Case probabilities

[MHordecki]

Hello everyone!

Let's say we've got OLLs (or PLLs, or whatever else). How do you calculate the probability for each case?

Are there any good papers on this?

Best regards,
MH

 

[Lucas Garron]

Take the probability base.
Divide by the number of symmetries.
An OLL that's 180 degrees symmetric is 1/2 (like double sune)
H-PLL is 1/4 (90-degree symmm)

(actually, non-symms create multiple cases, but it's about the same math)

If that's not enough, I can elaborate later.

 

[PCwizCube]

Probability Websites

Here are some sites that tell you the probabilities of OLL and PLL algorithms.

OLL:

• Macky's OLL Page
• Dan's OLL Pages

PLL:

• Macky's PLL Page
• Dan's PLL Pages

 

[AvGalen]

If you don't want to calculate you can simply check on this page: http://www.speedcubing.com/final_layer_print.html

 

[Lucas Garron]

Okay I have a bit more time...

Let's take the Sune/OLL (any standard orientation).
The chance of EO like the Sune (all correct) is 1/2*1/2*1/2 (last edge is dependent).
The chance of CO exactly like in the Sune is (1/3)^3 (last corner is dependent).

This makes 1/216, the base probability for OLL, and the chance of getting a Sune. But really, there are four Sunes (from AUF), so the chance of a Sune for OLL is 4/216 = 1/54.
The same applies (separately) to the mirror.

There are two unique AUFs for double Sune, so 2/216 = 1/108.
Four-edge-flip: 1/216

YOu can think of it as counting each possible case (counting AUF as distinct), and totalling to find the effective probability for a case.
Same applies to PLL, except /72.

 

[Lt-UnReaL]

How would you calculate the chances of an R-perm then? I come up with:

CP:
(1/4)^2 (2 correctly permuted corners)
(3/4) (Other corner is somewhere other than its correct spot, and the last corner is dependent)

EP:
Same thing as CP, so:
(1/4)^2
(3/4)

(((1/4)^2)(3/4)((1/4)^2)(3/4))*4 (4 AUFs)

Where did I go wrong? :/

 

[cmhardw]

How would you calculate the chances of an R-perm then? I come up with:

CP:
(1/4)^2 (2 correctly permuted corners)
(3/4) (Other corner is somewhere other than its correct spot, and the last corner is dependent)

EP:
Same thing as CP, so:
(1/4)^2
(3/4)

(((1/4)^2)(3/4)((1/4)^2)(3/4))*4 (4 AUFs)

Where did I go wrong? :/

Righty-R perm:
choose the two corners that are to be swapped, this chooses which edges need to be swapped. The corners must be adjacent, so this leaves 4 choices. The U layer can be AUF'd 4 ways so there are 16 righty R perms.

There are 288 possible permutations of the U layer (4!)^2 / 2 so that's 16/288 or 1/18. The lefty R perm, by a similar argument is also 1/18. So the combined chance for an R perm (righty or lefty) is 1/9.

--edit--
What it looks like in your calculation is you did not take into account that you are using conditional probabilities. For example after permuting 2 corners the chance that the remaining corners are not in the correct locations is 1/2 not 3/4. Also you wouldn't use (1/4)^2 for choosing the correct corners. I would imagine it this way. Choose the first "correct" corner. There are 4 choices for this. Then choose the other "correct" corner. This must be an adjacent corner to the first. There are 2 ways to choose this. This gives you 8 ways to choose the 2 correct corners. However, this is an overcount. If you choose the UFR corner, then the UFL corner, this is no different than choosing the UFL then the UFR. So you have doubly overcounted. Divide by 2 to find that there are 4 ways to choose the 2 correct corners.
--edit--

Chris

 

[joey]

I just guess.

 

[Lt-UnReaL]

How would you calculate the chances of an R-perm then? I come up with:

CP:
(1/4)^2 (2 correctly permuted corners)
(3/4) (Other corner is somewhere other than its correct spot, and the last corner is dependent)

EP:
Same thing as CP, so:
(1/4)^2
(3/4)

(((1/4)^2)(3/4)((1/4)^2)(3/4))*4 (4 AUFs)

Where did I go wrong? :/

Righty-R perm:
choose the two corners that are to be swapped, this chooses which edges need to be swapped. The corners must be adjacent, so this leaves 4 choices. The U layer can be AUF'd 4 ways so there are 16 righty R perms.

