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Probability Thread

TDM said:
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.
Click to expand...

I edited the original post. I was calculating a different scenario and didn't realize it. See the original post for the new question (to which the answer I gave is the right answer).
 
This gives me an idea. I am double posting because I think this one deserves its own post. Here are two more fun questions:

Question 1) Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the probability that none of the centers are in their correct orientations?

Question 2) Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the expected number of correctly oriented centers?
 
cmhardw said:
This gives me an idea. I am double posting because I think this one deserves its own post. Here are two more fun questions:

Question 1) Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the probability that none of the centers are in their correct orientations?

Question 2) Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the expected number of correctly oriented centers?
Click to expand...
0?
E: Just realised this is completely wrong... :fp
 
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A possibly simpler answer to
"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)?":

All centers are misoriented either 1, 2, or 3 turns clockwise. But the number of centers that are 2 turns clockwise must be even (i.e. 0, 2, 4, or 6); and there are two ways to choose each of the remaining centers. So out of 4^6/2 possibilities, this can be done in
\( {6 \choose 0}*2^6+{6 \choose 2}*2^4+{6 \choose 4}*2^2+{6 \choose 6}*2^0 = 365 \)
ways. Thus the probability is 365/2048.

And similarly, for
"Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the probability that none of the centers are in their correct orientations?", we have
Again, all centers are misoriented either 1, 2, or 3 turns clockwise. But now the parity is odd, so the number of centers that are 2 turns clockwise must be odd (i.e. 1, 3, or 5); and there are two ways to choose each of the remaining centers. So out of 4^6/2 possibilities, this can be done in
\( {6 \choose 1}*2^5+{6 \choose 3}*2^3+{6 \choose 5}*2^1 = 364 \)
ways. Thus the probability is 364/2048 = 91/512.
 
cmhardw said:
This gives me an idea. I am double posting because I think this one deserves its own post. Here are two more fun questions:

Question 1) Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the probability that none of the centers are in their correct orientations?

Question 2) Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the expected number of correctly oriented centers?
Click to expand...

Before looking at TDM's answer I got:
Q1: \( \frac{364}{2048}=\frac{91}{512} \)

Q2: \( \frac{3}{2} \)

Q2:
\( 0*\frac{364}{2048}+1*\frac{732}{2048}+2*\frac{600}{2048}+3*\frac{280}{2048}+4*\frac{60}{2048}+5*\frac{12}{2048}+6*\frac{0}{2048} \)

It seems this can also be calculated by reasoning that the expected number of oriented centers is 1.5 for a randomly scrambled 3x3x3 supercube. If you fix center orientation parity to even then the expected value of oriented centers is 1.5 centers (see my earlier post). Therefore the expected number of oriented centers if you fix odd orientation parity must also be also be 1.5 centers since odd and even parity both have probability 1/2
 
About "no centers solved": Ignoring edges/corners (i.e. ignoring parity) there are 3^6=729 cases with no centers solved. Fix an order of the six centers and interpret the cases as ternary numbers from 000000 to 222222. Adding 1 turns each even parity case except 222222 into an odd parity case and this is a one-to-one correspondence. So there's one more even parity case than there are odd parity cases, meaning the 729 split into 365 for even parity and 364 for odd parity.
 
43 Quintilian Possibilities, how many are solved?

This is kind of a weird question, but one that's been bugging me. I remember having a smiley face cube a long time ago, and I solved it, but, the centers were twisted upside down on 2 of the faces. I could never fix it, so it looked kind of goofy. If that were a standard cube with plain stickers, that would be solved. So I'm wondering how many solved possibilities there are among the 43 Quintilian possible permutations. If it's only one, then what about those other times when the centers might be twisted?

Also, I remember seeing special stickers for a cube that would simulate my smiley issue, where they had horizontal lines across all of the stickers, so not only would you have to solve, but also fix the twisted centers issue (and maybe other issues that I'm unaware of. First of all, what is that cube called, and does it have MORE than 43 quintilian possible combinations?
 
CyanSandwich said:
I think the 43 quintillion. combinations don't include spun centers. I'm basing that on the fact that they don't change anything on a Rubik's cube and the equation doesn't take it into account.
Click to expand...

Yeah, that's what I was thinking. So, 43 quintillion and only 1 solved. I guess guess rubik's cubes that aren't solved with spun centers must have much more than 43 quint.
 
cmowla said:
You guys are absolutely correct. In fact, there are (click me) solved positions for the 2x2x2 through 11x11x11 cubes, for example. The cubes which have stickers on the centers which makes them distinguishable (big cube centers) or helps you tell if they are rotated (fixed centers, like those on the 3x3x3) are called supercubes. So the answer to your question is 2048 (although 2048 is not counted as part of the 43 quintillion).
Click to expand...

Thank you. That will help me sleep at night :D:tu
 
DeeDubb said:
This is kind of a weird question, but one that's been bugging me. I remember having a smiley face cube a long time ago, and I solved it, but, the centers were twisted upside down on 2 of the faces. I could never fix it, so it looked kind of goofy. If that were a standard cube with plain stickers, that would be solved. So I'm wondering how many solved possibilities there are among the 43 Quintilian possible permutations. If it's only one, then what about those other times when the centers might be twisted?
Click to expand...

If you've ever wondered the same thing about the 20x20x20 cube, you can check out how many ways there are to rearrange a solved 20x20x20 cube and it still appear solved. :)
 
If I always use phasing (with ZZ), and after I use COLLs taken from ZZLLs, wich is the probability of a PLL skip? Thanks!
 
Saxsoprano said:
If I always use phasing (with ZZ), and after I use COLLs taken from ZZLLs, wich is the probability of a PLL skip? Thanks!
Click to expand...
You're preserving phasing, so the relative probabilities of the outcomes are the same as the probabilities of the EPLLs without including U perms.
H perm: 1/12 -> 1
Z perm: 2/12 -> 2
Solved: 1/12 -> 1
Chance of skip = 1/4
 
Just a silly thing - this weekend I got 3 PLL skips in three different rounds, all from the same alg x U R' U' L' U R U' r'?. I think I managed to predict the latter 2 skips! Any idea what the probability of this is from the 27 sighted solves I did this weekend? (Ignoring 2x2x2 because I know CLL)
 
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