Probability Thread

[mr. giggums]

Well the probability of getting a PLL skip after OLL is 1/72 (hence a 1LLL) but you will only use OLL when both corners and edges are misoriented which is 16/19 of the time. So 1/72*16/19=2/171. The chance of getting a OLL skip is 1/216 so a 1LLL. The chance of using COLL 1/8. Then after COLL the chance of edges permutated is 1/12. 1/8*1/12=1/96. Finally the chance of corners oriented 1/27 and after ELL the chance of corners permutated is 1/12. 1/27*1/12=1/324. Now all we have to do is add them all together so 2/171+1/216+1/96+1/324=.029828...
I'm sure I messed up somewhere but thats my answer.

 

[JeffDelucia]

Well the probability of getting a PLL skip after OLL is 1/72 (hence a 1LLL) but you will only use OLL when both corners and edges are misoriented which is 16/19 of the time. So 1/72*16/19=2/171. The chance of getting a OLL skip is 1/216 so a 1LLL. The chance of using COLL 1/8. Then after COLL the chance of edges permutated is 1/12. 1/8*1/12=1/96. Finally the chance of corners oriented 1/27 and after ELL the chance of corners permutated is 1/12. 1/27*1/12=1/324. Now all we have to do is add them all together so 2/171+1/216+1/96+1/324=.029828...
I'm sure I messed up somewhere but thats my answer.

I think thats right actually.. so about 3%

 

[LewisJ]

Just my thoughts on that idea - all the algs you need to learn for that add up to a significant portion of ZBLL! COLL + ELL + OLL + PLL = 40 + 30 + 57 + 21 = 148 algs. For a 3% chance...

 

[Feryll]

Bump. But I have a couple of questions:

1. What is the probability of getting a completed center face on a 4x4, so you skip the very first step?
2. What is the chance of getting 1 free paired pair while you were about to do the edge pairing (To start off with getting one, not multipairing accidentally)?
3. Free X-cross?
4. 2x1x1 block of anything on the 3x3?

These might have been on some website, but I can't find it.

 

[JeffDelucia]

Just my thoughts on that idea - all the algs you need to learn for that add up to a significant portion of ZBLL! COLL + ELL + OLL + PLL = 40 + 30 + 57 + 21 = 148 algs. For a 3% chance...

Yeah but then use that with EOline Petrus or just fridrich with edge controlling and your chances are much higher.

 

[Johannes91]

1. What is the probability of getting a completed center face on a 4x4, so you skip the very first step?

For a fixed center: (3/23)*(2/22)*(1/21) = 1/1771
For any center: I'm not sure, will maybe edit soon... [Edit: 1/296, see new post below]

2. What is the chance of getting 1 free paired pair while you were about to do the edge pairing (To start off with getting one, not multipairing accidentally)?

Chance of getting at least one free pair: 40.6204%.

Distribution: [[0, 0.593796], [1, 0.3092], [2, 0.0807901], [3, 0.0141328], [4, 0.00186375], [5, 0.000197873], [6, 1.76482e-05], [7, 1.36243e-06], [8, 9.39178e-08], [9, 5.5655e-09], [10, 4.17412e-10], [12, 3.16221e-12]]

3. Free X-cross?

Fixed cross and slot: 1/72,990,720.
Others: dunno

4. 2x1x1 block of anything on the 3x3?

Do you mean that either a corner-edge pair or an edge-center pair counts?

 

[Feryll]

1. What is the probability of getting a completed center face on a 4x4, so you skip the very first step?

For a fixed center: (3/23)*(2/22)*(1/21) = 1/1771
For any center: I'm not sure, will maybe edit soon...

2. What is the chance of getting 1 free paired pair while you were about to do the edge pairing (To start off with getting one, not multipairing accidentally)?

Chance of getting at least one free pair: 40.6204%.

Distribution: [[0, 0.593796], [1, 0.3092], [2, 0.0807901], [3, 0.0141328], [4, 0.00186375], [5, 0.000197873], [6, 1.76482e-05], [7, 1.36243e-06], [8, 9.39178e-08], [9, 5.5655e-09], [10, 4.17412e-10], [12, 3.16221e-12]]

3. Free X-cross?

Fixed cross and slot: 1/72,990,720.
Others: dunno

4. 2x1x1 block of anything on the 3x3?

