Probability Thread

[Discomantis]

What are the odds of getting a J perm in that one angle that looks like a PLL skip? Pictures are shown below of what that would look like.

 

[Cuber987]

Not sure if this was already discussed, but what is the probability of a 3x3 stage (or centers; edges) skip on big cubes?

 

[Jorian Meeuse]

Not sure if this was already discussed, but what is the probability of a 3x3 stage (or centers; edges) skip on big cubes?

Well on odd cubes it is 1 in 43 quintillion, on even cubes it is 1 in (4 x 43 quintillion).

What are the odds of getting a J perm in that one angle that looks like a PLL skip? Pictures are shown below of what that would look like.View attachment 22275View attachment 22276

2 possible cases where it looks like a skip, and 4 x 12 = 72 pll cases, so 2 in 72, so 1 in 36.

 

[OldSwiss]

What are the odds of getting a J perm in that one angle that looks like a PLL skip? Pictures are shown below of what that would look like.

How do you mean that? You always see 3 Colors on 2 sides...

 

[Discomantis]

How do you mean that? You always see 3 Colors on 2 sides...

I often think that I got a PLL skip when I get a J perm from that angle and DNF because of that. Also I found out that the odds of DNFing an average of 5 because of that occuring is approximately 1/137.

here's the calculator I used:https://www.wolframalpha.com/input?i=binomial+distribution+calculator&assumption={"F",+"BinomialProbabilities",+"x"}+->"1"&assumption={"F",+"BinomialProbabilities",+"n"}+->"5"&assumption={"F",+"BinomialProbabilities",+"p"}+->"1/36"

 

[xyzzy]

Looking into EO again…

Distribution of number of bad edges on all three axes:

Spoiler

(0, 0, 0): 1
(12, 12, 12): 1
(0, 0, 8): 3
(4, 12, 12): 3
(0, 0, 2): 48
(10, 12, 12): 48
(0, 0, 6): 48
(6, 12, 12): 48
(0, 0, 4): 108
(8, 12, 12): 108
(0, 2, 8): 576
(4, 10, 12): 576
(0, 2, 2): 1008
(10, 10, 12): 1008
(2, 10, 10): 1098
(2, 2, 10): 1098
(0, 8, 8): 3753
(4, 4, 12): 3753
(0, 2, 6): 4416
(6, 10, 12): 4416
(0, 2, 4): 6336
(8, 10, 12): 6336
(0, 4, 8): 6336
(4, 8, 12): 6336
(2, 2, 2): 12602
(10, 10, 10): 12602
(0, 4, 4): 14148
(8, 8, 12): 14148
(0, 6, 8): 14976
(4, 6, 12): 14976
(0, 6, 6): 22608
(6, 6, 12): 22608
(0, 4, 6): 29376
(6, 8, 12): 29376
(2, 2, 8): 31056
(4, 10, 10): 31056
(2, 4, 10): 38016
(2, 8, 10): 38016
(2, 6, 10): 95976
(2, 2, 6): 138636
(6, 10, 10): 138636
(2, 2, 4): 150336
(8, 10, 10): 150336
(2, 8, 8): 259056
(4, 4, 10): 259056
(2, 4, 8): 585792
(4, 8, 10): 585792
(2, 4, 4): 756576
(8, 8, 10): 756576
(2, 6, 8): 1147392
(4, 6, 10): 1147392
(2, 6, 6): 1415376
(6, 6, 10): 1415376
(4, 4, 4): 1483843
(8, 8, 8): 1483843
(2, 4, 6): 1829952
(6, 8, 10): 1829952
(4, 8, 8): 2536704
(4, 4, 8): 2536704
(4, 4, 6): 6295824
(6, 8, 8): 6295824
(6, 6, 6): 6527128
(4, 6, 8): 9836928
(4, 6, 6): 10465584
(6, 6, 8): 10465584

The denominator here is \( \frac{12!}{4!^3}\cdot2^{11}=70963200 \).

While the average number of bad edges is always 6 per axis, it's actually somewhat more common to have (4,6,6) or (4,6,8) than to have exactly (6,6,6).

The numbers of bad edges on two different axes can differ by at most 8, and this is the only constraint on the numbers of bad edges. IOW, of the sixty-five possible combinations \( \{x,y,z\} \) where \( 0\le x,y,z\le12 \) are all even and \( \max(x,y,z)-\min(x,y,z)\le8 \), all of them are actually attainable with legal cube positions.

