Probability Thread

[PoHos1]

what the **** so its wery big lucky

hele půjdeš na facebook můžeme pokecat

 

[AndersB]

OMG so lucky, that's almost as lucky as solving a 2x2 by only doing random moves!

 

[nqwe]

OMG so lucky, that's almost as lucky as solving a 2x2 by only doing random moves!

This is by far easier to "achieve".

 

[Bob]

This is by far easier to "achieve".

How so?

 

[mitch1234]

Probablilty of Getting God's Number for an Optimal Solution

I decided that I would do some simple calculations and figure this out for myself. Probablity for was rounded to two significant figures, so its a bit off but who cares about exact numbers. So here is what I have come up with:

Well there are 43,252,003,274,489,856,00 permutations for the cube. And all of those permutations all of them can be solved in 20 moves or less. No cube can be solved in >20 moves with an optimal solution. 20 moves is less likely an optimal solution than 8 is. So I decided I would calculate just how likely it is to happen, then put it in perspective.

For solving the cube in 20 moves there is a 3.0 x10^8 cases. These cases can not be solved in anything less than 20. So the odds of picking a random cube (you will need a lot of cubes for this) is ~1 / 144,173,344,248. Big number, well I told you would need a lot of cubes.

These cases practically don't exist, but we do have one really well know case of it, being the super flip.

Now to put it in perspective. There are about 7.056 billion people that live on earth currently. And if we had the entire population of earth get a cube and randomly scramble it a bit then all at once stop scrambling, we would need 20 more populations of the earth just to have 1 person have 1 cube that is exactly gods number.

Okay I'm done, time to sleep. There are probably some errors in this but just tell me so if you think there are, I'll gladly check my math.

 

[kinch2002]

Probablilty of Getting God's Number for an Optimal Solution etc

Thanks, that was interesting. Just wondering where you got the number 3 x 10^8 for the number of 20 move optimal cases from?

 

[Ranzha]

Thanks, that was interesting. Just wondering where you got the number 3 x 10^8 for the number of 20 move optimal cases from?

3×10^8 is the approximate figure from cube20.org, for one thing =)

 

[DrJohnFever]

What's the probability of taking your cube apart, putting the pieces back in completely randomly, and having the cube be solvable?

 

[Noahaha]

What's the probability of taking your cube apart, putting the pieces back in completely randomly, and having the cube be solvable?

1/12

1/3 chance corners are oriented, 1/2 chance of even PLL parity and 1/2 chance of edges flipped the right way.

 

[cuBerBruce]

Is there some reason this thread was moved (to Speedcubing Help/Questions)?

 

[ThomasJE]

Is there some reason this thread was moved (to Speedcubing Help/Questions)?

Probably because people are commonly asking many questions here; like the one I am about to ask.

After OLL/OCLL/WV etc., what is the chance of the corners being solved (counting H-perm as cornered solves, and discounting AUF's)?

 

[kinch2002]

Probably because people are commonly asking many questions here; like the one I am about to ask.

After OLL/OCLL/WV etc., what is the chance of the corners being solved (counting H-perm as cornered solves, and discounting AUF's)?

1/6
Think of it like this:
1. Pick any corner. Doesn't matter where it is because you're allowing AUFs (at least that's how I'm interpreting your question)
2. The corner to the right of it needs to be a specific piece out of the 3 corners left. So that's 1/3 chance.
3. The next corner needs to be a specific piece out of the the 2 corners last. So that 1/2 chance.
4. Both steps 2 and 3 need to happen for corners to be solved so it's 1/3*1/2=1/6

 

[ThomasJE]

1/6
Think of it like this:
1. Pick any corner. Doesn't matter where it is because you're allowing AUFs (at least that's how I'm interpreting your question)
2. The corner to the right of it needs to be a specific piece out of the 3 corners left. So that's 1/3 chance.
3. The next corner needs to be a specific piece out of the the 2 corners last. So that 1/2 chance.
4. Both steps 2 and 3 need to happen for corners to be solved so it's 1/3*1/2=1/6

Ah, thanks. Now I think of it, I could have done it myself, since PLL skip is 1/72, and EPLL skip is 1/12, so 72 divided by 12 is 6; therefore the chance of CPLL skip is 1/6.