Do you mean that either a corner-edge pair or an edge-center pair counts?

wohw, ur 2 smarte 4 me


I meant corner edge pair, but another question would be what are the chances of having no center-edge pair. These things have been in my mind forever

 

[Johannes91]

1. What is the probability of getting a completed center face on a 4x4, so you skip the very first step?

0.3379%, or approximately 1/296

This is very close to a simple estimate (\( 1-(1770/1771)^6 \approx 0.3383\% \)), so I believe it's correct.

Distribution, the first number is the number of faces with 4 centers of the same color and the second is the probability:

0 0.996621
1 0.00337051
2 8.6646e-06
3 2.51571e-08
4 1.131e-10
6 2.21766e-13

 

[That70sShowDude]

Yesterday I got 2 AUF-less PLL skips in a row.
What are the chances of this?

 

[Kirjava]

If you learned full OLL CLL PLL and ELL and when ever you got an all edges oriented OLL you used cll and whenever you got an all corners oriented oll you used Ell what would your probability of a 1lll be?

<+Kirjava> the probability doesn't change
<+Kirjava> unless you *know* that using a specific LL technique will give a skip for a specific senario
<+Kirjava> which no-one does/has/ever will

by the way, CLL isn't COLL.

 

[cmhardw]

If you learned full OLL CLL PLL and ELL and when ever you got an all edges oriented OLL you used cll and whenever you got an all corners oriented oll you used Ell what would your probability of a 1lll be?

<+Kirjava> the probability doesn't change
<+Kirjava> unless you *know* that using a specific LL technique will give a skip for a specific senario
<+Kirjava> which no-one does/has/ever will

by the way, CLL isn't COLL.

Based on Jeff's scenario the probably of a LL skip would change, and in fact it would be more likely. I think we can read COLL instead of CLL, which is what he probably meant. If you have an all edges oriented OLL and do a COLL alg, solving the permutation of the corners with respect to each other would increase the overall chance of a 1LLL.

Also, using ELL would affect the 1LLL chance when corners were both oriented and solved with respect to each other.

I'm about to head in to work, so I don't have time to work out the exact affect on the probability of a 1LLL, but it is certainly going to be more likely because of also knowing COLL and ELL.

Chris

 

[Kirjava]

Of course COLL would change the probability. But I decided to interpret the question as if it was CLL - as I think that question is more interesting.

Let me explain my reasoning behind believing that the probability doesn't change.

I don't like the idea of assuming OLL/PLL is the default method, unless you get a CLL skip in which case you use a subset of LL algs to finish in one look. Knowing OLL/PLL/CLL/ELL myself, I know that I use whichever group of algs will be faster for the first step. Therefore, for the last layer, no specific method is relied on and a specific one is chosen. When the choice of method is made, that choice dictates the probabilities of skipping.

Now, if we take the case of a CLL skip - the method chosen is CLL/ELL. The chances of this one look last layer is still 1/49.

For an OLL skip, the method you will use is OLL/PLL. Chances of this happening were 1/216.

I put forward that the chances of a LL step skip in any given solve is the same, since you have used a specific method to complete that LL and you inherit the probabilities from that method.

Of course, I understand what you're saying. I'm simply having difficulty consolidating it with what I'm saying ^_^.

On a side note, probability for skipping the second step does not change, as the probability is dictated by whatever method you use for the first step.

 

[cmhardw]

Of course, I understand what you're saying. I'm simply having difficulty consolidating it with what I'm saying ^_^.

Yes, but now think of it this way instead. I could actually do out the math of your chance to get a 1LLL, but it is not necessary to show that you have a higher probability of a 1LLL than someone using only OLL/PLL.

Consider the case of a CLL skip. You would surely notice this in your solving, perform the necessary ELL alg to solve the cube, and be done with the LL in 1 look. However, a OLL/PLL solver would see on the "X" OLL case and perform that, with 71/72 probability of still needing to execute a PLL case.

