Probability Thread

[goodatthis]

What is the probability of a L4E skip on 5x5, assuming freeslice is used.

Asking because it happened to me two days ago. The solve wasn't PB though

 

[cmhardw]

What is the probability of a L4E skip on 5x5, assuming freeslice is used.

Asking because it happened to me two days ago. The solve wasn't PB though

Using the midges as reference it should be
1/8! = 1/40320

If you used AVG or anything to try to partially build tredges throughout your solve, then the probability increases.

 

[MennoniteCuber1]

What are the most probable CLL cases for each orientation?

 

[kcl]

What are the most probable CLL cases for each orientation?

Anything not pure or diag?

 

[rybaby]

What is the probability of having a solved square in a scramble?

 

[Stefan]

What is the probability of having a solved square in a scramble?

1

(assuming you meant 3x3x3 or another puzzle with square stickers)

 

[rybaby]

1

(assuming you meant 3x3x3 or another puzzle with square stickers)

Why do you even waste your time with answers like these?
Now is there someone who would like to be helpful?

 

[cmhardw]

Why do you even waste your time with answers like these?
Now is there someone who would like to be helpful?

The probability of a solved face (not necessarily a layer) (3x3 square) is approximately: 1.3*10^(-7)

Clicky
Also clicky

 

[rybaby]

The probability of a solved face (not necessarily a layer) (3x3 square) is approximately: 2.2*10^(-12)

Clicky
Also clicky

How about a solved 2x2 square permuted properly? (Not necessarily aligned with centers)

 

[cmhardw]

How about a solved 2x2 squat permuted properly? (Not necessarily aligned with centers)

Oh, well you should have asked that in the first place

I will be away from my computer for a bit, but can work on this a little later in the day. This is a tricky question and will likely require very clever math, or a computer program to run an intelligent search search of case types. I think clever math should be able to do it, but there are others on here with a much more solid math background who would know better than I.

 

[ChickenWrap]

1

(assuming you meant 3x3x3 or another puzzle with square stickers)

I might not have invented my own BLD method, but at least I'm not an a**hole to random people on the internet.

 

[cmhardw]

I might not have invented my own BLD method, but at least I'm not an a**hole to random people on the internet.

The question was ambiguous, Stefan made an assumption (and not a bad one) about what the question was referring to.

How is that being an a-hole?

Stefan doesn't like ambiguous questions, and for good reason.

 

[rybaby]

Stefan made an assumption (and not a bad one) about what the question was referring to.

You must really undervalue my intelligence if you think that my original question would be answered by Stefans post.

 

[cmhardw]

What is the probability of having a solved square in a scramble?

1

(assuming you meant 3x3x3 or another puzzle with square stickers)

You must really undervalue my intelligence if you think that my original question would be answered by Stefans post.

Your question, as asked, was answered by Stefan's post. Your intended question was not answered by Stefan's post, because you did not ask your intended question.

That's the difference. Do you understand now why Stefan's post was not mean? He was, in fact, trying to teach you to ask more specific questions so that your intended questions get answered right away.

 

[cmhardw]

How about a solved 2x2 square permuted properly? (Not necessarily aligned with centers)

This is a rough estimation, but will at least give us a basic idea.

A 2x2x1 block is formed when a corner has 2 of it's adjacent edges correctly permuted, as well as the associated center piece.

The number of "positions around a corner" that have at least one 2x2x1 block is:
(3 choose 1)*10*2*4-((3 choose 2)*1)+((3 choose 3)) = 238

There are 24*22*20*6*4=253440 "positions around a corner"

This means that, in regards to one specific corner, the probability of having no 2x2x1 blocks around it is:
(253440-238)/253440 = 253202/253440

The approximate probability that at least one corner on a 3x3x3 would have at least one 2x2x1 is:
1-(253202/253440)^8 or approximately 0.75% chance.

This is a hack calculation, and is not the true value. This is my estimation to figure out roughly about what is this probability.

--edit--

I find it interesting that this probability is corroborated by:

8*(238/253440) which is also approximately 0.75%

This is the chance that one corner has at least one 2x2x1 block multiplied by the 8 corners.

 

[hawkmp4]

The number of "positions around a corner" that have at least one 2x2x1 block is:
(3 choose 1)*10*2*4-((3 choose 2)*1)+((3 choose 3)) = 238

There are 24*22*20*6*4=253440 "positions around a corner"

Could you elaborate on the details of those expressions/is thit correct?
24 possible orientations of the "target' corner?

FWIW I don't think that there's an issue of content, but one of tone. Stefan meant to teach, and OP didn't see that so it came off as abrasive. The dangers of text communication!

 

[cmhardw]

Could you elaborate on the details of those expressions/is thit correct?
24 possible orientations of the "target' corner?

I'm using inclusion/exclusion in a way that Stefan recently taught me.

In the way I am counting, I am picking an arbitrary corner and fixing it as my reference point. For clarity let's call the reference corner UFR. Around this corner are three adjacent edges, UF, FL, UR. There are also three center pieces that touch the corner on a diagonal (U, F, R) centers.

The number of "positions around a corner" that have at least one 2x2x1 block is:
(3 choose 1)*10*2*4-((3 choose 2)*1)+((3 choose 3)) = 238

(3 choose 1)*10*2*4 chooses which of the 2x2x1 blocks around the UFR corner to be "solved", then it permutes one of 10 possible edge pieces to the 3rd edge location around the UFR corner (the one that is not part of the 2x2x1 block), orients that recently permuted edge (2) and then there are 4 possible positions the centers can be in. This is because one center is part of the 2x2x1, but the others have 4 possible 90 degree rotations in their slice. This term calculates the number of cases where one 2x2x1 block exists. It is an overcount, and will be handled by the rest of the inclusion/exclusion procedure.

