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GMontag
Level 60
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So is my math wrong here, am I misunderstanding the combat formulas, or am I really just this unlucky? In the past day and a half, I've had 3 7v4 army battles against neutrals fail in strategic 1v1 16% luck games.

According to my understanding of how the luck works, you would have to roll either a 0, 1 or 2 out of 7 to not kill all 4 defending armies because 7*.6*.84 + 3*.16 = 4.008 which will always at least kill 4, with a .8% chance of killing 5.

So, the chance of failing with 7v4 is:

(.4^7 * (1-.528)) + (7*.6*.4^6 * (1-.688)) + (21*.6^2*.4^5 * (1-.848)) ~= 1/56

I can't have been in more than 20 of those 7v4 battles in that timespan, and 3/20 is nearly 10 times more often than 1/56. So am I wrong with my math here, or am I just on a bad run of variance?

Matma Rex
Level 12
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According to the Analyze window, the chances of succeeding 7v4 are ~98%, which matches your calculations. Probably just bad luck.

BumbleBee :)
Level 14
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but upto my normal practice, i found 9 vs 4 is more workable while 7 v 3 normally.

Duke
Level 5
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you're wrong.

7x4 fails 1 in 6 times with 16% luck.

7*.6 = 4.2 (with luck at zero -- you always win)

But with luck at 16%, 4.2 * .16 = .672, which yields a possible range of results from 3.528-4.872.

So .471 out of 1.344 is the percentage of time you have a remainder roll (the rest of the time you're over 4 on the first roll). That's 35%

The odds on the remainder roll are from .528 to .999 or .471 or 47%

Muliply the odds the together to get the possibility of missing you're roll: 16.5%.

Which is almost exactly 1 in every 6 attempts.

HTH

[中国阳朔]Chaos
Level 49
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If Duke is right, why does the analyse window gives it a 98% chance?
from experience, I think it's less than 1 out of 6, but maybe my total amount of 7x4 is not large enough.

GMontag
Level 60
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Duke, your math doesn't make any sense. You're counting all of the range from 3.528-4.872 as equally likely, which doesn't jive with the idea that each attacking army has a 60% chance of killing a defending army. My impression was that basically each army flipped a weighted coin and then you totaled up the kills, and only then multiplied by the luck factor, then rolled a percentile to deal with the remainder.

Knoebber
Level 54
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I have no idea why the anaylze window would be wrong.

Perrin3088
Level 44
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http://warlight.net/ViewHelp.aspx?s=LuckModifier

I questioned precisely how this works also.. and believe dukes math is slightly off as well

0% luck = 4.2 armies killed...
100% luck = IF(Rand()>=.6,1,0) repeated 7 times, armies killed...

any other factor of luck = (100% luck result-4.2)*(luck percentage)+0% luck percentage

so in the off chance you get 7 kills with 100% luck it would be..
(7-4.2)=2.8
2.8*.16=0.448
4.2+0.448=4.648 = highest possible kill result

and the furthest ranges have a much lower chance of occurring, as it is similar to playing a game of dice...

you have 7 rolls, the odds of 7 successes or 7 failures is much less then the chances of 3-4 successes...
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