Probability Thread

[ThomasJE]

What is the probability of one FACE being solved after a random scramble on a 2x2?

Don't know, but I've had a 2x2 scramble that had 2 opposite faces already solved (that left just PBL to do). I'll try to find that scramble.

 

[cuBerBruce]

What is the probability of one FACE being solved after a random scramble on a 2x2?

For at least one face being a solid color, I get 22654 chances in 3674160. That's approximately 1 chance in 162.

For exactly n faces being a solid color, I get the following distribution:

faces     count
  0       3651506
  1         22020
  2           597
  3            36
  4             0
  5             0
  6             1

 

[SweetSolver]

Very interesting. Thanks for that

 

[ncube]

What is the probability of the corners being permuted after Winter Variation?

 

[ben1996123]

What percentage of 2x2 states can be solved in 6 moves or less?

 

[mande]

Probability of at least 1 CE pair formed in a scramble?

 

[RNewms27]

What percentage of 2x2 states can be solved in 6 moves or less?

62360/3674160= 0.016972587~ 1.7%

 

[Rpotts]

What is the probability of the corners being permuted after Winter Variation?

1/6

 

[CubeRoots]

What is the probability of LL skip given last layer edges are oriented and two opposite edges are correctly permuted. (i.e. after zz-b phasing)

 

[nickvu2]

What is the probability of LL skip given last layer edges are oriented and two opposite edges are correctly permuted. (i.e. after zz-b phasing)

Based on this several posts back:

Probability that corners are correctly solved after f2l?
1/162, allowing AUF (1/648 if not allowing AUF)

So 1/648 if edges are solved, 1/1296 if edges are phased?

BTW, I've done 3,000+ solves with ZZ-b and I think I got 1 LL skip.

 

[JonWhite]

What is the probability that you answer this question correctly?
A) 25%
B) 50%
C) 25%
D) 33.3%

 

[Carrot]

What is the probability that you answer this question correctly?
A) 25%
B) 50%
C) 25%
D) 33.3%

0/4 ~ 0%

 

[IanTheCuber]

0/4 ~ 0%

Odder's sarcasm is awesome.

What is the propability that at Nationals, given there are about 200 people, that at least one would get a(n):
LL Skip?
PLL Skip?
OLL Skip?
Let's just say they all use Fridrich. I highly doubt it would happen, but it's a legitemate question. Right? RIGHT?

 

[Rune]

Odder's sarcasm is awesome.

What is the propability that at Nationals, given there are about 200 people, that at least one would get a(n):
LL Skip?
PLL Skip?
OLL Skip?
Let's just say they all use Fridrich. I highly doubt it would happen, but it's a legitemate question. Right? RIGHT?

Here "odds" is more proper than "probability".

 

[CubeRoots]

You can't answer that question. And thanks nickvu2. is that your main method?

 

[Julian]

You can't answer that question. And thanks nickvu2. is that your main method?

I think you can. You'd have to know what everybody is registered for, whether or not there are cutoff times (in which case you can't calculate it). You'd also have to assume that everyone will finish their solves, and that they're all using straight CFOP with no influencing techniques.

 

[Bob]

I think you can. You'd have to know what everybody is registered for, whether or not there are cutoff times (in which case you can't calculate it). You'd also have to assume that everyone will finish their solves, and that they're all using straight CFOP with no influencing techniques.

You'd also have to consider how many people will have the same solution. There are too many variables here.

 

[Julian]

You'd also have to consider how many people will have the same solution. There are too many variables here.

Oh, yeah, I didn't think about that.

 

[nickvu2]

You can't answer that question. And thanks nickvu2. is that your main method?

Yes, ZZ-b in my main method =)

 

[ThomasJE]

Odder's sarcasm is awesome.

What is the propability that at Nationals, given there are about 200 people, that at least one would get a(n):
LL Skip?
PLL Skip?
OLL Skip?
Let's just say they all use Fridrich. I highly doubt it would happen, but it's a legitemate question. Right? RIGHT?

1/72 for a PLL skip normally. So 200 x 1/72 = 2.777...
Same maths can be applied to OLL and LL skips.