CubeRoots
Veteran
sorry hould have quoted, It was tge question before i was talking about
you are right, you could answer that, in theory
Probability Thread
- Thread starter CubesOfTheWorld
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- cfop oll probability
sorry hould have quoted, It was tge question before i was talking about
you are right, you could answer that, in theory
kinch2002
Premium Member
Some assumptions:
300 averages of 5 are done (200 + later rounds)
All scramble/solve combos are independent of each other
All combos have a 1/15552, 1/72, 1/216 chance of the skips respectively. e.g. the scramble state doesn't affect your chances, and you don't do anything to force any skips.
i) ~9.2%
ii) ~99.9999999%
iii) ~99.9%
Bob
Premium Member
I certainly hope not. There's so much wrong with the math here that I don't know where to start.
Noahaha
blindmod
Seriously. That is the most cringeworthy post I've ever seen. Kill it with fire.
MTGjumper
Veteran
At first I thought the ellipsis indicated that you realised something was wrong.
Then I realised you were being serious.
SweetSolver
Veteran
What are the chances of solving a 2x2 in 2 moves after a random scramble? (I know it sounds weird but I was curious )
)
What are the chances of solving a 2x2 in 2 moves after a random scramble? (I know it sounds weird but I was curious
Chance of solving the 2x2 cube in 2 moves:
Within to two moves you can reach 54 states.
There are 3674160 possible configurations.
54 / 3674160 = 1 / 68040 = 0.00147%
Chance of solving the 2x2 cube in 2 moves or less:
There's one state defined as solved.
Within to one moves you can reach 9 states.
Within to two moves you can reach 54 states.
54 + 9 + 1 = 64
64 / 3674160 = 0.00174%
Quod erat demonstrandum.
ThomasJE
Premium Member
You can't get a recurring sign on a keyboard.
Eazoon
Regular
what is the probability of a 15 move solve that directly follows the fridrich method? feliks can do about 20 tps so it would be a sub-1 solve.
RNewms27
Veteran
Fridrich will not give a 15-move solve without skipping a large amount of the solve. The scramble will also need a sub-16 move optimal solution, not to mention the solution must be following Fridrich steps. Hardly possible.
Feliks has 7-8-9tps in solves. (R U R' U') over and over would not be a Fridrich solution.
Bob
Premium Member
I seriously doubt that it would be sub1. By the same logic, since an average Fridrich solve is 56 moves and he can turn 20 tps, all of his solves are sub-3.
Eazoon
Regular
First off, this is a probability thread, its hardly possible, not impossible. second, if he can do up to 9 tps, then could he get sub-3? the record fewest moves is 22 i think.
Rpotts
Veteran
The fewest moves record was not achieved with Fridrich. Faz's fastest reconstructed solves have 9+ TPS, but he can turn substantially faster in bursts.
A solve that directly follows the Fridrich method without skipping any steps can't really be less than 28 moves HTM.
1 moves cross
4 3 move pairs
F RUR'U' F'
A/U perm.
1 + 12 + 6 + 9 = 28
kinch2002
Premium Member
Actually IMO both 22 move WRs used Fridrich. Ofc steps were skipped, but that doesn't mean it's not Fridrich
Rpotts
Veteran
not cross - f2l fridrich.
freesyle F2L - LL skip ≠ Fridrich imo
freesyle F2L - LL skip ≠ Fridrich imo
Kirjava
Colourful
It wasn't freestyle F2L. I've heard that it was actually cross -> F2L.
Rpotts
Veteran
Oh really? I thought it was blockbuilding... Is the solve posted anywhere?
kinch2002
Premium Member
Here are the solutions. It's a little off-topic to be talking about whether this is Fridrich or not, so whatever. I'd say everything is standard Fridrich apart from from Jimmy starting with 2x2 block and that István kind of switched from green to yellow F2L, while cancelling a pair into the OLL. Reconstructions are posted by others from other threads
Jimmy:
Scramble : U L2 U' R2 B2 U L2 D2 R2 F2 R2 F L F' D2 L' U B' D R2 F'
Solution : U L U R2 U2 (bloc 222)
B D' R2 D' B R' (cross + second F2L)
B2 U' B U (3rd F2L)
D B2 D2 B' D2 B' D' (4th F2L + LL skip)
István:
scramble: D'R2DL2F2DLR2B'D2U2L2DF'L2UL2FU
Xcross on green: FD'FL'BR2
2nd f2l: RBR'B2DBD' (12 move with R2 R cancel)
and from here he should've done: U'BU'LU2L' (18 moves total, beaten the scramble by 1 move!)
but he did something like: U2BLU'LF'L'FU'L' (22 moves total)
Jimmy:
Scramble : U L2 U' R2 B2 U L2 D2 R2 F2 R2 F L F' D2 L' U B' D R2 F'
Solution : U L U R2 U2 (bloc 222)
B D' R2 D' B R' (cross + second F2L)
B2 U' B U (3rd F2L)
D B2 D2 B' D2 B' D' (4th F2L + LL skip)
István:
scramble: D'R2DL2F2DLR2B'D2U2L2DF'L2UL2FU
Xcross on green: FD'FL'BR2
2nd f2l: RBR'B2DBD' (12 move with R2 R cancel)
and from here he should've done: U'BU'LU2L' (18 moves total, beaten the scramble by 1 move!)
but he did something like: U2BLU'LF'L'FU'L' (22 moves total)
Rpotts
Veteran
Those are both closer to Fridrich than I had thought, still far "directly follows the Fridrich method," but yea, my bad Sheppz.
2.777... = 278% chance
On average, if there are 100 US Nationals, 278 of the US Nationals will have someone get a LL skip
...
learn probability before you try to answer a question in this thread, please
