Probability Thread

[Kit Clement]

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.

Is that not what he wanted though? He never specified that the two PLL skips came in a row.

 

[Christopher Mowla]

Is that not what he wanted though? He never specified that the two PLL skips came in a row.

It's kind of hard to tell exactly. Certainly what you gave him is what he asked for, but if I just got 2 PLL skips in a average of 12 and wanted to know the probability of that happening, I would want to know the exact probability of that particular situation (because otherwise to me it would be inaccurate). Therefore I just added on to what you did to make sure our friend's curiosity has been met.

 

[Stefan]

Got more time now...

Sorry, I made thinking/coding mistakes.

... meaning I thought the white and yellow crosses aren't independent, and not just because of the cases where both of them are solved, but also where one being unsolved also influences the other (like white center not being solved because a white edge is in the yellow layer, immediately ruling out the yellow center as well). But I was over-thinking it. Subtracting P(white and yellow) does handle it properly. And my coding mistake is funny, I used Python's Decimal class with many digits precision, but then used it too late in the process, after floats had already caused rounding errors. Should have just used WolframAlpha.

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

That's the easy part, the interesting part is counting, discounting and recounting the crosses properly, as seen in his new posts.

Is this a first?

Of course!

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

I confirm, though prefer integers and case counting instead of fractions and probabilities.
http://ideone.com/hetM2d

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

Well, we want to count each case with any solved cross exactly once. From my code/perspective:

c1 = number of cases with at least one solved cross
c2 = number of cases with at least two solved crosses
c3 = number of cases with at least three solved crosses
c6 = number of cases with at least six solved crosses

total = c1 - c2 + c3 - 10*c6

• The cases with exactly one solved cross are counted by using c1.
• The cases with exactly two solved crosses are counted twice by c1, so we have to discount them once (i.e. subtract c2).
• The cases with exactly three solved crosses are counted thrice by c1 and discounted thrice by c2, so we have to count them once again (i.e. add c3).
• The cases with exactly six solved crosses (*) are counted six times by c1, discounted 6c2=15 times by c2, and recounted 6c3=20 times by c3, so we have counted them 6-15+20=11 times and thus have to discount them 10 times (i.e. subtract 10*c6).
  (*) yes, there's just one such case

 

[Logiqx]

Thanks for sparing your time Stefan. It's nice to have a final resolution.

I was over-thinking it.

I started along those lines then decided there must be a simpler approach, easily visualised with a two-circle venn diagram. It just got a little harder to visualise with more than three circles!

That's the easy part, the interesting part is counting, discounting and recounting the crosses properly, as seen in his new posts.

I hope you enjoyed it. I haven't found it described elsewhere on this forum.

Of course!

My greatest cubing achievement, lol.

Well, we want to count each case with any solved cross exactly once. From my code/perspective

Excellent. Thanks for that explanation and I can now see my oversight.

I couldn't understand why I could get so close to the correct answer without using x*c2 and y*c3 and z*c6 , just 10*c6.

I'd overlooked how counting and discounting might end up with x=1, y=1 and z=10.

 

[Stefan]

I hope you enjoyed it.

I did, good stuff. Also thanks for your calculations, I might not have tried it without. And the mysterious 10 intrigued me, I didn't understand it at first, either.

 

[cubemaste r]

How many possible permutations does a 3x3x3x3 have?

 

[Jokern]

How many possible permutations does a 3x3x3x3 have?

I was trying to figure this out bymyself, but after some googling turns out I would be wrong. But anyway:
Short answer: about 10^120 (a few more than a 6x6x6)
long answer: http://www.superliminal.com/cube/permutations.html

 

[Saxsoprano]

Anyone knows the chance of an LL skip using always WV with ZZ method? Only curiosity.

 

[uberCuber]

Anyone knows the chance of an LL skip using always WV with ZZ method? Only curiosity.

