The Statistical Case for Combined Finals

[Ravi]

I have long been of the opinion that a five-solve trimmed average measures a speedcuber’s skill too unreliably to decide the winner of a competition. Recent events have shown this observation to be correct, especially after the tightening of the top group of competitors that has taken place over the last several years. One case that comes to mind is Breandan Vallance’s spectacular yet improbable 10.74 average in last year’s world championship. Another is last year’s US Nationals: although Rowe did have the best overall performance, his victory must be attributed at least partly to chance, as the three rounds of the competition actually produced three different winners. As I write this, the 2010 edition of US Nationals promises to be even more so, as none of the top three cubers in the first round managed to return to the podium in round 2.

I recently calculated some statistics to test just how uncertain averages can be. I started by assuming that any given speedcuber’s times are modeled fairly accurately by a normal distribution (bell curve)--this ignores factors such as lucky cases, DNFs, nerves, and mechanical slips*, but it should be close enough for our purposes. Then any speedcuber can be represented essentially by two numbers: his or her mean and standard deviation. The purpose of taking averages, of course, is to decrease the standard deviation while preserving the mean. The question: How much do standard deviations decrease when averages are used?

There are two ways to find out, and I somewhat regrettably chose the one that didn’t involve solving Rubik’s Cubes all night. First, I programmed the trimmed-averaging process into Mathematica (the input is a list of arbitrary length):

trim[x_] := Drop[Drop[Sort[x], -1], 1]
avg[x_] := Mean[trim[x]]

Now it’s time for some number-crunching. I had Mathematica calculate a million random averages using the default mean of 0 and standard deviation of 1 (a translation and a dilation away from our ideal cuber’s distribution), then calculate the standard deviation of the averages.

StandardDeviation[Table[avg[RandomReal[NormalDistribution[0, 1], 5]], {i, 1000000}]]

I tried this several times, generally getting results around 0.4765. This means that the standard deviation of your averages is slightly less than half of the standard deviation of your raw times; that is, if you ordinarily have a standard deviation of 1.4 seconds, the standard deviation of your averages will likely be around 0.7 seconds. For comparison, I tried a few other averaging schemes, with the approximate coefficients listed below:

single solve (mean of 1): 1.000
avg (median) of 3: 0.670
mean of 3: 0.577
mean of 4: 0.500**
avg of 5: 0.477
mean of 5: 0.447
mean of 10: 0.316
avg of 12: 0.295
mean of 3 avgs of 5: 0.275***
avg of 15: 0.262
mean of 15: 0.258
mean of 20: 0.224

You can experiment with this by adjusting the “5” in the code above and, for untrimmed means, changing “avg” to “Mean”.

What do these statistics mean? Well, averages of 5 are certainly more precise than single solves, but it is not difficult to improve further. An average of 12, for example, theoretically has only 62% of the standard deviation of an average of 5. Moreover, most major competitions already have three rounds of five solves each, so using combined finals for 3x3 could cut standard deviations nearly in half without requiring any additional time in the schedule. Of course, if more than five solves were used to calculate averages, it may be necessary to allow, say, two dropped times instead of one for the odd DNF. Or, in the interest of maintaining compatibility with existing average records, it might make more sense to simply rank competitors based on the mean of their averages throughout the competition, in the same way that a golf tournament is scored. This would have no adverse effects on scheduling, compatibility with past records, or (in my opinion) suspense, but it would make our competition results much more precise (coefficient = 0.275).

In more practical terms, let’s imagine a pair of hypothetical competitors, A and B. A and B both have standard deviations of exactly one second, but A is half a second faster than B on average. If they compete with a single solve in a one-on-one competition, then A has a 63.8% chance of winning. With trimmed averages of 5, A has a 77.1% chance of winning. With a mean of three averages of 5, as I suggested above, A has a 90.1% chance of winning****.

In the next few hours, the final round of this year’s US Nationals will take place. If, as I expect, no single person emerges as a clear champion, then I believe it will be time to seriously consider a more precise averaging scheme for our competitions.

May the best cuber win.




