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5 attacks 2 - looses too many times.: 11/17/2011 14:35:13

Nudge
Level 4
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Is this a bug? Should I post this as a bug?

I used to attack 4 vs 2 (having 5 armies and taking 4 of them to attack) thinking the 90+ percent chance was enough. I lost too many times. So I upped my attacks to 5 vs 2. (6 armies 5 of which attack the unowned territory.) More than 10 percent of the time I do not take over the territory. My opening move in a game is to attack 3 places with 5 armies. I fail to succeed on at least one of these opening attacks every 3rd game. And sometimes I fail on 2 of the attacks in the same game.

I think this is totally ridiculous and contrary to what the manual says. In the Guide about luck, it says 100 attacking will do anywhere from 50 to 70 at 75 percent luck.

50 % of 5 is 2.5 how am I only doing 1 sometimes?

60 % of 5 is 3 and should be a sure thing to kill 2 armies even at the high luck.
5 attacks 2 - looses too many times.: 11/17/2011 15:22:27

Nudge
Level 4
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Actually I finally remembered where I saw a table with this exact scenario, it was in the analyze function of the attack transfer window. I have not looked at that since my initial playing, but according to the graph, 5 vs 2 should win 99% of the time. And this is nowhere near my observed results.
5 attacks 2 - looses too many times.: 11/17/2011 16:12:22

cowmandude
Level 38
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You as a human have a bias in your perception of statistical distributions. If you think the calculations are off you will have to perform multiple trials to show this. I suggest you start a game on a big map and record the results of 200-300 attacks and then report them here(with the game number) if they are not what is represented in the graph.
5 attacks 2 - looses too many times.: 11/17/2011 16:21:24

Richard Sharpe
Level 59
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Cowman is right... you would need at least a hundred results to have anything of statistical value and even then it would be quite weak. The other aspect to consider would be to also count the number of times your 5 attacks 2 and all 5 survive... would be the opposite impact of luck since at least 1 should die based on the statistics.
5 attacks 2 - looses too many times.: 11/17/2011 17:02:29

00Dev
Level 55
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yeah! what sharpe said. if you only focus on the bad times, you won't remember the good times! =]

or then again, maybe you simply are the most unlucky person on warlight......had to be somebody! lol
5 attacks 2 - looses too many times.: 11/17/2011 17:05:43

Nudge
Level 4
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I do not know where to find the link, but I did 2 test games in multiplayer with my wife.

100 5 vs 2 attacks, 6 % failed iun both games.

this is closeR to the 10% I observe in normal play than the 1% claimed. Where do I find the links to post both games?
5 attacks 2 - looses too many times.: 11/17/2011 17:08:47

Nudge
Level 4
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5 attacks 2 - looses too many times.: 11/17/2011 17:18:37

00Dev
Level 55
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you're still not factoring in the times all 5 survived. you are focusing on your bad luck, and ignoring your good luck. Balance will be achieved, whether you realize it or not.
5 attacks 2 - looses too many times.: 11/17/2011 17:20:26

00Dev
Level 55
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(or you're the most unlucky guy on walight. like i said, had to be somebody. we can't rule that out)
5 attacks 2 - looses too many times.: 11/17/2011 17:38:21

Nudge
Level 4
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I am only looking at the one statistic. 5 vs 2 not killing 2. Sure there are cases when 5 takes no losses. 2 *.7 is only 1.4 so the likelihood of not hitting is much greater than the likelihood of not killing 2.
5 attacks 2 - looses too many times.: 11/17/2011 18:02:58

Ruthless
Level 36
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on 75% luck, you're looking at a 5v2 fail once every 20 times on average (5%).

your tests of 6% are very close to that average on both your games.

5v2 on the Analyze graph shows a % of 95% for me....not 99%
5 attacks 2 - looses too many times.: 11/17/2011 18:31:35

Richard Sharpe
Level 59
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Ruthless is right. I just looked at 5v2 is 95% on 75% luck (the default setting).

Makes sense. Think of it in terms of binomial distribution if you are familiar with the theory (gives the chance of X happening Y times)... given a 60% kill rate, 5 fails to kill 2 8.7% of the time. Add in the 75% luck modifier and that drops the percentage to around 6.5% (assuming I applied it correctly) which is exactly what you are seeing.

Attacking with 4 gives a 18% failure rate with 100% luck which drops to 13.4% failure rate at 75% luck (the analyze plot gives 12% failure rate, a margin that is well within reason).

So the stats are working as they should... may be off by a percent or two but that could be due to faulty application of the luck percentage by me.
5 attacks 2 - looses too many times.: 11/17/2011 18:43:05

Nudge
Level 4
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I just looked, the 99% came from a duel world game. For some reason the luck on that is set to 50%.
5 attacks 2 - looses too many times.: 11/17/2011 18:46:45

Richard Sharpe
Level 59
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Nudge, a lot of the single players have different luck values. For multi-player templates, I think only the 1v1 strategic and 2v2 auto-game have 16% luck... all others default to 75%.
5 attacks 2 - looses too many times.: 11/17/2011 19:02:34

Major Lazer
Level 29
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I can on wish you BLIRL
5 attacks 2 - looses too many times.: 11/17/2011 20:41:37

cowmandude
Level 38
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Someone with some training in statistics here. At 100% luck I would expect 5 to kill 2 with probability 0.91296 so initially it does seem that the analysis is giving you incorrect or imprecise information. But lets do some quick math just to be sure.

