OLL and PLL skip chances

Lt-UnReaL said:
I've never gotten an OLL skip in any speedsolve in the passed 5 months, and I solve about 30 times a day. ... How can these numbers be right? How can people get OLL/PLL skips all the time and I never get them... :(
Click to expand...
Very interesting. The chances of zero OLL skips for 5 months in a row, 30 solves a day is, 30 days in a month, is:

- 4,500 solves in a row
- OLL skip is 1/216 chance per solve
- non-(OLL skip) is 215/216 chance per solve
- single solve has a 215/216 chance of no OLL skip
- two solves in a row with no OLL skip = (215/216)^2
- three solves in a row with no OLL skip = (215/216)^3
- 4,500 solves in a row with no OLL skip is (215/216)^4500
- 0.000000000853472321
- 0.0000000853472321%
- 1 in 1,171,684,161 chance

A 1 in 1.2 billion chance... perhaps you should buy a lottery ticket? :P

Simulation of 881 billion (881,944,690,688) trials of 4500-solves:

OLL skips 0 = 230 / 881944690688 = 2.60787328761601E-08%
OLL skips 1 = 15416 / 881944690688 = 1.74795541747341E-06%
OLL skips 2 = 162271 / 881944690688 = 1.83992263589016E-05%
OLL skips 3 = 1097982 / 881944690688 = 0.000124495562090574%
OLL skips 4 = 5878114 / 881944690688 = 0.000666494629659204%
OLL skips 5 = 24749821 / 881944690688 = 0.00280627813300773%
OLL skips 6 = 86217883 / 881944690688 = 0.00977588321697837%
OLL skips 7 = 258301220 / 881944690688 = 0.029287689208549%
OLL skips 8 = 676006018 / 881944690688 = 0.076649479852603%
OLL skips 9 = 1575405550 / 881944690688 = 0.178628610913348%
OLL skips 10 = 3297845848 / 881944690688 = 0.373928873638025%
OLL skips 11 = 6274469723 / 881944690688 = 0.71143573845944%
OLL skips 12 = 10928087149 / 881944690688 = 1.2390898504616%
OLL skips 13 = 17562320340 / 881944690688 = 1.9913176557931%
OLL skips 14 = 26196827129 / 881944690688 = 2.97034807347885%
OLL skips 15 = 36456489384 / 881944690688 = 4.13364803586045%
OLL skips 16 = 47547253887 / 881944690688 = 5.39118318745234%
OLL skips 17 = 58338185419 / 881944690688 = 6.61472153922608%
OLL skips 18 = 67572855324 / 881944690688 = 7.66180192901743%
OLL skips 19 = 74133293942 / 881944690688 = 8.40566247801424%
OLL skips 20 = 77243172901 / 881944690688 = 8.75827857648795%
OLL skips 21 = 76616280920 / 881944690688 = 8.68719793077184%
OLL skips 22 = 72523103463 / 881944690688 = 8.22308975026826%
OLL skips 23 = 65657620246 / 881944690688 = 7.44464147686868%
OLL skips 24 = 56951977606 / 881944690688 = 6.4575452641562%
OLL skips 25 = 47419820300 / 881944690688 = 5.37673402886615%
OLL skips 26 = 37954497518 / 881944690688 = 4.30350087922089%
OLL skips 27 = 29254542807 / 881944690688 = 3.31704959686063%
OLL skips 28 = 21736639118 / 881944690688 = 2.46462610949485%
OLL skips 29 = 15592034123 / 881944690688 = 1.76791518647691%
OLL skips 30 = 10814643184 / 881944690688 = 1.22622691628922%
OLL skips 31 = 7260632098 / 881944690688 = 0.823252543460069%
OLL skips 32 = 4720558130 / 881944690688 = 0.535244236950678%
OLL skips 33 = 2976327270 / 881944690688 = 0.337473234027656%
OLL skips 34 = 1822461500 / 881944690688 = 0.206641246241679%
OLL skips 35 = 1083397451 / 881944690688 = 0.122841881405834%
OLL skips 36 = 626434696 / 881944690688 = 0.0710287960927938%
OLL skips 37 = 352151814 / 881944690688 = 0.0399290134311357%
OLL skips 38 = 193225312 / 881944690688 = 0.0219090056372204%
OLL skips 39 = 103108228 / 881944690688 = 0.0116910084145487%
OLL skips 40 = 53507456 / 881944690688 = 0.00606698544307344%
OLL skips 41 = 27216448 / 881944690688 = 0.00308595859665175%
OLL skips 42 = 13527912 / 881944690688 = 0.0015338730583487%
OLL skips 43 = 6620175 / 881944690688 = 0.000750633806167101%
OLL skips 44 = 3146212 / 881944690688 = 0.000356735749216389%
OLL skips 45 = 1441037 / 881944690688 = 0.000163393126033318%
OLL skips 46 = 651337 / 881944690688 = 7.38523636319978E-05%
OLL skips 47 = 280975 / 881944690688 = 3.1858573782083E-05%
OLL skips 48 = 121955 / 881944690688 = 1.38279646430961E-05%
OLL skips 49 = 51878 / 881944690688 = 5.88222827891058E-06%
OLL skips 50 = 21394 / 881944690688 = 2.42577570066334E-06%
OLL skips 51 = 8031 / 881944690688 = 9.10601320558443E-07%
OLL skips 52 = 2799 / 881944690688 = 3.17366840523357E-07%
OLL skips 53 = 1147 / 881944690688 = 1.30053506995459E-07%
OLL skips 54 = 438 / 881944690688 = 4.96629782598179E-08%
OLL skips 55 = 157 / 881944690688 = 1.78015698328571E-08%
OLL skips 56 = 2 / 881944690688 = 2.26771590227479E-10%