* I suspect that all of these factors, especially nerves, would mildly increase unpredictability and therefore all standard deviations. Also, the distribution is somewhat skewed in real life, since (for example) a speedcuber who averages 11 seconds is much more likely to screw up somehow and solve in 18 seconds than to get lucky enough to solve in 4 seconds. A generalized extreme value distribution is likely to provide a better model, but only slightly so.

** You may notice that the coefficient for a mean of n is 1/sqrt(n). This follows from the fact that for independent X and Y, stdev(X+Y) = sqrt[stdev(X)^2 + stdev(Y)^2]. Exact values for the standard deviations of trimmed averages are less obvious.

*** Mathematica code:
StandardDeviation[Table[Mean[Table[avg[RandomReal[NormalDistribution[0, 1], 5]], {i, 3}]], {i,
1000000}]]

**** Mathematica code:
N[CDF[NormalDistribution[0, Sqrt[2]], 0.5]]
N[CDF[NormalDistribution[0, Sqrt[2]*0.4765], 0.5]]
N[CDF[NormalDistribution[0, Sqrt[2]*0.275], 0.5]]
The factor of sqrt(2) comes from the fact that I did the calculation by subtracting two normal distributions centered on the origin and checking the probability of the signed difference being less than 0.5. Standard deviations of differences of normal distributions are the same as those of sums--see the double-asterisked item above.

 

[Kirjava]

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

 

[Carrot]

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

GO JAPAN!!

 

[riffz]

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

I've long been of this opinion. Moving on to the next round would be important because you would have more chances to set a better average.

 

[Ravi]

By the way, Rowe Hessler won in much the same way as last year: he did have the best overall performance, but could have lost, if his fourth-place finish in round 1 is any indication.

http://www.cubingusa.com/usnationals2010/results/index.php#7

 

[Lucas Garron]

An important fact to consider: In the earlier rounds of a tough competition, cubers often try to make it to the next round rather than a fast average.
Two DNFs are unacceptable, adding 2 seconds to your average can be.

In the finals, however, it's about the maximum speed and all its risks.

 

[Carrot]

An important fact to consider: In the earlier rounds of a tough competition, cubers often try to make it to the next round rather than a fast average.
Two DNFs are unacceptable, adding 2 seconds to your average can be.

In the finals, however, it's about the maximum speed and all its risks.

GOGO GARRON!

 

[Ravi]

An important fact to consider: In the earlier rounds of a tough competition, cubers often try to make it to the next round rather than a fast average.
Two DNFs are unacceptable, adding 2 seconds to your average can be.

In the finals, however, it's about the maximum speed and all its risks.

You have a point: It would be (and is) unfortunate for a serious competitor to be eliminated early because of DNFs. Of course, the same can happen in a final, but combined finals would increase the probability. At the same time, though, is there any reason for one average to be "worth more" than another in terms of ranking? I would argue that someone who averages 12, 11, and 13 is a better speedcuber than one who averages 14, 13, and 12.

Kir and riffz also have an interesting point, but I can't say I'm totally in favor of it. I think a great speedcuber is defined more by consistent skill than by occasional spikes of incredible speed--to use WC2009 as an example again, I'm more impressed by Tomasz Zolnowski's 11.09, 11.89, and 11.64 than by Breandan's 12.80, 12.20, and 10.74. Also, I would assume that a combined final would statistically point to the fastest competitor much more reliably than a "best of"-type format. (Of course, what you're talking about is far better than the straight best-of-3 used in WC1982. But the speedcubing world of 1982 was a totally different place.)

What do you guys think about the stats though? Does a theoretical improvement of 77.1% to 90.1% accuracy merit changing the rules?

 

[qqwref]

Ravi, there are two important factors that you are not considering. First: there is always a chance of a very nice time from a skip or a very bad time from a pop or similar problem. This even extends into averages as it is always possible in theory to get two DNFs (or bad times) or two skips (or very nice solves). Very bad times are more likely in a competition due to pressure, especially in a final like the one at Nationals 2010 where every competitor did their five solves in front of hundreds of people, the only one on stage in a silent and huge room. So the standard deviation approach doesn't do well. You can tell because your numbers give an average of 5 as varying LESS than a trimmed average of 5; this isn't true in real life as many competition averages are strongly hurt or helped by the removal of the best and worst times. So considering times to fall in a strict bell curve is an approach that may be easier but will not give a good model.