-((3 choose 2)*1) continues the inclusion/exclusion by subtracting the number of ways that there can be two 2x2x1 blocks solved around the UFR corner. There is, in fact, only one way and that is to have all three 2x2x1s around UFR solved forming a 2x2x2 block. However, for the inclusion/exclusion to work I have to count that there are (3 choose 2) ways to pick "two" 2x2x1s to be solved.

+((3 choose 3)) This continues the inclusion/exclusion by adding back the number of ways that three 2x2x1 blocks can be solved. There is only 1 way, and (3 choose 3) = 1.

Since we know that (3 choose 1)*10*2*4 overcounts the number of 2x2x1s when we have the special case of a 2x2x2, let's look at the 2x2x2 case.

(3 choose 1)*10*2*4 counts the 2x2x2 case three times, since there are three 2x2x1s forming the 2x2x2 block, and this term counts every way that a 2x2x1 can be formed.
-((3 choose 2)*1) subtracts the 2x2x2 block case three times (net is that this case has been counted 0 times). It subtracts it three times because there are three ways to look at exactly two sides solved on a 2x2x2.
+((3 choose 3)) adds back all cases where three 2x2x1s are solved. There is only one way, the special case where you have a 2x2x2. Now this special case has been counted a net total of 1 time, which is good because there is only 1 case.

For all cases where exactly one 2x2x1 block is formed (and there is not 2x2x2), these cases are all counted exactly once in the term (3 choose 1)*10*2*4. So this entire expression using inclusion/exclusion has counted every case with at least 1 2x2x1 block uniquely, and is not an overcount.

Once I have the number of cases using one corner as reference, I then view each corner as independent of all others (I view each corner "in its own bubble") and calculate the probability that every corner has no 2x2x1 around it. Take 1 minus that probability for the probability that at least one corner has at least one 2x2x1. This is not the true probability, as you cannot view each corner as completely independent ("in its own bubble"), but it will be a rough estimation of the true probability.

 

[Stefan]

He was, in fact, trying to teach you to ask more specific questions

This, thank you. I just demonstrated that the question as asked doesn't make sense and was hoping he'd clarify. I thought that was obvious.

2x2 square indeed seems rather difficult.

I find it interesting that this probability is corroborated by:

8*(238/253440) which is also approximately 0.75%

Yeah, for small probabilities, n*p and 1-(1-p)^n are almost the same:
http://www.wolframalpha.com/input/?i=8*p+vs+1-(1-p)^8+for+p+from+0+to+0.01
(This should show a comparison, a plot with the two functions. Reload if it doesn't (a moment ago that happened to me).)

Edit: Oh neat, this works as well: http://www.wolframalpha.com/input/?i=n*p+vs+1-(1-p)^n,+for+p+from+0+to+0.01,+for+n+from+2+to+10

The probability of a solved face (not necessarily a layer) (3x3 square) is approximately: 1.3*10^(-7)

Clicky

I think your "3 adjacent faces" term should end in 2^1, not 2^3.

Clicky

Just a quick question about this one. Why the "/2" in "4!/2*2^3"? The midges can be permuted all 4! ways.

 

[hawkmp4]

Thanks for the explanation and the link, Chris. Good stuff to know.

 

[Stefan]

I just realized you probably can't use inclusion-exclusion simply like that for the "3x3 face" problem. Inclusion-exclusion works by overcounting at first, then uncounting the cases unintentionally counted more than once, then recounting the cases now counted less than once, etc, until every case has been counted exactly once.

Consider the earlier "solved 7x7 edge among the last four" problem:

Counting all cases with at least 1 solved edge (because we haven't counted those with exactly 1 solved edge yet)...

• Counts all cases with exactly 1 solved edge exactly 1 time (each has now been counted 1 time).
• Counts all cases with exactly 2 solved edges exactly 2 times (each has now been counted 2 times).
• Counts all cases with exactly 3 solved edges exactly 3 times (each has now been counted 3 times).
• Counts all cases with exactly 4 solved edges exactly 4 times (each has now been counted 4 times).

Uncounting all cases with at least 2 solved edges (because we have double-counted those with exactly 2 solved edges)...

• Counts all cases with exactly 2 solved edges exactly -1 times (each has now been counted 1 time).
• Counts all cases with exactly 3 solved edges exactly -3 times (each has now been counted 0 times).
• Counts all cases with exactly 4 solved edges exactly -6 times (each has now been counted -2 times).

Counting all cases with at least 3 solved edges (because those with exactly 3 solved edges are at 0 again)...

• Counts all cases with exactly 3 solved edges exactly 1 time (each has now been counted 1 time).
• Counts all cases with exactly 4 solved edges exactly 4 times (each has now been counted 2 times). (marked red for later)

Uncounting all cases with at least 4 solved edges (because after 4-counting them, then (-6)-counting them and then 4-counting them, they're now overall double-counted)...

• Counts all cases with exactly 4 solved edges exactly -1 times (each has now been counted 1 time).

So you can see we counted, uncounted, counted and uncounted. In other words, we added the four numbers multiplied by +1, -1, +1 and -1, respectively. But that only turned out correct because of how "nicely" the countings and uncountings happened to "cancel out".

Now let's get back to the "3x3 face" problem.

For example when handling the "three adjacent faces" subcases of the "three faces" cases, you're *not* unintentionally 4-counting all "four faces" cases (like marked red above in the other problem) but only 2-counting the "four faces except two adjacent" subcases and even not counting the "four faces except two opposite" subcases.

Unless for some "miracle" it's still correct (like the naive +1, -1, +1, -1 turned out correct earlier), I'm afraid your already complicated expression is too simple.