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Probability Thread

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Chance to get an xx-cross with an LL skip on 3x3?
Depends on how you define the probability of getting an xx-cross. The chance of getting an LL skip with no influence is 1 in 15552. If you define getting an xx-cross as having 2 F2L pairs solved after you complete the cross, assuming that you didn't try to influence the pairs at all, the probability is a little less than 1 in 18816, but surely there is some influence here that significantly increases the probability. This would make the desired probability about 1 in 300 million.

@AlphaSheep This is INSANE math. I’m pretty good in math, how can I do this? And thats about right as I’ve got a 4 mover and 5 mover in about 2k solves.
It's not complex math; it's right there in the post.
 
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When I was in grade 3, I solved a layer of a rubik's cube... AND IT WAS COMPLETELY SOLVED. OK OK I kind of used a method, the steps were: 1.Solve a layer 2. Solve a different layer 3. Repeat step 2 until entire cube is solved. Kind of how you would solve a pyraminx as a beginner. But that is still incredible, probably one of the luckiest solves ever. What's the chance of something like that happening? How much layers would it take to usually solve that?
 
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Sounds almost like CFinity (but even less likely to work).

The chances of all 8 other edges and all 4 other corners being solved is:
1 / [(8! * 2^8 * 4! * 3^4) / (2 * 3 * 2)] = 1/1,672,151,040 = 0.000000000598 = 0.0000000598%

However, I think we can assume that you would be able to solve the cube if it were only a turn away (such as if the second or third layers were off by a single move). There are 16 cases like this, reducing your chances of reaching one of these to:
16/1,672,151,040 = 1/104,509,440 = 0.00000000957 = 0.000000957%
Basically, one in a hundred million.

If you wanted a 50% chance of having a solved cube (computed as a 50% chance of not having an unsolved cube, and assuming each layer is independent of each previous layer), you would need to do:
0.9999999904315 ^ n = 0.5
n ≈ 72,440,526 layers

(Although, since you had done some layers already, this might have somewhat affected the outcome of each trial, meaning they are not entirely independent. The average would likely be slightly less, but I'd guess it'd still hover around 70 million. Someone can correct me on that; it's probably a bit off. If anyone sees anything wrong with any of my math, feel free to point it out.)

In other words, you got VERY LUCKY! :)
 
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What is the probablity of having two 2x2 scrambles in a row that both have a completed adjacent-swap side?
Let's see...


So,
I have to calculate the probabilty of having two pieces next to each other, so around 1/7, then the same orientation to make a bar, so 1/3 for each piece, or 1/9, times the 1/7 is already 1/63 for a bar,

to have the next piece in for the adjacent swap would make it 1/1134,

then the last one would make it 1/17010

there are 4 different adjacent swaps for 1 face, which increases the chances to 4/17010, or 2/8505

having two of these in a row would make it 1/72335025, or 1/17010 ^ 2

so the answer is 1/72335025, or 0.000000138245615, or 0.0000138245615%


Remember, I might not be completely accurate, but this was my thought process, and correct me if I made any mistake.
 
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hm
6 colors to make the face
4 possible colors for the bar
24 possible permutations of the other corners
27 possible orientations of the other corners
that's 15,552 scrambles (funny, 1/15552 is the probability of LL skip on 3x3)

but that overcounts, since there are scrambles that have two completed opposite faces:
3 pairs of colors to make the faces
4 possible colors for each of the two bars
4 cuz AUF
that's 192

thus 15,360 scrambles possible.
total of 3,674,160 2x2 scrambles
(15,360/3,674,160)^2=(64/15309)^2=4096/234,365,481= about 1/57,218.
 
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Hello everyone!

I was wondering if anybody knew if there are PLLs or OLLs that were more common than others. For instance, I've been noticing quite more J perms than N perms, or more U perms than Z or H.

The OLL range is bigger, so if someone has noticed a variation on chance between different OLLs, I would like you to write it as a reply.

To make it clear, what I mean by this, is that a certain case as a bigger percentage chance of appearing.

Also I made a poll in whether you think change may vary between different cases.

Thank you and have a good day! :)

(PS: I'm sorry if someone made this thread before, but I could not find it.)
 
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Yes, the probabilities of each one are known. It has to do with symmetry; there's a pretty good explanation in this Reddit thread. Your findings make sense (especially if you consider a U-perm either of the two U-perms; same for a J-perm). You're 8 times more likely to get either U-perm than an H-perm, and 4 times more likely to get either U-perm than a Z-perm.

The Reddit thread links this page for OLL probabilities and this one for PLL probabilities.
 
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I am not positive if this has been asked yet or not, but what are the odds of your Ao5 and Ao12 being the exact same time. I ask because for the third time this week I have gotten the same average in Ao5 and Ao12 in three different events. One is 3x3 at 12.42, one is Pyraminx at 4.54 and the other was Skewb at 5.40. Thank you in advance for your help and answers.
 
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I am positive if this has bee asked yet or not, but what are the odds of your Ao5 and Ao12 being the exact same time. I ask because for the third time this week I have gotten the same average in Ao5 and Ao12 in three different events. One is 3x3 at 12.42, one is Pyraminx at 4.54 and the other was Skewb at 5.40. Thank you in advance for your help and answers.
Depends. This is impossible to calculate since its impossible to know what the underlying distribution of your solves are. If you are very consistent then the chances of this happening are really high. If you are so consistent that you usually get the same times every solve, then the chances of this happening are almost 100%. If you're very inconsistent, then the odds are still higher than you'd expect, because your times are distributed around your global average.
 
