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[WCA Regulations 2014] Mean of N for FMC

Do you support adding "Mean of 3" as an FMC format? (poll added December 10, 2013)


  • Total voters
    128

Robert-Y

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FM mo3 takes three hours which is a lot of time. I'm wondering if there's a way to determine the winner "sensibly" and save time. How likely is it to get two lucky FM solutions in a row? I haven't seen/heard of such a case before.

How about this format: Everyone does 2 solves and the winner is determined by finding the person with the best worst solve. I'm not sure if this has some name.

Example: Let's say these are the results for a final round of FMC:
1st: Shaniel Deppard: 28, 30
2nd: Yobert Rau: 29, 31
3rd: Mames Jolloy: 25, 40

In this case Shaniel is the winner, Yobert is second, Mames is third. Let's assume Mames got lucky with the 1st scramble. Even though she got a 25, it was not enough to let her rise above Shaniel and Yobert whose results are (arguably) more respectable, because of her worst solve. This seems sensible, right? In the case of a tie, I would suggest determining the winner by their best solve.

Currently I cannot see anything particularly wrong with this, but I haven't thought this through that much, admittedly.

Thoughts?
 

kinch2002

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FM mo3 takes three hours which is a lot of time. I'm wondering if there's a way to determine the winner "sensibly" and save time.
Have best of 1 in most comps. I really don't mind if some gets "lucky" at a UK comp and beats me. I suspect most other top solvers would feel the same.
Best of 3 in major champs.
I am totally cool with never having mean of 3 in UK comps.

Also, I'd like best of 3 to be a valid ranking format for competitions. You'd still get an official mean if you did all 3, but placings in that comp would be done by the best one. Then in non-major comps people don't HAVE to do all 3 to rank, but there is still an option for people to get a mean if they want.
 

porkynator

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How likely is it to get two lucky FM solutions in a row? I haven't seen/heard of such a case before.

Well, I have heard Vincent Sheu averages sup-30. So, yes, you can definitely get two lucky solves in a row.
The format you propose makes sense though.
But do we really want to avoid lucky winners at all costs? Luck is part of the game. Mo3 is a good format for the world ranking and an accurate way to measure your FMC level. If there isn't enough time for 3 FMC attempts, Bo1 or Bo2 is fine - if someone gets lucky, good for him!
 

okayama

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FM mo3 takes three hours which is a lot of time. I'm wondering if there's a way to determine the winner "sensibly" and save time. How likely is it to get two lucky FM solutions in a row? I haven't seen/heard of such a case before.

How about this format: Everyone does 2 solves and the winner is determined by finding the person with the best worst solve. I'm not sure if this has some name.

Example: Let's say these are the results for a final round of FMC:
1st: Shaniel Deppard: 28, 30
2nd: Yobert Rau: 29, 31
3rd: Mames Jolloy: 25, 40

In this case Shaniel is the winner, Yobert is second, Mames is third. Let's assume Mames got lucky with the 1st scramble. Even though she got a 25, it was not enough to let her rise above Shaniel and Yobert whose results are (arguably) more respectable, because of her worst solve. This seems sensible, right? In the case of a tie, I would suggest determining the winner by their best solve.

Currently I cannot see anything particularly wrong with this, but I haven't thought this through that much, admittedly.

Thoughts?

Personally, I like this idea (the best worst solve).
That is related to "competitor's number", same meaning as God's number.
I once suggested awarding the competitor who gets the best worst solve here.

My another thought is here:
https://www.speedsolving.com/forum/...The-FMC-thread&p=916703&viewfull=1#post916703
This way (Best of 1 in the 1st round, and Best of 1 in the final) can be employed in the current WCA regulation.
 

Sebastien

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I find it very weird to have this thread revived at this point, while the addition of the Mean of 3 format for FMC has turned out to have worked out way better than the vast majority had expected.

Besides my feeling that the Mean of 3 has definitely created a lot of new FMC enthousiasm around the world, here are some convincing numbers:

Competitions hosting FMC in 2014: 145 of 384 = 37,8%
Competitions hosting FMC in 2011-2013: 332 of 966 = 34,4%

FMC formats used in 2014:

FormatUsage
Best of 176 (51%)
Best of 215 (10%)
Mean of 358 (39%)

FMC formats used in 2011-2013:

FormatUsage
Best of 1243 (73%)
Best of 259 (18%)
Best of 332 (9%)

The concern about an FMC mean taking 3 hours has been expressed ever since this proposal was made, however I haven't seen this being an issue in practise a single time during this year. It was no problem to fit in 3 FMC attempts in at Euros and apparently neither at US Nats.

Further, I am still convinced that Mean of 3 is the best practical format for FMC.

One thing that I have said many times, which is not a strong argument but still, is that I find it quite useless to create a new WCA format for FMC, while there is an existing one (Mean of 3) that works very well for the event.

But let me be more precise: "Worst of 2" is definitely not better in my opinion than "Best of 2". It might be true that "Worst of 2" prevents FMC noobs to win because of a lucky solution, but "Worst of 2" certainly raises the randomness among the top FMCers. Take the following example:

Shaniel Deppard, Yobert Rau and Mames Jolloy are all Top FMCers and as those most of their solutions end up with insertions. As top FMCers, their insertions are always optimal.

Shaniel Deppard:
1st attempt: L3C in 22, 3 moves cancel => 27
2nd attempt: L4C in 19, 4 moves cancel => 31
Yobert Rau:
1st attempt: L3C in 22, 2 moves cancel => 28
2nd attempt: L4C in 19, 5 moves cancel => 30
Mames Jolloy:
1st attempt: L3C in 22, 1 move cancels => 29
2nd attempt: L4C in 19, 6 moves cancel => 29

I don't see while Mames Jolloy should win this competition just because of random insertion luck, while all 3 FMCers are obviously on the same level.

Meanwhile, another benefit of "Mean of 3" is that it mostly cancels out the insertion luck (that I assume to be randomly distributed).


How likely is it to get two lucky FM solutions in a row? I haven't seen/heard of such a case before.

The two first solves of Vincent's tied WR average have been LL skips.
 
Last edited:

c4cuber

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Have best of 1 in most comps. I really don't mind if some gets "lucky" at a UK comp and beats me. I suspect most other top solvers would feel the same.
Best of 3 in major champs.
I am totally cool with never having mean of 3 in UK comps.

Also, I'd like best of 3 to be a valid ranking format for competitions. You'd still get an official mean if you did all 3, but placings in that comp would be done by the best one. Then in non-major comps people don't HAVE to do all 3 to rank, but there is still an option for people to get a mean if they want.

you may take the sum of the 2 attempts and whoever is the lowest, gets the first prize.
 
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