World Championship 2007

[Pedro]

nah...makes no sense to me that someone who makes 2/2 wins when someone does like 19/20...

 

[AvGalen]

2/2 > 19/20. It is the easiest math there is. (1 > 0.95)

 

[normanbim]

that will be very hard stefan. he made 16 world records just this year. it took macky three years just to get 15 world records. i hope macky is coming. he is one of the reasons i started speedsolving and blindsolving. hail macky!

 

[AvGalen]

Stefan1,4,5,6 is just the same idea as 1,3,4,6: Don't do 6*4, but do (6/(1/4)). Don't do 4*6, but do ((4/(1/6)).
Solution: 4 / (1-(5/6))
Which one of the two has 2 solutions?

I just noticed I presented the riddle with the numbers 1, 4, 5, 6. I don't see a second solution with 1, 3, 4, 6.

normanbim: What were you responding to when you said this?

that will be very hard stefan

 

[Mark]

1,4,5,6 has two solutions: 4/(1-5/6) and 6/(5/4-1)

 

[AvGalen]

OK, same idea again (use 3 numbers to create 1/4 or 1/6)

 

[hdskull]

that will be very hard stefan. he made 16 world records just this year. it took macky three years just to get 15 world records. i hope macky is coming. he is one of the reasons i started speedsolving and blindsolving. hail macky!

that's because matyas competes in like 10 more events than macky does. all macky competes in is 3x3, OH, BLD, 2x2, 4x4, that's 5 events total.

EDIT: i just checked, matyas competes in 17.

 

[Stefan]

[that's because matyas competes in like 10 more events than macky does.

On the other hand, Matyas started much later, when the records were already a whole lot better than when Macky collected so many. Macky's last record is from 2005, Matyas only got one in 2006 and all others in 2007. I'm in the same boat, I got most of my records somewhat for free, because there was little serious competition back then. In retrospect I regret this, because I was a lot more motivated a while ago and now that the overall level has risen and I could try to improve accordingly (for example by learning the better methods for the magics), I feel a little fatigued, like I'm too old to try to get to the top again. I've already given up on the magics quite a while ago.

 

[Pedro]

2/2 > 19/20. It is the easiest math there is. (1 > 0.95)

I know the % is 'better' if you do 2/2 instead of 19/20, but that just doesn't seem fair for me...

now imagine some attempts 100 and gets 99 correct...that's some thousands of correct moves...

if someone does 2/2, that's something like 200-some moves...

isn't making 99/100 MUCH harder than 2/2? shouldn't someone who does that win?

 

[AvGalen]

I think you have to create a pretty complicated formula Pedro.

The following are ordered according to the current rules. Could you order them according to your ideas and then find a formula that matches that idea? (I have left the time factor out of the examples, that would complicated matters even more!)

9f16) For the 3x3x3 Cube: Multiple Blindfolded event the order in the results is based on:
Highest in the ranking are the competitors who solved all puzzles, among these competitors a larger number means better. For equal results the lower total time means better.
Next in the ranking are the competitors who solved at least one of the puzzles, among these competitors a larger number of solved puzzles means better, and after this a lower number of attempted puzzles means better. For equal results the lower total time means better.
Next in the ranking are the competitors who did not solve at least one puzzle, these competitors all finish at the worst position.

1000/1000
100/100
10/10
2/2
499/1000
99/100
99/1000
50/100
9/10
9/100
9/1000
8/9
6/7
4/5
2/3
1/2
1/10
1/100
1/1000
0/2, 0/10, 0/100, 0/1000

 

[Johannes91]

499/100

That's quite good.

 

[AvGalen]

Edited. I include this because I want to know if 50/100 or 499/1000 is considered better.

The only type of formula I can personally think of (besides the current rules which I think are fine) would be somethink like:
n points for each solved cube
-m points for each unsolved cube
Most points wins.

But it will be very hard/arbitrary to determine n and m and some unwanted effects might arise.

I think the current rules are good because they force you to do well if you want to be sure that you will win. 2/2 might sound easier than 9/10, but the current system is already pretty motivating to attempt a lot of cubes
1) If I attempt 2 and get 2/2 than my ranking still depends on someone else failure on 3/3 or 10/10
2) If I make 1 mistake and do 1/2 (50%) I will lose to someone that does 2/10 (20%)
3) If you attempt 100 cubes and all others attempt 5 you can take a gamble and pick the 5 easiest cubes (5%), memorize and solve only those and beat everyone that gets 4/5 or worse.
4) If you use trick 3 you can have 6*15 + 94*10 = All day to memorize and solve those 5 easieast cubes

 

[Pedro]

I think you have to create a pretty complicated formula Pedro.

The following are ordered according to the current rules. Could you order them according to your ideas and then find a formula that matches that idea? (I have left the time factor out of the examples, that would complicated matters even more!)

1000/1000-
100/100-
10/10-
2/2-
499/1000-
99/100-
99/1000-
50/100-
9/10-
9/100-
9/1000-
8/9-
6/7-
4/5-
2/3-
1/2-
1/10-
1/100-
1/1000-
0/2, 0/10, 0/100, 0/1000

well, for me it would be:

1000/1000 (of course )
499/1000
100/100
99/1000
99/100
50/100
10/10
9/10
9/100
9/1000
8/9
6/7
4/5
2/3
2/2
1/2
1/100
1/1000
0/2, 0/10, 0/100, 0/1000 (meaning all of them are at the last position )

no need for complicated formulas...

who solves more cubes wins
if 2 persons solve the same number, then the one who tried more wins...(99/1000>99/100;9/10>9/100>9/1000)

well, that's just what I think...

I don't think is fair in any way to give victory to someone who did 2/2 when another one did 999/1000...

 

[Stefan]

If you don't like the WCA rules, talk to the WCA. They have a forum for that.

Here's my suggestion from a while ago, inspired by the hour number discipline of the memorizers:

a) Allow a fixed amout of total time (e.g. one hour)
b) Count the number of correctly solved cubes, not the attempted ones.
c) Require fixed success rate (e.g. 2/3 of given cubes solved).

The current rules requiring 100% success rate stem from an era when three cubes was considered awesome. I actually talked with Ralf about two years ago and he considered blindsolving even one cube to be hard, which I think we now all agree is just wrong, as blindcubing is rather very easy. The higher the number of cubes gets, the more ridiculous it is to keep those old rules. Even the professional memorizers don't require perfection, in case of mistakes they just don't get full points (see the Scoring section in the page linked above).

 

[Pedro]

already posted there

(and I knew the wca forum, btw )

 

[gillesvdp]

The number of competitors for Multible Blindfolded has been raised...

WR for 2 cubes, here I come !!!

 

[Pedro]

nope...I'm coming for it hahaha

 

[gillesvdp]

haaa noooooo
I will do it sub2 (Matyas can do it )

 

[Pedro]

sub-2?O.o

then you'll beat me...I just did 2 cubes once...was like 4:39

 

[tim]

The number of competitors for Multible Blindfolded has been raised...

oh, really? i wish i would have registered for multi bld .