Probability Thread

[cuber314159]

If we assume I have done 100000 3x3x3 solves with full CFOP and no last layer influencing during F2L, what are the chances I have seen all the last layer cases?

I think it's probably easiest to do an estimate where we assume that the probability of getting each case is constant but obviously that isn't really the case

 

[brododragon]

You could use HARCS to do 100000 solves and see what the last layer is for them

 

[Q--]

If we assume I have done 100000 3x3x3 solves with full CFOP and no last layer influencing during F2L, what are the chances I have seen all the last layer cases?

I think it's probably easiest to do an estimate where we assume that the probability of getting each case is constant but obviously that isn't really the case

TLDR, 99.999997%
I may have completely messed up my math for someone like xyzzy to catch me on, but this is a large simplification of the problem:

Assume every case has a probability of 4/15552 of occurring (4x a last layer skip, or 1/3888).

The probability of NOT getting a specific case after 100000 solves is the probability of not getting that case, 3887/3888, times itself 100000 times.

You take the complement (1-answer) of this to find the chance of actually getting that case at some point in the 100000 solves, and multiply that by itself 3888 times (the number of cases) to get the chance of finding all of them in those solves, which is virtually 100%.

It seems odd that the odds are so high, but 100000 is a lot of solves so it makes sense. If I knew how many cases had each kind of symmetry I could give a real answer, and the odds would be a bit lower. This was a fun probability problem, so thank you.

 

[cuber314159]

TLDR, 99.99999997%
I may have completely messed up my math for someone like xyzzy to catch me on, but this is a large simplification of the problem:

Assume every case has a probability of 4/15552 of occurring (4x a last layer skip, or 1/3888).

The probability of NOT getting a specific case after 100000 solves is the probability of not getting that case, 3887/3888, times itself 100000 times.

You take the reciprocal of this to find the chance of actually getting that case at some point in the 100000 solves, and multiply that by itself 3888 times (the number of cases) to get the chance of finding all of them in those solves, which is virtually 100%.

It seems odd that the odds are so high, but 100000 is a lot of solves so it makes sense. If I knew how many cases had each kind of symmetry I could give a real answer, and the odds would be a bit lower. This was a fun probability problem, so thank you.

Interesting, I was expecting it to be a lot lower actually, like how the odds of having 365 people in a room all having different birthdays are astronomically low, but I suppose you are probably right.

 

[xyzzy]

You take the reciprocal of this

Complement (\( x\mapsto1-x \)), not reciprocal (\( x\mapsto1/x \)).

Haven't thought too carefully about it but your calculation seems correct. The final answer has two too many "9"s in it, though.

>>> (1-(3887/3888)**100000)**3888
0.9999999738084874

(the last four digits printed are wrong due to accumulated rounding error)

 

[kubesolver]

(1-(3887/3888)**100000)**3888

That's not fully correct calculation because the events aren't independent.
If you replace 100000 in the formula with 1 you will get positive result.

 

[xyzzy]

That's not fully correct calculation because the events aren't independent.
If you replace 100000 in the formula with 1 you will get positive result.

Right, yes, that is just an approximation due to non-independence. That part of the approximation gets more accurate as the number increases. (I'm not sure if there's a good way of calculating the exact answer without resorting to a very messy inclusion-exclusion calculation.)

 

[kubesolver]

one way to calculate is that expected number of unreached positions is (due to linearity of expectation)
3888 * ((1-(3887/3888)**100000)) = 0.00000067366
Which is the upper bound on probability of missing one.

 

[Q--]

Complement (\( x\mapsto1-x \)), not reciprocal (\( x\mapsto1/x \)).

Haven't thought too carefully about it but your calculation seems correct. The final answer has two too many "9"s in it, though.

>>> (1-(3887/3888)**100000)**3888
0.9999999738084874

(the last four digits printed are wrong due to accumulated rounding error)

Yeah, that’s what I meant and used in the calculation, good catch.

 

[ProStar]

TLDR, 99.999997%
I may have completely messed up my math for someone like xyzzy to catch me on, but this is a large simplification of the problem:

Assume every case has a probability of 4/15552 of occurring (4x a last layer skip, or 1/3888).

The probability of NOT getting a specific case after 100000 solves is the probability of not getting that case, 3887/3888, times itself 100000 times.

You take the complement (1-answer) of this to find the chance of actually getting that case at some point in the 100000 solves, and multiply that by itself 3888 times (the number of cases) to get the chance of finding all of them in those solves, which is virtually 100%.

It seems odd that the odds are so high, but 100000 is a lot of solves so it makes sense. If I knew how many cases had each kind of symmetry I could give a real answer, and the odds would be a bit lower. This was a fun probability problem, so thank you.

