The One-Answer Question Thread [closed]

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What possible reason could that be? I mean, why? What benefit do you gain from knowing that? It seems like such a random way to calculate a statistic about your times. I would much prefer to just know the mean then the average.

So another question, in race tournaments, when they say they want the average of your solves, are they asking to apply the same rule to your times (take out the best and worst then calculate the average.), or are they asking for the mean of it all?

Edit:
I think I understand why the average removes the best and worst now. It brings the times being averaged out closer to the median time, which I suppose makes it a more accurate reading of what you are likely to get. Hmm... I'll have to think more about this.
 
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My guess is that if you have one DNF , the mean for the session will always be DNF, regardless of the other times, ruining everything. Averages allow one DNF and assume it as the slowest time to be taken out.
In race tournaments they want your average, not your mean.
 
JF1zl3 said:
What possible reason could that be? I mean, why? What benefit do you gain from knowing that? It seems like such a random way to calculate a statistic about your times. I would much prefer to just know the mean then the average.

So another question, in race tournaments, when they say they want the average of your solves, are they asking to apply the same rule to your times (take out the best and worst then calculate the average.), or are they asking for the mean of it all?

Edit:
I think I understand why the average removes the best and worst now. It brings the times being averaged out closer to the median time, which I suppose makes it a more accurate reading of what you are likely to get. Hmm... I'll have to think more about this.
Click to expand...

In competitions the judges will write down your times on a little piece of paper and the average will be calculated and you're ranked based on that.
 
I think sub 15 is the first 'hard' step. I'm personally not much faster than you and am finding sub 15 to be quite illusive. Sub 20 can be achieved by just practicing, but I think realistically to break sub 15 you need to pinpoint your weaknesses and improve on them. Only then will you improve. But it's the same as all barriers, once you break it, you wonder why you ever found it difficult.
 
Re: The One Answer Question Thread

Hey! I just learned 2l oll/pll. I wanna learn full pll now. Is there a list of how often different pll cases come up? So i know which ones i just learn first :)
Or just the easiest to recognize/easiest algs :)

Sent from my GT-I9300 using Tapatalk 2
 
Lagom said:
Hey! I just learned 2l oll/pll. I wanna learn full pll now. Is there a list of how often different pll cases come up? So i know which ones i just learn first :)
Or just the easiest to recognize/easiest algs :)

Sent from my GT-I9300 using Tapatalk 2
Click to expand...

Dan's cubestation used to have the stats, but his website is down. You should learn Us, As, Ts and Ys, then slowly learn some more 2 corner and 2 edge swap like Fs, Rs and Js. Or whatever order you like, this is sort of what I first did. Gs can be a pain to recognize, but are quite worth it, since they do come up quite a lot.
 
Lagom said:
Hey! I just learned 2l oll/pll. I wanna learn full pll now. Is there a list of how often different pll cases come up? So i know which ones i just learn first :)
Or just the easiest to recognize/easiest algs :)
Click to expand...

I wouldn't worry too much about probabilities you will run into them all. Learn the ones that are longest to solve as 2-look PLL first.

E-perm was a joy to understand. Learn that one first if you don't already know it. It is longer but it is what a one look PLL should be, chaos to solved. Also H- and Z if they aren't already part of your 2look.

My 2 look PLL was corners first with an a-prem-a so I started 1-looking the cases that left me with a z-perm on the second look. I think that was F, Y, T's first. Then relearned both Us (sune and anti-sune) with both right and let hands so I could catch it and solve it wherever it came up without much cube rotation. Then I learned A-perm-B. Lastly I one-looked the ones that left me an H on the second look. H just doesn't come up too often anymore. I haven't actually gotten around to the g-perms and just do them as an A then a D but I know what direction each part is going to go so I'm not actually looking just doing a longer non-optimal alg. I need stop practicing this bad habit and seriously sit down a third time and try to learn these.
 
