• Welcome to the Speedsolving.com, home of the web's largest puzzle community!
    You are currently viewing our forum as a guest which gives you limited access to join discussions and access our other features.

    Registration is fast, simple and absolutely free so please, join our community of 40,000+ people from around the world today!

    If you are already a member, simply login to hide this message and begin participating in the community!

Probability Thread

AlphaSheep

Member
Joined
Nov 11, 2014
Messages
1,081
Location
Gauteng, South Africa
WCA
2014GRAY03
The probabilities aren't independent (e.g. if you have FR in the UB position and BR in the UR position, the matching corner position for both of those edges is UBR), so this is not exactly accurate, but it's pretty close to the correct value. I'm too tired to exert my brain, so I just did a simulation on a million random states and got this:

No pairs: 453642
1 pair: 433056
2 pairs: 106154
3 pairs: 7070
4 pairs: 78
Yeah, I just realised that I was thinking that the corner can only connect on one side, so I was completely wrong on that too. I'll have to rethink it. It's cool to have simulation results though because it means I know what the answer needs to be ;)
 

hagner

Member
Joined
Jul 26, 2016
Messages
68
Location
finland
back when i was doing lbl i got a l2l skip (last two layers) except for oll 23 otherwise everything was permutatet and oriented after first layer and i just did sexy moves to orient oll 23 and auf. what are the odds for this?

my times back then where around 1 minute, maybe even 1 and a half and i got a 33 second solve, but of course i didn't count as pb.
 
Joined
Apr 18, 2009
Messages
462
Location
San Diego, California
WCA
2007ESPI01
YouTube
Visit Channel
1) I'd prefer someone else to confirm my answers but I think the chances of one specific edge being paired with any 3rd layer corner in a specific orientation is 1/6 (1/3 chance of being oriented and 1/2 chance of being from the last layer). That means that the odds of at least one out of two specific edges being like that are 11/36, and for at least one out of 3 is 91/216 and for at least one of four specific edges, its 671/1296 or 51.77%.

2) Not sure I understand. Can you give an example?

3) Not sure how to work this out, because the state after 1 move is not completely independent of the state before that move. I think that the answer depends on how inefficient your solution is. The more moves you take, the more likely it is to happen, I'm sure.

4) Yes 1/81 is correct.

5) Yes, if the corner you insert is oriented then the chance of the rest also being oriented is 1/27. And if you don't influence permutation of the other corners during CLS, then yes, theres a 1/72 chance of a PLL skip. That means there's a 49/972 (about 5%) chance of skipping at least one of OLL or PLL and 1/1944 chance of skipping both. Same as nomal OLL+PLL for ZZ.
Thanks for your response. Surely, in your #5 response, the skipping of both CLS then PLL is not possible if the corner piece in DFR is a fixed LL corner. In other words, an LL skip after building the 2nd to last slot is not possible, even though it is possible to skip orientation, and be left with a TTLL case. Or you can be left with a CLS case after which you might get a PLL skip. But I don't think you can have both. But in this scenario, ~1/20 solves will have the LL done in one LS step via skip after CLS or execution of TTLL. That's useful to know THANKS!
 

wowitsbryce

Member
Joined
Mar 7, 2016
Messages
7
I just finished learning ELL. So, I started thinking, what are the chances of having the corners of the last layer solved? In other words, what are the chances of solving the first two layers and immediately get an ELL case? Does the probability change if partial edge control is used?
 

Berd

Member
Joined
May 25, 2014
Messages
3,847
Location
Nottingham
WCA
2014LONG06
YouTube
Visit Channel
I just finished learning ELL. So, I started thinking, what are the chances of having the corners of the last layer solved? In other words, what are the chances of solving the first two layers and immediately get an ELL case? Does the probability change if partial edge control is used?
Edge control won't affect the probablilliy of an CLL skip.
 

Torch

Member
Joined
Aug 6, 2013
Messages
1,441
Location
Austell, GA, USA
WCA
2014GOSL01
YouTube
Visit Channel
I have an interesting problem that someone might be able to help with; it's really more of a combinatoric problem than a probability problem, but here it is anyway:

If you have an icosahedron and and color each of the 30 edges red or blue randomly, how many possible unique colorings are there? This counts all rotationally symmetric positions as identical. This seems pretty difficult to me, but if someone could at least give me an estimate that would be great.
 

xyzzy

Member
Joined
Dec 24, 2015
Messages
2,296
If you have an icosahedron and and color each of the 30 edges red or blue randomly, how many possible unique colorings are there? This counts all rotationally symmetric positions as identical. This seems pretty difficult to me, but if someone could at least give me an estimate that would be great.
Such computations are usually done with the lemma that is not Burnside's.

We first need to list all the rotational symmetries of the icosahedron. There is 1 order-1 rotation (the identity, or "do nothing", rotation), 24 order-5 rotations (rotate around a vertex either 1/5 or 2/5 of a revolution), 20 order-3 rotations (rotate around a face) and 15 order-2 rotations (rotate around an edge).

Then for each rotation, we count how many ways there are to edge-colour the icosahedron such that applying the rotation doesn't change the colouring. Rather than just two colours, let's say we use n colours. For the order-1 rotation, that gives n^30 colourings; for the order-5 rotations, that gives n^6 colourings each; for the order-3 rotations, that gives n^10 colourings each; for the order-2 rotations, that gives n^14 colourings each. Adding these up, we get n^30 + 24n^6 + 20n^10 + 15n^14, which not-Burnside's-lemma says is exactly equal to the total number of rotations (60) multiplied by the number of rotationally distinct colourings.

Now we can substitute n=2 and get punch our calculator to get 17900160 as our answer, but also notice that the largest contributor to the above expression is by far the n^30 term, so we get a pretty good approximation just from dividing 2^30 (the number of colourings) by 60 (the number of rotational symmetries), which equals 17895697.067.

(I may or may not have made silly mistakes above, but you get the idea.)
 
Joined
May 17, 2015
Messages
341
Location
Lithuania
WCA
2015RIBI01
YouTube
Visit Channel
So how many cases are there on 2x2 when you have all of your yellow and white corners oriented, 3/4 of white layer is solved and the 4th piece of the white layer is on the top?

P.S. I don't know if this is the correct thread or not.
 
D

Daniel Lin

Guest
What's the chance of any scrambled cube having at least 1 twisted corner? (or 1-[no twisted corners])
ez pz

7*2/24-21*4/24/21+35*8/24/21/18-35*16/24/21/18/15+21*32/24/21/18/15/12-7*64/24/21/18/15/12/9+1*128/24/21/18/15/12/9/6

=44.34%
 
Top