This shows that your method has a higher chance of a 1LLL than OLL/PLL used only. This is equivalent to saying that if you did 1000 solves using your LL method, and a OLL/PLL solver did 1000 solves, then you are likely to have more 1LLL solves than the other person. No matter how you phrase it, you have an improved chance to get a 1LLL over a OLL/PLL only solver.

On a side note, probability for skipping the second step does not change, as the probability is dictated by whatever method you use for the first step.

Yes, your probability for skipping the second step does change, because you use two methods.

Consider CLL/ELL as your "base" method momentarily. You will "skip" ELL if you are solving, and after F2L you have an already oriented LL (but not permuted). In this case you have not yet done either CLL or ELL, but you would know to execute a PLL alg and thus skip the ELL step. Thus you have a higher chance to skip ELL than a CLL/ELL only solver.

Alternatively phrased you could imagine OLL and PLL as your "base." In this case you skip PLL in the case that all corners are not only oriented, but solved with respect to each other. In this case, rather than executing the "X" OLL, you would simply do the necessary ELL case to solve the cube. So your chance to skip PLL is higher than someone using only OLL/PLL alone.

Keep in mind that this result is very good for you! No matter how you consider it, your personal chance to skip the second step, or have a 1LLL in any form, is higher than that of anyone using either CLL/ELL alone or OLL/PLL alone.

Chris

 

[Kirjava]

Of course, I understand what you're saying. I'm simply having difficulty consolidating it with what I'm saying ^_^.

Yes, but now think of it this way instead. I could actually do out the math of your chance to get a 1LLL, but it is not necessary to show that you have a higher probability of a 1LLL than someone using only OLL/PLL.

Consider the case of a CLL skip. You would surely notice this in your solving, perform the necessary ELL alg to solve the cube, and be done with the LL in 1 look. However, a OLL/PLL solver would see on the "X" OLL case and perform that, with 71/72 probability of still needing to execute a PLL case.

My problem lies in only considering the two methods as a choice, instead of a combined system. I spoke to Johannes about this and he summed it up for me quite well;

22:15:03 < jlaire> but yeah, what I was saying, the probability of getting a 1LLL with a given system doesn't change if you learn additional systems, which I think is what you were getting at
22:15:28 < jlaire> but the overall probability goes up because a 1LLL with any one system gives you a 1LLL

This shows that your method has a higher chance of a 1LLL than OLL/PLL used only. This is equivalent to saying that if you did 1000 solves using your LL method, and a OLL/PLL solver did 1000 solves, then you are likely to have more 1LLL solves than the other person. No matter how you phrase it, you have an improved chance to get a 1LLL over a OLL/PLL only solver.

Of course, you are correct. I imagine working out the probability compared to OLL/PLL is actually quite difficult.

Here are some of my other thoughts about this, since I'm too lazy to type them out again;

21:57:05 < Kirjava> once you decide, you are bound by the probabilities of that system
21:57:36 < Kirjava> if you decide to use CLL/ELL because that would give you a skip
21:57:54 < Kirjava> the chances of that skip happening are the same as if you didn't know OLL/PLL at all
21:59:04 < Kirjava> I fully accept that I am probably completely wrong
21:59:23 < Kirjava> however, this explanation seems the most logical to me
21:59:31 < jlaire> if you only choose system X when it's easier than average, you can't really say that's the same as using system X every time
21:59:42 < jlaire> or if you choose the easiest of several systems
21:59:55 < Kirjava> for that particular solve it is
21:59:58 < Kirjava> not overall

And I came to some sort of conclusion;

22:09:06 < Kirjava> to me it seems like
22:09:20 < Kirjava> you're increasing the chances of getting a better LL
22:09:51 < Kirjava> yet the probibility of a skip is the same O_O
22:10:12 < Kirjava> but but
22:10:22 < Kirjava> there's a 1/216 chance of an OLL skip
22:10:31 < Kirjava> /or/ a 1/49 chance of a CLL skip
22:10:50 < Kirjava> these together must be higher than either one alone
22:11:21 < Kirjava> yet at the same time
22:11:31 < Kirjava> chances of either one happening are the same as they have always been

Thanks for giving me the clarity needed to understand why my reasoning was incorrect. Despite this, I still think there is some merit to it

On a side note, probability for skipping the second step does not change, as the probability is dictated by whatever method you use for the first step.

