Probability Thread

[guysensei1]

Just now, I was doing a solve, and after I finished 3 F2L pairs, I ended up with a sort of sideways J perm on the R face.

Whats the probability of that?

 

[cmhardw]

What's the probability of getting at least 1 LL skip in 15522 solves?

Getting at least 1 success in n trials with a probability of 1/n can be closely approximated by \( 1-\frac{1}{e} \) for large n.

 

[Sajwo]

What is the propability of getting 7 moves solution on skewb? FL 3 and LL 4?

Or getting pure U/H/Z perm?

 

[TDM]

How many solves would you have to do before the probability of having had at least one LL skip is more than 50%?

 

[cmhardw]

How many solves would you have to do before the probability of having had at least one LL skip is more than 50%?

For \( n \in \mathbb{N} \)

\( 1-\left(\frac{15551}{15552}\right)^n > \frac{1}{2} \)

Since n is a natural number this solution can be written:

\( n \geq 10780 \)

----------edit----------

How many solves would you have to do to get a 50% chance of seeing every last layer case? With AUF? W/O AUF? 90%?

I am really interested in this question. I haven't had time to really sit and think about this yet, but I really want to work on this question!

 

[Logiqx]

I got my first cross skip at the weekend using Prisma Puzzle Timer... no moves, no alignment required.

I've probably done about 5,000 timed solves and at most 10,000 solves in total.

I'm aware of the cross study but I decided to have a go at the maths myself. Have I got this right?

Probabilty of white cross = 1/190080
1/(24*22*20*18)

Probability of white / yellow cross ~ 1/95042
1/(24*22*20*18) + 1/(24*22*20*18) - 1/(24*22*20*18*16*14*12*10)
p(white cross) + p(yellow cross) - p(white cross + yellow cross)

Probability of any colour cross ~ 1/31704
I'm lazy and just looked at http://www.cubezone.be/crossstudy.html

In comparison to LL skip:
LL skip without needing AUF = 1/62208
LL skip needing AUF = 1/15552 - I've had one of these in an untimed solve

It's moderately interesting that they are in the same ball-park.

 

[ThomasJE]

How many solves would you have to do before the probability of having had at least one LL skip is more than 50%?

(15551/15552)^n <= 0.5 because that is the probabity of no LL skips.

I would use trial and error; but I can't right now. There's probably a quicker way though.

EDIT: Just worked it out on paper and with my phone calculator. 10780 solves are needed to have a 50% chance of at least 1 LL skip.

 

[uberCuber]

(15551/15552)^n <= 0.5 because that is the probabity of no LL skips.

I would use trial and error; but I can't right now. There's probably a quicker way though.

Chris Hardwick already posted the answer above, but, Logarithms:

\( \ln { \left( \frac{15551}{15552} \right)^n } < \ln{0.5} \)

\( n \ln {\frac{15551}{15552}} < \ln{0.5} \)

\( n \geq 10780 \)

 

[ThomasJE]

Chris Hardwick already posted the answer above, but, Logarithms:

\( \ln { \left( \frac{15551}{15552} \right)^n } < \ln{0.5} \)

\( n \ln {\frac{15551}{15552}} < \ln{0.5} \)

\( n \geq 10780 \)

Tapatalk sees three smilies. I can't see the fractions at all.

Did I get it right?

 

[uberCuber]

Tapatalk sees three smilies. I can't see the fractions at all.

Did I get it right?

Lol, yes, 10780 is the right number

 

[Stefan]

1/(24*22*20*18) + 1/(24*22*20*18) - 1/(24*22*20*18*16*14*12*10)
p(white cross) + p(yellow cross) - p(white cross + yellow cross)

This is not quite right.

 

[Logiqx]

This is not quite right.

Are you referring to the way I wrote it (somewhat lazy as it should really be 1/24 x 1/22 x 1/20, etc) or the final answer?

My calculation comes to the same result as the cross study, which if I understand correctly is 53759 / 5109350400.

My reasoning is that p(white cross) and p(yellow cross) intersect so I used the subtraction as a way of avoiding double-counting.

 

[Stefan]

Sorry, I made thinking/coding mistakes. It indeed gives the correct result, although I'll have to think about why it does. How would you do "any colour cross"?

 

[Methuselah96]

Sorry, I made thinking/coding mistakes. It indeed gives the correct result, although I'll have to think about why it does. How would you do "any colour cross"?

