Probability Thread

[joey]

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 OLL or a CLL case, always choose the OLL case.

What.. I'm confused.

 

[Kirjava]

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

Sidenote; While the aim is to increase the probability of a skip, this isn't the best technique for having a fast last layer.

 

[cmhardw]

@Joey: The part you quoted, and this is also in line with what Kirjava said, is that strictly from a math point of view, you can increase the chances of a skip in certain situations. The specific situation is that after having solved F2L, you are presented with a LL situation that can be interpretted as either an unsolved CLL case, or an unsolved OLL case. It cannot be considered as either a solved CLL or a solved OLL. In these specific cases you should always choose to do the OLL case if you want to increase your chances of a skip, with no exception. I changed the wording of the quote to make it a little more clear on this point.

Kirjava brings up a good point that this may not be the fastest case to execute, but it will be the method that increases your chances of a skip the most.

--edit--
Ok everyone, final corrections. I got too excited about the result and realized I had written down a number wrong in my reducing of a fraction. Basically, the chance of a skip is 2.3% as long as you follow the rule I mentioned in a previous post.

However, I found it interesting to look at what happens if you choose CLL or OLL with a perfect 50-50 split. In this case, the chance of a skip is 2.1% which is still higher than either method alone. So I would say, even if you are choosing CLL and OLL with equal chance, you still noticeably increase your chances for some form of skip during the LL.
--edit--

Chris

 

[Tim Reynolds]

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!

Let \( f(x) \) be this specific solver's probability of getting a solve of time \( x \). This is a discrete probability distribution since stackmats have limited precision. Then the probability is \( \sum_{x=0.00}^{\infty} f(x)^2 \), where the sum is taken over all numbers \( x>0 \) such that \( 100x \) is an integer. Maybe later tonight I'll try to write out a reasonable probability density and calculate this probability. But I'm kind of too lazy to do that.

The probability that any two consecutive times are the same out of an average of 12 is approximately 11 times this. Of course this overcounts a lot, but \( f(x)^2 \) is probably pretty small, so it's okay.

This also assumes each solve's probability function is independent of all the other times, which is in general a reasonably good assumption I would think.

 

[Robert-Y]

What are the chances of skipping the last two centres on a 5x5x5? (I'm not talking about a supercube)

 

[CuBeOrDiE]

What are the chances of skipping the last two centres on a 5x5x5?

I think it is : 1/ ((2^4 * 4!) * 2)
= 1/ (16 * 24 * 2)
= 1/ (32 * 24)
= 1/ ((32 * 20) + (32 * 4))
= 1/ (640 + 128)
= 1 / 768
But I am probably wrong ?

edit: whoops, should have been 1/((2^4*4!)^2)

 

[kinch2002]

What are the chances of skipping the last two centres on a 5x5x5?

I would say that you have a 4/8*3/7*2/6*1/5 chance of skipping all 4 xcentres. Same for +centres. So that gives (1/70)^2=1/4900 for skipping the whole last 2 centres

 

[Robert-Y]

What are the chances of skipping the last two centres on a 5x5x5?

I would say that you have a 4/8*3/7*2/6*1/5 chance of skipping all 4 xcentres. Same for +centres. So that gives (1/70)^2=1/4900 for skipping the whole last 2 centres

Thanks. Hmm... so skipping the last two centres on the 4x4x4 isn't that rare... (but I swear it hasn't happened yet for me!)

 

[mr. giggums]

@the 1 LLL debate: I came up with this answer at the top of ther page I don't know if it's right but I couldn't find anything wrong with it.

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.

 

[amostay2004]

What are the chances of skipping the last two centres on a 5x5x5?

I would say that you have a 4/8*3/7*2/6*1/5 chance of skipping all 4 xcentres. Same for +centres. So that gives (1/70)^2=1/4900 for skipping the whole last 2 centres

Thanks. Hmm... so skipping the last two centres on the 4x4x4 isn't that rare... (but I swear it hasn't happened yet for me!)

It isn't at all Happens quite often for me..though sometimes the last 2 centres need to be swapped

 

[kinch2002]

What are the chances of skipping the last two centres on a 5x5x5?

I would say that you have a 4/8*3/7*2/6*1/5 chance of skipping all 4 xcentres. Same for +centres. So that gives (1/70)^2=1/4900 for skipping the whole last 2 centres

Thanks. Hmm... so skipping the last two centres on the 4x4x4 isn't that rare... (but I swear it hasn't happened yet for me!)

It isn't at all Happens quite often for me..though sometimes the last 2 centres need to be swapped

Yeah so 1/35 that you'll get last 2 4x4 centres complete, but half of those times you'll need to swap them.

Btw it's 1/495 that you'll skip the 3rd-to-last centre, giving 1/(70*495)=1/34650 chance of last 3 centres skip.

