so today I was just cubing in the bus (yeah in a bus) and a woman was TOO SURPRISED when she saw that I could solve the cube in under 25 seconds. She asked if she could scramble the cube and ofc I said yes. She was scrambling it for like 2 mins. It was the easiest solve in my life. It was a one move away cross, an H perm and F2L SKIP ( sadly I wasnt timing ) She scrambled it again and I got an OLL SKIP ( wasnt timing again tho )
Probability Thread
- Thread starter CubesOfTheWorld
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Non-cubers always think that if they scramble the cube more it becomes harder to solve.
IsThatA4x4
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I hope you have more salt to take with this...
I don't know if you need to divide both by 2. If you take CP and EP together, only one of them matters for odd swaps. E.g. if we focus on corners first, we can allow any possible CPs, however looking at edges, there is now a 50/50 chance that the last 2 edges are swapped, which would be impossible. The case count for CP isn't affected by this I think (?).
As for what @Jaym-er said, what makes it so that I can't divide by 5? Just curious.
Because the way I see it, if we first calculate every possible "ZBLL" disregarding symmetry, any giving case's probability should be 1 / total case count / rotational symmetry. Some cases will have different symmetries, so it won't always be 5 but in the case of a skip it is. Where am I going wrong here?
ruffleduck
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Unlike 3x3, EP and CP are independant on megaminx. Odd swaps are impossible
IsThatA4x4
Veteran
I see! Interesting...
Good thing you had salt
What are the chances of getting almost 2 almost exactly same solutions from a random state scramble in 4 solves??
Generated By csTimer on 2022-06-20
avg of 5: 2.237
Time List:
1. (3.176) F2 R2 F U' R2 U' F' U R'
2. 1.884 F U F' R F' R2 U' R' F'
3. 2.094 U' F' R2 U2 F U2 F' U' R2
4. 2.732 R U R2 F' U F' R U2 R2 U'
5. (1.849) U2 F' U R U R2 U' F2 U2
Bold scrambles are the ones, they have the exact same EG-1 case and the solution is almost exactly same.
Generated By csTimer on 2022-06-20
avg of 5: 2.237
Time List:
1. (3.176) F2 R2 F U' R2 U' F' U R'
2. 1.884 F U F' R F' R2 U' R' F'
3. 2.094 U' F' R2 U2 F U2 F' U' R2
4. 2.732 R U R2 F' U F' R U2 R2 U'
5. (1.849) U2 F' U R U R2 U' F2 U2
Bold scrambles are the ones, they have the exact same EG-1 case and the solution is almost exactly same.
Zeke Mackay
Veteran
Not exactly probability but I've been stumped on calculating this.
How many cases are there for last bar of 7x7 last 2 centers?
If you aren't able to visualize it here's an explanation: One center has (at least) a 4x5 bar of center pieces built on left side and the other center has at least 20 correct center pieces and no more than 5 incorrect center pieces on it.
Ignoring AUF
How many cases are there for last bar of 7x7 last 2 centers?
If you aren't able to visualize it here's an explanation: One center has (at least) a 4x5 bar of center pieces built on left side and the other center has at least 20 correct center pieces and no more than 5 incorrect center pieces on it.
Ignoring AUF
Who better to answer this question than the madlad who genned algs for all 6×6×6 last bar cases…
Most cases do not have any AUF symmetry; the only ones that do are last bar skip and the one with two x-centres unsolved diagonally opposite each other.
The total number of cases (without reducing by AUF) is \( \binom51\binom51\binom51\binom62=1875 \). (The \( \binom51 \) factors correspond to the t-centres and obliques, while the \( \binom62 \) factor corresponds to the x-centres.) Reducing by AUF, all but three of these cases have no AUF symmetry, one has 4-way symmetry (skip) and two have 2-way symmetry (diag x-centres), so the number of cases up to AUF is \( 1872/4 + 1 + 2/2 = 471 \).
Your reasoning here is correct.
As pointed out by others (in the context of megaminx), this is off by a factor of 2 because only even permutations are possible.
Puzzlerr
Regular
no clue what i'm doing in a thread that involves math, but what are the odds of getting 3 pll skips in one average
sidenote, the last 2 plls skips were in a row
sidenote, the last 2 plls skips were in a row
Assuming no OLLCP/1LLL/etc., every individual PLL skip has a 1/72 chance of occurring; the chance of exactly three out of five solves having a PLL skip is
\( \binom53\left(\frac1{72}\right)^3\left(\frac{71}{72}\right)^2\approx0.000026\approx\frac1{38000} \)
(The probability of getting at least three PLL skips is about the same, since the chance of getting four or five PLL skips is much lower.)
