... embarassing for me but i have to admit after nearly 3K games: i still dont know exactly how the luck modifier works. Sure, i read WarLight Wiki too, but i still dont know exactly how the Luck Modifier (LM) influences the number of armies left after the attack. More precisely, i m talking about the number of attacking troops, NOT about the number of defending troops. The Wiki holds no information about that, it only tells me how the LM influences the outcome of an attack with respect of defending troops.

The luck modifier works the same in both directions.

Basically, it determines the possible variance from the expected outcome. If you have ten armies defending or attacking with standard kill rates, they are expected to kill 7 or 6 armies respectively. The luck modifier determines how far from that value the actual outcome can be.

So on offense with 16% luck, the expected outcome is 6 kills and minimum is 5.04 (6*(1-.16)). With 75% luck, the expect outcome is still 6 but the minimum is now 1.50

On defense with 16% luck, the expected outcome is 7 kills and minimum is 5.88 while at 75% it drops to 1.75.

In all cases, the decimal point is done separately to my understanding so this last instance gives 75% of 2 and 25% of 1 should the worst luck occur. Basically, the round number given is the minimum possible kills when either attacking or defending. (the same math can be used going up but it is bounded by the number of attacking armies so gets a bit more confusing)

Mosq, if you want, you can set up a game with me and we can go over the strategies involved with modified luck.

The number of armies left over after an attack is simply the number that the defenders didn't kill. For example, if you attack with 30 and the defenders kill 10, you'll have 20 left over to occupy the new territory.

To figure out how many the defenders will kill, multiply the number of defending armies by 0.7. For example, 20 defending armies will kill on average 14.

The luck modifier just determines how far off from 14 the actual result will be. In 0% luck, it will always be 14. With higher luck values, it can be higher or lower than 14, but 14 will always be the most likely result.

So do i understand correctly: the LM is with regard of defending / attacking troops uncorrelated?

An example: LM is 75%. "BLUE" territory, 18 troops, is attacked from "RED" territory with 14 attack troops. Bad luck for blue: After the attack BLUE and RED each have 7 troops left. But it could also be possible BLUE still has 15 troops left and RED loses all attack troops and has only 1 troop in the territory from which he attacked left? Or BLUE has 7 troops left and RED loses all attack troops?

For specific situations, you can always check by using the "analyze" option during an attack; And yeah, don't attack 15vs18 haha

Mosq, you are essentially correct.

Say you are playing on 100% luck meaning ten armies can kill anything from 0 to 10 (though over time it should average to 6). It is entirely possible that 18 attacking armies could kill zero while 15 defending armies could kill 15. Conversely, it is entirely possible that 18 attacking armies could kill all 15 defenders and those defenders kill 0 attackers.

Both scenarios are highly unlikely to play out but they are possible.

Doggy ... i sure have some experience with all this. Roughly estimated i performed several 100K attack on neutrals or opponents :-). The question was/is: Is the LM applied seperately in a random manner (influenced by the LM) on attacking and defending troops or if there is some correlation.

If there is no correlation, like it is with the Micro Luck randomizer, as i now suppose, then the range of results of such an attack is much wider than if there were a correlation. Bottom line: Almost everything is possible with such an attack with 75% LM.

Exactly. Hence people don't like that settings as much as lower luck.

To the best of my knowledge, the luck modifier is applied to the offence and the defence calculation of an attack completely independently; there is no correlation.

I'm not familiar with "the Micro Luck randomizer" (or, at least, I don't know it by that name).

Micro Luck randomizer i call the randomized rounding mode.

Ah okay, now I understand. That, too, has zero correlation with the other probabilities.

As far as I understand, Attacking 10 vs 10 with standard kill ratio and luck 50% will result with:

Attacker kill: 'default' value = 6.
After adding luck it will between 3 and 9 (6*(1+-0.5))

So the my question would be:
Are all values (3, 4, 5, 6, 7, 8, 9) equal with same probability or they are different. If so how to calculate them ?

ksz, all values would not be equal probability. They would form a bell curve with 6 being the most likely outcome and 3/9 being equally low. As for how to compute the odds, I am not certain but I believe it would be a binomial distribution to determine what the 'luck' figure was. Once determined, the luck figure is combined with the expected value according to the luck percentage.

Think of it as a series of coin tosses where heads is a successful attack and tails a failure. To get a 3 or 9 you would have to get all heads or all tails (10 in row is 1 in 1000 odds)... to get a 6 you just need 5 of each which is the expected outcome.

I have a question about luck, too.

I checked those graphs here:
http://wiki.warlight.net/index.php/Luck_Modifier#Analysis

and I'm still unclear in what range the luck modifier works. According to the graphs, for example, the number of armies killed when attacking with 100 (50% luck) can vary from 52 to 68.
If the formula was 60(1+-0.5) => the range should be from 30 to 90...but it's not that.


Also, according to previous posts, attacking with 1000 armies with 100% luck could end up (theoretically) with killing 0 armies, which to me doesn't sound right, because it's a huuuge deviation.



tl;dr in what range does the luck modifier works.

That formula is correct - you can kill 30 armies with 50% luck, 100 armies, and 60% kill rate, but it's extremely rare (40%^100) that it's near-zero.

Killing 51 armies is possible, but the chances of it are rare enough that it doesn't take a pixel on that graph, causing it to look like 0%.

---

100% is a special case - in that case, the expected value will not matter at all, and the kill amount can be any number.

The expected value is what keeps any other percentage, 0% to 99%, from killing 0 armies with 1000 in the case that someone has extreme misfortune.

thx for reply. So know i'm curious if anyone know(fizzer for sure:P) and would share this knowledge, how to calculate excat probabilty value for killing (at least) N armies when attacking X unit with army of Y. Assume that luck modifier is A and offense kill ratio is B. Looks like nice math task :D

Y is immaterial to the equation.

N = X*B*(1-A) with the answer of the right portion being rounded down to the nearest whole number.

A = .16, B = .60, N = 10; thus X = 20
On standard settings, you must attack with 20 to guarantee killing 10 defending armies.

A = .16, B = .60, N = 2; thus X = 4 (well known that 4v2 is 100% certainty)

A = .75, B = .60, N = 2; thus X = 14, you must attack with 14 to mathematically ensure victory.

Looking at your question again, the above solution is only partial... it determines the minimum armies killed by X attackers (or inversely, the minimum required to kill N defenders)

The probability of each would require using a binomial distribution.
http://en.wikipedia.org/wiki/Binomial_distribution

For example, use 4 attacking armies and 60% kill rate, 75% luck. Here are your possibilities:

2.6% of 0 successful attacks.
15.4% of 1 successful attacks
34.6% of 2 successful attacks
34.6% of 3 successful attacks
13.0% of 4 successful attacks
Rounding errors give a 100.2% total.

Then take the expected outcome (2.4 kills) and multiply by 1-luck (.25).
Multiple the successful attacks by the luck figure (.75)
Add these two numbers together.

2.6% of ending with 0.6
15.4% of 1.35
34.6% of 2.10
34.6% of 2.85
13.0% of 3.60

Now those remainders are the odds of rounding up or down.
1.04% of 0 kills (2.6% * 40%)
11.57% of 1 kill (2.6*60% + 15.4%*65%)
41.72% of 2 kills (15.4%*35% + 34.6%*90% + 34.6%+15%)
38.07% of 3 kills (34.6%*10% + 34.6%*85% + 13%*40%)
7.8% of 4 kills (13%*60%)

thanks a lot richard, that's exactly what i wanted to know :)