whereas, if you picked anything higher for your swing, you would still lose if you captured all of KCS's dice, but had only your 1 sided die left at the end. BUT HOW LIKELY IS THAT TO ACTUALLY OCCUR? In order to win the game with ONLY your 1 sided die left uncaptured, on the penultimate move, your opponent would have had to capture your second-to-last die and rolled a 1 [actually, since your 1 sided die is a trip die here, there are some other possibilities, I suppose -- but the basic point is still valid == the idea that you'd lose all your mid-to-large dice and yet manage to strike the death blow with your littlest die is somewhat unlikely. You are much MORE likely to have one large die left uncaptured after the first few moves that you try to protect and avoid rerolling to keep it out of
range of your opponent. So, if you do the calculation with An = 10 [your next largest die, aside
from the swing die itself], you come up with Ax <= 8 - 5 + 30/2 - 1/2 = 17.5 and setting the swing equal to 17 gives a much more powerful button -- and you still win if you manage to keep any one of your dice (except the 1-sided die). If you are lucky, the 17 will be out of reach of KCS in the beginning; and even if not, she'll have to take it out early and expose herself to your speed die or at least give you a chance to get one of your 10 or 12 sided die out of reach. This example is muddied a bit since the t1 doesn't count for initiative... and setting your Y to 4 will increase your chances of getting initiative from "unlikely" to "somewhat unlikely" [though if someone does a computation or simulation to get hard numbers here, I'll stand corrected]. Anyhow, the same basic principle holds for other buttons; while it's certainly true that I'd like to be able to win with as few dice-sides as possible, I ought to stop to consider what a realistic endgame between these buttons is likely to be. If I'm playing Yeti (10 20 30 30 X) vs. Crab (8 10 12 f20 f20) , the formula above (modified just to account for the fact that Crab has no swing die), tells me I want my X to be <= -20 +(3/2)*An - 1/2 but what is realistic? An = 10 doesn't work, X would have to be negative. An=20 yields X>=9.5, i.e. X=9 is best. An=30 would permit Xn<=24.5, i.e. X=20 would be work. So there are really two options here: set X=9 and try to protect any one of the three big dice 20 30 30; or set X=20 and try to either protect the two big dice 30 30, or keep ANY two dice at the end. It isn't obvious to me which is the better game, though I personally feel more comfortable with the second option.
