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One thing I always wondered...: 3/2/2018 02:53:12


MrTrolldemort 
Level 57
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Why is 16% the most common luck number for strategic templates?

What makes the number 16 so special? Why not make it 10% of 20% or something else?

I know it's not as big anymore as no luck is much more common but 16% has always had a large presence in ladder templates and whatnot. Anyone got a good explanation about this or is it just by convention that it's used?
One thing I always wondered...: 3/2/2018 03:11:08


Deadman 
Level 64
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EDIT: I was wrong. I'll wait for someone else to answer :p

Edited 3/2/2018 03:13:18
One thing I always wondered...: 3/2/2018 03:29:41

Fizzer 
Level 59

Warzone Creator
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It's the highest value where a 4 is guaranteed to take a 2 in weighted random.
One thing I always wondered...: 3/2/2018 03:53:17


Deadman 
Level 64
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Damn that's what I said! Then I created a single player game to verify and 4v2 seemed to work on 17% WR as well?



Edited 3/2/2018 03:56:47
One thing I always wondered...: 3/2/2018 04:07:33


Rikku 
Level 61
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I vaguely remember the idea that analysis attack graph for WR is only accurate within 1% so it could be more like 99.5 or whatever. Dunno if that is true or I am confusing it with something else could easily be the latter.
One thing I always wondered...: 3/2/2018 04:36:31

Fizzer 
Level 59

Warzone Creator
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Rikku is correct. The analyze graphs are an estimate and not 100% accurate.

Fun fact: When I first calculated the highest value that 4v2 would always succeed, I made a mistake and came up with 18%. For a while, this was the value used in strategic games, You can still see this if you dig back into old 1v1 ladder games.

4v2s would succeed almost every time at 18%. But maybe one in 100 games, one attack would have a 4v2 failure somewhere, and that player would rage.
One thing I always wondered...: 3/2/2018 04:36:48


knyte 
Level 58
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At 17% WR, you've got just below a 0.0005% chance of killing only one army. That's why it doesn't show up on the graphs.
One thing I always wondered...: 3/2/2018 05:00:37


Lord of Turnips
Level 59
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What's the specific formula for calculating how many armies kill how much based on the luck value?
One thing I always wondered...: 3/2/2018 05:08:07

TheUberElite
Level 42
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I might be misunderstanding how luck works but I thought 17% luck meant 4v2's failed .02048% of the time, still rarely, but much more common than knyte's figure.
One thing I always wondered...: 3/2/2018 05:19:49

TheUberElite
Level 42
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I'm not sure whether I'm right or Knyte is, but as I understand it:

A 4v2 is 4 separate attacks with a 60% success rate.

So long as one of them succeeds you have this:

Expected kill 2.4
Actual dice rolls 1

Difference 1.4 * .17 = .238
2.4-.238=2.162

Therefore always a success, since then WR comes into affect and rolls another dice to decide if itd kill 2 or 3 armies, 16.2% chance to kill 3, 83.8% to kill 2.

If you fail all the rolls though you get:

2.4 * .17 = .408
2.4-.408=1.992

Which I believe means a 0.8% chance to kill 1 army wenever you fail all 4 attacks.

The chance to fail 4 attacks being

.4 * .4 * .4 * .4, thus the total chance being .0256 * .008 = .0002048, or 0.02048%

Edited 3/2/2018 05:20:07
One thing I always wondered...: 3/2/2018 13:23:24


Rufus
Level 60
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TheUberElite is correct. You have 99.97952% chance on taking 4v2 on 17% WR.

By the way, even though 16% SR is a mimic of 16% WR, it still has some meaning -- it's the highest value where a tap is guaranteed.

Edited 3/2/2018 13:35:51
One thing I always wondered...: 3/2/2018 13:41:24


knyte 
Level 58
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Huh, I was just using a script made by Math Wolf some time ago that should be doing the same math as Nauz. I'll edit this later with details, but I probably just misused the script.
One thing I always wondered...: 3/2/2018 22:37:13


MrTrolldemort 
Level 57
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Thanks for the answer guys :)
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