I got 13 pll skips on my 2x2 in a row

[AvGalen]

I conclude that Stefan is deeply enjoying this thread

Then I conclude that you are right ^^

And I conclude that although you don't have sufficient evidence, you are probably right.

But I also suspect that Stefan isn't convinced that he is deeply enjoying just because of that sample of 2. He is more interested in finding out how to calculate the probability that you are wrong about him deeply enjoying.

(also, he would like to have deeply and enjoying defined)

 

[Stefan]

Thanks Herbert for the insight and confirmation of Johannes' result. I had suspected it would be something that complicated, I didn't see any easy way because of the heavy overlapping (e.g. solves 1-13 with solves 2-14 and 3-15 etc). And I wish my math teachers had been as knowledgeable. You make me wanna learn this stuff that I've heard about but never pursued.

Dene is right, Arnaud is mostly wrong, and needless to say, both increased the joy very much. Thanks guys.

 

[AvGalen]

Thanks Herbert for the insight and confirmation of Johannes' result. I had suspected it would be something that complicated, I didn't see any easy way because of the heavy overlapping (e.g. solves 1-13 with solves 2-14 and 3-15 etc). And I wish my math teachers had been as knowledgeable. You make me wanna learn this stuff that I've heard about but never pursued.

Dene is right, Arnaud is mostly wrong, and needless to say, both increased the joy very much. Thanks guys.

Well, what are the chances of having this many posts on a forum that has Pochmann-control activated and be always right?

I also share Stefans feeling that I want to know more about this stuff. Any good/fun sources for this, or are there just demotivating boring books :rolleyes:?

Found this source that might be useful for the topicstarter as well: http://www.webmath.com/lottery.html

 

[rahulkadukar]

I got a OLL,PLL and then another OLL skip and I thought that was lucky

 

[AvGalen]

I got a OLL,PLL and then another OLL skip and I thought that was lucky

It really was lucky (http://www.speedcubing.com/records/records-rules.html rule 3) to get any of those three skips. Getting them all in a row .....

 

[tfkscores]

how the hell did you find out what the chances are? im reapeating algebra 1 next year and am going to be a freshman. how did you figure this stuff out?

 

[Johannes91]

how the hell did you find out what the chances are?

Who? Herbert explained quite clearly how he did it. If you aren't familiar with the terms, look them up.

http://en.wikipedia.org/wiki/Markov_chain

If you can't understand something, read the relevant articles first.

The way I did it wasn't as elegant. The state after any number of solves can be represented with two natural numbers: n, the length of the current streak, and m, the length of the longest streak so far. Doing one solve gets us from (n,m) to (0,m) with probability 5/6, and to (n+1,max(n+1,m)) with probability 1/6. At the beginning, we are obviously at state (0,0) with probability 1. Keeping track of all different states and the distribution is fairly easy. Here's the code I used, it takes a few seconds to run.

im repeating high school algebra 1 so i have no idea what you all are talking about.
im reapeating algebra 1 next year

So what? Are you not allowed to learn outside school?

 

[tfkscores]

oh....my.....god....i have no idea what you just said lol. i suck at anything algebra related. apparently my brain isnt ready or something lol.

 

[Stefan]

Here's the code I used, it takes a few seconds to run.

No perl?

Thanks for that recurrence relation, I probably would've ended up with the same approach had I put more thought into it. Now I just quickly did it in perl and confirm both of your numbers.

View attachment 344

The graph shows the probability for at least 13 skips as a function of the number of trys. In good approximation it seems to be a linear function.

Small extra riddle: Is it linear? Explain your answer.

Herbert and Johannes are not allowed to answer, would be too easy for them.

 

[Rune]

I asked my computer to do it for me and it gave 4.700277693455794e-9 for the first one and 5.627572299115469e-9 for the second one (the rationals have quite large numerators and denominators).

Does that mean that you got an exact expression for the probability?

 

[moogra]

http://en.wikipedia.org/wiki/Binomial_probability

(100 C 13) * (probability of getting a PLL skip) ^13 * (probability of not getting a PLL skip) ^87 will be the probability of getting 13 PLL skips in 100.
If you want more, then you write a program
the general equation is
(totalnumber choose wantedNum) * (pllProbability)^wantedNum * (notPllProbability) ^ (totalNumber - wantedNum)
loop it from lower wantedNum bound to upper wantedNum

or just use binomialCdf on a TI-calculator (binomialpdf is 1 case)

 

[Stefan]

moogra, you might want to mention that you're answering a completely different question, the math noobs here might not notice that otherwise.

 

[Herbert Kociemba]

I asked my computer to do it for me and it gave 4.700277693455794e-9 for the first one and 5.627572299115469e-9 for the second one (the rationals have quite large numerators and denominators).

Does that mean that you got an exact expression for the probability?

With my Markovchain approach I am only able to find the expression for the second probability. The exact expression is

2836881009340468654612600338487002584885669527428222663495824930731/
504103876157462118901767181449118688686067677834070116931382690099920633856

but I do not think we get much insight from this.

 

[Jacco]

Sorry, I haven't read the whole thread but I think that the chance is way higher to hear somebody talking nonsense instead of in fact having 13 2x2 PLL skips in a row. No offense, just experience

 

[Tim Reynolds]

Small extra riddle: Is it linear? Explain your answer.

The expected number of runs of 13 grows linearly--this is a result of the fact that expected value is additive. This way of counting counts a run of 14 as 2 runs of 13. But the probability that there is at least one grows slower than that (P=E-something representing probabilities of >1 run), and slower than linearly.

Besides, if it's linear, then P->infinity. But P approaches 1 asymptotically. So unless it's linear on some interval, which I assert it's not, no.

 

[Swordsman Kirby]

http://i40.tinypic.com/rc1to6.gif

I wish I could get rid of the summation.

 

[moogra]

moogra, you might want to mention that you're answering a completely different question, the math noobs here might not notice that otherwise.

Oh I did not see the 13 skips IN A ROW. My answer is the probability of just getting 13 PLL skips in 100 solves.

 

[Stefan]

Besides, if it's linear, then P->infinity.

Yep, that's what I was after. I found it neat that one could tell the function isn't linear without understanding it at all (Markov chains huh?), solely by keeping in mind that its result is a probability. Background information ftw! Just like in another thread where someone computed the probability to win the lottery as about 1/10^50. If he had just considered *what* he wanted to compute, he should've realized that *how* he computed it must be wrong.

http://i40.tinypic.com/rc1to6.gif

False. You ignore cases like 13 skips followed by a nonskip followed by another skip.

 

[JLarsen]

ummm....im repeating high school algebra 1 so i have no idea what you all are talking about.

Dude, you're not the only one. Don't worry about it. Whenever I see probability, I see a long post of stuff, followed ussualy by Pochmann or Johannes correcting them, and whatever number I see at the end of their posts is the one I trust. Someone says how many cases, they do the math, I read the number at the end, I'm good.