Inequality describing improvement using a method

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I just had an idea about improvement. Being a mathy person, I turned it into an inequality. It states that the fastest you can get with a method is at least as fast as the best nonlucky solve that you have gotten. Of course, nonlucky is not strictly defined, but I would define it as a solve where you have to do every step, and each algorithm is at least medium long, not super easy (unless you intentionally influence it). My inequality is stated like this:

C(m) ≥ min(Nₘ)

Where: C(x) = the capability of method x (1/a where a is the best average you can get using x), m = the method that you use (more specific than just CFOP, Roux, ZZ, etc, I mean what algorithms you use and which LL cases you 2-look, 3-look, or 4-look, whether you know full CMLL, etc.), min(x) = the lowest value in the numerical set x, and Nₓ is all of your nonlucky solves using method x.

If this inequality is true, then it is definitely possible (maybe not easy) to try to get sub-13 or sub-12 with pure CFOP and no other algorithms. (I have gotten nonlucky sub-12 singles.) There are likely people who have gotten sub-12 with no extra alg sets, but I can't name any specifically..
 
If this inequality is true, then it is definitely possible (maybe not easy) to try to get sub-13 or sub-12 with pure CFOP and no other algorithms. (I have gotten nonlucky sub-12 singles.) There are likely people who have gotten sub-12 with no extra alg sets, but I can't name any specifically..


Lots of people could get sub-12 or even 10 with no extra alg sets. If feliks used PURE CFOP, no COLL or WV or easy ZBLLs or anything he could sub-10 still, I have no doubts about it. And I've talked to a guy on here once (I forget his username.. mozart maybe?) and I asked if he used any COLL or Multislotting or X-cross or anything and he said no. He was sub-10! He said he didnt find any of those things nessesary to get there (I wish I knew how he was so fast with pure CFOP :P)
 
It states that the fastest you can get with a method is at least as fast as the best nonlucky solve that you have gotten. Of course, nonlucky is not strictly defined, but I would define it as a solve where you have to do every step, and each algorithm is at least medium long, not super easy (unless you intentionally influence it).

I really do not like the term "lucky" or "unlucky" solves. What about solves where you slot one F2L accidentally while doing a cross? What about if each F2L case is easily inserted with a couple moves? What if the cross already has three parts completed? I count every solve as important as other, even if there is a OLL and PLL skip. I don't disqualify easy F2L pairs or cross, so I don't disqualify anything else as "lucky."
 
I feel this thread is unnecessary there are far too many variables that go into to calculating something like this all we know is the people that have practiced the most are the fastest that's just how it works... We cannot calculate the limits of a method although feliks has said he thinks it'd be possible to average somewhere in the 5 second mark we will never know until someone does it
 
I agree with the idea that, if you take your best solve with average-length algs and no skips, that should give you a sense of how good you can be on average (minus mistakes and bad lookahead and so on). I've used this in the past to get an idea of where I was at. I don't think it warrants making a ton of notation and writing a formula, though.

FWIW, I use Fridrich with no additional alg sets (I just try to orient LL edges during F2L) and I have a sub-10 avg12, and I don't even turn fast.
 
Guys, you are acting like this is a limit. If you read my original post, there is definitely a greater than or equal to sign there. All of you people saying things like "Feliks could easily be sub-10 with pure CFOP", I agree with you. Is sub-10 sub-12? Errr... yeah. 10 < 12. 8 < 12. 11 < 12. 10 < 13. I couldn't name any people specifically because Feliks just DOESN'T use pure CFOP. He uses lots of extra alg sets. By the logic of the inequality, a sub-5 global average is possible with the algorithms Feliks currently uses. I know I'm using Feliks as an example here, but plenty of other fast people could too. As long as their non-lucky PB is sub-5. If it isn't sub-5, then that is when the ≥ sign kicks in. Even if someone doesn't have a sub-5 non-lucky PB, they surely have a chance at averaging sub-5. The title of this thread contains the word Inequality, not Equation.

One more thing: It says in my original post that it may not be easy to get a sub-12 average using pure CFOP. I'm sorry, but this was a little bit unclear. It means that if you average around 15-16 seconds (about where I am right now) then it won't be very easy to get sub-12 with no extra algorithms. I never said impossible. For you fast people, I'm sure you can easily do that, but I wasn't really talking about you. This thread did need some clarification, and I'm sorry about that. Why don't you post about whether you think the inequality is true or not, and not whether you can get sub-10 times with 2-look OLL. Sorry about the confusion in the first post.
 
I really do not like the term "lucky" or "unlucky" solves. What about solves where you slot one F2L accidentally while doing a cross? What about if each F2L case is easily inserted with a couple moves? What if the cross already has three parts completed? I count every solve as important as other, even if there is a OLL and PLL skip. I don't disqualify easy F2L pairs or cross, so I don't disqualify anything else as "lucky."

Well, in the formula, it is a key factor (You won't get a OLL skip or something every time) but when it comes down to it, it is using your judgement. Was the F2L easy? Did I get easy LL cases?

Lots of people could get sub-12 or even 10 with no extra alg sets. If feliks used PURE CFOP, no COLL or WV or easy ZBLLs or anything he could sub-10 still, I have no doubts about it. And I've talked to a guy on here once (I forget his username.. mozart maybe?) and I asked if he used any COLL or Multislotting or X-cross or anything and he said no. He was sub-10! He said he didnt find any of those things nessesary to get there (I wish I knew how he was so fast with pure CFOP :P)

You are forgetting that the formula does not imply a limit, and that 10 < 12. Heck, anyone could get sub-12 with no extra alg sets. It is maybe not the easiest way to do it, but it is very possible.
 
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What does this even mean? Are you saying that I should one day be able to average 14 because I once had a 14 second single? That seems like a pointless oversimplification. Performance depends on many factors, of course.
 
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