The Statistical Case for Combined Finals

[macky]

Item 2 under Fairness is " (2) that the rules must apply equally to all competitors." Your point is still irrelevant to any of the Fairness items. Again, you're distorting Ravi's intention. You presented your concern as one involving the Fairness items, not "(Limited) forgiveness for outliers." If you meant to discuss "(Limited) forgiveness for outliers," I now understand why you brought up your concern. But I still disagree because of the arguments Pedro presented.

 

[reThinking the Cube]

But I still disagree because of the arguments Pedro presented.

Pedro failed to win the arguments you are choosing to agree with, and you can change your opinion at any time.
I will trust you to be *fair*.

For example:
Competitor A: (11), 11.2, 11.5, 11.6, (11.8) (same as above) [11.43 avg]
Competitor B: (9), 9.3, 11.4, (DNF), 15.8 (i.e. Pedro going-for-gold, and popping on 4th solve) [12.17 avg]

Which competitor's performance should allow them to advance into the next round, A or B?

Now, here is one from Chris Hardwick that I happen to also agree with:

I'd rather that the person who set the best average at the comp won, no matter what round it was in.

I'm not convinced that this identifies the best cuber at the tournament better than some other proposed methods in this thread.

For Fridrich solvers there is a \( \frac{15263}{15552} \) probability of a full step LL solve assuming no special techniques are used during LL beyond OLL and PLL. So the chance of getting at least two solves in a trimmed 5 cube average that are some form of LL skip is:

\( 1 - \frac{15263}{15552}^5 - 5\times \frac{15263}{15552}^4\times \frac{289}{15552} \approx 0.0033 \) or roughly 1/3% chance to get such an average.

Now, if you're good enough to make it through 3 rounds of 3x3x3 then the chance that at least one of your averages has at least 2 skips is approximately:

\( 1-(1-0.0033)^3 \approx 0.0099 \)

So each cuber who makes it to the final, and gets to do 3 averages, has an approximately 1% chance that one of their 3 averages has at least two skips. With 16 competitors making it to the final, the probability that at least one of those competitors gets at least one average with at least two skips is:

\( 1-(1-0.0099)^{16} \approx 0.147 \)

So this competition has an approximately 15% chance that the winner wins by luck, and not necessarily by skill.

A major assumption I am making here is that all of the 16 people in the final are good enough that getting at least two skips in a trimmed 5 cube average gives them a significant enough advantage to outclass the other cubers who would also make it to the final and not get such an average. I think this assumption, even if it might not be completely true now, will not be unreasonable within 5 years.

Another assumption is, of course, that other methods will contribute to this chance as well - some possibly more than this and some possibly less. This also does not include the chance for F2L pair skips, and the fact that people use advanced techniques to increase their chance of a skip in the LL. The real percentage of competitions won by luck would, I propose, be higher than this estimated chance.

Chris

Outlier Polluted Averages(OPA) - SUCK!

 

[Kirjava]

Personally, I don't have a problem with luck influencing results.

 

[reThinking the Cube]

Personally, I don't have a problem with luck influencing results.

Do you have a problem with having just 1 solve per round?

 

[Kirjava]

Yes, because that would be dumb.

 

[hawkmp4]

Personally, I don't have a problem with luck influencing results.

Agreed. If someone is able to take advantage of two easy F2Ls or two PLL skips or whatever in an average, that's just part of cubing. Good for them.

 

[reThinking the Cube]

Yes, because that would be dumb.

...luck?

 

[macky]

rTC, you ignored this part:

Item 2 under Fairness is " (2) that the rules must apply equally to all competitors." Your point is still irrelevant to any of the Fairness items. Again, you're distorting Ravi's intention. You presented your concern as one involving the Fairness items, not "(Limited) forgiveness for outliers." If you meant to discuss "(Limited) forgiveness for outliers," I now understand why you brought up your concern. But I still disagree because of the arguments Pedro presented.

And this:

For a cuber that performs 100% CONSISTENTLY (every solve is the same time), the median would be EXACTLY THE SAME AS THEIR AVERAGE. Consistent cubers won't LOVE the median as much as inconsistent cubers, but they also should have no good reason to HATE it. Seems pretty *fair* to me, and to argue otherwise would appear to be unnecessarily *misguided*.

It seems like you understand that your system supports a very inconsistent cuber (messes up to 2/5 of solves) over a perfectly consistent cuber whose median is 0.01 worse than this cuber's. Then we can't really do anything to change your mind.

I respect your opinion - "If you mess up twice in a round, well, too bad.", but it doesn't HAVE to be that way.

Yes, there are situations where speedcuber who should have qualified is eliminated because of two unfortunate mistakes. Pedro has already addressed this. We, and I'm sure the majority of speedcubers, understand this and still don't agree with you. A similar consideration came up in 2004 when we decided stop awarding an extra solve for a POP. One argument was that a POP followed by a bad solve that gets trimmed would mean 2 bad solves out of 6, both not counting. Many members in the community thought that 33% inconsistency (less than your %40) was already too much to go unpenalized.

