Czech Open 2007

[Erik]

Ok, lemme confess something. The first solve of the first round 3x3 was a very bad solve but I had a PLL skip

 

[Lt-UnReaL]

How does everyone get PLL skips so much in competition? I barely ever get them at all.

 

[gillesvdp]

Erik is not that lucky, he did not have a last layer skip in a competition yet !

 

[AvGalen]

I believe the chances of getting a PLL skip are 1/72 if you use regular OLL/PLL and 1/12 if you used F2L+edge control/COLL. That means that if there are 3 rounds of 5 solves each you have about a 15/72=20% chance of getting a PLL skip for regular PLL-skip.

This math is pretty rough (if you use F2L+EC/COLL your chance of having a PLL skip is not 15/12) but gives a good estimation

 

[Rama]

I only dream of getting a skip at the 2x2, it is funny to get a skip

 

[Lt-UnReaL]

If I solved the cube 30 times, everyday since January, that would be about 7000 solves. I've only had about 6 PLL skips in my life. I really don't see how a PLL skip comes 1/72 times. :/

 

[Erik]

I do have had quite some 3x3 solves in competition now so it's not that strange that I get a skip now. And since I know some extra stuff the probability for me to do the LL in 1 algorithm is a bit higher too

 

[Lt-UnReaL]

Wow, what a coincidence. I just got my new PB with a PLL skip, on my first solve today. 20.21, but it's a lucky solve so I won't really count it.

EDIT: Wtf? Another PLL skip in the same 12 solves? Well, what are the chances. 18.54 btw.

 

[joey]

If I solved the cube 30 times, everyday since January, that would be about 7000 solves. I've only had about 6 PLL skips in my life. I really don't see how a PLL skip comes 1/72 times. :/

Remeber that it is 1/72 chance for each solve on an induviual basis. So the total number of solves is irrelevant.

 

[gillesvdp]

Wow, what a coincidence. I just got my new PB with a PLL skip, on my first solve today. 20.21, but it's a lucky solve so I won't really count it.

EDIT: Wtf? Another PLL skip in the same 12 solves? Well, what are the chances. 18.54 btw.

That reminds me, what is the probability of getting 2 PLL skips in the same average of 5 ? (and in a competition moreover )

 

[fw]

That reminds me, what is the probability of getting 2 PLL skips in the same average of 5 ? (and in a competition moreover )

I think this can be modelled with a binomial distribution..

The probability is: (n over k) p^k (1-p)^(n-k)

where:
n is the number of solves (5)
k is the number of successes (2)
p is the probability of success (1/72 if Arnaud is right)
1-p is of course the probability of no success (71/72 if Arnaud is right)

which makes:
10 * (1/72)^2 * (71/72)^(5-2) =~ 1.8497 * 10^(-3)

which is about 0.185%.. This is the probability of getting EXACTLY 2 PLL skips (2 skips, 3 "non-skips").. Actually, you would have to add the one for 3 skips, 4 skips and even 5 skips to get the probability of "at least 2 skips"..

Hope this is right... Please correct me if I am wrong

EDIT: about 0.1876% for "at least 2 skips"

 

[gillesvdp]

So assuming there is only one average of 5 per competition, this should happen again in 540 competitions !


Thanks a lot