# Not a one in 12 chance to assemble the cube correctly?



## SenileGenXer (Sep 22, 2014)

We know there are 12 orbits of the cube. It is assumed that the probability of randomly assembling the cube and having a solvable puzzle is 1 in 12. 

Have we checked that? 

I was thinking of all the ways it's possible to misassemble the cube.

1) Clockwise corner twist (1 in 3 chance)
2) Counter-Clockwise corner twist (1 in 3 chance)
3) Edge flipped (1 in 2 chance)
4) Two edges swapped (1 in ? chance)
5) *Two corners swapped* (1 in ? chance)

The last part is a mistake you could make assembling the cube. Now because of 4-cycle moves like the T-Perm and F-Perm two corners swapped is indistinguishable from two edges. They will show up as the same PLL parity error. 

If you make mistake #4 *and* mistake #5 randomly assembling the cube they cancel each other out - there is no observable mistake. A cube assembled with two edges swapped and two corners swapped is solvable with a F-Perm/T-perm.

I was thinking that instead of 12 orbits there were 14 orbits but two pairs of orbits were conjoined. That is the solvable orbit is conjoined with the two edges and two corners swapped orbit. The two edges swapped orbit is conjoined with the two corners swapped orbit. I hypothesized that this produces a 2/14 chance of randomly assembling a solvable cube. 1/7 when simplified.

To test it I disassembled a cube into a brown paper paper bag, shook up the bag, reassembled the cube randomly and tried to see if it was solvable. Kept track of 50 random assemblies. If my theory was correct I should get 7.1 solvable cubes. If the conventional theory was correct I should get 4.1 solvable assemblies.

I got six.

Data available here.


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## Carrot (Sep 22, 2014)

for 4) and 5) combined you have these scenarios:
even corners + even edges = solvable
odd corners + even edges = unsolvable
even corners + odd edges = unsolvable
odd corner + odd edges = solvable
Each of above case has the same probability, hence 2/4 chance of an unsolvable permutation of the pieces.

also, number 1) and 2) are wrong.
there's 1/3 chance of getting corner orientation correct, all 2 corner twists can be made into 1 corner twists, therefore 2/3 for 1) and 3/3 for 2).

E: Also, your dataset only has 56% corner fails, and you KNOW it should be like 67%, therefore you are kind of bound to get more solvable cubes than expected. All in all, just no, it is 1/12.


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## SenileGenXer (Sep 22, 2014)

1/3 chance of getting a clockwise corner twist. 1/3 chance of getting a counter clockwise corner twist. 1/3 chance of getting it correct. More along the lines of what I was thinking. Post updated.

Even and odd edge/corners sounds right but are you sure about their probability?


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## Carrot (Sep 22, 2014)

SenileGenXer said:


> 1/3 chance of getting a clockwise corner twist. 1/3 chance of getting a counter clockwise corner twist. 1/3 chance of getting it correct. More along the lines of what I was thinking. Post updated.
> 
> Even and odd edge/corners sounds right but are you sure about their probability?



Yes I am.


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## kinch2002 (Sep 22, 2014)

In simplified terms,
1/3 chance the corner orientation is solvable
1/2 chance the edge orientation is solvable
1/2 chance the permutation is solvable
= 1/12


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## Hypocrism (Sep 22, 2014)

One of the permutation sets can be entirely ignored (because an odd permutation of edges or corners is not by itself impossible - see T perm). The other has a 1/2 chance of being solvable (given the edge permutation mod 2, the corner permutation must be the same mod 2).

Just like we don't count the orientation of all the corners when calculating 1/3 chance of corner orientation being solvable - we only need to take into account the "8th" corner because given the position of the other 7, it is the 8th which determines the cube's solubility.


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## Stefan (Sep 22, 2014)

SenileGenXer said:


> If you make mistake #4 *and* mistake #5 randomly assembling the cube they cancel each other out - there is no observable mistake.



There's not just no *observable* mistake, there's *no mistake*. You can't make mistake #4 *and* mistake #5.