Second: people's time distributions will change depending on a number of factors, such as tiredness (either from cubing all day or from just waking up), being warmed up, nervousness, how much they've been practicing that puzzle in the past few days, what cube they're using, whether they are distracted, and so on. However, not everyone is affected the same way, and not everyone will have the same factors affecting their average in a given round anyway. Because of this it's not accurate to talk about someone's mean and SD as a constant throughout the competition.

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

This.

It's for the last reason that I don't think it's quite fair to only rank people by their final round average. Taking a mean of a few averages could be good, but some people might be very negatively affected in one of the rounds and that would basically negate their chance of winning even if they totally destroyed in other rounds. So I like the idea of ranking by the best average obtained in the competition. That would allow people to do their best when they feel best, rather than expecting everyone to do their best at a given point; it would make previous rounds matter, instead of just being a "get under <x> seconds" task which is trivial for someone who could win; and it would make sure that anyone who leaves the competition with a world record average (and is thus, officially, the best ever at that event) is guaranteed to also have won that event.

 

[blade740]

One thing i'd like to point out is that while cubers may get different scrambles in preliminary rounds, all finalists have the same scrambles. Using the best of 3 rounds might mean that an event with easy scrambles in only one group of the first round might disadvantage cubers who are otherwise faster.

I really like the idea of using themean of your averages. The problem I stated still exists but is mitigated by averaging.

Ultimately, prizes of any sort, and the "winner" of a competition, are at the discretion of the organizers. I hope this will be taken into account in the future but I see no reason for a rule change.

 

[blade740]

One thing i'd like to point out is that while cubers may get different scrambles in preliminary rounds, all finalists have the same scrambles. Using the best of 3 rounds might mean that an event with easy scrambles in only one group of the first round might disadvantage cubers who are otherwise faster.

I really like the idea of using themean of your averages. The problem I stated still exists but is mitigated by averaging.

Ultimately, prizes of any sort, and the "winner" of a competition, are at the discretion of the organizers. I hope this will be taken into account in the future but I see no reason for a rule change.

 

[Ravi]

many competition averages are strongly hurt or helped by the removal of the best and worst times. So considering times to fall in a strict bell curve is an approach that may be easier but will not give a good model.

I think you are saying, to use the language of http://en.wikipedia.org/wiki/Kurtosis#Terminology_and_examples , that competition times tend to be leptokurtic (aka positive excess kurtosis): that most of the variance comes from occasional outliers rather than small, frequent variations. This is true to a point, but let's refer again to the statistics. I entered into Mathematica all of the statistics for everyone who made the finals (20 solves for each person, except for the two who DNFed earlier), and calculated kurtosis values, which are listed below in the order of final rankings:

1. Rowe Hessler -1.32
2. Phillip Espinoza -0.81
3. Andy Tsao -0.19
4. Andrew Kang +0.14
5. Michal Pleskowicz -1.18
6. Mitchell Stern -0.84
7. John Tamanas +0.59
8. Justin Adsuara +0.61
9. Weston Mizumoto +3.72
10. Patricia Li -0.71
11. Lucas Garron +2.72
12. Emily Wang -0.07
13. Erik Johnson +0.29
14. Timothy Sun +3.72
15. Dan Dzoan +11.23
16. Anthony Brooks +0.50

So nine of the sixteen did indeed have positive excess kurtosis as you predicted, while the other seven had negative excess kurtosis. For all of the four competitors with significantly positive (>1) excess kurtosis, it is easy to point to one or two solves which threw off the statistics massively--Dan's 34 in the final, Tim's 25 in the third round, Weston's 20 in the first round and 18 in the third, and Lucas's 22 in the first round--but which did not count in their respective averages. Thus it is only barely true that, as you argued, most real speedcubers' distributions in competition are more leptokurtic than a normal distribution, and that effect is almost entirely negated and even reversed in the process of trimming averages. Normal distributions aren't perfect models, but they're surprisingly accurate, especially when we're trimming the averages anyway.

