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Probability Thread

What is the chance of having zero misoriented edges on two axis while the third one has at least two misoriented edges on a cube that is in a random, solvable state?

P(edges oriented on all three axes) = 1/(2^11 ⋅ 12!/4!^3) = 1/70963200
P(edges oriented on F-B and L-R axes) = 1/(2^11 ⋅ 12!/(4!8!)) = 1/1013760
P(edges oriented on F-B and L-R, but not on U-D) = 1/1013760 − 1/70963200 = 23/23654400
P(edges oriented on exactly two axes ) = 3 ⋅ 23/23654400 = 23/7884800

Roughly 1 in 340000, in other words.
 
P(edges oriented on all three axes) = 1/(2^11 ⋅ 12!/4!^3) = 1/70963200
P(edges oriented on F-B and L-R axes) = 1/(2^11 ⋅ 12!/(4!8!)) = 1/1013760
P(edges oriented on F-B and L-R, but not on U-D) = 1/1013760 − 1/70963200 = 23/23654400
P(edges oriented on exactly two axes ) = 3 ⋅ 23/23654400 = 23/7884800

Roughly 1 in 340000, in other words.

Thank you very much! :)
 
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If you take all the solvable positions on the rubik's cube, count the EO from all 3 angles and then choose the EO with the lowest number of misoriented edges, (if multiple orientations have the same number, green front/red front/white front is the preference) what percent of the time wil each EO use rounded to at least 2 decimal places? (eg. green front has 6, red front has 4 and white front has 4, so red is chosen because red and white have the same number, but red is chosen before white because of the preference mentioned above.)

I hope someone is able to answer this challenging question. If you need more clarity, I will do my best to explain it better.
 
If you take all the solvable positions on the rubik's cube, count the EO from all 3 angles and then choose the EO with the lowest number of misoriented edges, (if multiple orientations have the same number, green front/red front/white front is the preference) what percent of the time wil each EO use rounded to at least 2 decimal places? (eg. green front has 6, red front has 4 and white front has 4, so red is chosen because red and white have the same number, but red is chosen before white because of the preference mentioned above.)

I hope someone is able to answer this challenging question. If you need more clarity, I will do my best to explain it better.

At first I was going to say this wasn't challenging, then I got stumped on how to count flipped edges in multiple orientations.

Define a reference orientation by assuming every quarter turn flips four edges. To count how many flipped edges there are on, say, the standard ZZ EO reference, "flip" the edges that are on the S slice, "flip" the edges that would be on the S slice on a solved cube, and then count. Repeat this for the M and E slices to get the numbers of misoriented edges on the three possible ZZ EO references.

Code:
from itertools import product
from collections import Counter

combs = Counter(tuple(sorted(a[:4]) + sorted(a[4:8]) + sorted(a[8:])) for a in product(range(3), repeat=12) if 4 == a.count(0) == a.count(1))
orients = [a for a in product(range(2), repeat=12) if sum(a) % 2 == 0]

axis_counts = [0, 0, 0]

for comb in combs:
    for orient in orients:
        flip_counts = [0, 0, 0]
        for i in range(3):
            flip_counts[ i] = sum((orient[j] + (j // 4 == i) + (comb[j] == i)) % 2 for j in range(12))
        axis_counts[flip_counts.index(min(flip_counts))] += combs[comb]

print(axis_counts)

The output is [37365070, 20481060, 13117070], id est the first choice gets chosen 52.654% of the time, the second choice 28.862%, and the third choice 18.484%.
 
At first I was going to say this wasn't challenging, then I got stumped on how to count flipped edges in multiple orientations.

Define a reference orientation by assuming every quarter turn flips four edges. To count how many flipped edges there are on, say, the standard ZZ EO reference, "flip" the edges that are on the S slice, "flip" the edges that would be on the S slice on a solved cube, and then count. Repeat this for the M and E slices to get the numbers of misoriented edges on the three possible ZZ EO references.