4,500 / 216 = 20.8333... so that's the number of OLL skips we would expect in 4,500 solves.
 
Last edited:
Stefan said:
Wrong! One OLL skip is *exactly* what you should expect.
Click to expand...
joey said:
You should *expect* one OLL skip, but it does not mean you *will* get that one OLL skip.
Click to expand...
That is not correct. To expect that a 1/216 chance OLL would occur only once in 216 solves means that there is a connection, and there's not. That type of thinking would be similar to expecting the numbers 1 through 20 to each appear exactly once each time a 20-sided dice is rolled 20 times in a row. That would be entirely unrandom, and also extremely improbable.

After 1.5 trillion (1,544,980,267,008) simulated trials of 216-solves each, the following counts of OLL skips showed up:

OLL skips 0 = 566065231316 / 1544980267008 = 36.638994257981%
OLL skips 1 = 571638686984 / 1544980267008 = 36.9997403326731%
OLL skips 2 = 284351838547 / 1544980267008 = 18.4048848143332%
OLL skips 3 = 93862260256 / 1544980267008 = 6.07530479581937%
OLL skips 4 = 23372499652 / 1544980267008 = 1.51280247075667%
OLL skips 5 = 4727938038 / 1544980267008 = 0.306019315518903%
OLL skips 6 = 816613350 / 1544980267008 = 0.0528559080939881%
OLL skips 7 = 125101642 / 1544980267008 = 0.00809729707695692%
OLL skips 8 = 17491555 / 1544980267008 = 0.00113215394225546%
OLL skips 9 = 2285378 / 1544980267008 = 0.000147922795442938%
OLL skips 10 = 281778 / 1544980267008 = 1.82382911948571E-05%
OLL skips 11 = 34683 / 1544980267008 = 2.24488304094439E-06%
OLL skips 12 = 3424 / 1544980267008 = 2.21620953556313E-07%
OLL skips 13 = 303 / 1544980267008 = 1.96119009718349E-08%
OLL skips 14 = 102 / 1544980267008 = 6.60202606972661E-09%

So the chances of getting exactly one OLL skip in 216 consecutive solves is about 37%, about 1 in 3, not 100%.

From any given consecutive 216 solves, you should expect to see one OLL skip (37.0%) or zero OLL skips (36.6%) about equally, and occasionally two OLL skips (18.4%) and three OLL skips (6.1%), and even four OLL skips (1.5%). The rest will occur far less often. Keep in mind this is the percentage chance of these occurring out of any 216 solves.

One or more OLL skips only happen about 2/3rds (63.4%) of the time in a given set of 216 solves, as 1/3rd of the time, there are no OLL skips.

Note: Actual calculated chance of zero OLL skips in 216 solves = (215/216)^216 = 36.70262219%, so the above simulation calculations have almost 3 significant digits of accuracy.
 
Last edited:
Let X be number of OLL skips in 216 solves. X~Bin(126,1/216)

E(X) = np = 216/216 = 1

So you expect to get exactly one OLL skip in 216 solves. This does not mean that after 215 solves, the probability of getting an OLL on the next solve is 1. Joey and Stefan are right.

(For what it's worth, if X~Bin(n,p), P(X=x) = nCx * (p)^x * (1-p)^(n-x), E(X) = np, Var(X) = np(1-p). This is a hell of a lot easier to do than write a program.)
 
