Probability Thread

[uberCuber]

I was going to write this in a PM so as not to go off-topic in this thread, but your inbox is full. :s

Who is fedora?

That's coming from a game started in a private cubing group on facebook. Each person participating is assigned another one as a target. Every single time a participant makes a post, whether it be in that facebook group, elsewhere on facebook, or even on speedsolving, the post has to contain a particular word (for the first round of this game quite awhile ago, the word was 'cube', and now this time it's 'fedora'). If a participant catches his target making a post that doesn't contain that word, he can post BLAMMO!, and the target is out. Last person left wins.

And there are a few other rules to make this actually work, but that's the gist.

 

[Stefan]

Oh. John has been to 40 competitions and is one of the fastest cubers, I couldn't imagine him asking those questions and it looked like he was quoting some noob to mock him or someone else using his account to ask them

Although, about his second question: Some OLLs in fact do give PLL skips more often than others (someone guess which).

 

[Logiqx]

for the how many times should I solve a cube before I have more than 50% chance of getting a PLL/OLL/LL skip: ceil(log(50%)/log((#-1)/#)) where #=15552 for LL skip, figure out how to use it

I've never really looked closely at the "my birthday" problem.

There is an easy approximation for cubers - 0.69314 * n ... rounding the result upwards. Remembering 69% is pretty easy.

I ran a few cubing numbers through Python using the original formula and my approximation:

import math
for n in [18, 54, 72, 216, 15552, 31704, 95042, 190080]:
    approx1 = int(math.ceil(math.log(0.5) / math.log((n - 1.0) / n)))
    approx2 = int(math.ceil(0.69314 * n))
    print '1 /', n, ':', approx1, 'solves (approx1) or', approx2, 'solves (approx2)'

Results - i.e. solves required for 50% chance of a specific event:

Spoiler

1 / 18 : 13 solves (approx1) or 13 solves (approx2) - J-Perm
1 / 54 : 38 solves (approx1) or 38 solves (approx2) - T-OLL
1 / 72 : 50 solves (approx1) or 50 solves (approx2) - PLL skip
1 / 216 : 150 solves (approx1) or 150 solves (approx2) - OLL skip
1 / 15552 : 10780 solves (approx1) or 10780 solves (approx2) - LL skip
1 / 31704 : 21976 solves (approx1) or 21976 solves (approx2) - CN cross skip
1 / 95042 : 65878 solves (approx1) or 65878 solves (approx2) - W/Y cross skip
1 / 190080 : 131754 solves (approx1) or 131753 solves (approx2) - W cross skip

 

[Carrot]

There are several approximations for the birthday problem but I like the two below. The second one is pretty easy to remember:

Spoiler

0.5 + sqrt(0.25 - 2 * N * log(0.5))

0.5 + 1.17741 * sqrt(N)

Here's a little Python script with a selection of values for N (re: cubing, computing and birthday):

import math
for n in [18, 54, 256, 365, 15552, 65536, 4294967296]:
    approx1 = 0.5 + 1.17741 * math.sqrt(n)
    approx2 = 0.5 + math.sqrt(0.25 - 2 * n * math.log(0.5))
    approx3 = math.sqrt(2 * n * math.log(2)) + (3 - 2 * math.log(2)) / 6
    print n, approx1, approx2, approx3

Results:

Spoiler

18 5.49532757142 5.52028868693 5.26427860676
54 9.15216115415 9.166596535 8.92111225942
256 19.33856 19.3451945187 19.1075113001
365 22.9943864654 22.9999431512 22.7633378354
15552 147.332043776 147.332897895 147.100997524
65536 301.91696 301.917380472 301.685916704
4294967296 77163.24176 77163.2432372 77163.0121865

It's right that your approximations are easier to remember etc, but they are approximations with errors The one I posted is an approximation with no known errors. (or at least no one has any instances and even if they did you could extend it using his paper to get it even more exact.)

 

[Tim Major]

Tipping point (over .5) for a dice is roughly 4 rolls.
(5/6)^x>0.5
Has to be 4ish rolls.

But logically, it should be a tipping point after 3 rolls.

So using my own maths I get 4 rolls, using intuition I get 3. Can someone explain?

The reason I ask is I explained the LL skip tipping point to a friend and we got in an argument and he suggested simplifying it to a die and I can no longer feel comfortable in my answer even though mathematically I'm right.

