02-03-2001 18:33									#1
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I'm having problems in balancing a scenario and I wonder if the problem is the formula I'm using to balance unit power. I have assumed that Attack factor * Hit points * Firepower = relative attack power (RAP). In playtest however the unit with higher RAP often fails to win against a unit with significantly lower DAP.

So the question is am I weighing HP and FP correctly?

	03-03-2001 02:53									#3
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William - you need to examine attack and defense separately from HP and FP. The stronger unit effectively gains a bonus equal to the difference between its strength and that of the weaker unit. For example, a catapult (6 attack) attacking a phalanx fortified on open ground (3 defense) effectively attacks at 9:3, rather than 6:3.

See this thread: http://apolyton.net/forums/Forum1/HTML/001761.html

	03-03-2001 08:17									#4
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Thank you both for your answers.

I now see the err of my math.

a/(a+d+.125) * Ahp * Afp / Dhp / Dfp = % chance the attacker will win

Do you agree Dave?

	03-03-2001 08:58									#5
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I remember that thread. DaveV did not agree with my views which appear at the end of the thread.

I still think that FP and HP cannot be lumped into the calculations with AP and DP. The AP/DP ratio calculates the odds of landing a blow on each round of combat. TO determine the odds of winning, you have to calculate the probability that the weaker party lands enough blows to kill the stronger party before it is killed itself. While it it true to say that the FP/HP ratio affects accidental behavior, it is relevant to note that when HPs are high enough, the probability of the weaker unit winning are always zero even if the stronger untis has only a slight advantage in FP/DP.

	03-03-2001 12:31									#6
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I worked out an example a while back but never posted it.

Say [4A 1D F2 H1] attacks [1A 2D F2 H1]. In this case, both attacker and defender will die if they are hit with five blows. From the defender's point of view, as long as he hits the attacker five times before he dies, he will win. The probability of this is:

defender lands five blows straight
(1/3)^5 = 0.41%

defender lands five blows but suffers one blow
(1/3)^5 x (2/3) x (5x4x3x2)/(1x2x3x4) = 1.37%

defender lands five blows but suffers two blows
(1/3)^5 x (2/3)^2 x (6x5x4x3)/(1x2x3x4) = 2.74%

defender lands five blows but suffers three blows
(1/3)^5 x (2/3)^3 x (7x6x5x4)/(1x2x3x4) = 4.27%

defender lands five blows but suffers four blows
(1/3)^5 x (2/3)^4 x (8x7x6x5)/(1x2x3x4) = 5.69%

note: this first component is the probability that the defender wins 5 rounds, the second component is the probability that the attacker wins X rounds and the last component is the number of ways it can happen.

Add them together, you get 14.48% which means the attacker has a 85.52% chance of killing the defender. So you see its not a matter of a simple formula, you'll typically need a spreadsheet to calculate the odds.

Actually I have my doubts about the way I calculated the last component. I have made a spreadsheet - if any qualified statistician is interested in checking it over. But the priciple of how HP and FP affect the odds is well illustrated I hope.

(edited this as I made a mistake in the calculation originally)
[This message has been edited by kobayashi (edited March 03, 2001).]
[This message has been edited by kobayashi (edited March 03, 2001).]
[This message has been edited by kobayashi (edited March 03, 2001).]

	04-03-2001 10:17									#7
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You are quite correct kobayashi, with the possible exception that there seems to be no allowance for a tie in your formula - which could be quite signicant in a low attack low defence combat.

It is important (for me) to bear in mind the purpose of the formula; to correct unit balance in my scenario. To achieve this goal what I need is to know the mean or average outcome not the angle of the probability bell curve. Please correct me if I am wrong, the mean outcome of both formulas is the same, no?

	04-03-2001 10:35									#8
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Hmmmm. Would this be the first scenario from the illustrious Mr. Keenan? Could he perhaps favor us with just a wee little hint as to the subject? Hmmmm?

	03-03-2001 23:11									#9
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Well lets see, using your formula on my example, we'd get

4/(4+2+.125)x1x2/1/2 = 65.3% which is a different answer.

(by the way I didn't catch how the 0.125 came about)

According to your formula, whenever the hp and fp of the attacker and defender are the same, you always get the same result since they cancel out each other.

According to my reasoning, the outcome will be different when the hp and fp change since they do not cancel out. When the hp/fp ratio increases (assuming they are still the same for both attacker and defenser) the chances that the superior unit (in terms of attack and defense factors)wins increases. As HP/FP increases to a certain point, even in the case of a A50 verses D49, the A50 will alway win.