He's just combining permutation and orientation of cross edges by distributing the 2 from a calculation that normally has them separate like: 2^4 x (12 x 11 x 10 x 9).
So then he would probably do a neutral cross like: 24 x 22 x 20 x 18 x 16 x 14 x 12 x 10 x 8 x 6 x 4 x 1 (not multiplying the last number by two because the orientation is already determined) derived from (2 orientation x 12 permutations) (2 orientation x 11 permutations)...(2 orientation x 2 permutation)(1 orientation x 1 permutation).

I made mistakes.

Is this a first?

 

[AFatTick]

This has most of the probabilities of pll and oll cases. http://badmephisto.com/badmephisto-speedcubing-method.pdf

 

[Logiqx]

How would you do "any colour cross"?

My approach would be to figure out the probability of all six crosses, then >=3 crosses (includes 6 cross probability), then >=2 crosses (includes 3 cross probabilities), then >= 1 cross (includes 2 cross probabilities).

I just had a quick go at it and my first result was slightly off... 1/31703.96091 instead of 1/31703.94349.

I can get the correct answer with a kludge in my calculation (re: chance of getting >= 2 or 3 crosses). The kludge uses the probability of six crosses but more times than I'd expected. I haven't got my head around why the kludge works yet. I might think about it again at lunch time...

I'd better do some actual work, tempting as it is to solve this little brain teaser!

 

[Logiqx]

Please forgive the strange notation. I've treated orientation and permutation as a single property.

I realise it could be written using factorials and distinguishing between orientation + permutation.

Six crosses - 1 case (11 edges correct, meaning the twelth is also correct)
1 / (24*22*20*18*16*14*12*10*8*6*4)

Five or Four crosses are not possible.

Three crosses - 20 cases
Two opposite crosses (10 edges, 12 cases) = 12 / (24*22*20*18*16*14*12*10*8*6)
+ Three adjacent crosses (9 edges, 8 cases) = 8 / (24*22*20*18*16*14*12*10*8)

Two crosses - 15 cases
Opposite crosses (8 edges, 3 cases) = 3 / (24*22*20*18*16*14*12*10)
+ Adjacent crosses (7 edges, 12 cases) = 12 / (24*22*20*18*16*14*12)

One cross - six cases
6 / (24*22*20*18)

Combining everything...
p(one cross)-(p(two crosses)-(p(three crosses)-10*p(six crosses))) = 1/31703.94349

This is the correct answer but it includes a kludge. I don't understand what the multiplication by 10 is doing!

I think it accounts for intersections of some of the three cross cases.

 

[RageCuber]

I just got two PLL skips in one avg12, no edge control or anything like that.
whats the probability?

 

[Kit Clement]

I just got two PLL skips in one avg12, no edge control or anything like that.
whats the probability?

Probability of PLL skip with no forcing skips - 1/72. Let X be the random variable for the number of PLL skips. This follows a Binomial(12, 1/72) distribution.

\( P(X = 2) = \binom{12}{2}(1/72)^2(71/72)^{10} = 0.011 \)

But you're probably more interested in cases of 2 PLL skips or more, as this probability does not include this.

\( P(X \geq 2) = 1 - P(X = 1) - P(X = 0) \)
\( = 1 - \binom{12}{1}(1/72)^1(71/72)^{11} - \binom{12}{0}(1/72)^0(71/72)^{12} = 0.0116 \)

 

[Christopher Mowla]

I just got two PLL skips in one avg12, no edge control or anything like that.
whats the probability?

I'm guessing that you wanted to know the probability for your specific event. If you just wanted to know the probability of getting 2 PLL skips in 12 solves in general, then Kit Clement's answer, \( \left( \begin{matrix}
12 \\
2 \\
\end{matrix} \right)\left( \frac{1}{72} \right)^{2}\left( \frac{71}{72} \right)^{10} \), is correct.

However, the probability of getting 2 PLL skips one after the other in 12 solves is different than if your first solve and 12th solve are PLL skips, for example.

So in general, for 12 solves, 2 of which are PLL skips,

The probability when there are x non-PLL skips in between the 2 PLL skip solves = \( \left( 12-\left( x+1 \right) \right)\left( \frac{1}{72} \right)^{2}\left( \frac{71}{72} \right)^{10},\text{ 0}\le x\le 10. \) [Link]