Code:
from itertools import product
from collections import Counter

combs = Counter(tuple(sorted(a[:4]) + sorted(a[4:8]) + sorted(a[8:])) for a in product(range(3), repeat=12) if 4 == a.count(0) == a.count(1))
orients = [a for a in product(range(2), repeat=12) if sum(a) % 2 == 0]

axis_counts = [0, 0, 0]

for comb in combs:
    for orient in orients:
        flip_counts = [0, 0, 0]
        for i in range(3):
            flip_counts[ i] = sum((orient[j] + (j // 4 == i) + (comb[j] == i)) % 2 for j in range(12))
        axis_counts[flip_counts.index(min(flip_counts))] += combs[comb]

print(axis_counts)

The output is [37365070, 20481060, 13117070], id est the first choice gets chosen 52.654% of the time, the second choice 28.862%, and the third choice 18.484%.

Thank you very much for the answer. You appear to have an amazing knowledge of cube theory. :)
 
At first I was going to say this wasn't challenging, then I got stumped on how to count flipped edges in multiple orientations.

Define a reference orientation by assuming every quarter turn flips four edges. To count how many flipped edges there are on, say, the standard ZZ EO reference, "flip" the edges that are on the S slice, "flip" the edges that would be on the S slice on a solved cube, and then count. Repeat this for the M and E slices to get the numbers of misoriented edges on the three possible ZZ EO references.

Code:
from itertools import product
from collections import Counter

combs = Counter(tuple(sorted(a[:4]) + sorted(a[4:8]) + sorted(a[8:])) for a in product(range(3), repeat=12) if 4 == a.count(0) == a.count(1))
orients = [a for a in product(range(2), repeat=12) if sum(a) % 2 == 0]

axis_counts = [0, 0, 0]

for comb in combs:
    for orient in orients:
        flip_counts = [0, 0, 0]
        for i in range(3):
            flip_counts[ i] = sum((orient[j] + (j // 4 == i) + (comb[j] == i)) % 2 for j in range(12))
        axis_counts[flip_counts.index(min(flip_counts))] += combs[comb]

print(axis_counts)

The output is [37365070, 20481060, 13117070], id est the first choice gets chosen 52.654% of the time, the second choice 28.862%, and the third choice 18.484%.
How about with two orientations instead of three?
 
How about with two orientations instead of three?

Good thing I posted the code instead of letting it get buried in my command history!

Just change flip_counts to be a two-element list and the range(3) to range(2); running the modified code snippet gives 67.1% for first choice and 32.9% for second choice.

(If you want to calculate this by hand, unlike the three-axis calculation, the number of cases here is small enough to be doable in maybe a few hours, but thank goodness computers exist.)
 
How many LL cases are there if there is a formed 2x1x1 block on U, either flipped or not?
I guess more than half of full LL.

Consider any particular edge. There are 2 possible corners that can connect to it. The chance of the first corner being on the correct side of the edge and oriented is 1 in 12. There's also a 1 in 2 chance that the first corner won't be touching the edge at all and 1 in 6 chance that the first corner is on the correct side but not oriented. In both of these cases there's a 1 in 9 chance of the second corner being on the correct side and oriented. That means the probability of having a 2x1x1 block involving the specific edge is 1/12+(1/2+1/6)*(1/9) = 17/108, or 15.74% of LL cases. Then you need to account for the fact that one of the other edges could be part of a 2x1x1 block, but my attention span ran out at about this point.
 
What is the probability that, in a random scramble, at least 1 corner edge pair is solved?

What is the probability that, in a random scramble, at least 1 corner edge pair is 1 move from solved?
 
What is the probability that, in a random scramble, at least 1 corner edge pair is solved?

What is the probability that, in a random scramble, at least 1 corner edge pair is 1 move from solved?

By solved do you mean both pieces in place?

I think the probability of a corner edge pair being formed is close to 1 - 1/e
 
What is the probability that, in a random scramble, at least 1 corner edge pair is solved?

What is the probability that, in a random scramble, at least 1 corner edge pair is 1 move from solved?

The 1 - 1/e figure I think is right as well, so like 63.2% ish.

1 move from solved *or* solved has to be pretty close to 100%, but I'm speculating.
 
By solved do you mean both pieces in place?

I think the probability of a corner edge pair being formed is close to 1 - 1/e

Sorry I meant formed, not necessarily solved.

Do you happen to how people arrived at that figure?
 
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