Best guess is that your formula applies in the case that hp/fp is very low. Something like how Newton's laws of motion are a subset of the theory of relativity.

It's been a pleasure having this discourse with you. I think your Barbarian Paper is an all time great.

	03-03-2001 23:40									#10
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Just a thought, but perhaps one of you math whizzes could gin up a "Unit Battle Outcome" tool. It would be a spreadsheet that would allow the developer to key in all the variables for an attacking unit and a defending unit. The results should should be a single number that gives the probability of victory for each side. Seems like this would be a very useful addition to the Scenario Developers toolkit.

I'd certainly be happy to host it on my web site, and I'm sure others would too!

	04-03-2001 10:59									#12
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The tying die! That makes the calculation infinitely more complicated. This brings to mind the time when I tried to determine the efficient frontier of...never mind that's irrelevant.

The point is, we don't need to have a formula, I can whip up a spreadsheet which simulates a single battle a hundred times and comes up with the average result. It will even have the average hit points left of the survivor. It will be huge though.

	04-03-2001 11:45									#13
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I'm sure that I don't know nearly as much about probability as anyone posting on this thread, but something tells me we don't know enough about the way a tie is handled by the program. I assume that ties don't occur. As I read the manual, the attack factor and the defense factor are expressed as a probability ratio. But either the attacker or the defender wins each round of combat. The amount of damage that is inflicted is determined by the FP, and the amount that can be sustained is determined by the HP. What is behind the assumption that ties actually occur?

	04-03-2001 17:07									#14
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Never thought much about having ties. Presumably this is easily tested out. Just have a 1 1 1 unit attack another and if it keeps dying, the tie idea must be true. Anyway, I have put together an excel spreadsheet (that assumes the tie is true)which does 100 simulations at a go. I will post it somewhere soon.

	04-03-2001 23:46									#15
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My understanding of the combat system is the same as Techumseh, so I don't understand about the die rolls and ties either. Where did that idea come from?

------------------
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	05-03-2001 03:29									#16
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The thread really got my attention and caused me to re-think my unit stats for future scenarios:

When you work with statistics you don't have to worry about ties since the calculation will assume an equal number of ties or split the ties evenly between the sides (Basically assign the damage evenly to both sides). How the game handles it is irrelevant IMHO since you are trying to calculate odds of each side winning, not what happens in the specific case of the odd tie. Assume the game "rolls" a number between zero and one with 3 decimal places then compares it to the odd level at which side A or B wins the toss (For example 0.500) there will only be one tie every 1000 rolls. I actually think the calc goes beyond 3 decimal places so there are probably no ties at all in the game.
I do not think the .125 bias on the denominator of the formula above applies. The odds should be a straightforward (a/a+d)

I made a spreadsheet using the BINOMDIST Excel function to assess the results of multiple rolls and the impact of HP and FP on the odds of winning. HP and FP are truly VERY important in the ultimate outcome of an attack. I also added calculations for the average damage sustained by the winning unit. In the process I actually made some (for me) shocking discoveries on how much the FP and especially HP impacts outcome.

For example a unit 11a,6d,2FP,3HP unit attacking a 15a,9d,2FP,2HP unit versus a 15a,10d,2FP,2HP unit attacking the same 15a,9d,2FP,2HP...

This is an example from my scenario under construction... the 15a represents the better, newer unit to attack with and the 11a unit represent an older unit but slightly more rugged than the 15a version...

Well amazing the older unit with 3HP wins 93.5% of the time versus only 86.9% for the newer unit!! The older does sustain an average of 16 points of damage, while the newer one only averages 12

Here is the classic shore bombardment conflict
The battleship
12a,12d,4hp,2fp
against Mech Infantry
6a,6d,3hp,1fp
Results: 100.00% victory, BB only sustains 8 pts of damage on average

I ran dozens of my units through it to see how they fare against eachother and got a lot of interesting insight. Thanks!

	05-03-2001 09:17									#17
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I must disagree. I originally did not know about the tie idea either. But to test it out, I made a simple scenario and used one warrior (1,1,1) to attack another. I repeated it 10 times. The attacking warrior would die about 80% of the time. If the game calculates the die roll to three decimal places, the odds would be 50/50. I can only conclude that Sid originally played board games with real dice and built in an exact simulation into Civ2.

