Probability Thread

[miniGOINGS]

One last thing: am I right?

Wow. I actually didn't understand your way of doing the calculations that well, and that's not how I would have done them, so I can't really tell at the moment.

 

[mr. giggums]

What about a edges skip? (on 3x3)

This is pretty easy when you think about it for a moment. You need the first edge in place - that's 1/12. Second edge must now go in place - that's 1/11. Etc...so you have 1/12*1/11*1/10...*1/2=1/12!. Then they all have to be orientated correctly (1/2 for each) but the last edge orientation is defined by the other 11, so you get (1/2)^11 for orientation

Final answer: 1/(12!*2^11) = 9.81*10^11

EDIT: I hope I got this right...I shouldn't be doing maths at uni otherwise...

I think you forgot that two edges can't be switched so you would have to divide that by 2 so it would be 4.91*10^11 I think.

 

[kinch2002]

What about a edges skip? (on 3x3)

This is pretty easy when you think about it for a moment. You need the first edge in place - that's 1/12. Second edge must now go in place - that's 1/11. Etc...so you have 1/12*1/11*1/10...*1/2=1/12!. Then they all have to be orientated correctly (1/2 for each) but the last edge orientation is defined by the other 11, so you get (1/2)^11 for orientation

Final answer: 1/(12!*2^11) = 9.81*10^11

EDIT: I hope I got this right...I shouldn't be doing maths at uni otherwise...

I think you forgot that two edges can't be switched so you would have to divide that by 2 so it would be 4.91*10^11 I think.

No I didn't, because they can be switched as long as the corners are free - just think of a J perm or something - only 2 edges switched. Thanks anyway

 

[mr. giggums]

What about a edges skip? (on 3x3)

This is pretty easy when you think about it for a moment. You need the first edge in place - that's 1/12. Second edge must now go in place - that's 1/11. Etc...so you have 1/12*1/11*1/10...*1/2=1/12!. Then they all have to be orientated correctly (1/2 for each) but the last edge orientation is defined by the other 11, so you get (1/2)^11 for orientation

Final answer: 1/(12!*2^11) = 9.81*10^11

EDIT: I hope I got this right...I shouldn't be doing maths at uni otherwise...

I think you forgot that two edges can't be switched so you would have to divide that by 2 so it would be 4.91*10^11 I think.

No I didn't, because they can be switched as long as the corners are free - just think of a J perm or something - only 2 edges switched. Thanks anyway

Oh I was assuming (I know I spelled it wrong) that the corners were already solved.

 

[miniGOINGS]

Kinch, I did my calculations the same way as yours (probably everyone else too...) 12*11*10*9*8*7*6*5*4*3*2 for EP, and 2^11 for EO.

479,001,600 * 2048 = 980,995,276,800

 

[kinch2002]

Yesterday I had a clock scramble where one side of edges (including centre piece) were all matched (but not all at 12 o'clock). The scramble program (as Maarten told me) generates a random position for the clock to be in and then figures out the generator for it. So, to have one particular side solved it would be (1/12^4) - just think about matching all 4 edges to the centre one at a time. So the probability of having at least one side like this is 1-(1-(1/12^4))^2=1/10368 approx. I've only done 1000 solves so I feel nice and lucky to get one

 

[Sir E Brum]

Probability of LL skip after F2L with edges oriented?

Probability of OLL skip after F2L with edges oriented?

 

[miniGOINGS]

EO'd LL Skip: 7 * 1/72 = 1/1944
EO'd OLL Skip: 1/27

 

[kinch2002]

Probability of LL skip after F2L with edges oriented?

Probability of OLL skip after F2L with edges oriented?

1. I assume AUF is allowed. Easy way: LL skip is 1/15552, so if EO is skipped already (normally 1/8) it's 1/1944.
Slightly harder way: You need to skip EP (doesn't matter where first edge is, so 1/3*1/2 = 1/6), CO (1/3*1/3*1/3 = 1/27) and CP (last 2 corners have no choice because I gave choice on edges so 1/4*1/3 = 1/12) = 1/6*1/27*1/12 = 1/1944
2. OLL skip is 1/216, and EO is normally 1/8 so this gives 1/27 - it's just CO skip really.

 

[Sir E Brum]

Probability of LL skip after F2L with edges oriented?

Probability of OLL skip after F2L with edges oriented?

1. I assume AUF is allowed. Easy way: LL skip is 1/15552, so if EO is skipped already (normally 1/8) it's 1/1944.
Slightly harder way: You need to skip EP (doesn't matter where first edge is, so 1/3*1/2 = 1/6), CO (1/3*1/3*1/3 = 1/27) and CP (last 2 corners have no choice because I gave choice on edges so 1/4*1/3 = 1/12) = 1/6*1/27*1/12 = 1/1944
2. OLL skip is 1/216, and EO is normally 1/8 so this gives 1/27 - it's just CO skip really.