Even more interestingly, because of the 4 choices of position for your 3rd centre, you have a 1/455 chance of skipping that, which is greater than the chance of skipping the 4th centre! This is all assuming that you have fixed colour for 3rd centre. Obviously if you don't mind which colour you go with then the chance of a skip is even greater (approximately 4*1/455)

 

[cmhardw]

@the 1 LLL debate: I came up with this answer at the top of ther page I don't know if it's right but I couldn't find anything wrong with it.

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...

Can you explain the part in bold more precisely? What exactly do you mean by "when both corners and edges are misoriented"? Do you mean that you don't have both corners and edges oriented at the same time? Because that would be 215/216 chance. I'm afraid I don't see how you got 16/19.

Chris

 

[blade740]

Thanks. Hmm... so skipping the last two centres on the 4x4x4 isn't that rare... (but I swear it hasn't happened yet for me!)

You probably assumed you skipped the 3rd and 4th (or 4th and 5th) centers instead. I'd assume that if the 5th and 6th were solved when you solved the 4th, at least one of them was solved before that.

 

[Rpotts]

@the 1 LLL debate: I came up with this answer at the top of ther page I don't know if it's right but I couldn't find anything wrong with it.

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...

Can you explain the part in bold more precisely? What exactly do you mean by "when both corners and edges are misoriented"? Do you mean that you don't have both corners and edges oriented at the same time? Because that would be 215/216 chance. I'm afraid I don't see how you got 16/19.

Chris

I think he's saying that you use OLL unless it's an OLL case that has either corners or edges oriented. Which he is claiming is 3/19 of time, when you would use COLL/ELL

 

[cmhardw]

I think he's saying that you use OLL unless it's an OLL case that has either corners or edges oriented. Which he is claiming is 3/19 of time, when you would use COLL/ELL

Oh ok, I gotcha. Mr. Giggums can you confirm this?
If this is the case, though, the chance to have either corners or edges (but not both) oriented is:
(1/27) * (7/8) + (26/27) * (1/8) = (7+26)/216 = 11 / 72

I did not allow for both corners and edges to be oriented, because this would constitute an OLL skip, which is handled in another case.

11/72 is approximately = 0.153
3/19 is approximately = 0.158

Hopefully that helps?
Chris

 

[mande]

What is the probability of not getting a single CE pair on a scramble?

 

[cmhardw]

Alright, one more post from me about the 1LLL debate, and then I'll let it rest. My previous calculation did not include another case type, and that should be the following:

Another way to get a 1LLL using the combined OLL/PLL and CLL/ELL method is to skip both the CLL and OLL steps, but not the full LL. This would essentially be getting an EPLL after F2L.

Also, I have to give myself one of these --> :fp for not doing the calculation a much easier way. It does not change the result by much, but this result should definitely be correct.

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

let q = the probability that, when having both an unsolved OLL and an unsolved CLL case after F2L, that you choose to execute a CLL alg.

Again, we are talking about the combined CLL/ELL and OLL/PLL method here. Case #1 refers to case #1 in my previous post.

P(1LLL) = 1 - P(case #1)
P(1LLL) = 1 - [(1283/1296) * q * (95/96) + (1283/1296) * (1-q) * (71/72)]
P(1LLL) = 1 - (q + 364372) / 373248
P(1LLL) = (8876 - q) / 373248

This is maximized when q=0 giving:

P(1LLL) = 8876 / 373248 = 2219 / 93312 = 2.4% approximately.

Pretty close to the same result, but if anyone was following along with the math I left out a case before, and I figured out the probability using a very long method when this shortcut existed the whole time.

Chris

 

[Zane_C]

This went in the accomplishement thread but it probably belongs here:
16.05, 15.26, (10.13)[PLL skip], 17.92, 16.16[OLL skip], (19.19), 16.91, 17.80, 18.33, 14.93[PLL skip], 11.70[LL skip], 14.69[PLL skip]

scrambles were:

Spoiler

1. 16.05 D' L' U R2 B2 R F D' F2 R U2 D F' L2 U' R B2 D R' F' R' F D2 U' L2
2. 15.26 U2 L2 R2 F R U2 F2 B' D2 U2 B' U B U F' L' U B' R2 D2 F' R' B' F L'
3. (10.13)[PLL skip] F2 U2 D L2 R2 B R B R B U' R B' F' R U' L' D B F' D2 U R2 F2 U'
4. 17.92 F L F2 B U' B D2 L F' R' L2 D' R2 D' B' F L D F' U2 R2 L U R2 D'
5. 16.16[OLL skip] R F B' D' R D2 L' B' F L U' L2 D' R' L2 F' R D' L D U' B' L B D
6. (19.19) B L B2 R F' B2 U' L U R2 F' L2 F R2 D' B U' D2 B L2 F2 D2 B' L2 R
7. 16.91 F B D' L2 B' F' U2 R2 D F2 U2 D' R2 F R2 L B U' F2 R2 D2 R F' L' U
8. 17.80 R L F' U B2 L D L2 F' D2 U' L' U' F' U' L2 D B R' D F' L' U2 D' R'
9. 18.33 L2 F' U2 L U' R U2 B2 U' B' L' F U' R2 B' R' U' B' R' L B F R D' R2
10. 14.93[PLL skip] B' L2 D' U' R L' B L U F R2 F2 L' F' U' F R2 U' D2 L' R' B2 R F' B'
11. 11.70[LL skip] U2 L2 U2 D2 L R2 B2 L' D2 U R2 U' D' L D2 U2 L B2 U L R F2 U2 F' B'
12. 14.69[PLL skip] U' F' D2 B2 F' R' L2 D2 F R B2 L F' B2 U' D L B' U2 L F B D U' L2