 

[Carrot]

Tipping point (over .5) for a dice is roughly 4 rolls.
(5/6)^x>0.5
Has to be 4ish rolls.

But logically, it should be a tipping point after 3 rolls.

So using my own maths I get 4 rolls, using intuition I get 3. Can someone explain?

The reason I ask is I explained the LL skip tipping point to a friend and we got in an argument and he suggested simplifying it to a die and I can no longer feel comfortable in my answer even though mathematically I'm right.

Roll with some dices and see for yourself...

jk.

Your friend is using the "expectance" instead.

It's true you are expected 0.5 6's in 3 throws, but the distribution just happens to be:
0: 125 cases
1: 75 cases
2: 15 cases
3: 1 case

So do an weighed average of that and you'll get 0.5 6's

But the question was 50% chance of getting at least one, which is the same as 50% chance of NOT getting one, then you suddenly have:
125/216 which is less than 50%, meaning you didn't hit the tipping point yet

 

[Logiqx]

It's right that your approximations are easier to remember etc, but they are approximations with errors The one I posted is an approximation with no known errors. (or at least no one has any instances and even if they did you could extend it using his paper to get it even more exact.)

That's fair enough... still close to within 1/4 of a solve on the cubing problems. One of my friends can probably do the second approximation to 5 decimal places in his head but he's a certified freak, lol.

 

[Jaycee]

Oh. John has been to 40 competitions and is one of the fastest cubers, I couldn't imagine him asking those questions and it looked like he was quoting some noob to mock him or someone else using his account to ask them

Although, about his second question: Some OLLs in fact do give PLL skips more often than others (someone guess which).

-It felt weird to be answering a question of his because frankly I'm still a "noob" and I definitely know who he is >_>

-The only thing I can think of is whatever OLL algorithm has the permutation of more common PLL (for example if your OLL algorithm does a G-perm would you be more likely to get a PLL skip than if it did an H-perm?)

 

[whauk]

Some OLLs in fact do give PLL skips more often than others (someone guess which).

I have no idea to explain this. Please enlighten us

 

[cmhardw]

Some OLLs in fact do give PLL skips more often than others (someone guess which).

-The only thing I can think of is whatever OLL algorithm has the permutation of more common PLL (for example if your OLL algorithm does a G-perm would you be more likely to get a PLL skip than if it did an H-perm?)

After reading Jaycee's post my guess would be a particular X-OLL alg that does a G perm. If you assume that the solver knows the reflections/inverses of that alg as well, then s/he could skip PLL 2/9 the time that OLL came up using AUF and reflection/inverse of the OLL alg (when necessary).

 

[Carrot]

After reading Jaycee's post my guess would be a particular X-OLL alg that does a G perm. If you assume that the solver knows the reflections/inverses of that alg as well, then s/he could skip PLL 2/9 the time that OLL came up using AUF and reflection/inverse of the OLL alg (when necessary).

yours would require learning multiple OLL's... If you take the 4 flipped edges you would due to symmetry be able to apply same alg from 4 different angles and therefore increase your chance of PLL skip by factor 4

Also, it seems all PLLs have same probability if you take AUF into account (correct me if I'm wrong)

 

[Stefan]

Spoiler: Why some OLLs give PLL skips more often than others

Because some OLLs occur more often than others.

 

[TDM]

After reading Jaycee's post my guess would be a particular X-OLL alg that does a G perm. If you assume that the solver knows the reflections/inverses of that alg as well, then s/he could skip PLL 2/9 the time that OLL came up using AUF and reflection/inverse of the OLL alg (when necessary).

If you're only using one alg, then you could use this and AUF before the alg to force a skip. If you're not including mirrors/inverses it doesn't have to be a G perm, just any PLL without rotational symmetry (N/H/Z).

Also, it seems all PLLs have same probability if you take AUF into account (correct me if I'm wrong)

Yeah, I think your right.

Spoiler: Why some OLLs give PLL skips more often than others

Because some OLLs occur more often than others.

Also a possible was of answering... although with Chris' method the least common OLLs, the ones with rotation symmetry, can be used to force a PLL skip more often, giving them an equal chance of a PLL skip, so you're not necessarily right.