Mmk. Tyvm.

 

[Aditya]

What is the probability of randomly turning the sides of the cube and solving it? lol. I wonder how you would calculate it.

 

[kinch2002]

Short and very approximate answer: 1/43 quintillion for each turn

Long answer:

Spoiler

Well I could take this question to mean 'what's the probability that a random state is the solved one?'. This is a fairly well known probability. There are approx. 43 quintillion (4.3*10^19) different states that the cube can be in (not counting positions that can only be achieved by taking pieces out).

This is found most easily by (8!)*(3^7)*(12!)*(2^11)/2. I'm sure there are other places on the forum that explain this probability as well, but do ask if you don't understand how I got those numbers!

In reality your question is slightly different to the one I answered because when the cube is scrambled you are usually at least 15 or so moves from being solved so the next 14 turns have 0 probability of solving it.

 

[Lucas Garron]

What is the probability of randomly turning the sides of the cube and solving it? lol. I wonder how you would calculate it.

I'm not sure. What is the probability of attending a competition?

Good approximation:
http://www.wolframalpha.com/input/?i=expected+value+geometric+distribution+p%3D1/(4.3e19)

 

[Aditya]

What is the probability of randomly turning the sides of the cube and solving it? lol. I wonder how you would calculate it.

I'm not sure. What is the probability of attending a competition?

Good approximation:
http://www.wolframalpha.com/input/?i=expected+value+geometric+distribution+p%3D1/(4.3e19)

Haha lol. How did you come up with that number? Wouldn't the probability of going to a competition be affected if you know that the person is a cuber?

 

[Lucas Garron]

What is the probability of randomly turning the sides of the cube and solving it? lol. I wonder how you would calculate it.

I'm not sure. What is the probability of attending a competition?

Good approximation:
http://www.wolframalpha.com/input/?i=expected+value+geometric+distribution+p%3D1/(4.3e19)

Haha lol. How did you come up with that number? Wouldn't the probability of going to a competition be affected if you know that the person is a cuber?

Yes, it would be. I'm just trying to get you to notice that your question was vague, and badly phrased, so that it almost sounds like it's not really answerable.

http://en.wikipedia.org/wiki/Geometric_distribution
It's a good approximation because the cube is almost completely random after a few dozen moves.

 

[mrCage]

I had a really interesting one come up a few days ago. After solving the cross and the first two pairs, all the remaining 12 pieces were in their correct locations, only 6 of them were misoriented.

The probability of having the last 12 pieces of the cube in the correct location after solving the cross and first two pairs is:

1 / [6! * 6!/2] = 1 / 259200

To give a comparison, the chance of a LL skip with no partial edge control, and with the possibility of AUF, is
1 / 15552

I thought it was pretty neat.

Chris

Maybe you're cubing too much?
**** happens - and chance happens...

Per

 

[Aditya]

What is the probability of randomly turning the sides of the cube and solving it? lol. I wonder how you would calculate it.

I'm not sure. What is the probability of attending a competition?

Good approximation:
http://www.wolframalpha.com/input/?i=expected+value+geometric+distribution+p%3D1/(4.3e19)

Haha lol. How did you come up with that number? Wouldn't the probability of going to a competition be affected if you know that the person is a cuber?

Yes, my question was phrased badly, and I am sorry if it confused anybody. The geometric distribution looks right, but coming from a Stanford guy, it HAS to be right lol jk.

 

[Bayamo Penguin]

"Easy Solve"

What is the probablity of an "easy solve" during a straight LBL (method included in the packaging of the Rubik's)
Cross, corner, middle layer, and top edges all solved..
then when orienting the top corners (URul/UruL), it all falls in to place, no mess, no fuss, no hassle

 

[riffz]

What is the probablity of an "easy solve" during a straight LBL (method included in the packaging of the Rubik's)
Cross, corner, middle layer, and top edges all solved..
then when orienting the top corners (URul/UruL), it all falls in to place, no mess, no fuss, no hassle

This is way to vague for someone to answer.

 

[Bayamo Penguin]

What is the probablity of an "easy solve" during a straight LBL (method included in the packaging of the Rubik's)
Cross, corner, middle layer, and top edges all solved..
then when orienting the top corners (URul/UruL), it all falls in to place, no mess, no fuss, no hassle

This is way to vague for someone to answer.

Vague in explanation? or Positionwise?