Last edited:
My simulation of 1.5 trillion trials, of 216-solves each, is legit. The numbers and results are real. If you expect to see exactly one OLL skip in any given set of 216 solves, then your expectations will not be met 63.0% of the time...

Any formula that predicts otherwise is either flawed or is using the word 'expect' to mean something different, which happens a lot in mathematics, so I am interested in knowing what your formulas mean (all of them) so I can understand your perspective. I think we're saying two different things correctly.
 
JasonD said:
My simulation of 1.5 trillion trials, of 216-solves each, is legit. The numbers and results are real. If you expect to see exactly one OLL skip in any given set of 216 solves, then your expectations will not be met 63.0% of the time...

Any formula that predicts otherwise is either flawed or is using the word 'expect' to mean something different, which happens a lot in mathematics, so I am interested in knowing what your formulas mean (all of them) so I can understand your perspective. I think we're saying two different things correctly.
Click to expand...

Expect.
http://en.wikipedia.org/wiki/Expected_value
We expect 1 OLL skip in 216 solves.
 
Where I think things get more interesting is that a lot of people partial edge control sometimes or always. This will change your OLL odds on an individual basis depending on the person's degree of partial edge control. I wonder if anyone here knows their true theoretical OLL skip percentage. I certainly don't know mine.
 
hawkmp4 said:
Expect.
http://en.wikipedia.org/wiki/Expected_value
We expect 1 OLL skip in 216 solves.
Click to expand...
Ok that's what I thought. We're talking about two different things. You were using "expect" in the mathematic sense for "expected value", which is a weighted average, which, of course, is 1. If this is what everyone meant by "expect", then they are right... but I doubt this is what they meant. If they meant "expect" in the non-mathematical (outside of probability theory) sense, then what's expected is that 2/3rds of the time there is not exactly one OLL skip, which is what I was trying to show. (Actually, what's expected is whatever you believe is expected, ha ha.) This is what I assume everyone meant, unless everyone here is a mathematician, and I'm out in left field on my assumptions. If so, my humble apologies.
 
Yeah. It's pretty clear to me from context (joey's comment about expecting 1 but not necessarily getting 1) and just from knowing Stefan and joey that they were also using it to mean expected value.
 
hawkmp4 said:
Yeah. It's pretty clear to me from context (joey's comment about expecting 1 but not necessarily getting 1) and just from knowing Stefan and joey that they were also using it to mean expected value.
Click to expand...
To be honest, I took their words to mean "I expect to get 1, but won't be surprised if I don't." Most people think of what to expect as the highest probability, and not as mathematical "expected value" (weighted average), and I thought that's all they were saying. I wanted to point out how far off such expectations would be. I didn't know the guys were mathematicians, and given "expect" is a common English word, I didn't follow their words in the mathematical way at all.
 
Possibility of getting x OLL skips in n amount of solves:
nCx * (1/216)^x * (215/216)^(n-x)
So getting 0 OLL skips in 4500(about 5 months) solves:
0C4500 * (1/216)^0 * (215/216)^(4500-0)=
(215/216)^4500
So getting no OLL skips in 4500 solves is about about 853472*10^-13% chance(0.0000000853472% chance)
 
Also note that this is not entirely chance based. You could make sure that you get an OLL or PLL skip (or both), but this would most likely be less efficient than solving the cube normally. In that case, you could expect to get it 100% of the time (like I said, this would most likely be less efficient and take a lot of brainpower and thinking).
 
WhatDoIPutHere said:
Also note that this is not entirely chance based. You could make sure that you get an OLL or PLL skip (or both), but this would most likely be less efficient than solving the cube normally. In that case, you could expect to get it 100% of the time (like I said, this would most likely be less efficient and take a lot of brainpower and thinking).
Click to expand...
you do realize this thread is 7 years old right?
 
You must log in or register to reply here.

Similar threads

Jteddy
Which Is Less Common?
Replies
6
Views
401
Wagnit
W
precompsubwr
Which do you think is better: PLL skip or OLL skip on 3x3
Replies
15
Views
2,039
CubeLite
CubeLite
A
What's this algorithm?
Replies
5
Views
1,419
pdilla
pdilla
Rubiks560
[video] Easy PLL Skip Cases for OLL
Replies
10
Views
5,133
Rubiks560
Rubiks560
Coolster01
Good Diagonal PLL Algs?
2
Replies
26
Views
5,202
IQubic
IQubic
Back
Top