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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.

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

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

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:

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. ;-)

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

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.

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.

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.

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.