 

[Stefan]

Also a possible was of answering... although with Chris' method the least common PLLs, the ones with rotation symmetry, can be used to force a PLL skip more often, giving them an equal chance of a PLL skip, so you're not necessarily right.

(I think you mean OLLs.) Yes, I very much like that idea. But I'd say you're doing more than just OLL then, as you're inspecting and reacting to the permutation. Also, the "solved OLL" case is one of those with rotation symmetry but you can't exploit that symmetry in that way.

 

[Rubiks560]

What is the probability of getting a PLL skip if you use OLLCP every solve?
And can you explain to me how you got the answer?

Nvm. 1/12 right?

 

[kcl]

What is the probability of getting a PLL skip if you use OLLCP every solve?
And can you explain to me how you got the answer?

I'm tempted to say it's 1/12 because that's what it is with COLL and COLL is a subset..

 

[TDM]

(I think you mean OLLs.) Yes, I very much like that idea. But I'd say you're doing more than just OLL then, as you're inspecting and reacting to the permutation. Also, the "solved OLL" case is one of those with rotation symmetry but you can't exploit that symmetry in that way.

Yeah, sorry I meant OLLs. And you can say that a G perm is a way of "solving the solved OLL", because technically it is as the OLL is solved after doing it I see what you're saying though.

What is the probability of getting a PLL skip if you use OLLCP every solve?
And can you explain to me how you got the answer?

Nvm. 1/12 right?

I'm tempted to say it's 1/12 because that's what it is with COLL and COLL is a subset..

Yes, the chance of an EPLL skip is 1/12. Chances of all EPLLs:
Ua: 4/12 = 1/3
Ub: 4/12 = 1/3
Z: 2/12 = 1/6
H: 1/12
Solved: 1/12

 

[cmhardw]

*This will seem to be a strange question. It includes edits based on me having miscalculated the original question, and instead calculating the answer to a different scenario. Everything is now corrected, and the answer given is for the question posed.*

Here's a fun question: Fully scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. What is the probability that none of the 6 centers are oriented correctly (as they are in the solved state)?

Answer below:

Spoiler

\( \frac{365}{2048} \)

\( \frac{{6 \choose 0}*4^5*2-{6 \choose 1}*4^4*2+{6 \choose 2}*4^3*2-{6 \choose 3}*4^2*2+{6 \choose 4}*4*2-{6 \choose 5}*2+{6 \choose 6}}{4^5*2} \)

--edit--

And a cool followup question.

Fully scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. What is the expected number of correctly oriented centers?

Hint: There is a short way and a long way to calculate this. Both give the same answer.

Spoiler

Easy way: 6 centers each with probability \( \frac{1}{4} \) to be solved. \( 6*\frac{1}{4}=\frac{3}{2} \) or 1.5 solved centers expected.

Long way: (calculation done before hand)
Number of states with:
0 centers oriented: 365
1 center oriented: 726
2 centers oriented: 615
3 centers oriented: 260
4 centers oriented: 75
5 centers oriented: 6
6 centers oriented: 1

\( 0*\frac{365}{2048}+1*\frac{726}{2048}+2*\frac{615}{2048}+3*\frac{260}{2048}+4*\frac{75}{2048}+5*\frac{6}{2048}+6*\frac{1}{2048}=\frac{3}{2} \)

 

[TDM]

Here's a fun question: Fully scramble a 3x3x3 supercube. What is the probability that none of the 6 centers are oriented correctly (as they are in the solved state)?

Answer below:

Spoiler

\( \frac{365}{2048} \)

\( \frac{{6 \choose 0}*4^5*2-{6 \choose 1}*4^4*2+{6 \choose 2}*4^3*2-{6 \choose 3}*4^2*2+{6 \choose 4}*4*2-{6 \choose 5}*2+{6 \choose 6}}{4^5*2} \)

Sorry, but I don't understand this... surely it would be (3/4)^6? Our answers aren't far from each other, but I just don't see how it can be anything else.

 

[cmhardw]

Sorry, but I don't understand this... surely it would be (3/4)^6? Our answers aren't far from each other, but I just don't see how it can be anything else.

Wow, you're right. Hmm... Now I am wondering what exactly it is I calculated, especially since the case count still adds up to 2048.

Hmmm.... I'm at work now, but I'll think on that more when I get home. It could just be coincidence and I might have just calculated nonsense.