Probability Thread

[CubeRoots]

I think you can. You'd have to know what everybody is registered for, whether or not there are cutoff times (in which case you can't calculate it). You'd also have to assume that everyone will finish their solves, and that they're all using straight CFOP with no influencing techniques.

sorry hould have quoted, It was tge question before i was talking about

you are right, you could answer that, in theory

 

[kinch2002]

What is the propability that at Nationals, given there are about 200 people, that at least one would get a(n):
LL Skip?
PLL Skip?
OLL Skip?
Let's just say they all use Fridrich. I highly doubt it would happen, but it's a legitemate question. Right? RIGHT?

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]

1/72 for a PLL skip normally. So 200 x 1/72 = 2.777...
Same maths can be applied to OLL and LL skips.

I certainly hope not. There's so much wrong with the math here that I don't know where to start.

 

[Noahaha]

I certainly hope not. There's so much wrong with the math here that I don't know where to start.

Seriously. That is the most cringeworthy post I've ever seen. Kill it with fire.

 

[MTGjumper]

1/72 for a PLL skip normally. So 200 x 1/72 = 2.777...

At first I thought the ellipsis indicated that you realised something was wrong.

Then I realised you were being serious.

 

[SweetSolver]

What are the chances of solving a 2x2 in 2 moves after a random scramble? (I know it sounds weird but I was curious )

 

[nqwe]

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]

At first I thought the ellipsis indicated that you realised something was wrong.

Then I realised you were being serious.

You can't get a recurring sign on a keyboard.

 

[Eazoon]

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]

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.

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]

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.

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]

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.

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]

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.

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]

The fewest moves record was not achieved with Fridrich.

Actually IMO both 22 move WRs used Fridrich. Ofc steps were skipped, but that doesn't mean it's not Fridrich

 

[Rpotts]

not cross - f2l fridrich.

freesyle F2L - LL skip ≠ Fridrich imo

 

[Kirjava]

It wasn't freestyle F2L. I've heard that it was actually cross -> F2L.

 

[Rpotts]

Oh really? I thought it was blockbuilding... Is the solve posted anywhere?

 

[kinch2002]

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)

 

[Rpotts]

Those are both closer to Fridrich than I had thought, still far "directly follows the Fridrich method," but yea, my bad Sheppz.

 

[aronpm]

1/72 for a PLL skip normally. So 200 x 1/72 = 2.777...
Same maths can be applied to OLL and LL skips.

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