Probability Thread

[whauk]

What is the probability of a first centre skip on 7x7?

there are 6 types of center pieces.
there are 24 pieces of each type.
so that makes 24!^6 possibilities.
if one center is solved there are 4!^6 possibilities for one certain center and 4!^6*6 if you want just some random center to be solved.
so thats (4!*6/24!)^6 in the end.
and thats 1.56*10^-130
not very likely

 

[That70sShowDude]

What are the chances of having none of the 8 f2l cubies in the last layer after solving the cross?

 

[TheTurtleman9]

On a 2x2x4, if solving center layers then outer, what is the probability of an outer layer skip (both of them)?

 

[cmhardw]

What are the chances of having none of the 8 f2l cubies in the last layer after solving the cross?

\(
\frac{(4!)^4}{(8!)^2} = \frac{1}{4900} \)

--edit--
Wow.. Interestingly, the probability of every single F2L piece ending up in the LL after finishing the cross is also \( \frac{1}{4900} \)

 

[Matt S]

What are the chances of having none of the 8 f2l cubies in the last layer after solving the cross?

\(
\frac{(4!)^4}{(8!)^2} = \frac{1}{4900} \)

--edit--
Wow.. Interestingly, the probability of every single F2L piece ending up in the LL after finishing the cross is also \( \frac{1}{4900} \)

That makes sense, because the number of corners and edges in the LL is equal to the number of each in the F2L once you subtract out the cross. So, having every F2L piece in the LL is equivalent to having every LL piece in the F2L, which is equivalent by symmetry to having every F2L piece in the F2L.

 

[cmhardw]

On a 2x2x4, if solving center layers then outer, what is the probability of an outer layer skip (both of them)?

There are 8 corner pieces remaining after the two inner layers have been solved, and all 8! permutations of those pieces are possible. There is no concept of corner orientation, so you can ignore this. Lastly, count how many "solved" states there are, allowing for the possibility of an AUF on either the U or D layers alone, or on both layers. Set this up as a fraction to calculate the total probability.

In case you haven't noticed, I'm purposefully not actually answering your question. The reason for this is that knowing how to calculate probabilities like this one is a good stepping stone for how to calculate the more complex ones. If my hints don't help, reply in this thread or PM me, but try to start from there and see if that helps.

Chris

 

[jaap]

What is the probability of a first centre skip on 7x7?

there are 6 types of center pieces.
there are 24 pieces of each type.
so that makes 24!^6 possibilities.
if one center is solved there are 4!^6 possibilities for one certain center and 4!^6*6 if you want just some random center to be solved.
so thats (4!*6/24!)^6 in the end.
and thats 1.56*10^-130
not very likely

This isn't correct.
For a particular type of centre piece, there are 24-choose-4 ways for them to be arranged so that the top face is correct, i.e.
24C4 = 24!/(20!4!) = 10626 ways
For all 6 types of top face centre pieces to be correct, that gives
1/(24C4)^6 = 1/ 1439523865867191634829376 = 6.95e-25
This is the probability that the top face centre is solved, regardless of the state of the other face centres.

Assuming for the moment that the six faces are independent and mutually exclusive, the probability of any of the six faces being solved is 6 times that, or about 4.17e-24. To calculate the real probability value you'd probably need to use the inclusion-exclusion principle, but the result won't be far off this approximation.

 

[TheTurtleman9]

On a 2x2x4, if solving center layers then outer, what is the probability of an outer layer skip (both of them)?

There are 8 corner pieces remaining after the two inner layers have been solved, and all 8! permutations of those pieces are possible. There is no concept of corner orientation, so you can ignore this. Lastly, count how many "solved" states there are, allowing for the possibility of an AUF on either the U or D layers alone, or on both layers. Set this up as a fraction to calculate the total probability.

In case you haven't noticed, I'm purposefully not actually answering your question. The reason for this is that knowing how to calculate probabilities like this one is a good stepping stone for how to calculate the more complex ones. If my hints don't help, reply in this thread or PM me, but try to start from there and see if that helps.

Chris

Would it be 1/8! because 8! is all of the permutations and orientations of the corners, and there's only 1 solved state?

Didn't put AUF in though. Would that be 1/(8!*16)?

 

[Rpotts]

This thread should be renamed, Ask Chris Hardwick a question about probability!

LOL

 

[cmhardw]

Would it be 1/8! because 8! is all of the permutations and orientations of the corners, and there's only 1 solved state?

Didn't put AUF in though. Would that be 1/(8!*16)?

Almost - and very close! The only thing to remember is that the final fraction that makes the probability for this type of problem is:

\( \frac{\mbox{number of "solved" states allowing for AUF and/or ADF}}{\mbox{Total number of possible states}} \)

Setup the fraction this way and you'll have your answer.

This thread should be renamed, Ask Chris Hardwick a question about probability!

LOL

Haha, I'm flattered, but at the same time there are a number of people on this forum whose knowledge of probability dwarf my own. I'm just interested in this type of finite probability, and I'm also a math teacher so I can't get away from math, it's too much fun!

Chris

 

[TheTurtleman9]

Oh,so , 16/8! ? 1/2520?

 

[cmhardw]

Oh,so , 16/8! ? 1/2520?

Yep!

 

[eastamazonantidote]

Stuff on ZBF2L

Awesome thanks!

 

[TheTurtleman9]

Oh,so , 16/8! ? 1/2520?

Yep!

Oh yay.

 

[Weston]

Whats the average move count for a CN 2x2x2 block ignoring the corner orientation?

 

[hawkmp4]

Whats the average move count for a CN 2x2x2 block ignoring the corner orientation?

That's a pretty complicated problem that's going to require a computer search, not just some probability calculations.

 

[Zane_C]

Just wondering what the probability of a complete middle layer edge skip is (LBL). Without any influence such as trying to preserve pairs when inserting corners.

 

[hawkmp4]

Just wondering what the probability of a complete middle layer edge skip is (LBL). Without any influence such as trying to preserve pairs when inserting corners.

1/8 * 1/7 * 1/6 * 1/5 for edge permutation...1/2^4 for edge orientation.
I believe. I could be wrong. It's really late.
so
1/26880

 

[Zane_C]

Thanks appreciated, just took a look and realised that's not too hard to figure out myself.

 

[hawkmp4]

You might multiply that by four for four orientations of the cube...but that's the general idea. If you read some posts by Chris Hardwick, you'll pick it up pretty quick. He's the man- at the math, and teaching it.