OLL/PLL vs. COLL/EPLL vs CLLEF vs CMLL

[Johannes91]

You're provided with the individual probabilities for the skips you want, so don't you just add the probabilities?

Using that logic, the probability that you won't skip either step is (26/27) + (71/72) > 1, so no, you can't just add them.

The correct probability that you skip Petrus OLL and/or PLL is 1 - (26/27) * (71/72) = 49/972 ≈ 5.0%.

Similarly for COLL and/or EPLL skip: 1 - (161/162) * (11/12) = 173/1944 ≈ 8.9%.

Someone good at math hopefully corrects me if I made a mistake (it's 5:36am here).

 

[irontwig]

You're provided with the individual probabilities for the skips you want, so don't you just add the probabilities?

Using that logic, the probability that you won't skip either step is (26/27) + (71/72) > 1, so no, you can't just add them.

The correct probability that you skip Petrus OLL and/or PLL is 1 - (26/27) * (71/72) = 49/972 ≈ 5.0%.

Similarly for COLL and/or EPLL skip: 1 - (161/162) * (11/12) = 173/1944 ≈ 8.9%.

Someone good at math hopefully corrects me if I made a mistake (it's 5:36am here).

Maybe he presumed or=xor instead of or=and/or. It's 4:52am here, yay!

 

[Johannes91]

Maybe he presumed or=xor instead of or=and/or. It's 4:52am here, yay!

Possibly, though I think and/or is more interesting practically. For completeness:

Skipping Petrus OLL xor PLL: (1/27) * (71/72) + (26/27) * (1/72) = 97/1944 ≈ 5.0%.
Skipping COLL xor EPLL: (1/162) * (11/12) + (161/162) * (1/12) = 43/486 ≈ 8.8%.

Is this correct?

 

[Lord Voldemort]

They are not mutually exclusive. My mistake.
So you were correct, irontwig, that I wasn't including and. :rolleyes:
Subtracting the "AND" possibility and the neither possibility corrects the probability, as well corrects the flaw that you proposed.

Hm... almost a 9% chance of a COLL/EPLL skip, or both means in most averages of 12, I''ll get 1. Maybe those extra 40 COLLs might be worth it.

 

[deco122392]

learning coll is always worth it =D its applicable for tons of methods =)

 

[Nukoca]

8.9%, eh? Not too bad... I hear COLL recognition is bad, but I'll get faster over time. Thanks everybody!

 

[deco122392]

Its not all to bad.
And naturaly, like everything else, it will get better with time.

Read this page:http://jmbaum.110mb.com/coll.htm

Mess around with it. Basically you Identify the oll then apply the corresponding algorithm.

 

[Lord Voldemort]

learning coll is always worth it =D its applicable for tons of methods =)

40 algorithms is a lot though.
I mean, I've learned 2 PLLs (I still haven't found a good N perm) and 36 OLLs, and it takes a while. I never really have a strict alg-learning routine, every once in a while I go "Hey! There's my alg sheet! I'll learn this OLL! Yay!"

 

[JLarsen]

Dude OLL PLL for Petrus is only 28 algs, and the oll's are super super fast because they're so short. I can sub 1.5 all of them.

The "classic" set up for petrus is COLL and EPLL.

 

[Lord Voldemort]

No, I was just saying that I am fairly slow at learning algs well.
I'm switching to ZZ, so I'm considering learning COLL.

I think I'll just learn the Pi, T, U, and L cases.
I can sub-1 Sune, Anti-Sune, and Double-Sune, so the speed probably won't be affected. For the Jason Baum page, on the L COLL page, are both cases dealt the same way, or is he giving you two views?

 

[deco122392]

The first moves: U, or U' or U2 are setups for the same algorithms because other wise there would be twice to memorize otherwise. But seeing as they are just reflections of each other you can solve them with a simple setup move and the same algorithm.

So yes, its the same case from different points of view and are delt with the same way.

(edit): wooooo 200th post!!!!

 

[Nukoca]

By the way, what is it that makes one case more likely than another?

 

[byu]

I'll use a simple example.

In EPLL, U-Permutation happens 2/3 of the time, Z permutation happens 1/6 of the time, and H permutation happens 1/12 of the time.

U-Permutation:
It can be done Clockwise and Counterclockwise, from 4 anges (2*4=8)

Z-Permutation:
It can be done from two angles (that give different results) (2)

H-Permutation:
It can be done from only one angle (1)

Solved state:
There is only one solved state (1)

8+2+1+1=12

So U-Perm is 8/12=2/3
Z-Perm is 2/12=1/6
H-Perm is 1/12
and solved state is 1/12

Remember, this is for EPLL only.

 

[deco122392]

haha that was quick, explanatory and to the point

 

[Nukoca]

I'll use a simple example.

In EPLL, U-Permutation happens 2/3 of the time, Z permutation happens 1/6 of the time, and H permutation happens 1/12 of the time.

U-Permutation:
It can be done Clockwise and Counterclockwise, from 4 anges (2*4=8)

Z-Permutation:
It can be done from two angles (that give different results) (2)

H-Permutation:
It can be done from only one angle (1)

Solved state:
There is only one solved state (1)

8+2+1+1=12

So U-Perm is 8/12=2/3
Z-Perm is 2/12=1/6
H-Perm is 1/12
and solved state is 1/12

Remember, this is for EPLL only.

I thank thee for thy care and honest pains.