(I'm not that great at math, so sorry if this is wrong)

Doesn't each LL case have different chances of appearing? Like, for example, an N-perm is way less likely to happen than a T-Perm, so you can't just say that you can do 21 solves and see all PLL cases (bad example but you get the general idea)

 

[Q--]

(I'm not that great at math, so sorry if this is wrong)

Doesn't each LL case have different chances of appearing? Like, for example, an N-perm is way less likely to happen than a T-Perm, so you can't just say that you can do 21 solves and see all PLL cases (bad example but you get the general idea)

Yes, that’s what I mentioned in the last paragraph about the symmetry. An N perm has 90 degree symmetry so it’s 1/72 as opposed to the 4/72 of a T perm. However, I couldn’t find a list of last layer cases by their symmetry so I simplified the problem.

 

[Romy4]

Got my fist ever last layer skip today (almost a pb cos my cross was rubbish), I’ve probably done under 5000 timed solves. What are the chances of getting a last layer skip? also how many last layer skips have people had? And how long have you been cubing to get that many last layer skips?

 

[kubesolver]

Got my fist ever last layer skip today (almost a pb cos my cross was rubbish), I’ve probably done under 5000 timed solves. What are the chances of getting a last layer skip? also how many last layer skips have people had? And how long have you been cubing to get that many last layer skips?

Eo skip is one in 8
Co skip is one in 27
Pll skip is one in 72
Roughly one in 15 thousands.

I got one in a speedsolve in so far which seems about fair considering number of solves I did in total

 

[Waffles]

I’ve had 3 Last Layer skips ever, have around 5000 3x3 solves on cstimer and hardly ever use it, and do like probably around 100 times solves every night. 1 LL skip was on 5x5 too (have around 150 solves in that session). I know I’m very lucky but I actually noticed the possible LL skip on the 2nd one and did a different pair solution to get it (also never gotten a no-AUF LL skip) and I believe I influenced the 5x5 LL skip by doing 2 adjacent pairs with a multi slot in between edge pairing stages.

 

[Cubing Forever]

However, I couldn’t find a list of last layer cases by their symmetry so I simplified the problem.

PLL: (Case(s): symmetry, probability)
A perms: no symmetry, 1/18 each
E: 180° symmetry, 1/36
F: no symmetry, 1/18
G perms: no symmetry, 1/18 each
H: 90° symmetry, 1/72
J perms: no symmetry, 1/18 each
N perms: 90° symmetry, 1/72 each
R perms: no symmetry, 1/18 each
T: no symmetry, 1/18
U perms: no symmetry, 1/18 each
V: no symmetry, 1/18
Y: no symmetry, 1/18
Z: 180° symmetry, 1/36

Too lazy to do it for OLL.

 

[Q--]

PLL: (Case(s): symmetry, probability)
A perms: no symmetry, 1/18 each
E: 180° symmetry, 1/36
F: no symmetry, 1/18
G perms: no symmetry, 1/18 each
H: 90° symmetry, 1/72
J perms: no symmetry, 1/18 each
N perms: 90° symmetry, 1/72 each
R perms: no symmetry, 1/18 each
T: no symmetry, 1/18
U perms: no symmetry, 1/18 each
V: no symmetry, 1/18
Y: no symmetry, 1/18
Z: 180° symmetry, 1/36

Too lazy to do it for OLL.

Thank you, but I meant all 3000+ last layer cases. Oll and Pll symmetry are fairly easy to find.

 

[yCArp]

Out of curiosity, I was wondering how many possible cases there are for F2L in one look. That means the entire F2L is solved in one algorithm, or in other words, solving all 4 pairs at the same time.

 

[xyzzy]

Out of curiosity, I was wondering how many possible cases there are for F2L in one look. That means the entire F2L is solved in one algorithm, or in other words, solving all 4 pairs at the same time.

Approximately \( \frac14\cdot(3^4\frac{8!}{4!})\cdot(2^4\frac{8!}{4!})=914457600 \) (914 million) up to rotation, roughly half that up to rotation and reflection (457 million), roughly a quarter up to rotation and AUF (229 million), roughly an eighth up to rotation, reflection and AUF (114 million).

 

[yCArp]

Thanks a lot. How about 3 pairs at a time? Does changing some of the relevant numbers in your formula get me the correct result?

 

[teh yoshi]

If you had all the stickers necessary to sticker a blank cube, but placed them randomly, what are the odds that it'll be solvable in the standard color scheme?

What are the odds that it'll be solvable in any color scheme?