1.29 and PTW: Closer but still not great

[Jon Shafer]

Vondrack, you have too much time.

JD, I'm hoping that if I whine enough to the whiners, they'll eventually quit whining. Of course the risk I run is making them whine more about life, so the whiners may end up worse than before.

[N. Machiavelli]

As a former 'whiner', the best thing you can do is hope that they eventually grow out of it. Whining back to them just gives them fuel for the fire

[Excelsior84]

Originally posted by vondrack
Of course, this is theory, as every single type of terrain gives the defender at least a small defense bonus, as others correctly pointed out.

This is exactly what The Civilization Calculator says (see http://www.columbia.edu/~sdc2002/civulator.html).

Yes, the approximation of 1 in 8 was with the 10% defense bonus for flat terrain.

Originally posted by vondrack
I tried to show how the calculation is done (it could be expressed in a more elegant way using combinatorics, but that would be difficult to understand for those that do not do much math), but if it is not as self-explanatory as it seems to me, just tell me and I will explain the formulas in more detail. You can figure out any combat result probability in this way.

Explain away, please. I think I figured out calculate it a while back from an old stat book, but I've now forgotten how it was determined, and further, where the stat book is.

[Jon Shafer]

Originally posted by vondrack
The probability of a regular archer succesfully fending off a regular swordsman:

1/64 + 9/4 * 1/64 + 54/16 * 1/64 = 1/64 * (1+9/4+54/16) = 106/16 * 1/64 = 0,1035 (10,3%)

Well, at least I was close. 10.3%, 1/12, same difference. It was just a rough estimate anyways. I don't have time to go out and do all that math. My mind is mush as it is right now anyways.

[Excelsior84]

So the Civ3 Demo game President is indulging in a little "fuzzy math?"

[vondrack]

Originally posted by Excelsior84
Explain away, please. I think I figured out calculate it a while back from an old stat book, but I've now forgotten how it was determined, and further, where the stat book is.

Well, it is like this:

A regular swordsman (attack 3) attacks a regular archer (defense 1). The probability that the archer wins one combat round, is 1:3. or 1 out of 4 (1/4). OTOH, the probability of the archer losing one combat round is 3:1, or 3 out of 4 (3/4).

Now, the archer has to win three different rounds, while letting the swordsman win no more than two. The formulas describe exactly that:

Should the archer win without a scratch, he has to win three consecutive rounds, not losing a single one (you multiply the elementary probabilities, that is the only thing I do not know how to explain - but I believe there is no need to, in your post above, you seem to understand that).

prob_no_scratch = 1/4 * 1/4 * 1/4 = (1/4)^3 = 1/64

However, the archer is allowed to lose one round, which means you can insert one lost round (probability 3/4) anywhere, but at the end (archer must win the last round to win the whole combat). There are three different ways of archer winning with losing one round, as he can lose in the first, second, or third round:

3/4 * 1/4 * 1/4 * 1/4 (prob of winning, but losing the 1st round)
1/4 * 3/4 * 1/4 * 1/4 (prob of winning, but losing the 2nd round)
1/4 * 1/4 * 3/4 * 1/4 (prob of winning, but losing the 3rd round)

The total probability of winning, but losing one round is a sum of the above three probabilities, that is:

prob_2hp_left = 3 * 3/4 * (1/4)^3 = 9/4 * 1/64.

You apply the same logic for the case when archers barely survives, losing two hitpooints:

1/4 * 1/4 * 3/4 * 3/4 * 1/4 (wins rounds 1,2,5, loses 3&4)
1/4 * 3/4 * 1/4 * 3/4 * 1/4 (wins rounds 1,3,5, loses 2&4)
3/4 * 1/4 * 1/4 * 3/4 * 1/4 (wins rounds 2,3,5, loses 1&4)
1/4 * 3/4 * 3/4 * 1/4 * 1/4 (wins rounds 1,4,5, loses 2&3)
3/4 * 1/4 * 3/4 * 1/4 * 1/4 (wins rounds 2,4,5, loses 1&3)
3/4 * 3/4 * 1/4 * 1/4 * 1/4 (wins rounds 3,4,5, loses 1&2)

Again, sum these partial possibilities up to get the total possibility that the archer will win the combat with just 1hp left:

prob_1hp_left = 6 * (3/4 * 3/4) * (1/4)^3 = 54/16 * 1/64

As there is no other way that would make the archer win over the swordsman, but winning with taking no, 1hp, or 2hp damage (taking an hp of damage twice at most, in any one, but the last round), summing up all the partial probabilities gives us the total probability of the archer killing the attacking swordsman:

1/64 + 9/4 * 1/64 + 54/16 * 1/64 =
1/64 * (1 + 9/4 + 54/16) =
1/64 * (16/16 + 36/16 + 54/16) = 1/64 * 106/16 = 10,35%

The trick is that you have to figure out all the possible combinations of lost and won rounds. This can be done in quite an elegant way using combinatorics, but I have to admit that I would have to look for a book, too...

