Modifiers for Attack/Defense

[Sieve Too]

quote: Originally posted by ruoxiaohai on 01-27-2001 01:33 AM Well, if you guys want to be anal, the pikemen ONLY get defensive advantages for unit numbers 16 to 21, I think.

Not too sure of this. If you leave both the Pikeman and, say, Knight in their respective slots, but increase their HP, the Pikeman loses the 50% defense multiplier.

quote: Also, many scenario designers have found that special unit "places" have advantages.

There are a number of these: Only a unit with the settling role in the Engineer slot works at double speed. Only a unit with the diplomacy role in the Spy slot, has the Spy's extra abilities. Same thing for Freight. Only the Partisan slot gives the extra attack strength vs 0 attack defenders. Only the Fanatics slot is a unit that can only be built under Fundamentalism. There's more.

[SlowThinker]

I am not sure if I have understood:

Mech. Inf. has 0,1,2...6 on his die (7 sides).
Artillery has 0,1 on his die (2 sides).

Is it OK?
(I thought there is just one 7-sided die, 6 sides own Mech. Inf., 1 side Artil.)

[DaveV]

As I understand it, the Mech. Infantry has a 48-sided die (0,1,...,47), while the artillery has an 8-sided die (0,1,...,7).

Here is a link to a thread detailing some calculations and experiments: http://apolyton.net/forums/Archives/...-1-000624.html

[Scouse Gits]

... and in the event of a draw the defender wins - thus a zero defence unit such as a barb king or diplo can (very) occasionally defeat an attacker.

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Scouse Git[1]

"CARTAGO DELENDA EST" - Cato the Censor
"The Great Library must be built!"
"A short cut has to be challenging,
were it not so it would be 'the way'." - Paul Craven

[SlowThinker]

I have read that thread quickly.

I hope I have understood it:

On defender's die	Nr. of attacker's wins	Nr. of defender's wins
0	47	1
1	46	2
2	45	3
...	 ...	Â
6	41	7
7	40	8
Total:	(40+47)*8/2	(1+8)*8/2

Generally:
Suppose m=max(a,d), n=min(a,d). Then wins are distributed among both units following way:  

	Attacker's wins (AW)	Defender's wins (DW)	Total number of attacks
a>=d	m-n/2-1/16	n/2+1/16	m
a<=d	n/2-1/16	m-n/2+1/16	m

 

In our example:  Artillery wins 1/2+1/16, Mech. Inf. wins 5 1/2-1/16.

(Edited: DaveV multiply these numbers by two: it looks more functional:
Artillery wins 1+1/8, Mech. Inf. wins 11-1/8;
11=6+5, 5 is the "bonus" for the stronger unit)

quote: Originally posted by DaveV on 01-29-2001 12:33 PM With the default numbers: the odds of mech. infantry winning each round are 11/12 (6 + 5 bonus vs. 1).Â

DaveV disregards sixteenths: Mech. Inf. wins 1 (or 1/2), Artillery wins 11 (or 5 1/2).

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We can multiply theses numbers  (AW and DW) by both relevant firepower and hit pointsand obtain two numbers representing the ratio how to divide wins among both units.

Using other words, we always use all the product f*h.  Single f and h factors are not interesting (they affect accidental behaviour of units only).
 

In our example:  Artillery wins (1/2+1/16)*4=2.25, Mech. Inf. wins (5 1/2-1/16.)*3=16.31 average outcome: Mech. Inf. holds 86% of hit points (0.86=1-2.25/16.31), Artillery is dying

Am I right?
[This message has been edited by SlowThinker (edited January 30, 2001).]
[This message has been edited by SlowThinker (edited April 10, 2001).]

[SlowThinker]

quote: Originally posted by DaveV on 01-29-2001 10:06 AM SlowThinker - I'm not sure I understand what you're saying. FP does make a difference for defenders, and HP for attackers.

I am sorry, there was a mistake in my last post. Look there again, I made a correction now (it is bold).