If you're using ZZ, EO is always solved, and if you're using WV, CO is always solved, so all that is left is PLL. An "LL skip" then would be equivalent to a PLL skip, which has a probability of 1/72

 

[Saxsoprano]

Ok, thanks!!

 

[JackJ]

So I celebrated my 18th birthday yesterday. My cousin also welcomed his first son yesterday. Additionally, another one of my cousins turned 10 yesterday. So are the odds of three specific people being born on the same day 1/(365^2) since the first person has a 100% chance of being born on random day of a 365 day calendar and the other people need two need to follow suit?

 

[Stefan]

So I celebrated my 18th birthday yesterday. My cousin also welcomed his first son yesterday. Additionally, another one of my cousins turned 10 yesterday. So are the odds of three specific people being born on the same day 1/(365^2) since the first person has a 100% chance of being born on random day of a 365 day calendar and the other people need two need to follow suit?

Apparently the days around September have a higher probability (at least in USA), so no. Also, don't forget February 29.

 

[JackJ]

Apparently the days around September have a higher probability (at least in USA), so no. Also, don't forget February 29.

I noticed this too. I think this happening may be more of a trend than a mathematical probability though.

Also, leap year shouldn't complicate the problem too much. Since my cousin was born in 2004, a leap year, the new equation should look something like this:

1/(365*366)

But hey, I don't know too much about math of this sort and wikipedia has an entire page dedicated to explaining just how complicated these types of problems can be.

 

[Stefan]

I think this happening may be more of a trend than a mathematical probability though.

What do you mean with "trend", and why wouldn't it be a mathematical probability?

This is one analysis I came across:
http://thedailyviz.com/2012/05/12/how-common-is-your-birthday/
http://thedailyviz.com/2012/05/18/how-common-is-your-birthday-pt-2/

Also, if you look at the source data at
https://docs.google.com/spreadsheet/ccc?key=0AjlIKRG8DtTqdDJ4S0FFSktma1dOWkhXZnRpV0tocmc#gid=0
then you'll see that births didn't just go up around September overall, but in every single one of those ten years.

 

[JackJ]

What do you mean with "trend", and why wouldn't it be a mathematical probability?

This is one analysis I came across:
http://thedailyviz.com/2012/05/12/how-common-is-your-birthday/
http://thedailyviz.com/2012/05/18/how-common-is-your-birthday-pt-2/

Also, if you look at the source data at
https://docs.google.com/spreadsheet/ccc?key=0AjlIKRG8DtTqdDJ4S0FFSktma1dOWkhXZnRpV0tocmc#gid=0
then you'll see that births didn't just go up around September overall, but in every single one of those ten years.

A quick Google search yields a good definition of trend: A general direction in which something is developing or changing.

I didn't say it wasn't a mathematical probability either, it just seemed like more of a trend to me. The graphs you presented me are, for the most part, uniform.

Assuming the date of conception between my parents and the other two couples was random, I think my approximation is pretty solid.

 

[Smiley]

I had a 2x2 scramble yesterday: U R' F R2 F R2 F R' U'
It is impossible to match up 2 colors in one move, how awesome is that???
anyway, it made me wonder what the chances are to start with no 2 colors matched up.

 

[Rune]

"I wouldn’t say that they hold it in over Christmas, but they force them out before New Years. Gotta get the tax deduction on the books".

 

[Stefan]

A quick Google search yields a good definition of trend: A general direction in which something is developing or changing.

Yeah but how is it developing/changing, and how does that make it less of a probability?

The graphs you presented me are, for the most part, uniform.

Not only are they not (although it's unclear what you mean with "for the most part" and what a uniform graph is), but they don't even appear to be based on a uniform distribution.

 

[Jaycee]

2 OLL skips in a row, both leading to EPLLs.

(1/216 * 1/6)^2 --- is that correct?

 

[TDM]

2 OLL skips in a row, both leading to EPLLs.

(1/216 * 1/6)^2 --- is that correct?

Yes.