Extreme WCA average/single ratios

The 20.90 20.90 20.90 is pretty suspicious. It's possible, but it seems much more likely that it was a mistake on the part of the judge or data entry person.
It is unlikely, but possible... I recently judged a similar thing at CMU (granted, it was Magic).

Stefan: Can you write this up for a few other common events, such as 3OH and 2x2?
I'd be interested to see this too.
 
The above was the ratio of currentRecordAverage/currentRecordSingle, here are the extremes for average/single with the single being in that average:

Code:
  avg  /  single =  av/si
112.86 /  16.46  =  [b]6.857[/b]  [url=http://www.worldcubeassociation.org/results/p.php?i=2009SHEP01]Daniel Sheppard[/url]

Oops. In case anyone is wondering, I was mucking around in the last solve :p

Aw, I was hoping maybe you were trying some sort of weird BLD solve.
 
Who is Shane Rowland?
world famous deaf rubiks

The 20.90 20.90 20.90 is pretty suspicious. It's possible, but it seems much more likely that it was a mistake on the part of the judge or data entry person.
It is unlikely, but possible... I recently judged a similar thing at CMU (granted, it was Magic).
Not even close. It's WAY easier for that to happen on Magic. 3x3 solves are all different :p
Yeah its pretty easy to do with magic. I got three 1.71s in the same average at nats last year.
 
It is unlikely, but possible... I recently judged a similar thing at CMU (granted, it was Magic).
Not even close. It's WAY easier for that to happen on Magic. 3x3 solves are all different :p
Interesting stats question then: given a Poisson distribution of solves, how much more likely is it to happen with a 1.51 average than a 21.05 average?
 
funny how nobody mentioned the current magic WR avg, but then again it's magic
 
It is unlikely, but possible... I recently judged a similar thing at CMU (granted, it was Magic).
Not even close. It's WAY easier for that to happen on Magic. 3x3 solves are all different :p
Interesting stats question then: given a Poisson distribution of solves, how much more likely is it to happen with a 1.51 average than a 21.05 average?

I'll give it a try, but I'm afraid I'll regret this :p

Let's assume µ=1,51 and 21,05 respectively (I don't know if this matters)
Let's also assume the intervals of outcomes are µ+-50%

We need a probability of getting 1,51 and 21,05, wich is 1/the number of outcomes. The number of outcomes are 151 and 2105.

λ for magic= 1/151= 0,0066
λ for 3x3= 1/2105= 0,00048

Lets say you get your most likely outcome, 1,51 on the first solve, now you need 2 identical times, gives n=2.

For magic: P(Y=2)= (0,0066^2*e^-0,0066)/2= 0,000022
For 3x3: P(Y=2)= (0,00048^2*e^-0,00048)/2= 0,00000012

Divide and you get about 183 times bigger chance to get two 1,51 in a row than two 21,05 in a row.

EDIT: Two in a row actually means three in a row, the first solve is x, and then you calculate the probability to get 2 more x.
 
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Johan: Will you do my math homework, pl0x? :)

Previous: There are enough ****heads in this world you can bring down with silly comments on a plastictoyforum, why choose a kid who is just having some fun?
 
Johan: Will you do my math homework, pl0x? :)

Previous: There are enough ****heads in this world you can bring down with silly comments on a plastictoyforum, why choose a kid who is just having some fun?

Someone who posted many fake videos, then said he would take them down, but never did.

Also someone who is believed to have cheated in comp..

He may be having fun, but it is kinda harmful.
 
I fully agree with you joey, that is not fun and ruins what a whole lot of people have put so much energy in. Hope he obides by the WCA rules in competition and general ethics when uploading videos :)
 
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