Talk:Luck

Thanks Cogniac, I've been wondering about this stuff for years now. :)

But still, this bit confuses me a little:

"Upon the death of the creature, a roll is made against the Luck of the character who did the most damage. If the roll is successful, both the amount of loot and the intensity of the properties on that loot are increased, after which the loot spawns."

So RAW, that means your Luck total only effects the chance of getting these bonuses, but has no effect on what these bonuses will actually be?

In that case, what are the bonuses themselves based on? A static figure, a random number? A percentage of the loot/intensities the creature would normally drop?

- Bomb Bloke 00:40, 1 March 2009 (UTC)

When I was writing that section and the sentences in the intro, I kept thinking that I needed to make it more clear that Luck gives you a chance at more and better loot, but does not directly affect what items you get and what properties appear on those items.
As to what actually happens when the loot itself spawns, I have no idea. To my knowledge, no dev team member has ever explained what happens after a Luck roll is successful, I.e. when the game rolls against your Luck and its successful, what happens next? We know that the amount and intensity of the loot is increased, but by how much? And what decides how much? I might send in a FoF question about it.
--Cogniac 02:16, 1 March 2009 (UTC)

Righto, reckon I'll do the same. More people ask, more chance of a response, I figure. - Bomb Bloke 02:33, 1 March 2009 (UTC)

Fun With Math

Bomb Bloke, concerning the Math behind Luck: I was going to make the argument that some readers would find the complete math workthroughs amusing, and that having it there didn't hurt anyone. But then I realized that it sort of was hurting people, because although the assumption that some people would find the complete workthroughs amusing wasn't an outright fallacy, if a user just wanted to see the relevant equations and forget the rest, they probably wouldn't be able to pick them out if they weren't particularly good at math. So I split the difference, and bolded the main equations. Furthermore, if JC ever gets the <math></math> tags working, the entire thing should look a lot cleaner.
--Cogniac 03:19, 15 March 2009 (UTC)

Very well. I still find it odd to include the "work back to L = C^1.8" steps.

You might like to include a "key" as to what the operators mean. Many people don't understand * and /, let alone ^ (granted, as you say, this'd be less of a problem with the math tags). Another idea might be to state the "final formulas" first up then show the workings underneath (for those people who really aren't interested in the process at all).

Also kinda worth pointing out is that the first equation could be extended like so:

L = C^1.8
ln(L) = 1.8 * ln(C)
ln(L) / 1.8 = ln(C)
e^(ln(L) / 1.8) = e^ln(C)
L^(ln(e) / 1.8) = C^ln(e)
L^(1 / 1.8) = C^1
L^(5 / 9) = C

- Bomb Bloke 03:54, 15 March 2009 (UTC)

e^(ln(L) / 1.8) = e^ln(C)
L^(ln(e) / 1.8) = C^ln(e)

What mathematical identity is that using?
--Cogniac 07:11, 15 March 2009 (UTC)

b^(m*n) = (b^m)^n

So:

e^(ln(L) / 1.8)
(e^ln(L))^(1 / 1.8)
L^(1 / 1.8)
L^(ln(e) / 1.8)

Er, the point I was getting at (and hoping to highlight by converting e^ln(C) to C^ln(e) on the other side of the equation) is that logs are redundant in this formula. Including them is sorta like adding 1000 to both sides in one step, only to subtract 1000 from each later on.

In fact the only identity required is this one:

y^z / x = y / x^1/z

That is to say, if x = y^z, then y = x^1/z. A fraction rule that is actually proven by the second formula you initially had on the page, so I won't bother extending it out.

- Bomb Bloke 07:03, 16 March 2009 (UTC)

I must've been incredibly tired when I looked at that, because I spent at least an hour looking for some crazy natural logarithm identity that would allow you to do

e^(ln(x) / y) -> x^(ln(e) / y)

with no intermediate steps, when in reality it was a simple exponent rule. Plus, I looked at the

C^ln(e) -> C^1

part and thought "Yeah, ln(e) = 1, that makes perfect sense." and then never noticed that everything happening on the right side of the equation was 100% useless redundancy.

I've removed the entire scientific calculator part, and just put the complete workthrough from L = C^1.8 -> L^(5 / 9) = C. Looking at it now, I don't even know why I wanted the scientific calculator part in there to begin with, as it was all an unnecessary obfuscation.
--Cogniac 08:10, 16 March 2009 (UTC)

Sorry, I did make that more of a riddle then it had to be.

- Bomb Bloke 10:52, 16 March 2009 (UTC)

Just Have To

I'm sorry but I just got to say how cool the calculators are! The chance to luck was inspired. Ceruleus 02:09, 19 March 2010 (UTC)