Probability Thread

[thejerber44]

How many combinations are possible with a 2-gen scramble?

 

[TheNextFeliks]

How many combinations are possible with a 2-gen scramble?

I think it should be (7!/2)*((3^6)/3)= 612,360 but not sure.

 

[cmhardw]

How many combinations are possible with a 2-gen scramble?

\( 7!*\frac{6!}{3!*2}*3^5=73483200 \)

You may find this page interesting.

 

[thejerber44]

\( 7!*\frac{6!}{3!*2}*3^5=73483200 \)

You may find this page interesting.

Thanks for the quick reply The website is awesome! A friend told me that he was doing some 2-gen solves (hand scrambles) and when he finished scrambling, the cube was a mere U2 from being solved. He's either really lucky, bad at scrambling, or lying

 

[cmhardw]

Thanks for the quick reply The website is awesome! A friend told me that he was doing some 2-gen solves (hand scrambles) and when he finished scrambling, the cube was a mere U2 from being solved. He's either really lucky, bad at scrambling, or lying

Yeah, odds are in favor of a poor scramble or a fib However, the truth is still possible if your friend got very lucky!

 

[StarOfDoom]

What would the prob. be of getting 3 LL skips in about 17k solves? (this is my situation)

 

[kcl]

What would the prob. be of getting 3 LL skips in about 17k solves? (this is my situation)

Depends if you use any influencing of last slot, WV, COLL, etc.

 

[Jaycee]

This is just my attempt at finding an answer to the question. I assume one of people more knowledgeable about this will give the correct at some point, but I think I'll try anyway.

Chance of an LL skip on any given single solve: 1/15552 = 0.0000643004115

Chance of not getting an LL skip on any given single solve: 15551/15552 = 0.9999356995885

3 solves had an LL skip out of 17000: (3(0.0000643))/17000 = something google can't calculate so I'll do this next thing instead

16997 solves did not have an LL skip: (16997(0.9999356995885)/17000 = 16995.9071 / 17000 (apparently google doesn't wanna go out that many decimal places)

16995.9071/17000 = chances of having 16997 non-LL-skips in 17000 solves = 99.9759241%

Inverse of that = 0.0240759% chance of having 3 LL skips in 17000 solves.

This is all assuming, of course that no influencing methods are used, which is probably not true. Oh well.

7 minutes later edit: Maybe not? Both things happened, (3 LL skips and 16997 non-LL-skips), so I think the probabilities get multiplied. In which case

2.40701035% chance of 3 LL skips in 17000 solves? Maybe. Eh. I stopped making sense a while ago.

 

[uberCuber]

I believe this should just be a binomial probability, right?

(17000 choose 3) * (1/15552)^3 * (15551/15552)^16997 = 7.3% chance of 3 LL skips in 17000 solves.

 

[GuRoux]

I believe this should just be a binomial probability, right?

(17000 choose 3) * (1/15552)^3 * (15551/15552)^16997 = 7.3% chance of 3 LL skips in 17000 solves.

That seems correct.

 

[StarOfDoom]

This is all assuming, of course that no influencing methods are used, which is probably not true. Oh well.

I do not do any influencing. I have enough trouble just finding and getting the f2l pairs inserted in the short amount of time, let alone partial edge control... so I don't think any influencing was involved.

 

[Jaycee]

Ah. In any case my math was wrong and uberCuber's was correct. Don't know why I was adding probabilities instead of multiplying them, that was pretty derp.

 

[LucidCuber]

Does anyone know the probability of getting 4 edges skipped after centres on a 4x4x4?

 

[cmhardw]

Does anyone know the probability of getting 4 edges skipped after centres on a 4x4x4?

The probability that at least 4 edge groups are paired after completing centers is (assuming you are not using Yau or any other approach to solve edges during the solving of centers):

P(at least 4 edge groups paired)=1-P(no edge groups paired)-P(1 edge group paired)-P(2 edge groups paired)-P(3 edge groups paired)

P(at least 4 edge groups paired) = \( 1-\frac{1}{24!}\sum_{i=0}^{3} \sum_{j=0}^{12-i} \binom{12}{i}*\frac{12!}{(12-i)!}*2^i*\binom{12-i}{j}*\frac{(12-i)!}{(12-i-j)!}*2^j*(-1)^{j}*(24-2i-2j)! \)

This is an approximately 0.208% chance.