 

[incessantcheese]

i apologize that i'm entering into your extremely analytical thread with more of an opinion/feel thing but -- trying to get rid of the risk/statistical variance in determining the winner isn't always a good thing. i think it's far more exciting to have it be possible for any person who's reasonably close to a winning average time to have a brilliant stroke of luck and nerves and win a competition - rather than it be a grind of who most deserves it. of course this will lead to some "undeserved" or "unfair" wins, but i think the winner of an event is just that - whoever managed to get into the last round and set a great average when things mattered the most.

 

[cmhardw]

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

 

[Pedro]

Yes, averages of 5 are not a really good indicator of people's skills. That's why we usually compare averages of 12 at home, since you have many many chances (as many as you want/can do) or even averages of 100.

But in competition you get at most 20 solves (usually 15 if you are a reasonably fast cuber) and giving people more solves would take too much time. Our competitions already take a lot of time (US Nationals was held over 4 days, but we usually have 2 and venue limitations and people who need to catch a flight and all that). Giving greater averages as rounds go on (5, then 7, then 9 solves or whatever) would be unfair to people who didn't qualify for, let's say, 0.01 and won't have the chance of a "better" measurement of their skills.

I agree with the point that, in competitions, it's about performing when it matters. In a World Cup of soccer, you can win all matches until the final by 5-0. If you lose the final by 0-1, you're not the champion. Maybe it wasn't "fair", but you knew that you should perform at that especific match and you won the other matches to qualify for that one. You don't get to play the final match if you fail to qualify, so who gets there made his way.

And since cubing, as other sports, takes into account pressure and nerves, I think it's fine to give the title to who performed best when they had to perform best.

The way it is now, people prepare themselves (or try to, at least) for the final, to do their best when it matters, so if we were to change the rules and use all solves, it would require a different preparation/way of thinking.

Also there is the problem that, for example, someone does really well in the 1st round but not so well in the 2nd, and doesn't qualify for the final. Suppose person A gets 10.00 and 14.00, respectively. Person B gets 11.00 and 12.00 and qualifies, but in the final he/she is nervous/tired/whatever and the average is 13.90. He/she will have a greater global average than person A, but he/she had more solves, so more chances to fail/succeed, and the comparison is not really valid in my view.

So how would we weight that? Who wins? Someone with a 12.00 average over 10 solves or someone with 12.01 average over 15 solves?

 

[Ravi]

First: there is always a chance of a very nice time from a skip or a very bad time from a pop or similar problem. This even extends into averages as it is always possible in theory to get two DNFs (or bad times) or two skips (or very nice solves). Very bad times are more likely in a competition due to pressure, especially in a final like the one at Nationals 2010 where every competitor did their five solves in front of hundreds of people, the only one on stage in a silent and huge room.

True in theory, again, but let's look at the statistics. Throughout this year's US Nationals, four people were eliminated because of multiple DNFs:

- one person with a DNF and four DNSes in the first round;
- one person with 22, 24, 27, and two DNFs in the first round;
- one person (Nicholas Sia) with 15, 16, 19, and two DNFs in the second round after a 14.23 average in the first; and
- one person (Toby Mao) with 15, 16, and three DNFs in the second round after a 15.78 average in the first.

Of the 80 solves in the final round, zero were DNFs (although Dan Dzoan’s 34.21 might as well have been one). Of the 160 solves in the third round, there was one DNF (by 31st-placer Chris Dzoan), and a total of four sup-20 solves. Of the sixteen finalists, only Dan Dzoan (in the first round) and Mitchell Stern (in the second) had any DNFs throughout the competition. That’s two out of 320. Conclusion: DNFs really aren't as much of an issue as they are usually imagined to be.

