Cuber A 1st round, 2nd round, 3rd round averages: 10.00, 12.50, 13.00
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Cuber A, perhaps, had a first round average with 2 skips (both PLL, one OLL and one PLL, both OLL, etc.). Based on Cuber A's global average of, probably, around 12.75 s/he received a significant advantage from the 2 skips over Cuber B who's global average is probably around 12.25 seconds.
First of all, let's be reasonable. There is no plausible way for a 12-13s performer to get a 10-second average based solely on luck. That would require something like the following:
13.00, 12.00, (5.00), (13.00), 5.00 -> 10.00
Of course, this assumes that the nonlucky solves actually averaged in the 12-13 range. Dropping this assumption, it is possible to conceive a slightly more plausible average like the following:
8.04, (7.89), 11.22, 10.74, (14.45) -> 10.00
but this still strikes me as highly unlikely, and at the very least it would require the cuber to perform much, much better than a 12-13s average would imply--not only on the 10 and 11, but also on the lucky solves, in order to make them that fast.
The question in question is this:
How much can two lucky solves affect an average, or a mean average? I believe the effect is much less than you imagine.
Suppose, for example, that any given OLL/PLL skip cuts a given speedcuber's times by exactly 2 seconds. (This may be generous or slightly conservative, depending on the speedcuber, OLL/PLL, reaction times, etc.) Then, on the rare occasion when the cuber gets two skips within the same average in competition, we can imagine that his average drops by roughly 0.67 seconds. (In fact, it would sometimes be a smaller difference than this, since the counting lucky time might have otherwise been the slowest solve and therefore dropped--but let's assume 0.67.) Then, assuming there are three rounds in this competition, that speedcuber's mean average would drop by 0.22 seconds.
There's a trade-off here: single averages would give a small chance of a 0.67-second error, while mean averages give three times the chance of a 0.22-second error. (I think we can agree that the probabilities of three lucky solves in an average or two in each of two averages are negligible.) So which side of the trade-off is better? That is, which side gives a greater likelihood of a luck-based upset?
You assumed, quite falsely, that all finalists are fast enough to win with a single counting lucky time. It seems more natural to assume that the probability of being within 0.67 of the lead is roughly three times the probability of being within 0.22. (Actually, I would venture to say that it is somewhat more than three times, since competitors tend to thin out at the head of the pack and bunch up behind.) EDIT: A quick look down the speedcubing.com homepage gives 35 competitions, in which a total of 15 speedcubers were within 0.67 seconds and 3 within 0.22. This is a small sample, but the stats do support my assumption. So although the mean-average system would give competitors three times the chance to get lucky twice, there would most likely be at least three times as many people close enough to win with a 0.67 boost compared to a 0.22 boost. (A rough guess gives an expected value of 15/35 * 0.0033 = 0.00141 competitors passing the leader by luck under the current system, and 3/35 * 0.0099 = 0.00085 under my system.) Therefore I conclude that
winning by luck is at least as unlikely, and probably more unlikely, under the mean average system compared to the single average system. In any case, the probabilities are on the order of a tenth of a percent, and we would be doing very well indeed if our competitions had anything near 99.9% accuracy.
Furthermore, I claim that lucky times are not only an
extremely minor issue in competitions, but that they actually fall well inside a typical cuber's bell curve. For example, in the last few weeks, I've recorded 268 official-rules solves at home. My overall average of 268 is 12.37 seconds, and my SD is 1.70. But of all 268 solves, the best single time is 9.07,
less than two standard deviations away from the mean. I don't recall whether the 9.07 was lucky, but there were several lucky times; they just weren't all that far from the mean. Theoretically, roughly 2.2% of times should be over two SDs away from the mean on each side of a bell curve. In my experience, virtually zero (in this case, exactly zero) solves are more than two SDs faster than average*. Conversely, there is much greater variance on the slower end of the distribution. What this means is that
lucky times are a far less important issue than a bell-curve approximation would indicate.
As my statistics in posts #1 and #23 showed, mean averages are significantly more accurate than single averages assuming normal distributions. As I have argued previously, freak outliers on the slow end rarely affect that accuracy due to their rarity. And as I showed here, lucky solves do not make a difference either. Thus I conclude once again that
mean averages provide a more accurate view of the relative speed of cubers in a competition than single averages do, not only assuming a normal distribution, but even when one considers unusually fast and slow times as well.
* Please tell me if this does not match your own experience.
a.s.a.p.
