Probability Thread

[mr. giggums]

I inspect for 1x2x1 block every scramble, and this just seems wrong to me...

(however I couldn't proove you wrong, or proove I am right for that matter. Just my many inspections)

I inspect for it too and it feels like two third. I'll do a test by doing random scrambles and get the average

1 = there is a 1x1x2, 0 = there isn't

1101101111100101001111011010111010011001101111111110111100111011011010111111110110000111001110000110

There was 100 scrambles 65 of them had at least one 1x1x2 pair. So that is 65% which is really close to my estimated 2/3. This isn't entirely accurate but it is close.

 

[Johannes91]

It's all joey's fault, he said that 15% seems reasonable. I suspected it might be wrong, I probably misremember what the code does.

Edit: Yepp, it only looks for ce-pairs that are connected _and_ solved with respect to centers.

 

[Robert-Y]

I tried 100 scrambles myself using http://www.ryanheise.com/cube/speed.html

and I got 68/100 with at least one 1x1x2 pair. Which of course, is very close to Mr Giggums' estimation of 2/3 too.

 

[cubersmith]

Chance of totally disassembling the cube and re-assembling it in a solvable position?

 

[cubernya]

That's well known. 1/12

1/2 for EO
1/3 for CO
1/2 for Permutation

 

[cubersmith]

That's well known. 1/12

1/2 for EO
1/3 for CO
1/2 for Permutation

Cool, can we use this idea to answer any question like that?
So if I said, if 5 corners and 4 edges are taken out, whats the chance of it being solvable?

 

[cubernya]

Assuming you still assemble it randomly (like blindfolded ) 1/12...as long as you take out 2 corners and 1 edge and assemble randomly it's 1/12

 

[cubersmith]

Assuming you still assemble it randomly (like blindfolded ) 1/12...as long as you take out 2 corners and 1 edge and assemble randomly it's 1/12

Haha thanks. I'm learning about probability. Could you tell me what ^ and * mean so I can understand this thread better?

 

[cubernya]

* is multiplication
^ is exponents

Example 5*2=10
5^2=25

 

[cubersmith]

* is multiplication
^ is exponents

Example 5*2=10
5^2=25

Explain the second one?

 

[cubernya]

You didn't learn that?

5^2 = 5*5
5^3 = 5*5*5
5^4 = 5*5*5*5

 

[cubersmith]

Oh, okay thanks. Yeah I did we just dont call it that in Scotland

 

[Chrisalead]

Probability of a OLL + PBL skip on the 2x2 ? (It happened to me solving with Ortega last week).

 

[kinch2002]

Probability of a OLL + PBL skip on the 2x2 ? (It happened to me solving with Ortega last week).

If you knew that the first layer permutation would be solved then you can't really call it a PBL skip. In which case, it's just a LL skip which is 1/162

 

[Godmil]

Oh, okay thanks. Yeah I did we just dont call it that in Scotland

*Phew, I thought it was just me who didn't recognise that name (despite knowing what it was).

 

[cubernya]

You probably were thinking of power?

 

[Chrisalead]

If you knew that the first layer permutation would be solved then you can't really call it a PBL skip. In which case, it's just a LL skip which is 1/162

Doh ! I should have noticed that ! Thanks ^^. Fun that the probability for a 2x2 LL skip is so high compared to a 3x3 LL skip ! 96 times more chances in fact !

 

[michaelfivez]

If you knew that the first layer permutation would be solved then you can't really call it a PBL skip. In which case, it's just a LL skip which is 1/162

With Ortega you don't know the permutation of the first layer? Or am I missing something?
A OLL+PBL skip is 1/162 * 1/6 then?

 

[MostEd]

i keep getting PLL skips on 7x7... while none on smaller size

 

[kinch2002]

With Ortega you don't know the permutation of the first layer? Or am I missing something?
A OLL+PBL skip is 1/162 * 1/6 then?

Surely you can plan the whole first face, and therefore would be able to 'predict' what the permutation of it will be?