As I argued previously, catastrophically slow times aren't a whole lot worse, especially when averages are trimmed. And, of course, lucky/unusually fast times are even less of an issue in general. Looking at the 318 solves collectively completed by the 16 finalists through the competition, we see that seven finalists suffered a single solve at some point which was over five seconds slower than their overall untrimmed average, but none had a solve over five seconds faster than average--the closest being Tim Sun's 9.40 compared to his 13.93 untrimmed (and therefore inflated by his third-round 25.33 solve) average.

your numbers give an average of 5 as varying LESS than a trimmed average of 5; this isn't true in real life

Well, you have a point. I'm beginning to wear myself out with all this statistical stuff (lol), but I'll let it suffice to say that real-life stats probably reflect your claim. Averages of 5 do vary slightly less than trimmed averages of 5 assuming perfect bell-curve distributions, presumably because the increased sample size (5 vs. 3) outweighs the likelihood of trimming a lopsidedly outlying time. However, this would probably change with the occasional twenties and thirties that happen in real competitions. There are very rarely two outliers in the same round, though--partly because of the rarity of the outliers, but also because people are more careful after having one--so using trimmed averages per round should fix nearly all major discrepancies. (This is part of the reason I suggested a mean of averages rather than a single trimmed average of 15 or 20.)

 

[Dene]

I don't think it should be changed.
The first reason is that people strategically cube to ensure they get to the next round (already mentioned).
The second reason is that one can mess up, but still make it to the next round. In usnats about a 16s average would get you into round 2. What if Rowe screwed up with a 16s average? Or a 14.5 average in the second round, and a 14 average in the third round? He was on a bad solve streak. But we all know that he is the best 3x3 cuber in the US, and he won and I think we all agree that he clearly deserved to win. If we did it your way, he wouldn't stand a chance.
The third reason is that it would be inconsistent with other speed sports. If Usain Bolt doesn't perform in the olympic finals then he misses out.

 

[Ravi]

Taking a mean of a few averages could be good, but some people might be very negatively affected in one of the rounds and that would basically negate their chance of winning even if they totally destroyed in other rounds.

It sounds to me like "very negatively affected in one of the rounds" is a euphemism for "not a consistently fast speedcuber." Am I missing something?

make sure that anyone who leaves the competition with a world record average (and is thus, officially, the best ever at that event) is guaranteed to also have won that event.

Personally, I have a little bit of scorn for world records--especially single solve records, but averages to an extent as well. Let's say there are two hypothetical speedcubers. One has been to dozens and dozens of competitions, and usually averages in the 11-12 range, but once had, say, two PLL skips in the same round which, combined with a streak of brilliance on the other two solves, gave him/her an 8.46 average and 6.70 single. The other hasn't competed in years, but shows up at one competition and averages 8.91, 9.10, and 9.02 in the three rounds, with a best single of 8.06. Which one would you say is the better speedcuber?

Yes, of course, averages don't generally vary that much. But they do vary. In my first post, I pointed out that taking averages reduces standard deviations (yes, assuming normal distributions) to 47.7% of those of single solves. Applying a "best of 3" format gives a coefficient of 74.8%; combining the two, we have a roughly 35.7% coefficient for "best of three averages of 5." This may be compared to 27.5% for "mean of three averages of 5" or 29.9% for "median of three averages of 5." (All of these are trimmed enough to remove the effects of nearly all outliers.) I also believe that (1) within reason, every solve should count, and count equally; and (2) best performance is less meaningful than average performance--I'm curious what the speedcubing community's consensus would be on these topics.

 

[qqwref]

I think you are saying [...] that competition times tend to be leptokurtic [...]

So nine of the sixteen did indeed have positive excess kurtosis as you predicted, while the other seven had negative excess kurtosis. [...] Thus it is only barely true that, as you argued, most real speedcubers' distributions in competition are more leptokurtic than a normal distribution, and that effect is almost entirely negated and even reversed in the process of trimming averages.

You're overanalyzing. With such a small sample size (20 solves) you can't expect most of the people to get a really bad or a really good solve, and that would explain why many of the people had a moderately negative calculated kurtosis. Instead of just looking at how many people had a positive kurtosis, and saying that 9 of 16 is indistinguishable from 50%, what if we analyzed all 16 people together and took the average kurtosis? I get +1.15, supporting my conclusion that a typical person would have a higher-than-average kurtosis. My whole point was that outliers are more common than a normal distribution would imply (but still pretty rare in most cases) so you can't expect every sample of 20 solves to include one of them.

