Probability Thread

[Ickathu]

What's the probability of each of the LL cases on pyraminx? I'm referring to the last layer that you get with Oka, not the LBL LL. This LL involves 3 edges and 3 centers.
Does that make sense?

 

[Schmidt]

Last 2 Center skip on 5x5?

So far: 1/Lifetime
I was just casually solving while my kids played on the playground, so I wasn't paying 100% attention to the solving, so when I had solved the first 4 centers I was turning the cube around to find the next unsolved centers. After a few x-y-z's I couldn't find any

 

[Bob]

So far: 1/Lifetime
I was just casually solving while my kids played on the playground, so I wasn't paying 100% attention to the solving, so when I had solved the first 4 centers I was turning the cube around to find the next unsolved centers. After a few x-y-z's I couldn't find any

I think the probability that you counted wrong is higher than the probability that you skipped 2 centers.

 

[JasonK]

What's the probability of each of the LL cases on pyraminx? I'm referring to the last layer that you get with Oka, not the LBL LL. This LL involves 3 edges and 3 centers.
Does that make sense?

Skip = 1/12
U-perm = 1/6
2flip = 1/4
Flipcycle = 1/2

 

[mDiPalma]

this question has likely been answered before, but advanced search has wierded out on me:

what is the distribution for the minimum amount of 2-generator [RU] moves it can take to solve a cube that CAN be solved 2-generator, without rotations? like if you perform an antisune on your cube, then the fewest moves it would take to be solved would be 7, and this state would be tallied in the 7 column.

does that make sense?

is this an impossibly hard question to answer? if it is, then don't bother.

 

[Stefan]

Look for "Analysis of the 3x3x3 <U, R> group" here: http://cubezzz.dyndns.org/drupal/text/fullcube.txt

 

[mDiPalma]

Thanks! I have some more questions!

What is the probability of a last layer skip on a 3x3 given that the cube is in a state that can be solved 2 generator <RU> and has the F2L completed?

How does this compare with a standard LL skip after standard F2L?

 

[Bob]

Thanks! I have some more questions!

What is the probability of a last layer skip on a 3x3 given that the cube is in a state that can be solved 2 generator <RU> and has the F2L completed?

How does this compare with a standard LL skip after standard F2L?

<RU> gen means that edge orientation is solved and corner permutation is solved.
P(OLL skip) = 1/27
P(PLL skip) = 1/12
P(LL skip) = 1/324

For a typical 3x3 solve, P(LL skip) = 1/15552, so it is 48 times more likely with <RU> gen.

Where did my numbers come from?
<RU> OLL: You can count them (4 each of S, As, L, Pi, U, T, 2 H, and 1 solved) or you can calculate them (3^4/3 [each corner has 3 orientations, but divide by 3 because only 1/3 of these is possible])

<RU> PLL: You can count them again (4 Ua, 4 Ub, 2 Z, 1H, 1 solved) or you can calculate them (4!/2 [4 positions for first edge, 3 for second, 2 for 3rd, and 1 for last edge, but divide by 2 because only 1/2 of all PLL are possible])

3x3 LL: P(OLL skip) = 1/216 and P(PLL skip) = 1/72. Multiply for your answer.

 

[F perm]

What is the chance of two N's in a row then a Z, then a H, then an E, in that order. It's all the unlikely PLL's. (This just happened to me!)

 

[ben1996123]

What is the chance of two N's in a row then a Z, then a H, then an E, in that order. It's all the unlikely PLL's. (This just happened to me!)

1/483729408, but it's not really that special since there are 4084101 different combinations of PLLs you can get in 5 solves, so its only really ~118 times rarer than most other combinations.

 

[Bob]

1/483729408, but it's not really that special since there are 4084101 different combinations of PLLs you can get in 5 solves, so its only really ~118 times rarer than most other combinations.

I think you made a mistake.

