Chances of getting a LL skip on a Megaminx?

[uriel rubik]

So that is the question, what are the chances of getting a LL skip on a Megaminx? Also, a PLL, and an OLL.

I've had a couple PLL, one OLL, and one LL. So I wanted to know, what are the odds of getting this.

I hope you can help.


Thanks!

 

[cmhardw]

LL skip is: 1/933120
OLL skip is: 1/1296
PLL skip is: 1/720

Chris

 

[hr.mohr]

What are the numbers for a 4 look LL? (OE, OC, EP, CP)

 

[Stefan]

What are the numbers for a 4 look LL? (OE, OC, EP, CP)

This is not rocket science. I suggest you think a bit.

 

[ajmorgan25]

LL skip is: 1/933120
OLL skip is: 1/1296
PLL skip is: 1/720

Chris

And I thought 3x3 LL skips were rare...

 

[uriel rubik]

Thanks Chris.

Wow, I tought they were more common. I guess I just was lucky.

This is not rocket science. I suggest you think a bit.

LOL

 

[EmersonHerrmann]

Hmmmmm...what is the chance of getting an LL skip on a 7x7 when using K7?

 

[puzzlemaster]

And to think. I've gotten an OLL skip on my megaminx before. I feel as if I just witnessed history.

 

[cmhardw]

This is not rocket science. I suggest you think a bit.

LOL

Now I regret answering the first question outright. Do you understand how these numbers are arrived at? The reason I ask is that I don't see why Stefan's comment is funny? (he means it seriously, and I would too)

Hmmmmm...what is the chance of getting an LL skip on a 7x7 when using K7?

I don't fully understand the steps of the method after solving the first 6x6x7 (assuming I am even thinking of the correct method) so I don't want to answer this. However I can tell you that the process is very simple. Count the number of "solved" states for a LL skip; count the number of possible states in the K7 method by the time you reach the last layer, and divide. Again, see Stefan's comment that I quoted in this same post.

Chris

 

[DavidWoner]

Hmmmmm...what is the chance of getting an LL skip on a 7x7 when using K7?

I don't fully understand the steps of the method after solving the first 6x6x7 (assuming I am even thinking of the correct method) so I don't want to answer this. However I can tell you that the process is very simple. Count the number of "solved" states for a LL skip; count the number of possible states in the K7 method by the time you reach the last layer, and divide. Again, see Stefan's comment that I quoted in this same post.

Chris

K4 on 7x7**

K4 is the name of the method, it doesn't change from puzzle to puzzle. Just like how reduction is still called reduction on a 7x7. The steps are still the same:

1. Opp centers
2. Block made of 3 edge groups and 3 corners in a single layer
3. Remaining centers without destroying block
4. finish that layer
5. finish everything but the layer opposite your first
6. CLL
7. ELL

[/Thom/Dan]

So for calculating the chance of a full LL skip on 7x7 when using k4 you only need to consider the 4 corners and 20 edge pieces.

I hope Chris meant to say 6x7x7, solving after forming a 6x6x7 just sounds unpleasant

Edit: On a more on-topic note, Chris Brownlee (UWR holder, did pretty much nothing but minx for a very long time) has only skipped 3/4 LL steps, and that was once. Not sure which 3, but if you are familiar with Chris than it should give you an idea of how rare a minx LL skip is (if numbers aren't enough).

 

[AvGalen]

Just use edge control to increase your chances "greatly"

 

[rahulkadukar]

Exactly whata re the chances iof getting a LL skip on a nxnxn cube when using LBL

 

[4Chan]

Exactly whata re the chances iof getting a LL skip on a nxnxn cube when using LBL

Very slightly better than if you were to use fridrich, since i assume theres a corner in the D layer.

However, i am not as brilliant as stefan nor chris, so i dont know how to find such numbers. D:

 

[trying-to-speedcube...]

Exactly whata re the chances iof getting a LL skip on a nxnxn cube when using LBL

Assuming you have solved 6 layers, all center pieces are solved.

The corner chances are constant on every cube.

An nxnxn cube has 4(n-2) edges in the last layer. There is a number of different edge pieces (4x4 has 1, 5x5 has 2, 6x6 has 2, 7x7 has 3 etc.) on each cube. You can calculate the number of different edge pieces by dividing n by 2, then subtracting 1. If you round it you have the number of different edges. In the last layer, there are 8 of each type, except for a middle edge piece, which only exists on odd cubes, of which there are 4.

Middle edge pieces have orientation and permutation, the other edge pieces only have permutation.

The number of possibilities for 8 edges is 8!, (parity exists), one of which is the solved state. That's 1/40320. The chance that the middle edges are solved is 1/((2^4)/2)*((4!)/4) (parity doesn't exist), which is 1/48. The chance that this is also solved in relation to te corners is 1/2 (H-perm is possible) so the chance that the middle edges are solved is 1/92.

The chance that the corners are solved is 1/((3^4)/3)*((4!)/6) (parity doesn't exist), which is 1/162.

Odd cubes:
_________1_________
162*92*(n/2-2)*40320

Even cubes:
________1________
162*(n/2-1)*40320

I think I have made a mistake somewhere, but I don't know exactly. I hope the more advanced cube theorists can point out my mistake.

Maarten

Edit: 500th post

 

[Stefan]

What are the numbers for a 4 look LL? (OE, OC, EP, CP)

This is not rocket science. I suggest you think a bit.

Oh well... 04mucklowd "reported" that, requesting to tell me "to stop being rude and post more thoughtful things".

I am deeply sorry. I promise I will never ever again encourage people to think on their own, pointing out that what they're asking for is easier than they might think so they might very well be able to answer it themselves and feel good about themselves and end up with better understanding. Again, very sorry.

 

[Kirjava]

Aahhahaha stefan you sarcastic bastard.

DavidWoner: Why so inaccurate?

Seems a good place to link this update.

 

[AvGalen]

@Stefan: Is this an episode of Twilight Zone or Candid Camera?

 

[uriel rubik]

@Chris: I didn't mean to say that Stefan wasn't being serious, I know he was. I just find funny the way he is a little bit rude, but respectful. Also it's funny his sarcasm.

I think the humor variates among the different countries.

@Kirjava: Thanks for the link, I was looking for a webpage with the algorithms.

 

[Stefan]

@Chris: I didn't mean to say that Stefan wasn't being serious, I know he was. I just find funny the way he is a little bit rude, but respectful. Also it's funny his sarcasm.

Yep, that was the intention.

Now about the serious part: Your question still hasn't been answered, so you still have the chance. What saddened me was that you didn't even try but just flat-out asked for the answer. This is rather easy intuitive math, plus you won't die if you show us your attempt and it's wrong.

 

[DavidWoner]

Aahhahaha stefan you sarcastic bastard.

DavidWoner: Why so inaccurate?

Seems a good place to link this update.

What do you mean "inaccurate?" I described exactly what you linked to, except I used terms that allowed it to be extrapolated for all bigcubes puzzles. Saying "make a 1x3x4" doesn't really apply when you are solving a 7x7.