In fraction form that would be about \( \frac{1}{481} \)

For those interested in the details, here is the distribution that this creates:

Spoiler

Each number represents the total number of all wing edge states that have the described characteristic:

exactly 0 edge groups paired: 368420070161105761075200
exactly 1 edge group paired: 191842564893283713024000
exactly 2 edge groups paired: 50126103570256861593600
exactly 3 edge groups paired: 8768672584653864960000
exactly 4 edge groups paired: 1156363172135829504000
exactly 5 edge groups paired: 122770256129792409600
exactly 6 edge groups paired: 10949790800019456000
exactly 7 edge groups paired: 845319706037452800
exactly 8 edge groups paired: 58271119441920000
exactly 9 edge groups paired: 3453103374336000
exactly 10 edge groups paired: 258982753075200
exactly 11 edge groups paired: 0
exactly 12 edge groups paired: 1961990553600

Notice that the summation correctly predicts that it is not possible to have exactly 11 edge groups paired, since having 11 edge groups paired means that the 12th is also paired.

Also notice that for exactly 12 edge groups paired, which has 1961990553600 positions that fit this characteristic, that this number is:
1961990553600=12!*2^12 which is exactly what we would expect.

Also notice that all of the numbers above sum to 620448401733239439360000 which is equal to 24!, which is the total number of wing edge permutations on the 4x4x4.

 

[LucidCuber]

Thanks for that. I haven't had a LL skip in a year and a half so I was due some luck.

 

[cmhardw]

Thanks for that. I haven't had a LL skip in a year and a half so I was due some luck.

Please take another look at my post above as I noticed an error in my calculation. I knew something looked wrong I had done this calculation before, and after checking my answer to my post above I noticed that I had forgotten something.

The chance of having at least 4 edges paired up after centers is far more likely than I calculated before. The chance is about 1/481.

See my post above for the corrected calculation. Also see the spoiler in my post above for a more detailed breakdown of all the cases, plus some verification of the result.

 

[Christopher Mowla]

Please take another look at my post above as I noticed an error in my calculation. I knew something looked wrong I had done this calculation before, and after checking my answer to my post above I noticed that I had forgotten something.

The chance of having at least 4 edges paired up after centers is far more likely than I calculated before. The chance is about 1/481.

See my post above for the corrected calculation. Also see the spoiler in my post above for a more detailed breakdown of all the cases, plus some verification of the result.

Interesting post Chris!

This reminded me of something. Knowing what you know about the structure of the 24! permutations of wing edges, would it be easy for you to confirm/find a counterexample to condition [2] in my conjecture in this post? The motivation for this, of course, is to be able to claim that at most 6 inner slices are required to pair the dedges (if the permutation is even. I didn't think it through, but it also might be true for odd permutations as well, but the purpose of it being an even permutation is, of course, for fewer moves).

I've been wanting to start a thread in puzzle theory about this conjecture, but I have been too busy with you know what. But if you might be able to confirm/prove condition [2] always can be met, then I definitely want to start a thread so that you can post your proof there!

 

[LucidCuber]

Please take another look at my post above as I noticed an error in my calculation. I knew something looked wrong I had done this calculation before, and after checking my answer to my post above I noticed that I had forgotten something.

The chance of having at least 4 edges paired up after centers is far more likely than I calculated before. The chance is about 1/481.

See my post above for the corrected calculation. Also see the spoiler in my post above for a more detailed breakdown of all the cases, plus some verification of the result.

Would it be possible to get each case (0 edges 1,2,3 edges etc) with a fractional probability? Those numbers are to big to comprehend at that level

 

[cmhardw]

Would it be possible to get each case (0 edges 1,2,3 edges etc) with a fractional probability? Those numbers are to big to comprehend at that level

Sure.

For clarity, all stats below are for the 4x4x4:

Probability of no edge groups paired after centers: \( \approx \frac{59}{100} \)

Probability of at least 1 edge group paired after centers: \( \approx \frac{41}{100} \)

Probability of at least 2 edge groups paired after centers: \( \approx \frac{1}{10} \)

Probability of at least 3 edge groups paired after centers: \( \approx \frac{1}{62} \)

Probability of at least 4 edge groups paired after centers: \( \approx \frac{1}{481} \)

Probability of at least 5 edge groups paired after centers: \( \approx \frac{1}{4609} \)

Probability of at least 6 edge groups paired after centers: \( \approx \frac{1}{52327} \)

Probability of at least 7 edge groups paired after centers: \( \approx \frac{1}{683837} \)

Probability of at least 8 edge groups paired after centers: \( \approx \frac{1}{10009627} \)

Probability of at least 9 edge groups paired after centers: \( \approx \frac{1}{167054487} \)

Probability of at least 10 edge groups paired after centers: \( \frac{1}{2377700325} \)

Probability of all 12 edge groups paired after centers: \( \frac{1}{316234143225} \)

 

[Deleted member 36445]

whats the probability that the 4 cross edges are oriented but not permuted, ie all on D layers but not correctly permuted?