I will present an example:
Artillery fights with Mech. Inf.

1. Civ2 offers four constants (a,d,h,f) for every unit:
Artillery (10a,1d, 2h,2f)
Mech. Inf. (6a,6d, 3h,1f)

2. You may translate theses four constants into two constants only (you may suppose h=1 and f=1 then):
Artillery (40a,4d)
40=10a*2h*2f; 4=1d*2h*2f
Mech. Inf. (18a,18d)

Suppose Mech. Inf. attacks Artillery : you can provide the standard deduction: Mech. Inf. will win 18 of 22 (=18a+4d) "internal" battles. It means that average outcome is: Artillery is dead, Mech. Inf. has 81,8% (18/22) hit points remaining.

[DaveV]

SlowThinker - I used to think exactly the same thing, but I've been convinced by others' testing (and my own) that there are differences between the various numbers. The combat odds are based on each side rolling an n-sided die, which results in a bonus to the stronger unit. So, to reconsider your example:

With the default numbers: the odds of mech. infantry winning each round are 11/12 (6 + 5 bonus vs. 1). Mech infantry needs to win 20 rounds, so will probably take 2 hits = 4 hitpoints.

With "unified" numbers: the odds of mech. infantry winning a round are 8/9 ( 18 + 14 bonus vs. 4). Mech infantry needs to win 10 rounds, so will probably take 1 hit = 1 hitpoint.

[SlowThinker]

Sorry, my last post is indigested. I am rebuilding it now.

[SlowThinker]

I have rebuilt the post now.

[DaveV]

quote: Originally posted by SlowThinker on 01-29-2001 07:03 PM DaveV disregards sixteenths: Mech. Inf. wins 1 (or 1/2), Artillery wins 11 (or 5 1/2).

Yes, sorry, this was an approximation that overstates the attacker's strength. It provides a rough estimate of relative strengths without having to compute fractions. I use this formula in games so I don't continually have to resort to my calculator.

quote: Originally posted by SlowThinker on 01-29-2001 07:03 PM Am I right?

I think so!

[SlowThinker]

This bad explanation lies somewhere on Apolyton:

quote: "FIRE!" Making War in Civilization II by Marc Fisher The actual equation used to resolve each combat round is: __a__ (a+d) where: a = attacker's attack rating d = defender's defense rating The result is a fractional number (percentage). A random number is generated, and if the result is less than the percentage, the defender loses hit points equal to the attacker's firepower rating. The reverse happens if the random number is higher - the defender's firepower rating is subtracted from the attacker's hit point rating

I trusted to that...
[This message has been edited by SlowThinker (edited January 30, 2001).]

[DaveV]

Oops, after further thought I have to disagree with your "expected result" calculation. The question is, how much damage do we expect the mech infantry to sustain while destroying the artillery? I think the answer is a straight ratio of the combat strengths:

mech infantry damage/artillery damage = 16.31/2.25

artillery damage = 100%; mech infantry damage = 2.25 / 16.31 = 14%; so I expect the mech infantry to survive with 86% of hitpoints. A small difference, I know...

[SlowThinker]

Of course, it was my mistake. A big difference in a general case.
I corrected it, somebody may consider that post useful.

Note:
If we disregard sixteenths, then weaker unit makes 1/(2*m/n-1) of damage.
For large ratio m/n, weaker unit looses a half of his power ("-1" in the expression looses his significance).

[SlowThinker]

I have a presentiment that attacker has a 50% chance to get vet status, defender has a 100% chance.

Is it OK?

[Scouse Gits]

SlowThinker - I concur tests I have made would tend to suggest that the attacker will win vet status (without Sun Tzu) 50% of the time - I do not have any stats for the defender


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Scouse Git[1]

"CARTAGO DELENDA EST" - Cato the Censor
"The Great Library must be built!"
"A short cut has to be challenging,
were it not so it would be 'the way'." - Paul Craven