Probability Thread

[Discomantis]

What are the exact odds of getting one or more four move 2x2 scrambles in your first 2x2 round? (

 

[cubenerd74]

What are the exact odds of getting one or more four move 2x2 scrambles in your first 2x2 round? (

According to Wikipedia, there are 1847 out of 3674160 positions that can be solved optimally in 4 moves(assuming HTM). There are also 385 scrambles that are illegal according to wca regs, meaning 3671928 out of 3673775 legal scrambles are NOT 4 movers. That gives a \( \left(\frac{3671928}{3673775}\right)^5\approx99.7\% \) chance of no 4 movers in 5 solves, so the answer is about 100-99.7=0.3%.

 

[Swagrid]

What are the odds of a CN standard CFOP cross being solved, where being offset by a D/D'/D2 counts as solved

Essentially a pre-built cross, but not aligned.

I tried to work this out myself and got 1/7925.98587297, ofc rounding to 1/7926 for simplicity's sake.

However I am very bad at maths and have most likely gotten something wrong.

 

[FenTheGoat]

What are the odds of a CN standard CFOP cross being solved, where being offset by a D/D'/D2 counts as solved

Essentially a pre-built cross, but not aligned.

I tried to work this out myself and got 1/7925.98587297, ofc rounding to 1/7926 for simplicity's sake.

However I am very bad at maths and have most likely gotten something wrong.

I believe it would be (4!*2^4)Which equals 1/768. I could be extremely off, but I’ll explain my reasoning.

You have to have 4 edges in the right spot (4!). You need 4 edges oriented correctly (2^4). Then you multiply them and get 1/768.

 

[qwr]

I believe it would be (4!*2^4)Which equals 1/768. I could be extremely off, but I’ll explain my reasoning.

You have to have 4 edges in the right spot (4!). You need 4 edges oriented correctly (2^4). Then you multiply them and get 1/768.

4 edges in the right spot is not 1/4!. There are 12 edges on a cube if I counted correctly, and you need them to be in the right order relative to each other so 4! doesn't work. This also only considers one color cross.

 

[Davian]

F2l skip possibility

 

[IsThatA4x4]

F2l skip possibility

// EP = 12!/4!
// EO = 2^8
// CP = 8!/4!
// CO = 3^4
1/(12!/4! x 2^8 x 8!/4! x 3^4)
= 1/695,280,402,432,000 (non-CN)
or 1/115,880,067,072,000 (CN)

Please tell me if I am wrong as always

 

[Davian]

I dont understand Hieroglyphics

 

[IsThatA4x4]

I dont understand Hieroglyphics

12! = 12 x 11 x 10 x [...] x 1
2^8 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

 

[qwr]

I dont understand Hieroglyphics

wait until you find about generating functions

 

[Bruce MacKenzie]

What are the odds of a CN standard CFOP cross being solved, where being offset by a D/D'/D2 counts as solved

Essentially a pre-built cross, but not aligned.

I tried to work this out myself and got 1/7925.98587297, ofc rounding to 1/7926 for simplicity's sake.

However I am very bad at maths and have most likely gotten something wrong.

The number of arrangements of a particular set of four edges:

24 * 22 * 20 * 18 = 190,080

For each face, four of these meet your criteria and there are six faces. So:

4 * 6 / 190,080 or 1 in 7,920

The odds would be slightly less than this since cubes with more than one solved cross are counted more than once in the above.

 

[Jason Tzeng]

What's the odd of getting a scramble with two 2×2×1 block?

(Got this scramble few minutes ago:
R2 B2 L2 D' B2 D2 L2 F2 R2 F2 R2 U2 R' F L2 B' L D' R' D' R' )

 

[Jorian Meeuse]

What's the odd of getting a scramble with two 2×2×1 block?

(Got this scramble few minutes ago:
R2 B2 L2 D' B2 D2 L2 F2 R2 F2 R2 U2 R' F L2 B' L D' R' D' R' )

That scramble is really good, easy 1 look F2L on orange. I'm a white-yellow solver and still managed to get 6.06 on a colour I never do and a cube I never use

Can't help you with the probability rn, maybe I'll try tomorrow if it isn't super hard

 

[randomness...]

possibility of me understanding any of the numbers on this thread?

 

[Discomantis]

possibility of me understanding any of the numbers on this thread?

100%

There is going to be at least one thing you understand in this thread

 

[randomness...]

100%

There is going to be at least one thing you understand in this thread

ive gone through the thread on multiple occasions, it all flies right over my head

 

[Bruce MacKenzie]

What's the odd of getting a scramble with two 2×2×1 block?

(Got this scramble few minutes ago:
R2 B2 L2 D' B2 D2 L2 F2 R2 F2 R2 U2 R' F L2 B' L D' R' D' R' )

For a set of two corners and four edges the number of permutations is given by:

(24 * 21) * (24 * 22 * 20 * 18) = 95,800,320

So for the example you give the odds are 1 in 95,800,320. With rotation and mirror symmetry there are 48 equivalent patterns of this type.

