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

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#### Johannes91

##### Member
But what is the probability of getting 13 PLL skips in a row somewhere in let's say 100 solves?
Exactly 13 or >=13?
I meant the second version, but now I'm interested in both (and btw I don't know the answers).
I don't know an easy way to calculate it on paper, so 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).

I guess you're more interested in the how than the what, though.

#### Rune

blah;200162 Exactly 13 or >=13?[/QUOTE said:
I meant the second version, but now I'm interested in both (and btw I don't know the answers).
You can, of course, always find the answer(s) here:
"An Introduction to Probability..." William Feller, Volume 1, 1968

#### dougbenham

##### Member
I would lol if we found out this guy was using CLL (or COLL whatever you call it) and didn't even realize it

#### panyan

##### Member
the probability of that is 1/6^13 = 1/13060694016
if you did that thirteen times then it comes to 1/169000000000 chance, the chance of winning the lottery is 1/(1.4x10^51). That is an impressive pll skip scenario if it did happen

#### Rune

"the chance of winning the lottery is 1/(1.4x10^51). "

Think it over!

#### tfkscores

##### Member
ok this isnt bs i actually did do it and i should've turned my camera on. it doesnt matter if you believe me cause i wouldnt even believe it. lol hope i didnt upset any1

#### TomZ

##### Member
In an average of N, the streak of 13 can start at N-13 'points'. So the probability of getting a streak of 13 in an average of N is 1/6^13*(N-13). That means to have just a 1% chance you need to do 13 MILLION solves.

The streak of 13 or more is a bit trickier. That requires adding all the possible streaks from 13-N together, giving us the following sequence:

Sum(X from 13 to N) 1/6^X*(N-X)

I don't know of any algebraic way to calculate that (I'm not even sure it exists), but using some computer programming I generated this graph:

Pretty darn unlikely if you ask me.

#### Stefan

##### Member
So the probability of getting a streak of 13 in an average of N is 1/6^13*(N-13).
Wow is that wrong. And not only because you just said it's impossible to get 13 PLL skips in 13 solves.

The streak of 13 or more is a bit trickier. That requires adding all the possible streaks from 13-N together, giving us the following sequence:

Sum(X from 13 to N) 1/6^X*(N-X)
This would be wrong even if you plugged in the correct formula for the "=X" case. Btw, "13 to N" would've been clearer than "13-N".

Pretty darn unlikely if you ask me.
Just to clarify: I don't care whether he actually got that streak. I'm here solely for the probability riddle he caused.

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#### brunson

##### Member
Hmm, I wonder if this approach would work:

Calculate the probability of 13 PLL skips in 100 solves (not too hard), then multiply by the number of permutations of 100 solves (100!), and divide by the number of permutations that have those 13 skips adjacent to one other.

Computation left as an exercise for the student. ;-)

Actually, I'll hack at it when I'm not at work.

Edit:
Someone critique this...
Code:
 100(1/6)^13*100!
--------------------
13!(100-13)!(100-13)
Where:
100(1/6)^13 = P(13 skips in 100)
100! = permutations of 100 solves
13! = permutations of the 13 skips
(100-13)! = permutations of the non skips
(100-13) = the number of places in the 100 solves where your streak could occur

Actually, that result is greater than 1, so there's a problem in there somewhere.

Edit2: My denominator doesn't force the skips to be adjacent

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#### tfkscores

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

#### Stefan

##### Member
Calculate the probability of 13 PLL skips in 100 solves (not too hard), then multiply by the number of permutations of 100 solves (100!), and divide by the number of permutations that have those 13 skips adjacent to one other.
But what about all the PLL skips other than those 13 of the streak, somewhere else in the 100 solves?

#### brunson

##### Member
Calculate the probability of 13 PLL skips in 100 solves (not too hard), then multiply by the number of permutations of 100 solves (100!), and divide by the number of permutations that have those 13 skips adjacent to one other.
But what about all the PLL skips other than those 13 of the streak, somewhere else in the 100 solves?
Good point. Even if my calculation was correct it would only be the chance of *exactly* 13 skips and all in a row.

#### panyan

##### Member
"the chance of winning the lottery is 1/(1.4x10^51). "

Think it over!
oh woops, it is 49! not 42! (i have never played the lottery )

#### Stefan

##### Member
"the chance of winning the lottery is 1/(1.4x10^51). "

Think it over!
oh woops, it is 49! not 42! (i have never played the lottery )
You're... not exactly getting closer.

I suggest you follow Rune's advice and think it over. Independent of how you arrived at your number, do you not realize how tiny it is? Like I said earlier, it would pretty much mean that nobody ever won the lottery. If every person had lived since the big bang and filled out 1 lottery ticket every second, the probability that anyone ever won would still be almost zero. You gotta be kidding me.

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#### Herbert Kociemba

##### Member
The probabiltiy is about 5.627572299*10^-9 to get at least 13 PLL skips in a row within 100 which as almost exactly 73.5 times more than 1/6^13.

I do not think there is a closed expression for the result. You can describe the process by a Markov Chain with 14 states (for 0 to 13 skips in a row) and a 14x14 transition matrix. The 100 th power of this matrix gives the desired result.

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.

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