 

[cuBerBruce]

Probably because people are commonly asking many questions here; like the one I am about to ask.

And people like me will probably ignore this thread from now on.

 

[Smiles]

What is the probability of all edges being oriented in a ZZ solve before starting...
with a fully fixed colour scheme?
being fully colour neutral?

I got a 3 move cross on white with all the edges oriented (qqtimer scramble). I wasn't doing ZZ, but it was interesting that I never turned F or B during the solve.

 

[ThomasJE]

What is the probability of all edges being oriented in a ZZ solve before starting...
with a fully fixed colour scheme?
being fully colour neutral?

I got a 3 move cross on white with all the edges oriented (qqtimer scramble). I wasn't doing ZZ, but it was interesting that I never turned F or B during the solve.

Fixed colour scheme is 1/2048/~0.05% (source), but full colour neutral; I have no idea.

 

[JDogg]

OLL Case Probabilities

I've searched around the forum and Google and can't seem to find the right answer here. Are the probabilities typically listed for all the OLL cases wrong? I did the math and something isn't adding up right. There are several places that list the probabilities for each of the 58 cases (57 + the 1 solved case), like the OLL document on this page. Cubefreak shows the same probabilities and I've seen the same numbers listed in a couple other places too.

Of the 58 cases:
2 of them have a 1/216 probability.
4 of them have a 1/108 probability.
52 of them have a 1/54 probability.

With common denominator of 216, that's 2/216 + 8/216 + 208/216, which equals 218/216, or 100.926%. There's an extra 1/108th there. Those probabilities can't possibly be right.

Is there something obvious I'm missing here? Does anyone actually know how those probabilities are calculated? I assume it's a calculation based simply on the combination of the number of edges that are flipped and then the number of corners that are rotated a certain way (which is why the majority of cases end up being equally likely). But I wouldn't know where to begin trying to calculate them myself.

 

[whauk]

4 of them have a 1/108 probability.

i counted 5. and with 51 having 1/54 probability it should turn out correctly.

on calculating these: one OLL with a certain AUF has 3^3 possibilities for CO (every corner has 3 and the last one is fixed) and 2^3 possibilities for EO (same). that turns out to be 216 OLL cases with a certain AUF. however we dont mind AUF and consider one OLL "equal" with another if U,U',U2 makes them exact same cases. most OLLs look different when doing U,U',U2 so they actually have a chance of 4/216=1/54. (because all 4 possibilities of {(none),U,U',U2} were counted as different cases before) then again there are some that look identical after U2 and therefore have a chance of 2/216=1/108. and even some cases look identical after U,U' and U2 so they have a chance of 1/216.

 

[Sa967St]

Without ignoring AUF, there are 3^3=27 possible corner orientations and 2^3=8 edge orientations. The common denominator comes from 27x8=216.

Within these 216 cases, most of the OLL cases appear four times, once for each y rotation. These are the OLLs that have a 4/216=1/54 probability of occurring. The ones that appear two times have a 2/216=1/108 probability and the ones that appear only once have a 1/216 probability.

There are five OLLs (1, 21, 55, 56, 57) that have a 1/108 probability and two (20, solved) that have a 1/216 probability.

Essentially, the probabilities depend on their reflections. The ones that are 1/108 look the same when you do a y2 and the ones that are 1/216 look the same when you do y, y2 and y'.

 

[JDogg]

i counted 5. and with 51 having 1/54 probability it should turn out correctly.

on calculating these: one OLL with a certain AUF has 3^3 possibilities for CO (every corner has 3 and the last one is fixed) and 2^3 possibilities for EO (same). that turns out to be 216 OLL cases with a certain AUF. however we dont mind AUF and consider one OLL "equal" with another if U,U',U2 makes them exact same cases. most OLLs look different when doing U,U',U2 so they actually have a chance of 4/216=1/54. (because all 4 possibilities of {(none),U,U',U2} were counted as different cases before) then again there are some that look identical after U2 and therefore have a chance of 2/216=1/108. and even some cases look identical after U,U' and U2 so they have a chance of 1/216.

Jeez. I triple-checked my math by counting up the probabilities on 3 different websites and missed the 1/108 for the single dot case on all 3 sites. So to answer my question - yes, there was something obvious I was missing.

Interesting about the calculations though. It works different than I thought, but it's still neatly simple.