	05-03-2001 10:31									#18
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To understand the "tie roll" concept please read this thread: http://apolyton.net/forums/Forum1/HTML/001761.html

Nemo: We are certain that the (a/a+d) formula is not complete. If it were then zero defence units would never inflict damage, yet they do.

	05-03-2001 11:13									#19
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William, I have never seen a zero DF unit kill an attacking unit of any strength. I think damage caused by a zero defense unit is an AI "cheat", and only affects full strength attackers. Of course, I could be wrong.
[This message has been edited by techumseh (edited March 05, 2001).]

	05-03-2001 11:57									#20
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Techumseh: Would you would like to see a zero defence unit win a battle? Attack a diplomat with a warrior at 1/3 strength. I just ran a quick test (on diety) and the worrior lost five out of five.

Nemo: Thanks for introducing me to binomial distribution! I knew there must be a simple way to calculate the probability.

	05-03-2001 12:05									#21
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Well, I said I could be wrong. Now go to bed!

	05-03-2001 12:15									#22
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Including the "tie factor" the chance of the defender winning is equal to

BINOMDIST(ROUNDUP(Ahp*10/Dfp)-1,ROUNDUP(Dfp*10/Ahp)+ROUNDUP(Dhp*10/Afp)-1,(a*8)/((a+d)*8+1),TRUE)

a = attacker's attack
d = defender's defence
Dhp = Defender's hit points
Dfp = Defender's firepower
Ahp = Attacker's hit points
Afp = Attacker's firepower
[This message has been edited by William Keenan (edited March 05, 2001).]

	05-03-2001 12:26									#23
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Good Night!

	05-03-2001 12:33									#24
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Binomial distribution has two parameters: n (number of trials) and p (probability for success in one trial). The expected number of successes out of n trials is np, and the stadard deviation for the expected number is sqrt(np(1-p)). If np>5, then the normal distribution is a good approximation for the Binomial distribution.
I'm not sure if it is useful to you guys.

	05-03-2001 12:42									#25
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BINOMDIST(ROUNDUP(Ahp*10/Dfp)-1,ROUNDUP(Dfp*10/Ahp)+ROUNDUP(Dhp*10/Afp)-1,(a*8)/((a+d)*8+1),TRUE)

The maximum number of times the attacker could be hit by the defender without being destroyed.


BINOMDIST(ROUNDUP(Ahp*10/Dfp)-1,ROUNDUP(Dfp*10/Ahp)+ROUNDUP(Dhp*10/Afp)-1,(a*8)/((a+d)*8+1),TRUE)

The maximum possible number of combat rounds.


BINOMDIST(ROUNDUP (Ahp*10/Dfp)-1,ROUNDUP(Dfp*10/Ahp)+ROUNDUP(Dhp*10/Afp)-1,(a*8)/((a+d)*8+1),TRUE)

The percentage chance the attacker will score a hit. If you don't want to include the "tie factor" just leave out the +1.


The chance of a warrior successfuuly defending against an attacking warrior would be BINOMDIST(9,19,47%,TRUE) or 60.26%

	05-03-2001 15:15									#26
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This thread gets my vote for best thread of the year. Good Night.

	05-03-2001 19:31									#27
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Wow! This is all too mathematical to me. For me the bottom line is that the Microprose documentation is wrong! They may have conceived it that way, but when the programmers got done with the combat system, it came out different.

------------------
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	05-03-2001 20:49									#29
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Wow! This thread certainly grew since I last stuck my nose in. I think I can answer a couple questions asked earlier.

The odds of an attacker winning one round must be expressed as two separate cases. Case 1, A <= D, the odds are (A-.125)/(2*D); case 2, A > D, the odds are 1-((D+.125)/(2*A)).

Mark Wagner wrote a battle odds calculator that I think is correct. It can be downloaded at http://www.geocities.com/SiliconVal...loads_Page.html

	06-03-2001 06:43									#30
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I agree that I did not take into account the special defense/attack bonus related to shore bombardment/ships caught in port/2x defense/air defense etc...

Where does this .125 come in? Does it just give the defender the victory when there is a tied die roll? If it truly is designed as a die roll are we talking about a 6 sided die? As in Roll A (1-6) x a > Roll D (1-6) x d for the attacker to win? This would give the 1a vs 1d battle a 21-to-15 edge to the defender but not 80-20?

I will do some more testing with actual combat situations... I think it is also critical to test Human-vs-AI and AI-vs-Human because I am willing to bet the results will be completely different!!