Hit Points and Firepower

[Captain Nemo]

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!

[kobayashi]

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.

[William Keenan]

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.

[techumseh]

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).]

[William Keenan]

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.

[techumseh]

Well, I said I could be wrong. Now go to bed!

[William Keenan]

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).]

[techumseh]

Good Night!

[Xin Yu]

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.

[William Keenan]

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%

[kobayashi]

This thread gets my vote for best thread of the year. Good Night.

[Gothmog]

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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"There is no tiddle-taddle nor pibble-pabble in Pompey's camp."

[CyberChrist]

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

[DaveV]

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

[Captain Nemo]

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!!