Probability Thread

[Bayamo Penguin]

Maybe this is clearer?: This is the position/ situation: F2L complete. Top edges correcly positioned. You still need to position and orient the top corners, and you are using URul, UruL to position them. This is one of those simple LBL solves many people begin with. Then as you complete the algorithm they do fall in to the correct slot, and fall in to correct orientation. I rememebr it would happen when i first started learning, and in fact, some tutorials mention that this may happen. I just was wondering how often that happened.

 

[Johannes91]

Skipping orienting corners has probability 1/27, or about 3.7%.

 

[eastamazonantidote]

I want to know the chances for each X-VHF2L case. On Jason Baum's site, this would be cases: 1, 2, 3, 4, 5, 6, 11, 12, and 16. Assuming complete randomness, what are the chances for each case? If it's easier to just do ZBF2L in general I can work out the VHF2L chances from that. Is this even possible to compute accurately?

 

[qqwref]

I'd help you out but I don't really understand X-VHF2L. How are you reducing to these cases? Is it an automated process or something like "oh I just intuitively pair" (in which case the probability of each case cannot be computed)?

 

[eastamazonantidote]

I'd help you out but I don't really understand X-VHF2L. How are you reducing to these cases? Is it an automated process or something like "oh I just intuitively pair" (in which case the probability of each case cannot be computed)?

Unfortunately I do my pairing intuitively. I was worried intuition would hurt the odds. You could pair up algorithmically, but I doubt there would be any advantage there. Alright, how about ZBF2L? I can generate the probabilities for VHF2L from the ZBF2L cases.

 

[Gold_A]

What is the probability of a parity on one layer of the square 1, after having it in cube form and every piece in it's respective layer?

 

[Lucas Garron]

What is the probability of a parity on one layer of the square 1, after having it in cube form and every piece in it's respective layer?

1/2, under reasonable assumptions.

 

[JeffDelucia]

If you learned full OLL CLL PLL and ELL and when ever you got an all edges oriented OLL you used cll and whenever you got an all corners oriented oll you used Ell what would your probability of a 1lll be?

 

[CuBeOrDiE]

If you learned full OLL CLL PLL and ELL and when ever you got an all edges oriented OLL you used cll and whenever you got an all corners oriented oll you used Ell what would your probability of a 1lll be?

1/22 I assume, because OLL would be taken care of, and there are 22 PLL cases, one of which is the solved cube. but there is something wrong with your question; if you have all corners oriented and did ELL, you would still have to solve them because ELL orients edges AND corners. correct me if I'm wrong

 

[JeffDelucia]

If you learned full OLL CLL PLL and ELL and when ever you got an all edges oriented OLL you used cll and whenever you got an all corners oriented oll you used Ell what would your probability of a 1lll be?

1/22 I assume, because OLL would be taken care of, and there are 22 PLL cases, one of which is the solved cube. but there is something wrong with your question; if you have all corners oriented and did ELL, you would still have to solve them because ELL orients edges AND corners. correct me if I'm wrong[/QUOTE

I don't think thats right =/

 

[CuBeOrDiE]

why not?

 

[eastamazonantidote]

why not?

Because the chances of PLLs occurring is not equally spread. Just look at badmephisto's PLL page for the probabilities. The chance of a PLL skip is 1/72.

 

[CuBeOrDiE]

hmmm... could easily be true, but I wonder why?

edit: how did he get that probability, by many trials or by some kind of mathematics?

 

[Johan444]

hmmm... could easily be true, but I wonder why?

edit: how did he get that probability, by many trials or by some kind of mathematics?

Depends of how many different cases of each PLL you can get. Take i.e. T-perm, the corner swap can be on right, left, back and front side of the up layer. That makes it four different T-perms. A PLL skip only has one case, it's 4 times rarer.

 

[Meep]

Does anyone know the probability of getting a last 4 edges skip in 5x5 reduction?

 

[CuBeOrDiE]

hmmm... could easily be true, but I wonder why?

edit: how did he get that probability, by many trials or by some kind of mathematics?

Depends of how many different cases of each PLL you can get. Take i.e. T-perm, the corner swap can be on right, left, back and front side of the up layer. That makes it four different T-perms. A PLL skip only has one case, it's 4 times rarer.

oh, i didn't think of it that way. thanks!

 

[CubesOfTheWorld]

hmmm... could easily be true, but I wonder why?

edit: how did he get that probability, by many trials or by some kind of mathematics?

Depends of how many different cases of each PLL you can get. Take i.e. T-perm, the corner swap can be on right, left, back and front side of the up layer. That makes it four different T-perms. A PLL skip only has one case, it's 4 times rarer.

H-perm also has one case. J and R perms are like the T-perm. You can use the same algorithm in four different cases.

Whenever I get an H-perm, I think, "Man... Why couldn't that be a PLL skip? It has the exact same chance..."

EDIT: J and R's are not like the T-perm. One kind of R or J is like the T-perm.

 

[LewisJ]

Does anyone know the probability of getting a last 4 edges skip in 5x5 reduction?

My thoughts say 1/8! (1/40320)

Pretend the last 4 midges don't matter; then, the 8 wings must all be in the correct spot. The 8 wings can be permuted in any possible way (8!) but only one has all the edges solved.

 

[JeffDelucia]

Whats the probability of having LL edges oriented? LL corners?

 

[miniGOINGS]

Whats the probability of having LL edges oriented? LL corners?

Edges: 1/8
Corners: 1/27