But also the probability of getting three PLL skips in some average-of-5 increases to 100% asymptotically as you do more solves, and if you count rolling averages (as most people do for unofficial solves), this probability also increases faster.
ottozing
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(Apologies if the statistics/probabilities for this request are already documented and I've somehow missed it)
For FMC solving I'm curious to know the % chance of the following EO cases occurring on a random scramble
1. The probability of getting a 4 bad edges case solvable in 3 moves or less
2. The probability of getting an 8 bad edges case solvable in 4 moves or less
3. The probability of getting a 6 bad edges case solvable in 4 moves or less
I basically want to know how often 2 moves to 4 bad & 1 move to 6 or 8 bad could yield a 5 move or less EO using NISS
For FMC solving I'm curious to know the % chance of the following EO cases occurring on a random scramble
1. The probability of getting a 4 bad edges case solvable in 3 moves or less
2. The probability of getting an 8 bad edges case solvable in 4 moves or less
3. The probability of getting a 6 bad edges case solvable in 4 moves or less
I basically want to know how often 2 moves to 4 bad & 1 move to 6 or 8 bad could yield a 5 move or less EO using NISS
GodCubing
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That one dude who exposed the WCA for not using random state did this I think it was pochman, but I will have to check. I know he analysed the frequency of 12 flips, but doing the other EO cases wouldnt' be that hard, but in the context of FMC it would be CN EO, aka EO on all 3 axis so I suppose this is useless (I doubt pochman analysed EO on all 3 axis). I could do fixed orientation by myself but that wouldn't be relevant
ottozing
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Fixed orientation would be fine I think since for the purposes of what I'm looking for, I want to know how common these outcomes are on a single axis. I don't really care what's going on with all 3 axis at once
Knowing the data for these 3 scenarios would give me a good idea of how much better it is to check 2 moves to 4 bad followed by NISSing VS checking 1 move to 8 or 6 bad. I already suspect that 2 moves to 4 bad followed by NISSing is more likely to produce a 5 move EO than 1 move to 6 bad based on the current EO data I have, but I'm less sure of how it compares to 1 move to 8 bad given it has more cases in the 2-4 move range
mencarikebenaran
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are you a mathematician?
I mean I majored in mathematics but that calculation is something any high schooler (11th/12th grade?) should be able to do.
My computer tells me the answers are:
1. 32/99 = 32.3%
2. 13/45 = 28.9%
3. 52/231 = 22.5%
Full distribution:
2 bad edges: [0,0,40,22,4,0,0]
4 bad edges: [2,24,134,276,56,3,0]
6 bad edges: [0,0,8,200,556,160,0]
8 bad edges: [0,1,20,122,284,68,0]
10 bad edges: [0,0,0,0,0,54,12]
12 bad edges: [0,0,0,0,0,0,1]
Read this as: the i'th number in the row is the number of positions with that many bad edges whose minimal solution takes i moves.
ottozing
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Thanks for doing this, as well as posting the additional data!
Knowing this now, I'll have to think about how I want to approach NISS EO. On the one hand, 2 moves to 4 bad certainly performs slightly better than 1 move to 8 bad & quite a bit better than 1 move to 6 bad. On the other hand, Doing NISS EO's with only one turn on one of the sides of the scramble makes it easier to check other EO's for DR triggers
This is also not factoring in 1-2 moves to 2 bad which is very likely to give 4-5 move EO's, though these tend to only work well when the other axis don't have very many bad edges when looking for short solutions to DR minus trigger
I have a lot to think about
How many cases are there when you have a belt and oriented yellow and white sides? I'm including parities (M2 U2 M2) and excluding AUFs and rotation symmetry (z2/x2). I'm currently working on a method and I'm genning some algs, I want to know how many exactly I have to gen. I know its going to be a lot though
IsThatA4x4
Veteran
Using the batch solver gives...
It doesn't because it's taking to long to calculate
Fine I'll do it myself
Every piece is oriented, so we only care about permutation.
There are 8 corners and 8 edges to consider:
the corners it's 8!, for the edges it's 8! Again
Since we can't only swap two edges, we divide by 2
So it should be 8!^2 / 2 = 812851200 cases excluding AUF and symmetry
Good luck genning 800 million algs, I'll see you at the end of time
Edit: I see we're not counting AUFs, this would get pretty complicated because each case has different symmetries, but dividing by 4 would get pretty close, which is still around 200 million cases
Last edited:
ruffleduck
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Sounds like PBL but on 3x3. Should be the same number as sq1: 21^2 = 441
edit: you were pretty vague with your terminology ("oriented" relative to what?) and I may have misinterpreted what you are looking for.
