I'm in a game with luck set to 20%. I attacked a neutral 1 with 2 and lost.

The result says I had -1.20 luck. How is that possible with 20% luck setting? 1.2 x .2 = .24. So the worst result I should be able to get with a 20% luck factor is 1.2 - .24 or .96. I would have needed to fall to .49 (or below) to fail to kill the one army and take the territory.

Or am I missing something?

As it happens failing to take that 1 neutral resulted in me missing a +62 army bonus the next turn.

.96 will get rounded up to 1 96% of the time. 4% of the time, it will be rounded down to 0.

So there are effectively two rolls then. The first to establish the percentage odds and then a percentile roll to deal with the remainder.

In my scenerio the first roll could bring any result between .96 and 1.44. Then the second roll would take whatever the remainder is and apply that chance out of a hundred. To end up with 0 I had to fall between .96 and .99 on the first roll (a 4 out of 48 chance given the range of numbers between .96 and 1.44, or roughly 6%) and then, assuming the worst result of .96 exactly, I would need to roll 1-4 out of 100 on the second roll, a 4% chance. So the combined chances are somewhere between .06% and .24%. So I'd fail that attack an average of .15% of the time or 1 in 667 times.

That makes sense I guess. thanks for the answer.

Yep! Check out the help page under "Luck Modifier" for a full explanation.

BTW: That first 6% should've been 8%.

To be 100% sure you win any attack/defense you just have to exceed the remainder. Hence the default luck setting should be 15%. Of course, I'll always take 0% too.


To kill 1, an attack with 2 would be 1.2 -- so if luck is 15% or less (.18), you'll never fail. An attack with 1 would fail 88% of the time (because of the replacement factor - if attacking a stack, you'd fail 40% the time).

To kill 2, an attack with 4 would be 2.4 -- so if luck is 15% or less (.36), you'll never fail. An attack with 3 would fail 20% of the time.

To kill 3, an attack with 6 would be 3.6 -- so if luck is 15% or less (.54), you'll never fail. An attack with 5 would fail 7.5% (I think).

To kill 5, an attack with 9 would be 5.4 -- so if luck is 7% or less (.378), you'll never fail. An attack with 8 would fail 20% of the time.

etc. etc.

Which is why we need more tournament and FFA games with lower luck factors!!

.06 and .24 are the chances? I just want to clarify this is correct, because I've now failed 2 attackers on 1 defender twice in the same 20% luck game, and I have likewise failed this attack a few other times recently.

I want to make sure that something didn't get altered or whatever, because that seems to be an absurdly high number of times to fail an attack that has a .06-.24% chance of happening.

my math was slightly off -- it's .08% to .24% or an average of .16% or 1 in every 625 attacks.

I just played a game with luck at 0% where I lost 6 of my first 10 attacks of 3 on 2. The odds were a straight 80% on each attack and I lost 6 out of 10 times or 3 times the expected rate. Unlikely yes, but not out of the realm of possibility.

I'm sure I notice more when the luck is off the expected result against my favor a lot more.

"Randoming remainders is important to WarLight since it prevents players from exploiting the system. If results were rounded instead, it would make micromanagement of your armies very beneficial which reduces the overall fun of the game. For example, in this fictional scenario it would be a waste to attack with 3 armies since it would have the same effect as attacking with 2 except it puts an additional army of yours at risk and takes that army away from other benefit. By randoming remainders, this makes every army beneficial and removes the need to micromanage."


your trying to counter the reason he uses percentages instead of rounding

I for one, prefer a higher luck factor, so that every battles outcome is not predetermined, as that is highly unrealistic.. the greater force generally wins, but not always, best determined by luck factor

and Duke, as i understand luck, it would not be 2v1 is .96-1.44...

the difference is .24 to the lower number true, but the higher is shorter.

there are 3 luck chances... none killed, 1 killed, 2 killed...

