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Old March 2, 2001, 09:33   #1
William Keenan
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Hit Points and Firepower
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?

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Old March 2, 2001, 17:26   #2
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quote:

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


That should be your question. Your problem is that you are weighing the attack factor too greatly. I'm no mathematician, and I haven't read much of the civ2 instruction manual, but I have tested around greatly with HP and FP and I can relate my observations. HP= hit points (obviously). A unit with 2hp will take twice the amount of damage to kill than a 1hp unit. And so a 4hp unit will take twice as much to kill as a 2hp, and 4x's as much as a 1hp unit. FP= the amount of damage done with each hit. The attack factor vs defense factor, with consideration of terrain, veteran status and fortresses, calculates the odds of which a hit is scored. If the total attack value of 1 unit is 5, and the total defense value of the 2nd unit is 6, that just means that the chances of the 2nd unit scoring a hit is greater.

Now, playing around with all these values, I can use HP and FP to get certain roles out of my units. I began experimenting with this in my Danube scenario, so you might want to look at the unit values there. I will take 2 units that I want to be roughly equal in overall strenght. But to give them the same statistics would be boring. What I would do is give one unit higher attack and defense values, and give the other unit higher firepower. Say the unit with the higher attack and defense is a swordsman, and the higher firepower is the archer. Now if the values are adjusted to the right levels, you can have some very interesting combat experiences. You the results of combat involving the archer will vary more so than the swordsman. The swordsman will produce consistant results, while an archer may go down without scoring a hit, or kill the opponet without taking a hit. It gets really cool when an archer faces a swordsman. Say they battle on non defensive terrain like plains or grassland. If the unit values are proper, the archer will have the advantage and usually win, regardless if it attacks or defends. But on defensive terrain, the swordsman will have the advantage. If he is the defending unit, his odds are siginificantly greater at winning the battle than if the archer is defending. This is because the swordsman has the higher values and his defense will be increased more so than the archer. And this increase will surpase the bonus the archer gets with firepower.

Sometimes I will give a unit a great amount of firepower and very low attack to make things very interesting. A unit like this can easily win a battle without taking a hit, or lose a battle without dealing a hit, and score anywhere in between. The results can not be easily predicted. I often use this with assassin type units or siege engineers/weapons. It does take a lot of experimentation to set the unit values to the results you want. I pride myself on being one of the only scenario creators that greatly tampers with these values to make combat more interesting. So you can look at almost any of my scenarios and use those unit values as refrences.

I hope that helped.

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Old March 2, 2001, 17:53   #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
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Old March 2, 2001, 23:17   #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?
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Old March 2, 2001, 23:58   #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.
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Old March 3, 2001, 03:31   #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).]
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Old March 3, 2001, 14:11   #7
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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.
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Old March 3, 2001, 14:40   #8
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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!
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Old March 3, 2001, 16:32   #9
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quote:

Originally posted by Kull on 03-03-2001 12:35 PM
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?


Hmmmm ... No!

It will be ready for release very soon Kull.

quote:

Originally posted by kobayashi on 03-03-2001 01:11 PM
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.


Well, to tell you the truth I don't understand your formula well enough to apply it.

quote:

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


That comes from the tieing die roll going to the defender. It's in the thread that Dave links above.

quote:

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.


I don't know much about statistical math but the experience that I have had leads me to believe that your formula is most accurate proposed thus far.

quote:

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


Thanks, glad you liked it. It only took 3 * 10 ^ 10 hours to research. I wish I had that kind of time now! A new version of the paper will be released soon. It's jam packed with tons of new information about barbarians, and includes contributions from Kull and SlowThinker.
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Old March 4, 2001, 01:17   #10
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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?
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Old March 4, 2001, 01:35   #11
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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?
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Old March 4, 2001, 01:59   #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.
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Old March 4, 2001, 02:45   #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?
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Old March 4, 2001, 08:07   #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.
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Old March 4, 2001, 14:46   #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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Old March 4, 2001, 18:29   #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!
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Old March 5, 2001, 00:17   #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.
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Old March 5, 2001, 01:31   #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.
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Old March 5, 2001, 02:13   #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).]
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Old March 5, 2001, 02:57   #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.
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Old March 5, 2001, 03:05   #21
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Well, I said I could be wrong. Now go to bed!
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Old March 5, 2001, 03:15   #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).]
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Old March 5, 2001, 03:26   #23
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Good Night!
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Old March 5, 2001, 03:33   #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.
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Old March 5, 2001, 03:42   #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%

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Old March 5, 2001, 06:15   #26
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This thread gets my vote for best thread of the year. Good Night.
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Old March 5, 2001, 10:31   #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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Old March 5, 2001, 11:28   #28
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quote:

Originally posted by Captain Nemo on 03-04-2001 05:29
PMHere 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


Unless the manual is wrong about this then the FP of both attacker and defender is reduced to 1 when it comes to shore bombardments of any kind (sea to land attacks).

Also Air units with a turns in air of 1 defending a city have their DF increased to 4 times the normal value when defending against attacks from air units with a turns in air of 2 or more.

Just thought I'd remind you to enter those facts into your equations
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Old March 5, 2001, 11:49   #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/SiliconVall...oads_Page.html
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Old March 5, 2001, 21:43   #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!!
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