Probability Thread

[cubenerd74]

Does anyone know the odds of, after solving your third pair, you get a fourth pair skip into an OLL skip?

1/10(Edge solved)x1/15(Corner solved)x1/216(OLL skip)=1/32400

 

[Deleted member 74176]

I know this is a very common question, but what are the odds of a 4-mover in 2x2? I've done about 12k solves and am yet to get one

 

[OtaMota]

1 in 2000.

 

[Jorian Meeuse]

I know this is a very common question, but what are the odds of a 4-mover in 2x2? I've done about 12k solves and am yet to get one

It's around 1 in 2000 (1 in 1989.26... to be more precise). I don't know what your level of 2x2 has been throughout those 12k solves, but realize that not all 4 movers are easy to spot. It is very possible that you have had 4 movers without realizing

Try this: R U R' U' R' F R F' x2 R U R' U' R' F R F'
Spot the 4-mover.

Try U2 R U' L' U R' U' L U x2 R U' L' U R' U' L
Spot the 4-mover.

Notice how, despite these 4-movers definitely being possible to spot, they are also relatively easy to miss if you aren't paying full attention.

 

[OtaMota]

Pocket Cube - Wikipedia

en.wikipedia.org

 

[xyzzy]

I don‘t know how easy this would be to calculate, but what is the probability of having the same EP on both sides of a Sq1

Since nobody correctly answered this yet:

There are 10 possible EPs per face, with these probabilities:
Skip: 1/24
H: 1/24
Oa: 1/24
Ob: 1/24

Z: 1/12
Opp: 1/12

Ua: 1/6
Ub: 1/6
Adj: 1/6
W: 1/6

If you don't solve parity before EP, the probability is 4/24^2 + 2/12^2 + 4/6^2 = 19/144 ~ 13.2%.
If you do (e.g. with CSP or CPP), the probability doubles to 19/72 ~ 26.4%.

(If you treat the two O perms as the same and the two U perms as the same, then the probabilities increase to 55/288 ~ 19.1% without parity pre-solved, 55/144 ~ 38.2% with parity pre-solved.)

 

[Caden Fisher]

Since nobody correctly answered this yet:

There are 10 possible EPs per face, with these probabilities:
Skip: 1/24
H: 1/24
Oa: 1/24
Ob: 1/24

Z: 1/12
Opp: 1/12

Ua: 1/6
Ub: 1/6
Adj: 1/6
W: 1/6

If you don't solve parity before EP, the probability is 4/24^2 + 2/12^2 + 4/6^2 = 19/144 ~ 13.2%.
If you do (e.g. with CSP or CPP), the probability doubles to 19/72 ~ 26.4%.

(If you treat the two O perms as the same and the two U perms as the same, then the probabilities increase to 55/288 ~ 19.1% without parity pre-solved, 55/144 ~ 38.2% with parity pre-solved.)

Thank you!

 

[Isaiah Scott]

What is the probability of a PLL skip on a megaminx?

 

[cubenerd74]

What is the probability of a PLL skip on a megaminx?

If you count cases with AUF, then it would be 5/(5!(CP)x5!(EP)/5(Rotational symmetry)/2(Parity))=1/288=0.3472%

 

[Isaiah Scott]

Thanks!

 

[Jorian Meeuse]

Since nobody correctly answered this yet:

There are 10 possible EPs per face, with these probabilities:
Skip: 1/24
H: 1/24
Oa: 1/24
Ob: 1/24

Z: 1/12
Opp: 1/12

Ua: 1/6
Ub: 1/6
Adj: 1/6
W: 1/6

If you don't solve parity before EP, the probability is 4/24^2 + 2/12^2 + 4/6^2 = 19/144 ~ 13.2%.
If you do (e.g. with CSP or CPP), the probability doubles to 19/72 ~ 26.4%.

(If you treat the two O perms as the same and the two U perms as the same, then the probabilities increase to 55/288 ~ 19.1% without parity pre-solved, 55/144 ~ 38.2% with parity pre-solved.)

Lol thanks for the correction
I hope it was past midnight when I wrote my stupidly oversimplified answer otherwise that would be very embarrassing haha

 

[Weston_Todd]

What's the probability of getting a CN cross skip?

 

[cubenerd74]

What's the probability of getting a CN cross skip?

the probability of getting a cross skip on a certain color is 1/24*1/22*1/20*1/18=1/190080. Since there are 6 colors, the answer is 6/190080=1/31680=0.003157%.

 

[Jorian Meeuse]

the probability of getting a cross skip on a certain color is 1/24*1/22*1/20*1/18=1/190080. Since there are 6 colors, the answer is 6/190080=1/31680=0.003157%.

Actually, the probability is ever so slightly smaller because there are scrambles with multiple crosses solved. It doesn't change a lot, but it matters a bit. For future reference, you can use this link (thanks to @qwr for telling me this):

CubeZone - Cross study

 

[OtaMota]

How many algs in an alg set do you need to know so that in 50% of your ao5s, all five solves are from that alg set. What about 10% instead of 50%?

 

[FenTheGoat]

How many algs in an alg set do you need to know so that in 50% of your ao5s, all five solves are from that alg set. What about 10% instead of 50%?

Depends on the alg set. For most alg sets, like pll, it’s not an even split on the chances you’ll get a certain case (for example, most plls are a 1/18 chance, some are a 1/36, and the N perms are 1/72)

 

[OtaMota]

That is true. Okay so I guess I will only use ZBLL in this example. I am aware that depending on the probability the case appears, you may need to learn more algs to get 50% of ao5s full ZBLL. What is the min-max number of ZBLLs you would need to know to have a 50% chance of having full ZBLL during an ao5?

 

[TheSpeedSkewber]

Just interested after seeing this on JDAW6's Progression, what's the chance of getting a LL Skip and a PLL skip in the same Mo3?

 

[cubenerd74]

Just interested after seeing this on JDAW6's Progression, what's the chance of getting a LL Skip and a PLL skip in the same Mo3?

Assuming the two are in different solves, that would be 1/15552(LL skip)*1/72(PLL skip)*6, which is about
0.0005%.

 

[JDAW6]

Assuming the two are in different solves, that would be 1/15552(LL skip)*1/72(PLL skip)*6, which is about
0.0005%.

Bro that is so low I am just lucky l9l