Probability Thread

[Rpotts]

This went in the accomplishement thread but it probably belongs here:
16.05, 15.26, (10.13)[PLL skip], 17.92, 16.16[OLL skip], (19.19), 16.91, 17.80, 18.33, 14.93[PLL skip], 11.70[LL skip], 14.69[PLL skip]

scrambles were:

Spoiler

1. 16.05 D' L' U R2 B2 R F D' F2 R U2 D F' L2 U' R B2 D R' F' R' F D2 U' L2
2. 15.26 U2 L2 R2 F R U2 F2 B' D2 U2 B' U B U F' L' U B' R2 D2 F' R' B' F L'
3. (10.13)[PLL skip] F2 U2 D L2 R2 B R B R B U' R B' F' R U' L' D B F' D2 U R2 F2 U'
4. 17.92 F L F2 B U' B D2 L F' R' L2 D' R2 D' B' F L D F' U2 R2 L U R2 D'
5. 16.16[OLL skip] R F B' D' R D2 L' B' F L U' L2 D' R' L2 F' R D' L D U' B' L B D
6. (19.19) B L B2 R F' B2 U' L U R2 F' L2 F R2 D' B U' D2 B L2 F2 D2 B' L2 R
7. 16.91 F B D' L2 B' F' U2 R2 D F2 U2 D' R2 F R2 L B U' F2 R2 D2 R F' L' U
8. 17.80 R L F' U B2 L D L2 F' D2 U' L' U' F' U' L2 D B R' D F' L' U2 D' R'
9. 18.33 L2 F' U2 L U' R U2 B2 U' B' L' F U' R2 B' R' U' B' R' L B F R D' R2
10. 14.93[PLL skip] B' L2 D' U' R L' B L U F R2 F2 L' F' U' F R2 U' D2 L' R' B2 R F' B'
11. 11.70[LL skip] U2 L2 U2 D2 L R2 B2 L' D2 U R2 U' D' L D2 U2 L B2 U L R F2 U2 F' B'
12. 14.69[PLL skip] U' F' D2 B2 F' R' L2 D2 F R B2 L F' B2 U' D L B' U2 L F B D U' L2

Can someone please calculate this? I tried, but I'm probably wrong.

Calculate what? The probability of getting full step, full step, PLL skip, full step, OLL skip, full step, full step, full step, full step, PLL skip, LL skip, PLL skip, in that order?

so full step probability is (215/216)x(71/72) = 15265/15552 ≈ 98.1545%
PLL skip probability is (215/216)x(1/72) = 215/15552 ≈ 1.3824%
OLL skip probability is (1/216)x(71/72) = 71/15552 ≈ .4565%
LL skip probability is (1/216)x(1/72) = 1/15552 ≈ .00643%

btw it all adds up to 15552/15552 so i think my calcs are correct

(15265/15552)x(15265/15552)x(215/15552)x(15265/15552)x(71/15552)x(15265/15552)x(15265/15552)x(15265/15552)x(15265/15552)x(215/15552)x(1/15552)x(215/15552)

6.808388 x 10^-13

.0000000000006808388

 

[cmhardw]

so full step probability is (215/216)x(71/72) = 15265/15552 ≈ 98.1545%
PLL skip probability is (215/216)x(1/72) = 215/15552 ≈ 1.3824%
OLL skip probability is (1/216)x(71/72) = 71/15552 ≈ .4565%
LL skip probability is (1/216)x(1/72) = 1/15552 ≈ .00643%

btw it all adds up to 15552/15552 so i think my calcs are correct

(15265/15552)x(15265/15552)x(215/15552)x(15265/15552)x(71/15552)x(15265/15552)x(15265/15552)x(15265/15552)x(15265/15552)x(215/15552)x(1/15552)x(215/15552)

6.808388 x 10^-13

.0000000000006808388

This is the probability if he must get the exact same kind of skips, in the exact same order that he did during the solve. If you allow the exact same kind of skips, but in any order, you would have to multiply your probability by:
12! / (7! * 3!) = 15840

(6.808388 x 10^-13) * 15840 = approximately 1 x 10^-8
or
0.00000001 probability allowing the exact same scheme of skips to happen in any order.

Chris

 

[Rpotts]

Yeah i was calculating it based on johans question, those skips in the same order. Good to know it was correct, if the question is in the same order.

 

[cmhardw]

Yeah i was calculating it based on johans question, those skips in the same order. Good to know it was correct, if the question is in the same order.

Yes, your math is good for the situation your were calculating.

Chris

 

[DanHarris]

Chris, you are my hero.

 

[TeddyKGB]

Me and a friend were racing yesterday, he's quite a bit slower than me so he solves one cube and I'll solve two, during one particular race, both my cubes had not only the same OLL case but also the same PLL case, just wondering what are the odds of that happening?

 

[joey]

Could be 1/15552^2 = 1/241864704.

 

[cmhardw]

... both my cubes had not only the same OLL case but also the same PLL case, just wondering what are the odds of that happening?

Should be \( \sum_{j}\sum_{i} [P(ith OLL)*P(jth PLL)]^2\ \)

I honestly don't have the programming know-ho to do this quickly :-s But given a list of the OLL and PLL probabilities it can definitely be done.