There are 288 possible permutations of the U layer (4!)^2 / 2 so that's 16/288 or 1/18. The lefty R perm, by a similar argument is also 1/18. So the combined chance for an R perm (righty or lefty) is 1/9.

--edit--
What it looks like in your calculation is you did not take into account that you are using conditional probabilities. For example after permuting 2 corners the chance that the remaining corners are not in the correct locations is 1/2 not 3/4. Also you wouldn't use (1/4)^2 for choosing the correct corners. I would imagine it this way. Choose the first "correct" corner. There are 4 choices for this. Then choose the other "correct" corner. This must be an adjacent corner to the first. There are 2 ways to choose this. This gives you 8 ways to choose the 2 correct corners. However, this is an overcount. If you choose the UFR corner, then the UFL corner, this is no different than choosing the UFL then the UFR. So you have doubly overcounted. Divide by 2 to find that there are 4 ways to choose the 2 correct corners.
--edit--

Chris

I re-read your post several times and I just can't understand what you mean. :/

 

[Lucas Garron]

I re-read your post several times and I just can't understand what you mean. :/

Don't try to do specific stuff.
For any PLL:

*There are 4 (even) possibilities for the UFR corner.
There are 3 (even) possibilities for the UBR corner.
There are 2 (even) possibilities for the UBL corner.
There is 1 (even) possibilities for the UFL corner.

There are 4 (even) possibilities for the UF edge.
There are 3 (even) possibilities for the UR edge.
There are 2 (even) possibilities for the UB edge.
There is 1 (even) possibilities for the UL edge.

Number of cases for, say, R-perm:
*4, for color permutation equivalence. (R-perm y' = R-perm, even if they permute differently colored edges)
2, for parity dependence
s, the number of symmetries, for AUF

Makes (4*2*s)/(4*3*2*1*4*3*2*1) = s/72.

In the case of R-perm, s=4 (s=8 if you count mirrors).
For Z, it's s=2, for H, s=1.

*Note, for most computations like OLL and PLL, you can throw out the starred lines, by fixing a piece WLOG.

 

[blade740]

I just guess.

I just ask lucas.

 

[Lt-UnReaL]

I re-read your post several times and I just can't understand what you mean. :/

Don't try to do specific stuff.
For any PLL:

*There are 4 (even) possibilities for the UFR corner.
There are 3 (even) possibilities for the UBR corner.
There are 2 (even) possibilities for the UBL corner.
There is 1 (even) possibilities for the UFL corner.

There are 4 (even) possibilities for the UF edge.
There are 3 (even) possibilities for the UR edge.
There are 2 (even) possibilities for the UB edge.
There is 1 (even) possibilities for the UL edge.

Number of cases for, say, R-perm:
*4, for color permutation equivalence. (R-perm y' = R-perm, even if they permute differently colored edges)
2, for parity dependence
s, the number of symmetries, for AUF

Makes (4*2*s)/(4*3*2*1*4*3*2*1) = s/72.

In the case of R-perm, s=4 (s=8 if you count mirrors).
For Z, it's s=2, for H, s=1.

*Note, for most computations like OLL and PLL, you can throw out the starred lines, by fixing a piece WLOG.

This is in a way I understand. Just one question, though. What do you mean by parity dependence?

 

[Lucas Garron]

This is in a way I understand. Just one question, though. What do you mean by parity dependence?

Well, I was being sneaky and simple. I was implicitly corresponding every parity case to a non-parity. It should actually be like this:

...
**There is 1 possibility for the UB edge.
There is 1 possibility for the UL edge.
...
[Take out the line you bolded]
...
(4*s)/(4*3*2*1*4*3*1*1) = s/72


**This is because of permutation parity restriction, which you hopefully understand (parity(EP) == parity(CP)).

 

[Kenneth]

Why so complicated?, for PLL; there are 6 corner combos 4x adjacent, diagonal and solved. Overlay the edges, first edge 4 places, second 3 and the others are dependent (4! / 2!). So PLL probability is 6 * 4 * 3 = 1:72