 |
| |
|
|
Edward
|
|
debeest,
I don't think Xin Yu's formula
strongUnit+(strongUnit-weakUnit) vs. weakUnit
is accurate; it's just more accurate than a raw
combatValue*hitpoints*firepower
formula.
Xin Yu's formula is going in the right direction in that the difference in the two units' strengths is more important than straight numbers would lead one to believe. However, according to Marquis de Sodaq's combinatorial equation as I understand it, the combat values are more important than hitpoint & firepower values. Any realistic simplification of the combinatorial formula has to treat combat value and firepower/hitpoints differently.
My guess is that an approximation of strength would be more like
(combatValue^2)*firepower*hitpoints
where combatValue is the unit's natural attack or defense value times all the modifiers (vet, terrain, etc.). Warning!!! I'm not saying the particular formula above is a good or correct approximation, just that a good approximation needs to (like the above formula) treat combat value differently than it treats the other two traits.
|
|
| |
|
|
|
Xin Yu
|
 |
Emeryville, CA, USA
Jan 1970
time: 21:12
|
|
Strong+(Strong-Weak)/Weak is the odd for the strong unit to win one round. If we consider HP and FP then the full formula is:
Strength Ratio=[Strong+(Strong-Weak)]*HP_Strong*FP_Strong/[Weak*HP_Weak*FP_Weak]
Roughly speaking, if Strength Ratio>1, then the strong unit wins, if <1, then the weak unit wins.
Example: 'a=8 hp=1 fp=3' unit attack 'd=3 hp=2 fp=1' unit behind city walls on forest:
Strong=3*3*1.5=13.5, HP_strong=2, FP_strong=1 (defending unit);
Weak=8, HP_weak=1, FP_weak=3 (attacking unit).
Strength Ratio=[(13.5+5.5)*2*1]/(8*3*1)=38/24=19/12. Defense unit wins.
|
|
| |
|
|
|
debeest
|
|
Um, ok. I finally made time to sit down and study the calculations a bit and I can see that the complicated single-round formula produces the same results as Xin Yu's simple single-round formula, although I haven't delved into it enough to see why the seemingly unrelated formulas equate.
However, if I understand it correctly, Xin Yu's formula ignores the tie-to-the-defender advantage. For example, suppose a 3-strength unit attacks a 2-strength unit. Xin Yu's formula says that the exact probability per round is (3 + (3-2))/2, or 4/2. The two-dice method says (3*8) * (2*8) = 384 possible combinations. Of those, the weaker unit gets a higher score 15*(15+1)/2 = 120 times. The scores are equal 16 times. The stronger unit gets a higher score the other 248 times. If the tie scores were split evenly, the totals would be 256 and 128, Xin Yu's ratio. So, Xin Yu's formula (somehow) gives the exact probability of each unit's winning a single round, EXCEPT for a modest-to-negligible defender advantage. Is that right, Xin?
And then, of course, the combinatorials formula kicks in and the stronger unit whups ass very disproportionately, because the randomness reduces toward zero the more hit points (divided by firepower) are involved in the battle.
|
|
| |
|
|
|
Marquis de Sodaq
|
 |
... no, a Marquis.
Jul 2000
time: 23:12
|
|
Debeest, in your example of 3 vs. 2, the small p (single round odds) is:
Stronger attacker: p = ((Attack * 8) - 1)/(2 * (Defend * 8), or,
p = 23/32, or almost 3/4. Xin Yu's 2/3 (4:2) is similar, but far from exactly the same.
"So, Xin Yu's formula (somehow) gives the exact probability of each unit's winning a single round" - It looks like what Xin's formula does is not find the single round odds, but approximate the big P. In this case, at least, Xin Yu's formula gives the ratio of favorable possible outcomes of the battle.
If this applies generally to any combat, it could be a good simplification to use. Obviously, testing Xin's number against P with a slew of different combat ratios would show if this is true. If so, small p and Xin's number would be opposite sides of the same coin.
------------------
"There is no fortress impregnable to an ass laden with gold."
-Philip of Macedon
|
|
| |
|
| |
|
|
|
Edward
|
|
Yes. As Marquis de Sodaq pointed out...
Xin Yu's formula is an inexact ('though quite good) guess as the victor of a whole battle. It does NOT factor in the defender tie advantage.
Marquis de Sodaq's recently posted formula is exactly correct for the victor of a single mini-combat round. It does (as it must) factor in the defender tie advantage.
Yes, they give similar results, but then again, all of the proposed formulas give somewhat similar results. (A good reason for one to use a formula that's inaccurate but simple for one to calculate on-the-fly.)
|
|
| |
|
|
|
Xin Yu
|
|
Emeryville, CA, USA
Jan 1970
time: 21:12
|
|
quote:
Originally posted by debeest on 05-02-2001 03:35 AM
So, Xin Yu's formula (somehow) gives the exact probability of each unit's winning a single round, EXCEPT for a modest-to-negligible defender advantage. Is that right, Xin?
|
Yes! The defense bonus diminishes as attack and defense (adjusted for terrain, etc.) factor increases. Roughly speaking, the chance for a tie is
minimum (attack, defense)/(8* attack * defense).
Which means, if the attack factor is weaker than the defense factor, then the chance for a tie is
1/(8 * defense);
if the attack factor is stronger than the defense factor, then the chance for a tie is
1/(8 * attack).
If you want an exact formula, the attack side should substract, and the defense side should add, half of this tie value, or minimum (attack, defense)/(16* attack * defense), from one-round odds (not the odds ratio given by my formula, but the odds. Odds ratio= win/lose, odds=win/(win+lose), they are different).
How big is the chance of a tie? A warrior against a warrior, both non-vet, the defender on grass unfortified will give 1/16, or 0.0625; if either of the two is a vet then the chance gets down to 1/24, or 0.0417. I think we can ignore it if the chance of tie is below 0.01 (happens when either the attack or defense (adjusted for bonus) is 7 or up), unless the attack factor and defense factor are equal.
|
|
| |
|
|
|
DaveV
|
 |
USA - EDT (GMT-5)
Jan 1970
time: 00:12
|
|
quote:
Originally posted by Edward on 05-02-2001 12:27 PM
Yes. As Marquis de Sodaq pointed out...
Xin Yu's formula is an inexact ('though quite good) guess as the victor of a whole battle.
|
Arrrgh. No, it's the odds for a single round. A slight advantage in the Xin Yu formula will result in a much greater advantage in a battle.
quote:
It does NOT factor in the defender tie advantage.
|
If you're going for accuracy, subtract 1/8 from the attacker and add 1/8 to the defender. So in the example above, 3 strength unit attacking 2 strength unit, the odds are 3.875:2.125, or 31/48, exactly the same result as the other formula. I think it's easier to calculate without the fractions, and close enough if you bear in mind that the defender has the advantage in close contests (see your point below).
quote:
Yes, they give similar results, but then again, all of the proposed formulas give somewhat similar results. (A good reason for one to use a formula that's inaccurate but simple for one to calculate on-the-fly.)
|
For a long time I used a formula similar to debeest's, and was surprised when catapults beat musketeers and musketeers had trouble beating fortified pikemen on a river. The Xin Yu formula explains such results, and gives me a much better idea of what's going to happen in a battle. The math is no more demanding than the debeest formula, IMHO.
|
|
| |
|
| |
|
| |
All times are GMT. The time now is 05:12.
Apolyton Time is 00:12. |
top of page
|
| archivepost |
|
Forum Rules:
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts
|
HTML code is ON
vB code is ON
Smilies are ON
[IMG] code is ON
|
|
|
|
|
|
26-04-2001 02:51
|