In a 100% luck situation we are sampling from X~Binom(n=5,p=.6) where the events {2,3,4,5} are success and {0,1} are failure.

With 75% luck we are taking a sample from Y=RandRound(L(x-np)+np) where RandRound(x) is a function which evaluates to floor(x) with prob 1-(x-floor(x)) and ceil(x) otherwise.

(NOTE:This is my interpretation of http://wiki.warlight.net/index.php/Luck_modifier If I'm wrong please let me know.)

Note that we can convert an event in X into an event in Y(and indeed warlight does this all the time). So the question is, are the same set of events in X still considered success? If they were we would assume the same probability of success.

Infact we really only have to test the edge cases. Using a simple calculator we see that:

If x=2 then y=round(2.25)=SUCCESS regardless of rounding method.
If x=1 then y=round(1.5)= 50% SUCCESS 50% depending on rounding method.

Since thus we conclude that P(SUCCESS in Y)=P(SUCCESS in x+P(x=1)/2)=.95136 which is very close to the approx 6% you observedi n your trial.

Thus I think we should conclude not that there is a bug in the random generator, but that there is a bug in the analysis tool.
5 attacks 2 - looses too many times.: 11/17/2011 20:46:06

cowmandude
Level 38
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Ahh I had start typing that post awhile ago and then went back to doing real world work and didn't see that you had posted that the luck was off in the game you were playing. It turns out the my analysis agrees with what the map gives on 75% luck and everything is working fine.

Carry on.
5 attacks 2 - looses too many times.: 11/17/2011 20:52:32

Richard Sharpe
Level 59
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Cowman,

Wouldn't the x=0 case contribute to that percentage you quoted? Its only a 1% chance (0.4^5) which when added to your 4.864% failure rate gives us 5.888%... exactly what nudge witnessed.
5 attacks 2 - looses too many times.: 11/17/2011 23:03:15

cowmandude
Level 38
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That case has already been accounted for. My calculation is as follows

.95136=1-(5*.6*.4^4/2+.4^5)

That's 1-(P(x=1)/2+P(x=0)
5 attacks 2 - looses too many times.: 11/17/2011 23:45:28

Perrin3088
Level 44
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The analysis tool uses a brute force calculation btw.. it will run the attack X times and gives a graph showing the possibilities..
it will regularly give different answers to the same possibility due to the fact that it does *imho* brute force enough.. afaik it only does 100 attempts, maybe 1000, either way is still not reliable enough to be given exact accuracy, as it can regularly be off by as much as 2%
5 attacks 2 - looses too many times.: 11/18/2011 00:30:03

Richard Sharpe
Level 59
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Cowman, why the divide by 2? The rest all makes sense to me.

.95136=1-(5*.6*.4^4/2+.4^5)
5 attacks 2 - looses too many times.: 11/18/2011 08:03:32

Qi
Level 55
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with 75%:

- if there is a spot i consider very important, i attack with 7
- if there is a spot i consider vital, i attack with more than 7

i've seen a 9 or 10 attack fail to take a 2 once. 7 very rarely doesn't take the 2.
5 attacks 2 - looses too many times.: 11/18/2011 09:03:34

Perrin3088
Level 44
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it would require 14 armies attacking, at 75% luck, to have 100% chance to win killing 2

13*.6=7.8
7.8-(7.8*0.75)= 1.95

minimum armies killed would be 1.95, thus s light chance to not kill 2, using 13
5 attacks 2 - looses too many times.: 11/18/2011 09:30:17

Qi
Level 55
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i think fizzer wrote somewhere about rounding up when doing these calculations. maybe that means 13 kills 100% too
5 attacks 2 - looses too many times.: 11/18/2011 11:21:31

Perrin3088
Level 44
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I'm fairly certain that all subsets are not rounded.. ie .95 armies killed = 95% chance of success... but we haven't had a debate involving something like this for months, so it is probably buried somewhere I won't bother to find, lol
5 attacks 2 - looses too many times.: 11/18/2011 18:00:03

Math Wolf
Level 63
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Something is off here.

In the first, he has a -7 cumulative offensive luck, in the second one he has -13. These values are too big (I could calculate how improbable it is).

I tried it myself in single player and I got -6.4
5 attacks 2 - looses too many times.: 11/18/2011 18:02:58

cowmandude
Level 38
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Divided by two because if x=1 there is a 50% chance that y=2 and a 50% chance that y=1.

Your minimum kills would be the minimum of possible events in Y when x=0. This would imply(from the formula in a previous post) minKills= floor(np(1-L)). Fixing p=.6 and L=.75 this means minKills = floor(.15*n).

Thus from this I would agree with Perrin's analysis.

The wiki suggests that in the event of a fractional kill, that kill takes place with the probability of the fraction. I.E. 1.95 kills results in 1 kill 5% of the time and 2 kills 95% of the time.
5 attacks 2 - looses too many times.: 11/18/2011 18:12:57

Richard Sharpe
Level 59
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Mathwolf... the huge negative makes sense given the context. He is only ever making the same attack with a high chance of success. As such there is never an opportunity to have positive luck (safe players don't get positive luck by the very fact they are safe).

He would have to repeatedly attack 3 v 2 or something similar in order to see positive luck impact the game.
5 attacks 2 - looses too many times.: 11/18/2011 18:14:40

Richard Sharpe
Level 59
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Makes sense cowman... had completely forgotten about that aspect. Was trying to figure out where it came from forgetting about the luck aspect and instead simply doing the binomial distribution consideration.