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?

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

Edited 3/2/2018 03:13:18

It's the highest value where a 4 is guaranteed to take a 2 in weighted random.

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

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.

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.

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.

What's the specific formula for calculating how many armies kill how much based on the luck value?

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.

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

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

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.

Thanks for the answer guys :)