 

[Owen]

This went in the accomplishement thread but it probably belongs here:
16.05, 15.26, (10.13)[PLL skip], 17.92, 16.16[OLL skip], (19.19), 16.91, 17.80, 18.33, 14.93[PLL skip], 11.70[LL skip], 14.69[PLL skip]

scrambles were:

Spoiler

1. 16.05 D' L' U R2 B2 R F D' F2 R U2 D F' L2 U' R B2 D R' F' R' F D2 U' L2
2. 15.26 U2 L2 R2 F R U2 F2 B' D2 U2 B' U B U F' L' U B' R2 D2 F' R' B' F L'
3. (10.13)[PLL skip] F2 U2 D L2 R2 B R B R B U' R B' F' R U' L' D B F' D2 U R2 F2 U'
4. 17.92 F L F2 B U' B D2 L F' R' L2 D' R2 D' B' F L D F' U2 R2 L U R2 D'
5. 16.16[OLL skip] R F B' D' R D2 L' B' F L U' L2 D' R' L2 F' R D' L D U' B' L B D
6. (19.19) B L B2 R F' B2 U' L U R2 F' L2 F R2 D' B U' D2 B L2 F2 D2 B' L2 R
7. 16.91 F B D' L2 B' F' U2 R2 D F2 U2 D' R2 F R2 L B U' F2 R2 D2 R F' L' U
8. 17.80 R L F' U B2 L D L2 F' D2 U' L' U' F' U' L2 D B R' D F' L' U2 D' R'
9. 18.33 L2 F' U2 L U' R U2 B2 U' B' L' F U' R2 B' R' U' B' R' L B F R D' R2
10. 14.93[PLL skip] B' L2 D' U' R L' B L U F R2 F2 L' F' U' F R2 U' D2 L' R' B2 R F' B'
11. 11.70[LL skip] U2 L2 U2 D2 L R2 B2 L' D2 U R2 U' D' L D2 U2 L B2 U L R F2 U2 F' B'
12. 14.69[PLL skip] U' F' D2 B2 F' R' L2 D2 F R B2 L F' B2 U' D L B' U2 L F B D U' L2

Can someone please calculate this? I tried, but I'm probably wrong.

 

[Johan444]

This went in the accomplishement thread but it probably belongs here:
16.05, 15.26, (10.13)[PLL skip], 17.92, 16.16[OLL skip], (19.19), 16.91, 17.80, 18.33, 14.93[PLL skip], 11.70[LL skip], 14.69[PLL skip]

scrambles were:

Spoiler

1. 16.05 D' L' U R2 B2 R F D' F2 R U2 D F' L2 U' R B2 D R' F' R' F D2 U' L2
2. 15.26 U2 L2 R2 F R U2 F2 B' D2 U2 B' U B U F' L' U B' R2 D2 F' R' B' F L'
3. (10.13)[PLL skip] F2 U2 D L2 R2 B R B R B U' R B' F' R U' L' D B F' D2 U R2 F2 U'
4. 17.92 F L F2 B U' B D2 L F' R' L2 D' R2 D' B' F L D F' U2 R2 L U R2 D'
5. 16.16[OLL skip] R F B' D' R D2 L' B' F L U' L2 D' R' L2 F' R D' L D U' B' L B D
6. (19.19) B L B2 R F' B2 U' L U R2 F' L2 F R2 D' B U' D2 B L2 F2 D2 B' L2 R
7. 16.91 F B D' L2 B' F' U2 R2 D F2 U2 D' R2 F R2 L B U' F2 R2 D2 R F' L' U
8. 17.80 R L F' U B2 L D L2 F' D2 U' L' U' F' U' L2 D B R' D F' L' U2 D' R'
9. 18.33 L2 F' U2 L U' R U2 B2 U' B' L' F U' R2 B' R' U' B' R' L B F R D' R2
10. 14.93[PLL skip] B' L2 D' U' R L' B L U F R2 F2 L' F' U' F R2 U' D2 L' R' B2 R F' B'
11. 11.70[LL skip] U2 L2 U2 D2 L R2 B2 L' D2 U R2 U' D' L D2 U2 L B2 U L R F2 U2 F' B'
12. 14.69[PLL skip] U' F' D2 B2 F' R' L2 D2 F R B2 L F' B2 U' D L B' U2 L F B D U' L2

Can someone please calculate this? I tried, but I'm probably wrong.

Calculate what? The probability of getting full step, full step, PLL skip, full step, OLL skip, full step, full step, full step, full step, PLL skip, LL skip, PLL skip, in that order?