Your system would be great if all mistakes people ever made were timer failures. Competitors should then always perform around their true potential except for errors outside his control, so the median would be the best measure. In reality, most slow times are the competitor's fault to some degree. POPs and even corner twist DNFs are mistakes that can been avoided by better preparing the cube.

 

[Kirjava]

You're implying that I shouldn't have a problem with a round with 1 solve because I don't mind luck influencing results, right?

I thought I told you to stop trolling.

 

[reThinking the Cube]

rTC, you ignored this part:

:fp

 

[macky]

RTTC DE MACKY = UR TOO GUD DR MAN TOO GUD = 73 CUAGN DE MACKY \SK E E E

 

[Kirjava]

I think you broke macky.

 

[hawkmp4]

I lol'd. Poor macky.

Only fools would argue with RTC.

 

[Pedro]

Dude, you still failed to argument against my example where a better cuber would be out for 0.01 if we used the median time...is it fair?

A: 10, 10, 11.5, 11.6, 12
B: 11.2, 11.3, 11.4, 15, 20

also, did someone read this?

To summarize: someone with a bad 2nd round would have minimal chances to win, in a strong competition (like needing a sub-9 average, dependind on the other people results, if we used Ravi's suggestion of taking the mean of 3 averages to determine the winner.

 

[reThinking the Cube]

RTTC DE MACKY = UR TOO GUD DR MAN TOO GUD = 73 CUAGN DE MACKY \SK E E E

Too funny. My eyes starting watering. Reminded me of something -

http://www.youtube.com/watch?v=PbcctWbC8Q0

 

[hawkmp4]

Dude, you still failed to argument against my example where a better cuber would be out for 0.01 if we used the median time...is it fair?

A: 10, 10, 11.5, 11.6, 12
B: 11.2, 11.3, 11.4, 15, 20

also, did someone read this?

To summarize: someone with a bad 2nd round would have minimal chances to win, in a strong competition (like needing a sub-9 average, dependind on the other people results, if we used Ravi's suggestion of taking the mean of 3 averages to determine the winner.

I read it, and I agree. He may have read it, but he's not interested in intelligent conversation anymore, just trolling.

 

[Pedro]

I read it, and I agree. He may have read it, but he's not interested in intelligent conversation anymore, just trolling.

No, I don't mean him, I mean the rest (specially Ravi).

 

[reThinking the Cube]

Dude, you still failed to argument against my example where a better cuber would be out for 0.01 if we used the median time...is it fair?

A: 10, 10, 11.5, 11.6, 12
B: 11.2, 11.3, 11.4, 15, 20

also, did someone read this?

To summarize: someone with a bad 2nd round would have minimal chances to win, in a strong competition (like needing a sub-9 average, dependind on the other people results, if we used Ravi's suggestion of taking the mean of 3 averages to determine the winner.

I read it, and I agree. He may have read it, but he's not interested in intelligent conversation anymore, just trolling.

No. I asked him my question first, and he has been avoiding answering me. Lighten up please.

For example:
Competitor A: (11), 11.2, 11.5, 11.6, (11.8) (same as above) [11.43 avg]
Competitor B: (9), 9.3, 11.4, (DNF), 15.8 (i.e. Pedro going-for-gold, and popping on 4th solve) [12.17 avg]

Which competitor's performance should allow them to advance into the next round, A or B?

 

[Daniel Wu]

Dude, you still failed to argument against my example where a better cuber would be out for 0.01 if we used the median time...is it fair?

A: 10, 10, 11.5, 11.6, 12
B: 11.2, 11.3, 11.4, 15, 20

also, did someone read this?

To summarize: someone with a bad 2nd round would have minimal chances to win, in a strong competition (like needing a sub-9 average, dependind on the other people results, if we used Ravi's suggestion of taking the mean of 3 averages to determine the winner.

I read it, and I agree. He may have read it, but he's not interested in intelligent conversation anymore, just trolling.

No. I asked him my question first, and he has been avoiding answering me. Lighten up please.

For example:
Competitor A: (11), 11.2, 11.5, 11.6, (11.8) (same as above) [11.43 avg]
Competitor B: (9), 9.3, 11.4, (DNF), 15.8 (i.e. Pedro going-for-gold, and popping on 4th solve) [12.17 avg]

Which competitor's performance should allow them to advance into the next round, A or B?

A obviously.

 

[reThinking the Cube]

For example:
Competitor A: (11), 11.2, 11.5, 11.6, (11.8) (same as above) [11.43 avg]
Competitor B: (9), 9.3, 11.4, (DNF), 15.8 (i.e. Pedro going-for-gold, and popping on 4th solve) [12.17 avg]

Which competitor's performance should allow them to advance into the next round, A or B?

A obviously.

Why you no likeee B.Pedro?:confused: