Probability Thread

[xyzzy]

Please correct me if I missed something

Non-independence across the edges.

Your answer is probably correct to an order of magnitude, but this type of question ("probability of skipping last n whatevers out of N total whatevers") is nigh impossible to answer completely correctly.

Consider the following similar question: let's say I'm a CFOP user and I don't do any multislotting. What's the probability of skipping the last slot (out of four slots, total) when solving F2L? The naïve answer is 1/15 (corner) × 1/10 (edge) = 1/150, but this is wrong. Last slot skips are much rarer than that, and more insidiously, it also depends on F2L algs used. This is not method-independent, so there's no canonical answer. The crux here is that I'm asking for a last slot skip. This is only possible if the solution to the third slot affects multiple slots (*), and even then, if it's a solution that breaks down into two triggers (e.g. R U' R' L' U L), then the first trigger already solves a different slot.

 Twizzle link 

// solving the orange-green pair
R U' R' // red-green is the third slot
L' U L // orange-green is the last slot

(*) or if you entirely skip F2L after cross, which, you know

Same problem shows up in edge pairing. What's the probability of skipping L4E on a 5×5×5, assuming I use freeslice for F8E? Naïve calculation suggests 1/8! = 1/40320, but there are factors that can decrease and other factors that can increase this value. If, during F8E, I prioritise using edge pieces that are already in the active slice and in the correct orientation to be paired up (i.e. ones where I don't need to use a flip alg), that means that what's left at the end of F8E is substantially less likely to be in compatible orientations (because the matching ones have already been used up). On the other hand, if I prioritise inserting edges to make them have matching orientations, that increases the likelihood of getting free edge pairs/triplets when I slice the centres back at the end of F8E.

(This is an actual thing I do for edge pairing on big cubes, btw. Not purely hypothetical.)

The original question by @OtaMota asked about L4E on 4×4×4. To answer this, we first need to know how the first eight edges were paired in the first place. Was it with freeslice? Was it with chain pairing (in which case, L4E skip is only possible if all of edge pairing was skipped)?

Edit: lmao, of course I wrote this before (January 2017): https://www.speedsolving.com/threads/probability-thread.20384/post-1212367

 

[OtaMota]

Non-independence across the edges.

Your answer is probably correct to an order of magnitude, but this type of question ("probability of skipping last n whatevers out of N total whatevers") is nigh impossible to answer completely correctly.

Consider the following similar question: let's say I'm a CFOP user and I don't do any multislotting. What's the probability of skipping the last slot (out of four slots, total) when solving F2L? The naïve answer is 1/15 (corner) × 1/10 (edge) = 1/150, but this is wrong. Last slot skips are much rarer than that, and more insidiously, it also depends on F2L algs used. This is not method-independent, so there's no canonical answer. The crux here is that I'm asking for a last slot skip. This is only possible if the solution to the third slot affects multiple slots (*), and even then, if it's a solution that breaks down into two triggers (e.g. R U' R' L' U L), then the first trigger already solves a different slot.

 Twizzle link 

// solving the orange-green pair
R U' R' // red-green is the third slot
L' U L // orange-green is the last slot

(*) or if you entirely skip F2L after cross, which, you know

Same problem shows up in edge pairing. What's the probability of skipping L4E on a 5×5×5, assuming I use freeslice for F8E? Naïve calculation suggests 1/8! = 1/40320, but there are factors that can decrease and other factors that can increase this value. If, during F8E, I prioritise using edge pieces that are already in the active slice and in the correct orientation to be paired up (i.e. ones where I don't need to use a flip alg), that means that what's left at the end of F8E is substantially less likely to be in compatible orientations (because the matching ones have already been used up). On the other hand, if I prioritise inserting edges to make them have matching orientations, that increases the likelihood of getting free edge pairs/triplets when I slice the centres back at the end of F8E.

(This is an actual thing I do for edge pairing on big cubes, btw. Not purely hypothetical.)

The original question by @OtaMota asked about L4E on 4×4×4. To answer this, we first need to know how the first eight edges were paired in the first place. Was it with freeslice? Was it with chain pairing (in which case, L4E skip is only possible if all of edge pairing was skipped)?

Free slice.

 

[CubeRoot]

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.

what's the avg move count and distribution of fixed axis EO?

 

[ruffleduck]

Non-independence across the edges.

I see. So 1% is the upper bound, and assumes the solver has no bias to order or approach in solving edges

 

[caleb_g]

i got 4 EP skips in a row on megaminx on the weekly competition today, i believe thats 1 in 12,960,000... which is insane. someone correct me if im wrong

(i use 4LLL, so once i get to LL i just orient edges, and then permute them, but they were already done 4 times in a row)

 

[Exmestify]

What is the probability of getting 2 pll skips in a row without auf on 3x3?

 

[FenTheGoat]

What is the probability of getting 2 pll skips in a row without auf on 3x3?

Chance of a pll skip is 1/72, one with out auf is 1/288. And 2 in a row would be (72*4)^2. That is

1/82,944 or the equivalent of getting about 5 LL skips

 

[caleb_g]

Chance of a pll skip is 1/72, one with out auf is 1/288. And 2 in a row would be (72*4)^2. That is

1/82,944 or the equivalent of getting about 5 LL skips

getting 5 LL skips in a row would be (15,552)^5 which is 1 in 909,769,216,957,750,444,032

also i know you probably didnt mean in a row but im silly and like huge numbers

 

[Exmestify]

and the fact that this is the third time this has happened to me in 1 year and i have 1 ll skip (no auf lol)

 

[THE_V01DL0RD]

this isn't probability but how many combinations are on a 2x2x3?

 

[An Animating Cuber]

this isn't probability but how many combinations are on a 2x2x3?

Rephrase it to “what is the probability I get a specific scramble on a 2x2x3” and now it is about probability

 

[tom0989123]

This is just a thread for people to post interesting possibilities of a case. Like, it is a 1/72 chance for a PLL skip. Enjoy.

dude is that right because I once got a PLL skip three times in a row be-for I gave my GAN 356 away
like it didn't want me get get rid of it so In it's own way it was giving me PLL skips

 

[THE_V01DL0RD]

dude is that right because I once got a PLL skip three times in a row be-for I gave my GAN 356 away

1 in 373248

 

[tom0989123]

1 in 373248

hah no way!!

 

[Skewby_Doo]

I was doing some cubedesk 1v1 and got 4 f perms in a row. Something similar happened to me a few months ago in OH, but this was weird.

 

[Discomantis]

I was doing some cubedesk 1v1 and got 4 f perms in a row.

That's a 1 in 104976 chance

 

[THE_V01DL0RD]

what are the odds of you knowing how to solve a cube

 

[Arcanist]

what are the odds of you knowing how to solve a cube

1/20 is the general number given

 

[FenTheGoat]

what are the odds of you knowing how to solve a cube

Let’s say there’s 10x amount of people that know how to solve a cube than have been to a comp. There’s about 200k people who’ve been to a comp, multiply by 10 you get 2M. 2M/8B= 1/4,000 people

 

[Arcanist]

Let’s say there’s 10x amount of people that know how to solve a cube than have been to a comp. There’s about 200k people who’ve been to a comp, multiply by 10 you get 2M. 2M/8B= 1/4,000 people

I think it is much higher than that.