 

[Discomantis]

How many solves on average would it take to get all PLL's at least once? (Including a PLL skip.) I'm wondering this because I'm close to having gotten all PLL's officially. In case anyone doesn't know, here are the odds of each PLL:
H perm, Nb perm, Na perm, PLL skip: 1/72 chance each
Z perm and E perm: 1/36 chance each
All other PLL's: 1/18 chance each

 

[sDLfj]

How many solves on average would it take to get all PLL's at least once? (Including a PLL skip.) I'm wondering this because I'm close to having gotten all PLL's officially. In case anyone doesn't know, here are the odds of each PLL:
H perm, Nb perm, Na perm, PLL skip: 1/72 chance each
Z perm and E perm: 1/36 chance each
All other PLL's: 1/18 chance each

I think it would take 72 solves for there to be a 50% chance of having gotten them all, which basically means it's "supposed" to take 72 solves to get them all.

 

[xyzzy]

I think it would take 72 solves for there to be a 50% chance of having gotten them all, which basically means it's "supposed" to take 72 solves to get them all.

That's definitely wrong. Why do you think there's a 50% chance of getting every case after 72 solves?

 

[qwr]

I think it's a variation of https://en.wikipedia.org/wiki/Coupon_collector's_problem
but idk how to do the calculations since there are different probabilities each

 

[IsThatA4x4]

How many solves on average would it take to get all PLL's at least once? (Including a PLL skip.) I'm wondering this because I'm close to having gotten all PLL's officially. In case anyone doesn't know, here are the odds of each PLL:
H perm, Nb perm, Na perm, PLL skip: 1/72 chance each
Z perm and E perm: 1/36 chance each
All other PLL's: 1/18 chance each

I don't know if this is the correct way to do this, there are definitely much smarter people than me on the forum who can correct me if I'm wrong:

Let's start by asking how many solves it would take to get the 4 1/72 PLLs (H, Na, Nb, Skip). Using the binomial distribution, we can see that for the probability of 4 successes (p=1/72) to be >0.5, we need 265 solves.
For the 5 1/18 PLLs, we need 264 solves for P>0.5;
For the 2 1/36 PLLs, we need 61 solves for P>0.5

So, if we do 265 solves to have good odds of getting all 4 of the 1/72 PLLs, we will also have good odds of getting all the 1/18 PLLs, and it is almost impossible we have not gotten the 2 1/36 PLLs by now. So, after 265 solves, it is more likely than not that we have received every PLL.

 

[sDLfj]

That's definitely wrong. Why do you think there's a 50% chance of getting every case after 72 solves?

My logic was that since all of the solves are independent, it would take 72 solves to get an H perm (or any others like it), and by the time that was accomplished you should have gotten 1 of each N perm, a skip, 2 Z perms, etc. But it seems like there are smarter people that know what they're talking about, lol.

 

[qwr]

Let's start by asking how many solves it would take to get the 4 1/72 PLLs (H, Na, Nb, Skip). Using the binomial distribution, we can see that for the probability of 4 successes (p=1/72) to be >0.5, we need 265 solves.
For the 5 1/18 PLLs, we need 264 solves for P>0.5;
For the 2 1/36 PLLs, we need 61 solves for P>0.5

I don't know how you derived this. The PLLs are not independent in terms of how likely it is to see a new one. That's why I linked the coupon collector's problem

 

[SpeakCuba]

Does anyone here know the probability of getting a PLL skip 3 solves in a row?... Happened to me multiple times.

 

[Discomantis]

Does anyone here know the probability of getting a PLL skip 3 solves in a row?... Happened to me multiple times.

1 in 373248 chance. (1/72 chance raised to the third power.)

EDIT: Inputted number from calculator incorrectly

 

[fdskljgrie]

whats the probability of getting double parity on each solve in a 4x4 average of 5?

 

[Discomantis]

whats the probability of getting double parity on each solve in a 4x4 average of 5?

1/4^5=1/1024

 

[OldSwiss]

whats the probability of getting double parity on each solve in a 4x4 average of 5?

For me it's around 100% o_O

 

[THE_V01DL0RD]

while I was doing 2x2, my scramble had one side done already
i accidentally deleted that solve
what are the odds of me getting a scramble with a side solvedd again?

 

[Discomantis]

while I was doing 2x2, my scramble had one side done already
i accidentally deleted that solve
what are the odds of me getting a scramble with a side solvedd again?

This might be wrong but I'm just going to give it my all:
So I'm going to count positions here. If you count last layer cases including different AUF's, you have 4 possibilities for the first corner's permutation, then 3 for the next, then 2 for the next, and 1 for the last. Then, you can orient 3 of them in 3 different ways. So, the number of last layer cases are 4x3x2x3^3, or 648. Then, I will calculate the number of possible face cases there are. So, there are 4 possibilities for the first corner's permutation on the face, then 3 for the 2nd, then 2 for the third, then 1 for the last one. Multiply these together and you get 24 side cases. Multiply that by 648 and you get 15552 cases. Multiply this by 6 because there are other possible face colors and you have 93312 cases. Divide this by the number of positions on a 2x2 and you have a 1 in 39.375 chance. If anyone knows that this is wrong please correct me.