there is 1 default chance. 1.2

if luck gives you 0, then your chance is 20% closer to 0 then to 1.2 which is .24 difference
but if luck gives you 1, then your luck difference is 0.04 so you have a 1.16 kill rate
"1.2-1=.2*.2=.04"
and if luck gives you a 2, then your luck difference is 0.16 giving you a 1.36 as opposed to the predicted 1.44
"2-1.2=.8*.2=0.16"


so on a 2v1 with 20% luck you have a chance to get .96 1.16 or 1.36
as i understand it

so the odds of failing both those rolls is .4*.4= .16

which is a 16% chance, not a 6% chance

then you need to fail the 4% chance afterwards

which gives you, as i understand it *and i could be way off base*
10.24% chance to lose 2v1, so closer to 1 in 10

hmm.. i think i messed up that last part by a digit...
0.16*0.04 0.0064 which would be closer to 1 in 157

Since there's only 1 army to kill it's irrelevant whether you would kill 2.

There's no additional factor to apply once you arrive at .2075% odds.

Let me try and make it clearer (this is with a 20% luck factor):

Roughly 8% odds that you end up with 96, 97, 98 or 99 (out of range between 96 and 144, that's 4 out of 48 or 8.3%) (if it's 100 or above, congrats you killed the 1 defender).

Then between 1% and 4% odds that you end up rolling a 1, 2, 3 or 4.

Therefore the range of possibilities where you fail is between (8.3% x 1%) and (8.3% x 4%) or (.083%) and (.332%).

The average of those two possibilities is .2075%.

Divide 100 by .2075 and you end up with 1 in 481.


Impaller - For it to happen twice in one game to you is unlikely, unless you've made hundreds of 2x1 attacks.



[I initially thought the odds on the first roll were 6% so the odds dropped when I changed that 8%, they dropped slightly again when I used exact numbers instead of rounding]

hmm...

i thought the combat system worked differently then that.

with 0 luck is 1.2 kills, thus 1 killed and 20% chance of 2 killed, with luck, it rolls each attack with a 60% chance *base* to kill, and then takes the result of those rolls, plus the base result *1.2* and then averages them based on the luck percentage...

I made about 10 2x1 attacks that game, if that.

http://warlight.net/App.aspx?GameID=1085788 Turn 5 and 10. Cost me a potentially crucial bonus on turn 5.

Anyway, the reason I even bothered to mention it is because this has been happening to me very frequently lately. I had it happen to me 3 times in a 2 day stretch a few weeks back. I really don't recall this ever happening prior to a month or so ago, and it's happened at least 6-7 times in the last month. Perhaps I was just really lucky earlier on and have been really unlucky lately, or perhaps I never really paid attention to it earlier and now I do, but I wanted to make sure something hadn't changed that I wasn't aware of.

Also, perhaps the base for strategic games can be made to 15% or less luck to remove this from ever happening? I know you like Warlight to have some bit of variance (I don't disagree, necessarily), but this very small amount of luck that almost never comes up but is devastating when it doesn't seem to offer anything positive. Also, could it at least be changed so that it doesn't list the attack as 100% likely to succeed using the analyze tool? If there is any chance of failure, then there isn't a 100% chance to succeed.

I am also experiencing whacky luck probability. I've been tracking all my 3x2 attacks for the last 5 games and am between 50% and 60% successful. I should be 80% successful (luck is set to between 0-20%). That's a sample of only around 50 attacks, but a big deviation from the expected result.

I think 3x2 is 72% or something similar on 20% luck, but close to 80% on 0% luck.

I guess I understand that point Imp.

3x.6 = 1.8, 1.8x.2 = .36, range = 1.44-2.16, the median is still 1.8, but you're point is that you disregard the range above 2.0 as immaterial to the result?

odds result of 1st roll is between 1.44 and 1.99 = 55/72 or 76.3% (odds you'll have a remainder roll). Then the median result in the range between .44 and .99 is .715 (odds you'll kill 2 on remainder roll). We know the result has to be higher than 71.5% because 23.7% of the time you don't even have a remainder roll.

Therefore overall odds that you'll kill 2 instead of 1 would then be 100-(76.3*.285) = 78.25%.

I think, or am I getting that wrong?

youre experiencing problems with the fact that it is extremely hard to program a random number generator that is "random"

I think the more or less incomplete Help about this leads to all the misunderstandings.