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basic question thread: 8/28/2016 00:00:22

nateTEKtonik
Level 25
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Alright, so I'm getting use to the game but there is one question I've had for awhile that if I got an answer to I think it would really up my game and also help some other people.

Say that the offensive kill rate is 60% and the defensive is 70% (Which are both the standard). I'm going to attack a neutral space that has 2 armies in it. What is the exact number it will take to take the space and the more important question is how do you calculate how much it would take (say the neutral space was 5 instead of 2)?

The game I'm on now also has a luck modifier of 16%. A little more insight into that would help to.

Thanks in advance!!
basic question thread: 8/28/2016 00:37:02


DerWyyy
Level 53
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2 nuetral armies means u should attack with 3 minimum, unless the kill rate is jacked. 4 is 100% safe. typically u can go a little under two times what the neutral armies are ex-

if theres 10 neutral armies, 17,18,19 will probably do it, 20 is the 100% safe. unless the rate is jacked up
basic question thread: 8/28/2016 00:44:56


Apollo
Level 56
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60% Kill Rate,

x=number of armies you are attacking with

x*0.6=number of armies it will kill



although warning that ^^^^^^^^^ is usually for 0% luck SR, it also works for 16% but its less safe, it can happen that you miss a attack

Edited 8/28/2016 00:45:09
basic question thread: 8/28/2016 00:48:54


master of desaster 
Level 64
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To take over a territory, only the attack killrate is relevant.

Let's say we got 60% attack killrate, straight round. To kill 2 armies, you need to attack with 3, cause 3*0.6=1.8 (rounded 2).
5 defender need 8 armies to be taken over. 8×0.6= 4.8 (rounded 5)

On the other hand, 2 defender kill always 1 army from the attacker, cause 2*0.7=1.4 (rounded 1, defense killrate 70%)
That xan be important if you defend only a territory, cause with 5 defenders 5*0.7=3.5 (rounded 4) you kill as many attackers as with 6 armies 6*0.7=4.2 (rounded 4).

On 0% weighted random it gets calculated the same way. The only difference is, that at an attack with 3 (example above, 1.8 armies killed), you kill at least 1 army. The second army gets killed to a percentage of 80%.

Sadly i don't exactly know how killrates on 16% wr get calculated. Inputs on that would be welcome
basic question thread: 8/28/2016 00:51:27


Cata Cauda 
Level 57
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^^^
Kind of what Wyyy said, but I would like to add that the attack-anaylse tool is really really helpful for beginner regarding that question.

Edited 8/28/2016 00:51:47
basic question thread: 8/28/2016 01:29:29


Semicedevine
Level 59
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but Wyyy didn't even answer the question... <_<
basic question thread: 8/28/2016 01:37:47


Semicedevine
Level 59
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alright, pull up a google calculator or whatevsss

this is the type of sht I do 25/7 strat lulz:

1) put in a number, multiply it by 0.6, and that's how much damage you'd do if you were ATTACKING with that many armies

2) put in a number, multiply it by 0.7, and that's how much damage you'd if you were DEFENDING with that many armies

(with "damage" being how many armies you're gonna kill)

oh and almost always you're gonna get a decimal number... just round it to a whole number the regular way and that's how it is for a game with 0% luck factor and straight round (regular settings)

but for 16% luck modifier.... umm...

https://www.warlight.net/wiki/Luck_Modifier

Edited 8/28/2016 01:39:21
basic question thread: 8/28/2016 14:55:14

nateTEKtonik
Level 25
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Perfect guys! Thank you very much!
basic question thread: 8/28/2016 15:19:42


GiantFrog
Level 58
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Sadly i don't exactly know how killrates on 16% wr get calculated. Inputs on that would be welcome


the luckmodifier tells you the weight thats been given to the outcome of the randomly generarted number of armies you kill with your attc/defense.

Those kills are calculated with a binominal distribution:
each attacing/defending army has a chance of attckillrate/defensekillrate to kill 1 enemy army, resulting in (p over k)*(p^k)*(1-p)^(n-k) chance to kill EXACTLY (if you search for the chance of killing at least k armies, you need to sum up all those who kill at least k) k armies (k in [0,n]), where p is the offenskillrate/defensekillrate.
The random outcome you got will then be weighted with the luckmodifier. the rest (1-luckmodifier) will be the expected valuce (#attcers*attckillrate, equivalent for defense):
(random value)*luckmodi+(1-luckmodi)*(expected value) is your outcome.
the outcome can now, as always, be rounded using WR or SR, same as usual.

if you are intrested in when an attc, using luckmodi, is 100% safe, all you need to do is assume all your armies failed to kill an enemy (chance for that is (1-killrate)^(#attacers) ).
Meaning (1-luckmodi)*(expected value) rounded (WR or SR) has to be at least as high as the number of armies you want to kill.

Edited 8/28/2016 15:52:23
basic question thread: 8/28/2016 15:48:25

Krzysztof 
Level 65
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To take over a territory, only the attack killrate is relevant.

He's noob, he can't play, never trust him.

Attacking one defender with one army, off kill ratio 60%
-deff kill ratio: 50% or above: you fail
-deff kill ratio: less than 50%: you take over attacked territory :P

Edited 8/28/2016 15:48:50
basic question thread: 8/28/2016 16:03:46


Perrin3088 
Level 44
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Krz that's only true on straight round.
on WR it's 70% chance you lose and kill nothing if it has 1 army standing
then after that there is a 60% chance you will kill the 1 army.
so overall you have about an 18% chance of success 1v1 standard rules
basic question thread: 8/28/2016 16:04:54


DerWyyy
Level 53
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idk how to caculate, sorry. I just tell you what i do, attack with a little less than two times the # of armies the enemy has
basic question thread: 8/28/2016 16:17:48


Perrin3088 
Level 44
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and on WR, the luck modifier is based on the difference between the standard kill rates and the ends.

so for 16% luck, 3 armies kills 1.8 - (1.8*0.16) to 1.8 + ((3-1.8)*0.16)
so 1.512 to 1.992 armies..
which is why 16% became the default at one point, since it is above 1.5 armies killed at it's lowest
basic question thread: 8/28/2016 16:25:58


Perrin3088 
Level 44
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and the only difference between WR and SR is that on SR the 1.512 would be a guaranteed 2 killed, while on WR it would only give 51.2% chance of a kill on the second one.
if you are playing WR and you want 100%, then the bottom value should be higher than the defenders.
(4 * 0.6) - ((4*0.6)*0.16) = 2.016 = guaranteed win against default 2 neutrals.
at one point the default value was 20% to make it more strategic, but people raged at the slight chance of 4v2 failure - (4 * 0.6) - ((4*0.6)*0.2) = 1.92 at bottom.
basic question thread: 8/28/2016 16:31:50


GiantFrog
Level 58
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which is why 16% became the default at one point, since it is above 1.5 armies killed at it's lowest


i dont see that being a reason for 16%WR, maybe 16%SR. 1.992 highest may be intresting for WR.
basic question thread: 8/28/2016 16:41:38


Perrin3088 
Level 44
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yeah, after writing the post afterwards I came to the conclusion it was initially made 16% for the guarantee 4v2, and it just worked for the guarantee SR 3v2
basic question thread: 8/28/2016 16:42:15


Perrin3088 
Level 44
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I can't remember which order the changes were precisely.. it was quite some time ago, ahah
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