Probability Thread

[jaap]

You might multiply that by four for four orientations of the cube

No. That is only done with LL calculations because an asymmetrical case can occur in four orientations and therefore represents 4 distinct positions. Rotating the cube as a whole does not change the probabilities in this case because the middle-layer-solved position is fully symmetrical about the y axis.

 

[That70sShowDude]

Probability of a 2x2 scramble containing no "2 bars"?

 

[hawkmp4]

You might multiply that by four for four orientations of the cube

No. That is only done with LL calculations because an asymmetrical case can occur in four orientations and therefore represents 4 distinct positions. Rotating the cube as a whole does not change the probabilities in this case because the middle-layer-solved position is fully symmetrical about the y axis.

Right! Of course. If I multiplied that by four, that would be calculating the probability of all the middle layer edges being in the middle layer, and having correct permutation relative to each other, yes?

 

[Christopher Mowla]

Probability of Having Inner-Layer Odd Parity on Big Cubes

I am not very good at probability, but I think I might have just calculated that the probability percentage of having inner-layer odd parity for a cube of degree n equal to:

\( \left( 1-\frac{1}{2^{\left\lfloor \frac{n-2}{2} \right\rfloor }} \right)\times 100 \)


Here is a link to the formula in wolfram.

According to this formula, any cube size beyond 15 has a 99._ percent chance!

The cube sizes under that have the following:

5X5X5 50.0%
7X7X7 75%
9X9X9 87.5%
11X11X11 93.75%
13X13X13 96.875%
15X15X15 98.4375%


Is this right?

 

[Rinfiyks]

I might have just calculated that the probability percentage of having inner-layer odd parity

Having what? :confused:
Are you referring to parity when you end up with 2 edges left to complete during reduction?

 

[Lucas Garron]

I was just hoping someone could verify the formula.

What's there to verify? Of course it's correct.

Probability of no parity: 2^-(#orbits)

 

[ariasamie]

After which OLL is there more chance of getting a PLL skip, or the 1/72 for PLL skip is the same for all of the OLLs?

 

[~Adam~]

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

I had a similar but more improbable situation arise. 1st and 2nd F2L pair inserted leaving all remaining pieces in their correct position only leaving 3 corners to be oriented.

Can anyone please tell me the odds of this occurring or better (Solved or 2 edges need flipping) after inserting 2 F2Ls?
Thanks

 

[hawkmp4]

After which OLL is there more chance of getting a PLL skip, or the 1/72 for PLL skip is the same for all of the OLLs?

Orientation is independent of permutation when you use OLL. 1/72.

 

[~Adam~]

After which OLL is there more chance of getting a PLL skip, or the 1/72 for PLL skip is the same for all of the OLLs?

I would assume the probability is the same for all OLLs.

edit - I got Ninja'd =(

 

[XXGeneration]

What are the chances that when you solve your normal cross, an x-cross is also done? Like when you do your normal cross, you manage to do an extended cross inadvertently?

 

[Stefan]

What are the chances that when you solve your normal cross, an x-cross is also done?

About 1%, if I remember correctly.

 

[Johannes91]

What are the chances that when you solve your normal cross, an x-cross is also done?

About 1%, if I remember correctly.

Correct, 1.036368...%.

pairs cases
0 3619921801
1 37715396
2 192606
3 596
4 1
+ 3657830400

Source: http://pastebin.com/yvMyxHGM

 

[ben1996123]

What are the chances that when you solve your normal cross, an x-cross is also done?

About 1%, if I remember correctly.

Correct, 1.036368...%.

pairs cases
0 3619921801
1 37715396
2 192606
3 596
4 1
+ 3657830400

Source: http://pastebin.com/yvMyxHGM

So you are 1000 times more likely to get a corners skip than an F2L skip after cross? lol.

EDIT: Some more probabilities that I would like to know (sorry if some have already been posted):

1. 2x2 first layer skip
2. separation skip on 3x3xn where n is 2 or 4+
3. Centers skip on 4x4
4. Megaminx star skip
5. First center skip on 11x11

 

[StachuK1992]

I'm aware that this is the probability thread, so this isn't entirely relevant, but I was wondering if anyone could clear this up from a while ago.

Spoiler

So. CPLL+ELL would be...how many algs? I want to say 84, 21 PLLs per 4 edge cases (all, adj, opp, none)

But Thom then said 6*48=288.

So then I looked again, and thought

21 (regular PLL)
21 (all edges flipped)
Those only get a set of PLL each, since the angle doesn't matter.

21*2 (opp edges)
One set where flipped on L/R, one on U/B.

21*4 (adj edges)
One set per AUF.

21 + 21 + 42 + 84 = 168.

Is this correct?

Spoiler

#Non-EPLL PLLs=22-4=18 (Don't forget pure flips)

ELL: 29
Regular PLL: 18-1=17 (Pure flip=Solved)
PLLEF: 18 (Like CLLEF)
Opp flip: 2*18-2=34 (N-perm overcount)
Adj flip: 4*18-1-6=65 (E&N-perm overcount)
Sum: 19+17+18+34+65=153

Please correct me if this is wrong.

I'm not sure if it's right, but I sure do like the numbers better.
However, recog will probably deter me from learning this, but I can at the very least learn ELL and see what I feel like from there.

Thanks.

 

[Lucas]

So you are 1000 times more likely to get a corners skip than an F2L skip after cross? lol.

EDIT: Some more probabilities that I would like to know (sorry if some have already been posted):

1. 2x2 first layer skip
2. separation skip on 3x3xn where n is 2 or 4+
3. Centers skip on 4x4
4. Megaminx star skip
5. First center skip on 11x11

Mmm... let's see... correct me if I am wrong:

1. For one color: 24/(8! / 4! * 3^4) = 24/136080
For any color: 6 times powered? : 1 - (1 - that above)^6 = 0.00106
3. 4!^6/24! = 3.08 * 10^(-16)... a bit difficult
5. 1 / (24! / 4! * 20!)^20 = 1/3.37*10^80 = 2.97 * 10^(-81)

 

[irontwig]

What's the probability to have these blocks already built:
a: 2x1x1
b: 2x2x1
c: 2x2x2

I assume you can then just half the probabilities if you use non-colour-neutral block F2L.

 

[Erzz]

Would it be possible to create a 3 dimensional graph with one variable as corners solved, one as edges solved, and another as probability that the rest of the cube would be solved? Or something similar.

What I mean is, if someone covered the whole cube except the centers and one edge, and you solved that one edge, the probability of the cube being solved after being uncovered. For an LL skip for example, you would solve 8 edges and 4 corners (F2L), the point on the plane where those two meet would be the probability of a LL skip. Or am I too tired to be thinking about math?

If this is possible, what would the equation be?

 

[eastamazonantidote]

Would it be possible to create a 3 dimensional graph with one variable as corners solved, one as edges solved, and another as probability that the rest of the cube would be solved? Or something similar.

What I mean is, if someone covered the whole cube except the centers and one edge, and you solved that one edge, the probability of the cube being solved after being uncovered. For an LL skip for example, you would solve 8 edges and 4 corners (F2L), the point on the plane where those two meet would be the probability of a LL skip. Or am I too tired to be thinking about math?

If this is possible, what would the equation be?

Sounds awesome. Really awesome. I'll get on it at some point, though I'm sure people will beat me to it. It won't be an equation but a series of points (I don't feel like working equations out right now). Get back to me in a week and I should have something worked out.

 

[Erzz]

On a related note, what is the probability of a full skip after solving the cross?
Also, the probability of a LL skip without AUF.