The above mentioned method is pretty much using the elementary math and common sense (while doing exactly what the fancy combinatoric formulas do, just expressing it in a primitive way. Hence, I am able to think it up every time without the book...

Originally posted by Trip
Vondrack, you have too much time.

After reading Excelsior's post, I thought... oh, my, it so easy... but then, I realized I was not able to put the formula together off the cuff... which sorta pissed me off, as I used to be good at math... I took a piece of paper and the remains of my brains, knowing I must be able to figure it out... and I was, which soothed me a bit... well, the rest was just that I wanted to show the world how smart I am...

[Dinner]

Originally posted by JoeDaddy715
it's funny, trip always whines about people whining.... Isn't that whining too???

Whining about whining is the worst sort of whining...

[vondrack]

Originally posted by Dom Pedro II
Excelsior, i was in fact wondering if increasing the hitpoints would perhaps balance out the system a bit more. Thank you very much. I'll try it.

They will not balance it, they will change it so that the units with higher attack values will win more often over units with lower defense values... effectively decreasing randomness (means it will not be balanced any better, but it will better suit your taste).

Adding hitpoints does increase the probability that the unit with the higher combat-effective value will win. A veteran swordsman is more likely to kill a veteran archer than a regular swordsman is to kill a regular archer. And an elite swordsmen has even greater chance of defeating an elite archer. By increasing hitpoints to, say, 50, the unit with the higher combat-effective value will (for all practical purposes) ALWAYS win.

[Destroyer]

I thought the original post made some excellent points. While I love civ3, it should not have had civ2s better features stripped out.

Scenario building is the thing which has etched civ2 into legend.

[Excelsior84]

Sorry you had to write all that out! I didn't mean for you to explain the *logic,* that was simple, I just wanted the elegant formula. I once *had* such a formula, but I've forgotten it now... Hrm...

*hunts for that book*

[Excelsior84]

Originally posted by vondrack
Adding hitpoints does increase the probability that the unit with the higher combat-effective value will win. A veteran swordsman is more likely to kill a veteran archer than a regular swordsman is to kill a regular archer. And an elite swordsmen has even greater chance of defeating an elite archer. By increasing hitpoints to, say, 50, the unit with the higher combat-effective value will (for all practical purposes) ALWAYS win.

Yup, exactly right. It's just that the really low numbers like 3 make it far random than is human nature to accept. I *like* the more powerful unit to win more often than it does, others don't. Civ2 was far less random than Civ3 with regards to combat outcomes because the *minimum* HP of a unit in Civ2 was 10.

[vondrack]

Originally posted by Excelsior84
Sorry you had to write all that out! I didn't mean for you to explain the *logic,* that was simple, I just wanted the elegant formula. I once *had* such a formula, but I've forgotten it now... Hrm...

*hunts for that book*

Oops...
Sorry to offend your intelligence...

The formula actually just looks elegant, but when you want to calculate its value, you will end up with something very similar to what I posted... because the fancy combinatoric notation has to be rewritten into simple multiply/add operations... But I guess I will also have a look for the book... just to remind myself of the times when all that was everyday food for me...

[Excelsior84]

Yes! I got it!

The probability is (sorry for the formatting, as it looks hideous):

((5!)/((3!)((5-3)!)) * ((.25)^3)((.75)^2)) + ((5!)/((4!)((5-4)!)) * ((.25)^4)((.75)^2)) + ((5!)/((5!)((5-5)!)) * ((.25)^5)((.75)^1)) = 90/1024 + 15/1024 + 1/1024 = 106/1024 = 0.103515625...

Or, in words, the combination of 5 taken 3 at a time times success probability of .25 to the 3rd power times the failure rate of .75 squared plus 5 taken 4 at a time times success probability of .25 to the 4th power times the failure rate of .75 plus 5 taken 5 at a time times success probability of .25 to the 5th power times the failure rate of .75 to the 0th power (1).

[notyoueither]

Originally posted by Oerdin


Whining about whining is the worst sort of whining...

No, it's not... Not when this board was 75% (possible exaggeration) whining about Civ3 and flaming Firaxis. It got to the point that Civ3 General was as good as VC country for anyone actually interested in playing the game.

BTW, I don't think DP is really whining. More like lamenting for what is not, and what is unlikely to be.

[Coracle]

Originally posted by CyberShy
Dom Pedro, it's a game.

It's more than that. It's a buggy disappointment. A BIG one.