P(N) = 1/36
P(Z) = 1/36
P(H) = 1/72
P(E) = 1/36

P(N-N-Z-H-E) = (1/36)^4 * 1/72 = 1/120932352

The number of "different combinations" depends on your definition of what a combination is. Is Ua the same as Ub? Since the OP says only "N" instead of Na and Nb, I assume he groups them together. Therefore, there are only 13 "different" permutations (U, H, Z, A, E, J, T, R, F, G, V, N, and Y) so the number of combinations is 13^5 = 371,293. (I'm ignoring PLL skip as a case because I'm lazy.)

The original sequence of permutations is rarer than the expected combinations, but not 1/118. The PLLs that the OP got were N, Z, H, and E, which makes up 7/72 of all PLL cases. To only get one of these PLLs each time and not any of the others will occur on average (7/72)^5, which is about 1/115125. The most common combinations will involve one of the other PLLs, which account for 8/9 of PLL cases. "Most other combinations" will have only PLLs from that group, and the probability of getting 5 PLLs from that group is (8/9)^5 ~ 5/9, which makes the OP's combination about 63,886 times rarer than "most other combinations."

....I think.

 

[F perm]

I think you made a mistake.

P(N) = 1/36
P(Z) = 1/36
P(H) = 1/72
P(E) = 1/36

P(N-N-Z-H-E) = (1/36)^4 * 1/72 = 1/120932352

The number of "different combinations" depends on your definition of what a combination is. Is Ua the same as Ub? Since the OP says only "N" instead of Na and Nb, I assume he groups them together. Therefore, there are only 13 "different" permutations (U, H, Z, A, E, J, T, R, F, G, V, N, and Y) so the number of combinations is 13^5 = 371,293. (I'm ignoring PLL skip as a case because I'm lazy.)

The original sequence of permutations is rarer than the expected combinations, but not 1/118. The PLLs that the OP got were N, Z, H, and E, which makes up 7/72 of all PLL cases. To only get one of these PLLs each time and not any of the others will occur on average (7/72)^5, which is about 1/115125. The most common combinations will involve one of the other PLLs, which account for 8/9 of PLL cases. "Most other combinations" will have only PLLs from that group, and the probability of getting 5 PLLs from that group is (8/9)^5 ~ 5/9, which makes the OP's combination about 63,886 times rarer than "most other combinations."

....I think.

Yeah, this seems right. Super rare!
I think it was two Na's in a row.
The average sucked though

 

[nqwe]

I think you made a mistake.

P(N) = 1/36
P(Z) = 1/36
P(H) = 1/72
P(E) = 1/36

P(N-N-Z-H-E) = (1/36)^4 * 1/72 = 1/120932352

This is the chance of gettting these permutations, but not getting them in a specific order

 

[Bob]

This is the chance of gettting these permutations, but not getting them in a specific order

What would be the calculation for these permutations in the specified order, then?

 

[nqwe]

Sorry, I was wrong..
1/120932352 * 5! is the chance of getting these permutations in any order, which includes N-N-Z-H-E. I thought 1/120932352 would give you the chance of getting the permutations in any order.

 

[Sajwo]

I just get 6 R perm in a row (all PLLs was on right hand). What's the chance fot something like that?

 

[JasonK]

I just get 6 R perm in a row (all PLLs was on right hand). What's the chance fot something like that?

For it to happen in those specific six solves:
Chance of R(b)-perm = 1/18
(1/18)^6 = 1/34012224

But for it to happen in any random set of six consecutive solves, it's a lot higher.

 

[Bob]

For it to happen in those specific six solves:
Chance of R(b)-perm = 1/18
(1/18)^6 = 1/34012224

But for it to happen in any random set of six consecutive solves, it's a lot higher.

...and also more likely if it could have been any permutation 6 times.

 

[PoHos1]

4x4 lucky..

hey all .. I have mabye a big lucky because I solve rubiks cube 4x4 and first solve is with PLL skip and lucky second solve is with pll skip and wery lucky 3. solve is agein witch pll skip
so how its possibly ?
how is big % to get 3 times.. ?

 

[vd]

Probability of PLL skip on 3x3 is 1/72. On 4x4, there is parity possible, so probability is just 1/144. So getting 3 PLL skips in the row on 4x4 has probability of 1/144*144*144 = 1/2 985 984. Very unusual i guess.