I get the feeling this thread is a hoax, especially with the 50 assemblies experiment. Way too much work for a way too small sample size. Hard to believe you actually did that


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## AvGalen (Sep 22, 2014)

I like how he checks theory with reality, but "his math" of "12 becomes 14, so 1/12 becomes 2/14" makes no sense to me.
also....that samplesize.
Has anyone ever done a computer simulation where all pieces are put randomly and then checked for solvability? That I would expect a few million states to be checked in a second (and the 1/12 to be clearly showing)


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## TMOY (Sep 22, 2014)

AvGalen said:


> I like how he checks theory with reality, but "his math" of "12 becomes 14, so 1/12 becomes 2/14" makes no sense to me.


The flaw in his reasoning is pretty obvious: he's assuming his 14 differentevents all have the same probability of happening, which unfortunately is not the case.


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## qqwref (Sep 22, 2014)

SenileGenXer said:


> I was thinking that instead of 12 orbits there were 14 orbits


wat



SenileGenXer said:


> If my theory was correct I should get 7.1 solvable cubes. If the conventional theory was correct I should get 4.1 solvable assemblies.
> 
> I got six.


Pretty sure that's not enough data to reject either null hypothesis.


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## SenileGenXer (Sep 22, 2014)

AvGalen said:


> I like how he checks theory with reality, but "his math" of "12 becomes 14, so 1/12 becomes 2/14" makes no sense to me.
> also....that samplesize.
> Has anyone ever done a computer simulation where all pieces are put randomly and then checked for solvability? That I would expect a few million states to be checked in a second (and the 1/12 to be clearly showing)



My thinking was looking at the orbitals. I don't know if anyone has named and numbered them but I think they work out like this:

1) You can assemble it correctly.
2) you can assemble it with a corner twist (+1)
3) you can assemble it with a CC corner twist (or -1)
4) You can assemble it with an edge flip
5) Edge flip plus corner twist
6) Edge flip plus CC corner twist
7) You can assemble it with a PLL error
8) PLL error and a corner twist
9) PLL and a CC corner twist
10) You can assemble it with a PLL error and and edge flip
11) PLL error plus edge flip plus corner twist
12) PLL error plus edge flip plus CC corner twist.

These are the orbitals generated by the 1/3 * 1/2 *1/2 math. the 12 states that exist. That's the result but that's not how a cube is assembled. I was thinking instead of PLL error there were a differentiation between corner PLL error and edge PLL error as you assemble it and before it's solved. A proto-orbital with both edge and corner parity was conjoined to the solved orbital - 2 chances to assemble it correctly instead of only 1/12 results.

In the cold light of morning I can't figure out where the 14 came from. If I make a list of proto-orbitals there are 24. 2/24 = 1/12


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## Christopher Mowla (Dec 18, 2014)

en.wikipedia.org/wiki/Rubik%27s_Cube said:


> 519 quintillion[SUP][28][/SUP] possible arrangements of the pieces that make up the Cube, but only one in twelve of these are actually solvable. This is because there is no sequence of moves that will swap a single pair of pieces or rotate a single corner or edge cube. Thus there are twelve possible sets of reachable configurations, sometimes called "universes" or "orbits", into which the Cube can be placed by dismantling and reassembling it.


Clearly only one third of corner orientation _positions_ are reachable, but if we twist a single corner we can reach all other 3^8 - 3^7 possible corner orientation positions with legal moves.

Therefore, shouldn't there just be (2)(2)(2) = 8 "cube universes instead of (2)(2)(3) = 12 "cube universes"? It seems there has been a confusion of the orientation _operations_ with the orientation _positions._


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## Stefan (Dec 18, 2014)

cmowla said:


> if we twist a single corner we can reach all other 3^8 - 3^7 possible corner orientation positions with legal moves.



If we twist a single corner *one way* we can reach *other 3^7* possible corner orientation positions with legal moves.

Instead of the cube, imagine the integers, and instead of turning, imagine adding or subtracting 3. Then you have *three* "universes":
{..., -6, -3, 0, 3, 6, ...}
{..., -5, -2, 1, 4, 7, ...}
{..., -4, -1, 2, 5, 8, ...}
If you're in one "universe" (i.e., at one number in that "universe"), you can make moves (add or subtract 3) as much as you want, you stay within that "universe". Just like the three corner orientation "universes".


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## Christopher Mowla (Dec 18, 2014)

Ah, thanks Stefan. For some odd reason, I never tested this thoroughly to see why it was true.


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