Yes, your probability for skipping the second step does change, because you use two methods.


Consider CLL/ELL as your "base" method momentarily. You will "skip" ELL if you are solving, and after F2L you have an already oriented LL (but not permuted). In this case you have not yet done either CLL or ELL, but you would know to execute a PLL alg and thus skip the ELL step. Thus you have a higher chance to skip ELL than a CLL/ELL only solver.

Alternatively phrased you could imagine OLL and PLL as your "base." In this case you skip PLL in the case that all corners are not only oriented, but solved with respect to each other. In this case, rather than executing the "X" OLL, you would simply do the necessary ELL case to solve the cube. So your chance to skip PLL is higher than someone using only OLL/PLL alone.

I'd say that for the former you would be using the OLL/PLL system and CLL/ELL system for the latter. This is because I don't consider CLL/ELL as an extension to OLL/PLL or vice versa, I consider them seperately.

You cannot consider CLL/ELL to be the method used when you know both systems and get an OLL skip, since you automatically use the OLL/PLL system as it's the best one to use in that situation, and the probability of a PLL skip is still 1/72.

However, I think this is less to do with probability and more to do with perspective.

Keep in mind that this result is very good for you! No matter how you consider it, your personal chance to skip the second step, or have a 1LLL in any form, is higher than that of anyone using either CLL/ELL alone or OLL/PLL alone.

This doesn't mean much when my execution is terrible

 

[ariasamie]

someone please answer this guy:

Avg 29.52

30.13
(25.58)
26.97
28.83
29.02
(33.44)
29.46
31.44
30.47
29.34
29.78
29.78

What are the odds of the last two being exactly the same?

this was originally posted on "NEW" Race to Sub-30!

 

[JackJ]

At Minnesota Open 2010, Chester Lian got 4 pyraminx LL skips in a row. What are the odds of that?

 

[Rpotts]

someone please answer this guy:

Avg 29.52

29.78
29.78

What are the odds of the last two being exactly the same?

this was originally posted on "NEW" Race to Sub-30!

Probably depends on the solvers SD

EDIT: I enjoyed reading every last bit of the chris/kirjava debate.
How does CLL/ELL and OLL/PLL overlap affect probabilities, if the discussion is still open.

 

[eastamazonantidote]

Assuming a completely random last slot + last layer, what are the chances for each ZBF2L case? I have been thinking about this and I can figure out how many cases intuitively but I can't come up with probabilities.

 

[cmhardw]

EDIT: I enjoyed reading every last bit of the chris/kirjava debate.
How does CLL/ELL and OLL/PLL overlap affect probabilities, if the discussion is still open.

The reason it affects probabilities is that the two methods have some overlapping situations.

Considering your solving method to be OLL/PLL and nothing else, you have the following possibilities for the LL.

1) You must perform an OLL case, followed by the fact that you must perform a PLL case.
2) You skip the OLL step, but must still perform the PLL case.
3) You must perform the OLL case, but you skip the PLL step.
4) You skip both the OLL and PLL steps.

Now for CLL/ELL only and nothing else you have the following possibilities:
1) You must perform the CLL case, followed by the fact that you must perform the ELL case.
2) You skip the CLL step, but you must perform the ELL case.
3) You must perform the CLL case, but you skip the ELL step.
4) You skip both the CLL and ELL steps.

So now let's consider the method: OLL/PLL or CLL/ELL. The probabilities are affected because you have some overlap of the cases between methods at the OLL step.

1) You have either a CLL or OLL case, but neither case was skipped. If you execute CLL, then followup with ELL. If you execute OLL, then follow up with PLL.

2) You have either a CLL or OLL case, but neither case was skipped. You perform a CLL case, then the ELL step is skipped.

3) You have either a CLL or OLL case, but neither case was skipped. You perform a OLL case, then the PLL step is skipped.

Ok the next steps are overlap steps.
4) You skip CLL, but do not skip the full OLL. You perform the correct ELL case to solve the LL.

5) You skip OLL, but do not skip the full CLL step. You perform the correct PLL to solve the LL.

6) You skip the LL (with the possibility of a necessary AUF to solve the cube).

