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Probability Thread

cmhardw

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#5
I had a really interesting one come up a few days ago. After solving the cross and the first two pairs, all the remaining 12 pieces were in their correct locations, only 6 of them were misoriented.

The probability of having the last 12 pieces of the cube in the correct location after solving the cross and first two pairs is:

1 / [6! * 6!/2] = 1 / 259200

To give a comparison, the chance of a LL skip with no partial edge control, and with the possibility of AUF, is
1 / 15552

I thought it was pretty neat.

Chris
 

Lucas Garron

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#6
Before someone says it, it is a 1/1 chance of me being stupid, or a 1/1 chance of this thread being a failure.
Saying that doesn't make your thread more legitimate.

I changed the title, and let's see if we can salvage it:

If you keep doing random alignments and twists on a Square-1, you will hit the square-square shape 2/1839 of the time.
 
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#8
not cubing related, but interesting

if you buy a lottery ticket on a tuesday, you are more likely to die before the results are announced than you are to win on the wednesday
 
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#9
Before someone says it, it is a 1/1 chance of me being stupid, or a 1/1 chance of this thread being a failure.
Saying that doesn't make your thread more legitimate.

I changed the title, and let's see if we can salvage it:

If you keep doing random alignments and twists on a Square-1, you will hit the square-square shape 2/1839 of the time.
I just have to say this:

WOLFRAM ALPHA IS AMAZING

That is all.

not cubing related, but interesting

if you buy a lottery ticket on a tuesday, you are more likely to die before the results are announced than you are to win on the wednesday
That's actually quite funny. But how exactly were the odds of dying calculated?
 
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TheMachanga
#12
My friend (who is not very good at cubing) was solving his 4x4x4 and he got a 3x3x3 skip NO JOKE. He finished pairing up and after that, he had to do a D' and a U2 THAT'S IT.

This might be the luckiest solve you could have on a 4x4x4!
He scrambled it him self...most likely. It was most likely a crappy scramble.
 
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DaijoCube
#14
My friend (who is not very good at cubing) was solving his 4x4x4 and he got a 3x3x3 skip NO JOKE. He finished pairing up and after that, he had to do a D' and a U2 THAT'S IT.

This might be the luckiest solve you could have on a 4x4x4!
He scrambled it him self...most likely. It was most likely a crappy scramble.
After solving centers then edges, you most likely can't keep track of the scramble. Scramble was legit.
 
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#16
I want to know the chances of COLL cases and ZBLL cases. Cride5 has tried several times (maybe only once?) to explain this to me with complicated language. I might know what is going on now and I will try to explain.

So for OLL, the chance of Pi, Sune, Anti-Sune, T, U, and L is 1/54 while H is 1/108 and a skip occurs every 1/216 (these statistics are pretty well known). Just taking these cases for EO (setting 1/54 to be the standard):

6 * chance + 1 * .5 chance + 1 * .25 chance = 1

4/27 is the standard chance of occurance.

So the chance of OCLL occurrences after EO:
T: 4/27
U: 4/27
L: 4/27
H: 2/27
Pi: 4/27
Sune: 4/27
Anti-Sune: 4/27
Skip: 1/27

Now for COLLs. PLL probabilities are well known, so I can use those for my data.

Corners correctly placed: 1/18 + 1/18 + 1/36 + 1/72 (U,U,Z,H) = 11/72
Diagonal corner swap: 1/72 + 1/72 + 1/36 + 1/18 + 1/18 (N,N,E,V,Y) = 1/6

Now the others should be divided evenly amongst themselves, if my reasoning is correct (no reason you should trust this). Which means every other case should have a nice 49/432 chance of occurrence. What? Hmmm. Weird number. I guess I'll post it with PLL chances:

T
Correct: 11/486
Diagonal: 2/81
Left: 49/2916
Right: 49/2916
Top: 49/2916
Bottom: 49/2916

U
Correct: 11/486
Diagonal: 2/81
Left: 49/2916
Right: 49/2916
Top: 49/2916
Bottom: 49/2916

L
Correct: 11/486
Diagonal: 2/81
Left: 49/2916
Right: 49/2916
Top: 49/2916
Bottom: 49/2916

H
Correct: 11/972
Diagonal: 1/81
Left: 49/5832
Right: 49/5832
Top: 49/5832
Bottom: 49/5832

Pi
Correct: 11/486
Diagonal: 2/81
Left: 49/2916
Right: 49/2916
Top: 49/2916
Bottom: 49/2916

Sune
Correct: 11/486
Diagonal: 2/81
Left: 49/2916
Right: 49/2916
Top: 49/2916
Bottom: 49/2916

Anti-Sune
Correct: 11/486
Diagonal: 2/81
Left: 49/2916
Right: 49/2916
Top: 49/2916
Bottom: 49/2916

Skip
Aa: 1/486
Ab: 1/486
E: 1/972
F: 1/486
Ga: 1/486
Gb: 1/486
Gc: 1/486
Gd: 1/486
H: 1/1944
Ja: 1/486
Jb: 1/486
Na: 1/1944
Nb: 1/1944
Ra: 1/486
Rb: 1/486
T: 1/486
Ua: 1/486
Ub: 1/486
V: 1/486
Y: 1/486
Z: 1/972

I postulate that ZBLL has the same chance for every edge permutation. There are 12 possible edge permutations and COLL + EPLL generates each edge permutation equally often. Therefore the chance of each ZBLL case is that of the chance of the respective COLL case divided by 12. I don't want to have to post all of those probabilities.

One last thing: am I right?
 

kinch2002

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#20
What about a edges skip? (on 3x3)
This is pretty easy when you think about it for a moment. You need the first edge in place - that's 1/12. Second edge must now go in place - that's 1/11. Etc...so you have 1/12*1/11*1/10...*1/2=1/12!. Then they all have to be orientated correctly (1/2 for each) but the last edge orientation is defined by the other 11, so you get (1/2)^11 for orientation

Final answer: 1/(12!*2^11) = 9.81*10^11

EDIT: I hope I got this right...I shouldn't be doing maths at uni otherwise...
 
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