However there are a number of different ways of having two 2x2 blocks. For example one could have

F' R F' B U' B' L2 B L F U' B' U F' U B2 L'
or
F R U' R D' L' U R' U' F' L U' F R' F U' D R F'

I haven't counted up all the distinct ways one could have two solved 2x2 blocks but they would all have similar probabilities.

ADDENDUM

After playing around with a cube editor, I composed the following ten ways of having two 2x2 patches on two adjacent cube faces and had not exhausted all the possibilities. And then there are all the cases where the 2x2 patches are on opposite faces. I'd guess there are probably in excess of 20 types in total. Each type would then have 48 symmetry equivalent patterns. So, I'd estimate that the odds of getting two 2x2x1 blocks with a random scramble would be something like 1 in 100,000.

Type 1
U B U' D' F2 L' F D' R U L F' L' F' U D L

Type 2
U R' F D' L B U B D R F L' F' B' U B' D R' F R'

Type 3
L' F R' B R2 U' D F' U' F' R F' R' D' R B' R' L

Type 4
U R2 U B R' U' D' L F D2 B' L' B R L U' L2 B'

Type 5
R' U R2 D L D' F' B D' B' D F' B D' F D' F' B' R L

Type 6
R' U R F D R D' R B' R' D' F2 R B' R' F' D' L' D L

Type 7
L2 D F R L U' F' L' F2 D B' R B U D' F' U

Type 8
L U' D' F' L F D' R' L U' R' F R2 U B2 L F L

Type 9
U R2 U F D R' F B L' D' B2 D2 R' B' R F'

Type 10
F' U L' U R D' B D L' B R' U' B' U' F' B L' F' R'

 

[Swagrid]

What's the odd of getting a scramble with two 2×2×1 block?

(Got this scramble few minutes ago:
R2 B2 L2 D' B2 D2 L2 F2 R2 F2 R2 U2 R' F L2 B' L D' R' D' R' )

side note, crazy scramble. acn link

y' x' L2 D2 R D' R U' R // XXXCross
L' U2 L y' U2 R U' R' // 4th
R' U2 R U R' U R U' R' U' R U' R' U2 R // ZBLL

can cancel 4th into ZB for 28 mover.

 

[Enirolf]

I don't know the probability, but in my pr 3x3 ao5 I got a Uperm 4 times. On the first two solves of the average, I got the same Uperm with the same auf

 

[Bruce MacKenzie]

For a set of two corners and four edges the number of permutations is given by:

(24 * 21) * (24 * 22 * 20 * 18) = 95,800,320

So for the example you give the odds are 1 in 95,800,320. With rotation and mirror symmetry there are 48 equivalent patterns of this type.

However there are a number of different ways of having two 2x2 blocks. For example one could have

F' R F' B U' B' L2 B L F U' B' U F' U B2 L'
or
F R U' R D' L' U R' U' F' L U' F R' F U' D R F'

I haven't counted up all the distinct ways one could have two solved 2x2 blocks but they would all have similar probabilities.

ADDENDUM

After playing around with a cube editor, I composed the following ten ways of having two 2x2 patches on two adjacent cube faces and had not exhausted all the possibilities. And then there are all the cases where the 2x2 patches are on opposite faces. I'd guess there are probably in excess of 20 types in total. Each type would then have 48 symmetry equivalent patterns. So, I'd estimate that the odds of getting two 2x2x1 blocks with a random scramble would be something like 1 in 100,000.

Type 1
U B U' D' F2 L' F D' R U L F' L' F' U D L

Type 2
U R' F D' L B U B D R F L' F' B' U B' D R' F R'

Type 3
L' F R' B R2 U' D F' U' F' R F' R' D' R B' R' L

Type 4
U R2 U B R' U' D' L F D2 B' L' B R L U' L2 B'

Type 5
R' U R2 D L D' F' B D' B' D F' B D' F D' F' B' R L

Type 6
R' U R F D R D' R B' R' D' F2 R B' R' F' D' L' D L

Type 7
L2 D F R L U' F' L' F2 D B' R B U D' F' U

Type 8
L U' D' F' L F D' R' L U' R' F R2 U B2 L F L

Type 9
U R2 U F D R' F B L' D' B2 D2 R' B' R F'

Type 10
F' U L' U R D' B D L' B R' U' B' U' F' B L' F' R'

I sat down and wrote some code to count the number of ways to have two solved 2x2x1 blocks on adjacent faces. On a face there are four 2x2x1 blocks which may be turned four ways to give 16 patterns. For two adjacent faces the blocks may be combined in 144 ways which are disjoint. That is the blocks do not intersect. Applying the 48 rotational and mirror symmetries gives 1728 ways to apply these patterns to the cube as a whole. Many of the original patterns have symmetry or are symmetry conjugates of one another so one does not get the full 48 times the number of patterns.

So, 1728 / 95,800,320 = 1 in 55,440 chances of getting two solved 2x2x1 blocks on adj. faces of the cube in a random scramble.