--------

Ok so now to convince you that the method of combining both methods together has a higher probability of 1LLL I am going to actually calculate the probability of skipping at least one step in each method.

Now for OLL/PLL methods the probability of skipping either OLL, or PLL, or possibly both, is:
1 - (71/72)*(215/216) = 287 / 15552

Now for CLL/ELL methods.
The probability of skipping either CLL, or ELL, or possibly both, is:
1 - (161/162) * (95/96) = 257 / 15552

Ok now for OLL/PLL and CLL/ELL combined method.
The chance for either a 1LLL or a full LL skip is the chance of getting any of either cases: 2, 3, 4, 5 or 6 from my list of the combined method above.

------
Lemma to help for cases 2 and 3.

We need to know the probability of getting a LL that has neither a OLL nor CLL skip. This is:
1 - (1 / 216 + 1 / 162 - 48 / 62208)
1 - (288 / 62208 + 384 / 62208 - 48 / 62208)
1 - (624 / 62208)
1283 / 1296
------

------
Lemma for case #4

4) You skip CLL, but do not skip the full OLL.

This probability is:
(1/162) * (7/8) = 7 / 1296
------

------
Lemma for case #5

5) You skip OLL, but do not skip the full CLL step.

This probability is:
(1/216) * (5/6) = 5 / 1296
------

So the probability of a 1LLL (or LL skip) with the CLL/ELL or OLL/PLL combined method is:

P(1LLL) = P(case #2) + P(case #3) + P(case #4) + P(case #5) + P(case #6)

P(1LLL) = (1283/1296) * (1/96) + (1283/1296) * (1/72) + (7/1296) * (95/96) + (5/1296) * (71/72) + (1/15552)

P(1LLL) = 115 / 3456

(115 / 3456) > (287 / 15552) > (257 / 15552)

OR

in prettier form:

3.3% (CLL/ELL or OLL/PLL) > 1.8% OLL/PLL > 1.7% CLL/ELL

So the chance is nearly double that you get a 1LLL (or 0LLL) compared to either OLL/PLL method alone, or CLL/ELL method alone.

--edit--
Actually, I just realized that this depends on the frequency (call this q) that you choose between cases 2 and 3 as I list them for CLL/ELL or OLL/PLL methods. I will take a look at this more in depth when I get home tonight. But 3.3% is NOT the exact answer using both methods, because of the choice you have between cases 2 and 3. I will fix this tonight.
--edit--

Chris

 

[cmhardw]

--edit--
Actually, I just realized that this depends on the frequency (call this q) that you choose between cases 2 and 3 as I list them for CLL/ELL or OLL/PLL methods. I will take a look at this more in depth when I get home tonight. But 3.3% is NOT the exact answer using both methods, because of the choice you have between cases 2 and 3. I will fix this tonight.
--edit--

Ok I fixed it, and I have a more in depth analysis.

So let's look at the probability of a 1LLL given that you use CLL/ELL and OLL/PLL combined method.

Let q = probability that, given the choice being executing an OLL or a CLL case at the last layer, that you choose to continue the solve by solving the CLL case.

P(1LLL) = q * P(case #2) + (1-q) * P(case #3) + P(case #4) + P(case #5) + P(case #6)

P(1LLL) = q * (1283/1296) * (1/96) + (1-q) * (1283/1296) * (1/72) + (7/1296) * (95/96) + (5/1296) * (71/72) + (1/15552)

P(1LLL) = (8571-1283q) / 373248

And this probability is maximized when q=0. So now that q=0 the probability of a 1LLL using this combined method is:

P(1LLL) = 8571/373248 = 2857/124416 = 2.3% chance

------------

Final Analysis of CLL/ELL and OLL/PLL combined method

This method is best utilized, with a 2.3% chance of a one look last layer or LL skip combined, as long as you always obey the following golden rule:

1) Whenever presented with a last layer that gives you a choice between an unsolved OLL case or an unsolved CLL case, always choose to continue by executing the OLL case.

As long as you obey this rule 100% of the time, you will have a 2.3% chance at a 1LLL or a LL skip combined. This is greater than the 1.8% chance using OLL/PLL alone, and both are greater than the 1.7% chance using CLL/ELL alone.

I hope I have cleared this up.

Chris