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What is the probability of a cross skip?
TL;DR: I think 1/31,680. I'm not positive.

Many of us know that the chance you'll get an OLL skip in any given solve is 1/216; for PLL skip, 1/72; and for a full LL skip, 1/15,552. But what is the chance you will get a cross skip in a solve? I've never heard of anyone ever getting one. I tried to figure out the chance. Here's what I came up with.

The cube has 43,252,003,274,489,856,000 positions (12! x 8! x 2^11 x 3^7 / 2). How many of these positions have a cross on them? First of all, in a cross, all of the corners are completely irrelevant. We can start our calculation by using the number of positions of the corners, which is 88,179,840 positions (8! x 3^7). For there to be a cross, four edges matter and eight don't. For the four edges that matter, the cross edges, there are six options, because the cross could be on any of the six faces. The eight edges that don't matter could be permuted 8!/2 ways. This is because there are eight places to put the first irrelevant edge (because four spots are already occupied by the cross edges), seven places to put the second (because now there is a spot occupied by an irrelevant edge), six places for the third, and so on. The /2 at the end is because we have already chosen a position for the corners, and it's impossible to switch just two pieces on a cube. There are 2^7 ways that the eight irrelevant edges could be oriented, since the orientation of the eighth edge depends of the orientation of the other seven.

This means that there are, in total, 8! x 3^7 x 6 x 8! x 2^7 / 2 = 1,365,277,881,139,200 cube states that have a cross on them. Dividing this number by the total number of 3x3 positions gets us the chance that any random scramble will have a cross on it, thus, a cross skip. Ultimately this comes to 1/31,680, which is very close to half as likely as a last layer skip.

All of this is my own work (except figuring out 43 quintillion). I'm not positive that this is correct. I think my formula may not properly address cube states with multiple crosses on them. Any thoughts are welcome.
 
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Your calculation is very interesting, and I feel it is quite accurate. It does make me think though, since I use LMCF instead of CFOP, I too am looking for 4 pieces in the correct position as the first step (a full EG1 face, four corners with the color, any permutation). However, according to my calculation the probability of skipping the EG1 face is 1544 - 1. I have had it happen many times. In both cases we need 4 pieces pre-solved. In one case the odds are 30,000+ against, in the other case 1500+ against. The same thing happens when we look at F2L vs. E2L pairs, in terms of the chance of skipping the first pair. In CFOP you must pair an edge with its exact corner. In E2L, any two solved edges count as a pair and any two edges can be solved as a pair. So again the chance of a skip is way higher.
 
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Consider only permutation and orientation of the edges (we need not worry about the parity of their permutation). There are simply 12!*2^11=980995276800 possible solvable states of the edges. A cross necessitates having 4 edges solved, leaving 8!*2^7 possibilities. With 6 colors, this is 30965760, 1/31680 of our sample set. Of course, as you mentioned, this does not account for the unlikely possibility that there are multiple crosses.

Let's do this:
2 crosses:
Adjacent crosses (RU): 12 pairs of faces*(5 unimportant edges)!*2^4=23040
Opposite crosses (UD): 3 pairs of faces*(4 unimportant edges)!*2^3=576

3 crosses:
(RUD): 12 sets of faces*(2 unimportant edges)!*2^1=48
(RUF): 8 sets of faces *(3 unimportant edges)!*2^2=192

4 crosses:
(RUFD): 12 sets of faces*(1 unimportant edge)!*2^0=12
(RULD): 3 sets of faces*(0 unimportant edges)!*1 (we are guaranteed that EO is valid here since all of the edges are solved)=3

5 crosses:
6 sets of faces*0!*1=6

6 crosses:
1.

Note that every case with at least 4 crosses has 6, but we ignore this fact for the purposes of counting correctly.

There are (sort of) 30965760 states with 1 cross. We have double counted the states with 2 crosses, so we subtract these. However, that means that each case with 3 crosses has been counted 3-3=0 times, so we add these back. Etc.

Our final count is 30965760-(23040+576)+(48+192)-15+6-1=30942374. This is roughly 1/31704 of all possible states.
 
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Your calculation is very interesting, and I feel it is quite accurate. It does make me think though, since I use LMCF instead of CFOP, I too am looking for 4 pieces in the correct position as the first step (a full EG1 face, four corners with the color, any permutation). However, according to my calculation the probability of skipping the EG1 face is 1544 - 1. I have had it happen many times. In both cases we need 4 pieces pre-solved. In one case the odds are 30,000+ against, in the other case 1500+ against. The same thing happens when we look at F2L vs. E2L pairs, in terms of the chance of skipping the first pair. In CFOP you must pair an edge with its exact corner. In E2L, any two solved edges count as a pair and any two edges can be solved as a pair. So again the chance of a skip is way higher.
The reason for the higher probability of a skip is that the pieces of the EG face (not necessarily EG-1, correct?) need only be solved relative to each other (and not even that since the permutation doesn't matter). Let's compute the probability of that for fun:

1 face:
6 colors*6 permutations of solved layer*27 orientations*24 permutations=23328

2 faces:
adjacent: 12 pairs of faces*4 permutations of pieces in solved layers*3 orientations*2 permutations=288
opposite: 3 pairs of faces*6^2 permutations of pieces within layers*4 AUF=432

3 faces:
(RUF): 8 sets*1 (cuz the cube must be solved)=8
(RUD): 12 sets*4 permutations=48

4 faces:
15 sets of 4*1 (cuz the cube is solved)=15

5 faces:
6

6 faces:
1

Total=23328-(288+432)+56-15+6-1=22654
This is roughly 1/162 of the 3674160 possible states.
 
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