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Lagom said:
Hey! I just learned 2l oll/pll. I wanna learn full pll now. Is there a list of how often different pll cases come up? So i know which ones i just learn first :)
Or just the easiest to recognize/easiest algs :)

Sent from my GT-I9300 using Tapatalk 2
Click to expand...
16 of the 21 algs have the best probability of happening (1 in 18).
I personally taught my friend these PLL's first (in this order):
T, Jb, Ja, F, Y, Na
That's because the algs I have for them are very similiar:

T = R U R' U' R' F R2 U' R' U' R U R' F'
Jb = R U R' F' then most of the T perm until the cube is solved
Ja = F U' then T perm without the first 4 moves then the last 4 moves of T perm again
F = R' U' F' then T perm with the last move (F') replaced by U R
Y = F R U' R' U' then J perm but stop in the middle of the R2 and do an F'
Na = R U R' U then J perm then U' R U' R'

That's 6 algs learned by just learning the T perm and executing it with slight modifications
 
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I am looking to buy a Witeden 2x2 cube version 1, however I don't want to buy off a Chinese website as shipping takes ages and I am not a patient person ;) I found a variety of European sites that sell them, but none of them specify whether it's a v1 or a v2, so I was wondering if there is a way you can determine which version it is just by the photo.

EDIT: This would be the best site for me, is it possible to see which version it is? http://www.amazon.fr/Cube-Magique-2...f=sr_1_1?s=toys&ie=UTF8&qid=1361547370&sr=1-1
 
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Wassili said:
I am looking to buy a Witeden 2x2 cube version 1, however I don't want to buy off a Chinese website as shipping takes ages and I am not a patient person ;) I found a variety of European sites that sell them, but none of them specify whether it's a v1 or a v2, so I was wondering if there is a way you can determine which version it is just by the photo.

EDIT: This would be the best site for me, is it possible to see which version it is? http://www.amazon.fr/Cube-Magique-2...f=sr_1_1?s=toys&ie=UTF8&qid=1361547370&sr=1-1
Click to expand...

Yes. The WitTwo II ships out in violetish box while the WitTwo I ships out in orangeish box. From that picture alone, you can't really tell.
 
I meant to make this an edit of my last night's post, but at around 1:00am (Central Time), the website would not load. Maybe it was site maintenance Probably just my internet. I dunno. Either way this is my new question:

Alright, in my editing I have discovered a slight problem. So if the way to determine an average is to remove the single worst and single best time and then calculate the mean of the remaining times, and a certain race competition has 2 sets of scrambles, and that same race is asking for the average of both of them together (into one measurement of time), am I supposed to do
a) take the averages of the 2 sets, add them together, then divide by two? (I believe this finds the Mean of the 2, not the average)
or
b) look through all of my solve times for both sets and find the single worst, and single best time and remove both of them, then take the remaining times and calculate the mean of them?

Example
Set 1:
15.32
30.45
37.84
34.56
30.67
Average of set 1: 31.89 (Calculated by omitting 15.32 and 37.84. Then adding #2,3,4 and dividing that by 3.)

Set 2:
47.56
32.47
39.56
36.76
36.03
Average of set 2: 35.09 (Calculated by omitting 32.47 and 47.56. Then adding #3,4,5 and dividing that by 3.)

Now to find the Average of both of these sets combined I have the following results based on which case I stated earlier is used:
a) 33.49 (Calculated by adding 31.89 and 35.09 and then dividing by 2)
or
b) 34.79 (Calculated by omitting 15.32 and 47.56. Then adding #2,3,4,5 of set 1 and #2,3,4,5 of set 2 and dividing that all by 8.)
 
The One Answer Question Thread

sneaklyfox said:
Actually I think it's option A. But really, the best thing is to ask whoever is running the race in question. That way, everyone entering the race should be doing it the same way, whichever way it should be.
Click to expand...

Or maybe ill just post both. Ill post an Ao2 and a Mo2. In my opinion, the more statistics, the better. :)
 
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