Probability Thread

Probability of Feliks getting 3x3 single WR back?

Why u ask? Who cares? Congratulate collin instead of wishing it was feliks. I hate it when ppl do this. (ex: cubing world andrew ricci how to get fast at 3x3 and ppl complained about hot it wasnt feliks.) He isnt the only fast cuber by the way. Respect plz for collin.
 
Why u ask? Who cares? Congratulate collin instead of wishing it was feliks. I hate it when ppl do this. (ex: cubing world andrew ricci how to get fast at 3x3 and ppl complained about hot it wasnt feliks.) He isnt the only fast cuber by the way. Respect plz for collin.

Yeah I know Faz isn't the only fast cuber ofc. Collin does deserve it and congrats but I'm rooting for faz to get single wr back. :P
 
Couldn't you do it as this: P = 1 - (215/216)(71/72) = 1 - 15265/15552 = 287/15552? Just seems more intuitive to me.

I just got an OLL skip on my 4x4. Is that a 1/432 chance or am I doing something wrong?
 
Yeah, the probability of not getting parity + getting an OLL skip is 1/432. However, I could argue that getting an OLL skip after parity should also count (this case for most people) bringing the probability back down to 1/216 (this assumes parity is solved by flipping the one miss oriented edge and not just a random one).
 
What's the probability of not getting EPLL (PLL skips are counted as EPLLS) in an average of 50?
Assuming every solve path goes through EPLL (2-look PLL, COLL/EPLL, etc):

EPLL skip: 1/12
not EPLL skip: 11/12
not EPLL skip 50 times: (11/12)^50 = 0.01289947263

EDIT: misread the question. My answer is for not skipping EPLL in 50 solves.
 
Assuming every solve path goes through EPLL (2-look PLL, COLL/EPLL, etc):

EPLL skip: 1/12
not EPLL skip: 11/12
not EPLL skip 50 times: (11/12)^50 = 0.01289947263

EDIT: misread the question. My answer is for not skipping EPLL in 50 solves.

I was assuming 1 look PLL.
 
What's the probability of not getting EPLL (PLL skips are counted as EPLLS) in an average of 50?

Getting an EPLL is equivalent to having CP solved, which has a 1/6 probability.
The probability of not having solved CP 50 times in a row (assuming no COLL or anything like that) is (5/6)^50 = 0.00010988481... or about 1/9100
 
Couldn't you do it as this: P = 1 - (215/216)(71/72) = 1 - 15265/15552 = 287/15552? Just seems more intuitive to me.

I just got an OLL skip on my 4x4. Is that a 1/432 chance or am I doing something wrong?
You could do it that way too.
Yes, it's 1/432.
What's the probability of not getting EPLL (PLL skips are counted as EPLLS) in an average of 50?
Depends how much OLLCP you know :p
 
Since it seems that misscrambles happen way too often, because of scramblers not properly checking the scramble, I was thinking about writing some application to check if a picture of a scrambled cube matches the scramble.

One issue I stumbled upon is the fact that people have weird colorscheme's. Therefor, it's only possible to check for "patterns", for example (top side):
[A, A, B]
[C, D, A]
[B, A, E]
Where A, B, etc. could be any color, since we don't know what colors the cube has anyway. But we do know that (UBL, UB, UR, UF) all have the same color.
Since the application needs a picture of a cube, we have 3 sides to check (or, 27 stickers, minus 6 stickers used as "reference")

My question:
For a scramble and a "pattern", constructed like described before, what is the probability that the pattern seems to match the scramble, but the cube was actually misscrambled? Or, in other words, how trustworthy is this method? :)

Sorry if anything is unclear.
 
My question:
For a scramble and a "pattern", constructed like described before, what is the probability that the pattern seems to match the scramble, but the cube was actually misscrambled? Or, in other words, how trustworthy is this method? :)

Sorry if anything is unclear.

I don't understand what's wrong with the current method. You know the orientation you scrambled the cube in, and you know that say, yellow on the scrambled image should be black on the physical cube.

If this is for a human to check, would they insert the colours into the scrambler beforehand? Or would they have to read it as a bunch of letters? If you're inserting the colours yourself, it'd be hard to implement in competitions and Prisma Puzzle timer already does it if it's for personal use. If you meant for somebody to read it as letters, it's harder to recognise than just knowing that one colour may correspond to a different colour.

Do you mean mis-scrambling by missing or adding one or two moves, or any scrambled state that aren't the one it needs to be? A 3-cycle on the D face might be hard to recognise if a human is just glancing at the cube, but at the same time, it wouldn't be possible to get to that state by only mistaking two moves. Also, you should put an example of how much of the cube is properly checked, since different people check different amounts. Try something like checking the U and F face, since I think most people check those.
 
If this is for a human to check, would they insert the colours into the scrambler beforehand? Or would they have to read it as a bunch of letters?

The human doesn't check anything, because clearly humans suck at checking scrambles :)
In some further stage, it should be possible to take a picture of the cube with your smartphone and have the app say "Yeah, I think this matches the scramble" or "You misscrambled".
 
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