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Probability Thread

Julian

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I like this. Difficult, but a rough maximum to start the thinking: Assume a 20 move "naive" scramble. Probability that moves 11-20 are the inverse of 1-10 = 1 in 18^10 = 1 in 3.57 Trillion

The actual probability is likely way higher than this, as there's so many possibilities, e.g. where the scramble has junk like R L R' L' in it. I'm guessing by a factor of a hundred.

So, it *could* happen, given millions of cubers doing hundreds of thousands of solves in a lifetime. And I want to see the dude's face when it does. He'd be all like "Holy crap!"
This is wrong because scambles aren't just random strings of letters. First, a random combination is selected, then a scramble is generated that leads to that combination.
So the probability is one in ~43 quintillion.
 

Julian

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Sorry for the double post, but I'm writing this from my iPod at school, and when I tried to edit my last post, It wouldn't let me scroll though it to get to the end.
anyway...
What the probability of a relative cross solved on any colour?
How many combinatons are there when scambling using <R,U>?
After solving a 2x2x3 block in Petrus, how many combinations are there left?
 

Kynit

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After solving a 2x2x3 block in Petrus, how many combinations are there left?

6 corners to permute (6!) * 3 orientations for all but one corner (3^5) * 7 edges to permute (7!) * 2 orientations for all but one edge (2^6) / Inability to get PLL parity (2)

=~ 28 billion - does this sound right?
 

Johannes91

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Is there any possibility that with a random scramble the cube will solve itself? how about being 1 move away? or a u2?
Yes.

To get a probability, you need to define "random scramble". For random state the answer is obvious. But for a random move sequence of 20 or more moves the exact answer seems practically impossible to determine (feel free to prove me wrong!).

What the probability of a relative cross solved on any colour?
I'm not 100% sure what you mean by relative, but see http://www.cubezone.be/crossstudy.html.

How many combinatons are there when scambling using <R,U>?
6*5*4 * 3^5 * 7! / 2 = 73,483,200

After solving a 2x2x3 block in Petrus, how many combinations are there left?
6! * 3^5 * 7! * 2^6 / 2 = 28,217,548,800
 

cuBerBruce

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After the OLL is finished, what is the probability that the PLL is in the right place (no U, U', U2, y, y', y2 needed to apply the alg) ??

1/4: there are 4 ways to AUF, and 1 is right.

I think the question perhaps needs a little clarification.

Generally a PLL has 4 angle cases and 4 AUF cases. For example, for the A-Perms, the algs I use require same color corner stickers facing me. If not, I have to rotate the cube, or at least the U layer so that same color corner stickers are facing me. I now have the correct angle case. After executing the A-Perm alg, I still might not have the U layer aligned properly with the bottom two layers. This can be handled by an AUF after the algorithm or by rotating the bottom two layers before the alg. If the Schmidt only cares about if the angle case is correct, the answer would be a little larger than 1/4. This is because the N-Perms and H-Perm only have 1 angle case, and Z -Perm and E-Perm have 2 angle cases. I view PLL skip as having 1 angle case and 4 AUF cases.

If we assume there is no attempt to influence the PLL case, and we only care about having the U layer correct for applying the alg, and don't care about if there will need to be an AUF or not, then the probabilty I get is 11/36 (or 85/288 if you want to consider AUF adjustments on PLL skips to matter).

If the question is for having to do no adjustment for either the angle case or for the AUF case, the probability becomes (11/36)*(1/4) = 11/144.

Of course, there are a number of assumptions here, such as you're not trying to influence the PLL case and you only use one particular alg for each PLL case (always applied from the same angle) regardless of which angle or top/bottom alignment situation you might get.
 

Marcell

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I'm not sure I get this - sorry if it's just me. Anyway: By stickers do you mean edge + corner stickers as well? Why is x between 0 and 10? What do you mean ignoring twisted corners - we leave them there twisted? And what about flipped edges?
 

riffz

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I'm not sure I get this - sorry if it's just me. Anyway: By stickers do you mean edge + corner stickers as well? Why is x between 0 and 10? What do you mean ignoring twisted corners - we leave them there twisted? And what about flipped edges?

I mean only corners. And yes, completely ignoring twisted corners as if they are solved.

x is 10 or less because that is the maximum number of stickers you would have to shoot to in order to solve the corners. It's intuitively pretty clear. Assuming you start with the buffer piece solved, There can at most be 3 cycles you have to break into. Two of these will be of length 2 and the other of length 3, and you will have to shoot to 3 stickers to solve a 2-cycle and 4 to solve a 3-cycle. 3+3+4=10
 

cuBerBruce

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What is the probability in a BLD solve that, ignoring twisted corners in place and using a fixed buffer, you will have to shoot to x stickers? (0 <= x <= 10)

I mean only corners. And yes, completely ignoring twisted corners as if they are solved.

x is 10 or less because that is the maximum number of stickers you would have to shoot to in order to solve the corners. It's intuitively pretty clear. Assuming you start with the buffer piece solved, There can at most be 3 cycles you have to break into. Two of these will be of length 2 and the other of length 3, and you will have to shoot to 3 stickers to solve a 2-cycle and 4 to solve a 3-cycle. 3+3+4=10

What about four 2-cycles?
And does it matter if the buffer is not in a cycle?

Anyway, I used GAP to get the distribution of cycle structures for permutations of 8 objects.
Code:
Even Permutations
{ 7 }		5760
{ 6, 2 }	3360
{ 5 }		1344
{ 5, 3 }	2688
{ 4, 4 }	1260
{ 4, 2 }	2520
{ 3, 3 }	1120
{ 3, 2, 2 }	1680
{ 3 }		 112
{ 2, 2, 2, 2 }	 105
{ 2, 2 }	 210
{  }		   1

Odd Permutations
{ 8 }		5040
{ 6 }		3360
{ 5, 2 }	4032
{ 4, 3 }	3360
{ 4, 2, 2 }	1260
{ 4 }		 420
{ 3, 3, 2 }	1120
{ 3, 2 }	1120
{ 2, 2, 2 }	 420
{ 2 }		  28
 
Last edited:

toastman

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Awesome stats! Thanks dude.

So, if my maths are correct, 16065/40320 or 39.8% of corner solves do *not* require breaking into a new cycle. The rest do. I hate when that happens. I have to go back and count all the corners by putting my fingers on them.

if my maths are correct, according to his signature, Amostay2008 "pwns" on 3360/40320 or 8.3% of corner solves.
 

cuBerBruce

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So, if my maths are correct, 16065/40320 or 39.8% of corner solves do *not* require breaking into a new cycle.

To get 16065, I see that you added up all the cases where there is at most one cycle. I note that if you use a fixed buffer, that buffer may not always be part of a cycle. For example, take the 7-cycle case. There are 5760 such permutations. 7/8 of them or 5040 will have the buffer in the cycle, and in the remaining 720 cases, the buffer piece is in its correct place (not part of the cycle). If you consider it to be "breaking into a new cycle" in this case when the buffer is not in a cycle to start with, but you still have a cycle to solve, then you have overcounted the number of cases of "not breaking into a new cycle."

Under these assumptions, the number of cases I calculate reduces to 13700 cases, or just a little under 34%.
 
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