Could be 1/15552^2 = 1/241864704.

Joey I think you are on the right track with this. I think you could argue that a really good estimate to the probability would be 1/15552. You could say that the probability that you get a LL case on the first solve is 1. Then the probability that you get the matching case to that on the second solve is 1/15552. I don't know if the above sum equates to this, and if it does that would be really cool!

Chris

 

[Lucas Garron]

Should be \( \sum_{j}\sum_{i} [P(ith OLL)*P(jth PLL)]^2\ \)

o=(Table[RotateRight[#,i]&/@#,{i,4}]//Tally//Length)&/@OLLs;
p=Length/@GatherBy[CanonicalizeLL[StringJoin[#1]]&/@Tuples[{PLLGeneratorCPs,PLLGeneratorEPs}],CubeLL[PerformNewPLLInverse[#]]&];
((o/216)^2//Total)*((p/72)^2//Total)

28073/30233088 ≈ 0.000928552

I don't know if the above sum equates to this, and if it does that would be really cool!

Of course it doesn't. What kind of math teacher are you?

Think of it this way for OLL only: Instead of \( (1/58)^2*58 \), most of the probabilities are just a slight bit larger, and a few of them are quite a bit smaller than 1/58. So, the actual probability for a repeated OLL/PLL/both is always lower than such an estimate. But of course not by much.

 

[DavidWoner]

Yeah i was calculating it based on johans question, those skips in the same order. Good to know it was correct, if the question is in the same order.

David Cramer would be proud.

 

[Rpotts]

Yeah i was calculating it based on johans question, those skips in the same order. Good to know it was correct, if the question is in the same order.

David Cramer would be proud.

yay

but wait, doesn't Cramer kinda dislike speedcubing?

 

[Carson]

I am looking for a list of probabilities for each of the OLL's. Does anyone know where I can find this?

 

[oprah62]

I am looking for a list of probabilities for each of the OLL's. Does anyone know where I can find this?

Jessica's site or badmephistos(I think).

Btw what are the chances of getting 2 solves with the exact same LL and same time? I don't think the time probability can be calculated though. Probably (disregarding time), the same as ^.

edit:here
http://www.ws.binghamton.edu/fridrich/Mike/orient.html

 

[Carson]

I am looking for a list of probabilities for each of the OLL's. Does anyone know where I can find this?

Jessica's site or badmephistos(I think).

Btw what are the chances of getting 2 solves with the exact same LL and same time? I don't think the time probability can be calculated though. Probably (disregarding time), the same as ^.

Badmephisto's site has PLL probabilities, but not OLL. I did find it on Macky's site though.

 

[CubesOfTheWorld]

I am looking for a list of probabilities for each of the OLL's. Does anyone know where I can find this?

Jessica's site or badmephistos(I think).

Btw what are the chances of getting 2 solves with the exact same LL and same time? I don't think the time probability can be calculated though. Probably (disregarding time), the same as ^.

edit:here
http://www.ws.binghamton.edu/fridrich/Mike/orient.html

The probability of getting the same time cannot be calculated.

 

[ben1996123]

What is the probability of a first centre skip on 7x7?

 

[eastamazonantidote]

What is the chance of each ZBF2L case occurring? I asked this before but the Hardwick/Kirjava debates took over.

 

[trying-to-speedcube...]

The probability for the both-pieces-stuck-in-slot cases is 1/25 total, and I'm pretty sure this is equally distributed. There are 6 such cases, so the probability for any of those 6 is 1/150.

Now for the edge-stuck-in-slot cases. The probability of an edge being stuck in the slot but not the corner is 1/5*4/5=4/25. There are 6 cases, so the probability for each of those is 4/150.

Same goes for corner-stuck-in-slot cases. 1/5*4/5=4/25. There are also 6 cases, so also 4/150 for all of those.

Then there's the biggest group, where both corner and edge are in the U-layer. The chance of that happening is 4/5*4/5=16/25. There are 24 cases total for this. For ease, let's say the corner is always in UFR. Then there are 4 places where the edge can be, all with 2 orientations. These are all equally distributed. 16/25/24=4/150.

Summary: All the cases have a 4/150 chance of occuring, except for the both-pieces-stuck-in-slot cases, which have a 1/150 chance of occuring.

As for the edge orientation, the cases are always equally distributed when there is at least 1 of the F2L pieces in the U-layer. If not, there is an AUF possible. If the F2L edge is wrongly oriented, the cases have equal probabilities. If the F2L edge is correctly oriented though, the 2 adjacently flipped edges have a probability of twice as high as the 2 oppositely flipped edges.

Going by this list:
In the first group every single case has a probability of 4/150/8=1/300.
In the second group, the first case has 2 cases with a probability of 1/300, the second, third and fourth case have 4 cases with a probability of respectively 1/1200, 1/300, 1/600 and 1/1200 for the skip again. The cases in the fifth and sixth cases all have probabilities of 1/300. The rest of the cases also all have a probability of 1/300 of occuring.

 

[irontwig]

I am looking for a list of probabilities for each of the OLL's. Does anyone know where I can find this?

Looks the same after y: 1/216
Looks the same after y2: 1/108
The rest: 1/54

 

[Rpotts]

Well put