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 [...] .147

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

Interesting point. But there's always luck involved in cubing, and nobody has enough luck on a competitive event to win - or even place - without already having a considerable amount of skill. As Ravi pointed out, a significantly slower person still has a decent chance of winning if SDs are high enough; different scrambles will be of different difficulties when some competitors are opposite or fully color neutral and some even use other methods; OLL and PLL skips are almost completely random; and even veteran cubers can have hardware issues (or gravity issues!) that can spoil a time regardless of their skill.

Essentially, I don't think it is the goal of a competition to determine the best cuber (mainly because this is really hard), but to determine the cuber who delivered the best performance. To me, this goal is best served by giving the top prize to the person who got the best average in any round. If you want to determine who is really objectively fastest I suggest the following:
- First round, every competitor does up to 5. Get rid of anyone who did not either achieve a counting single which was at least as good as the winning average, or come within (say) 2 seconds of it. The idea here is to prune it down to the people who could plausibly be the best.
- Have everyone left do 5 more solves. Now we have 10 per person.
- From now on, we use a trimmed average of every solve done so far with 1/5 of the solves (rounded down) trimmed off from each side. Cutoff the person with the slowest average, then do one more solve. Continue with that procedure until only one person is left, and that person is likely to be the best cuber there, in terms of average.


EDIT

It sounds to me like "very negatively affected in one of the rounds" is a euphemism for "not a consistently fast speedcuber." Am I missing something?

I don't mean that someone has a high SD or messes up a lot, I mean that someone is physically or mentally not in their best condition when a certain round takes place. I was once at a competition where the organizer forgot the scorecards where we were staying and a group of people (myself included) had to drive back and fetch them. When we got there we were pretty much yelled at for holding the competition up and had to do the 3x3 first round immediately. I was not warmed up, annoyed, and nervous, and of course I did badly (enough in fact to not make the next round). Are you going to say this is my fault for not being consistently fast? Cubers are people, not sources of statistical noise.

Personally, I have a little bit of scorn for world records--especially single solve records, but averages to an extent as well. Let's say there are two hypothetical speedcubers. One has been to dozens and dozens of competitions, and usually averages in the 11-12 range, but once had, say, two PLL skips in the same round which, combined with a streak of brilliance on the other two solves, gave him/her an 8.46 average and 6.70 single. The other hasn't competed in years, but shows up at one competition and averages 8.91, 9.10, and 9.02 in the three rounds, with a best single of 8.06. Which one would you say is the better speedcuber?

The second one is better. But records are not about judging who is better: they are about what the best recorded performance is. This is obvious for single solve records but I guess people don't realize it applies to average records as well. The exact same situation happens in every other situation where records are kept, and it is important to realize that asking "who is best" is a much more complicated question than asking "what is the best result in the database" - and that therefore determining the answer will require more work on your part.

 

[Ravi]

And since cubing, as other sports, takes into account pressure and nerves, I think it's fine to give the title to who performed best when they had to perform best.

I am very uncomfortable with the idea that the winner of a competition should be based on anything except who solves Rubik's Cubes fastest.

Also there is the problem that, for example, someone does really well in the 1st round but not so well in the 2nd, and doesn't qualify for the final. Suppose person A gets 10.00 and 14.00, respectively. Person B gets 11.00 and 12.00 and qualifies, but in the final he/she is nervous/tired/whatever and the average is 13.90. He/she will have a greater global average than person A, but he/she had more solves, so more chances to fail/succeed, and the comparison is not really valid in my view.

So how would we weight that? Who wins? Someone with a 12.00 average over 10 solves or someone with 12.01 average over 15 solves?

I was thinking that everyone who didn't advance to the next round would be out of the tournament. Qualification for rounds could even be based on cumulative averages/means of averages/etc. I think that would make the situation you're talking about very unlikely, albeit possible I guess. I don't know. I'm open to suggestions other than means of averages (maybe longer final rounds?)--I just think that it's a bit of a joke to declare one speedcuber better than another based on a competition whose format furnishes such enormous margins of error in comparison to the differences between its competitors.