ilikecubing
Member
Probability of a PBL skip on 2x2?
probability that corners are solved after finishing F2L on 3x3?
probability that corners are solved after finishing F2L on 3x3?
This is wrong because scambles aren't just random strings of letters. First, a random combination is selected, then a scramble is generated that leads to that combination.I like this. Difficult, but a rough maximum to start the thinking: Assume a 20 move "naive" scramble. Probability that moves 11-20 are the inverse of 1-10 = 1 in 18^10 = 1 in 3.57 Trillion
The actual probability is likely way higher than this, as there's so many possibilities, e.g. where the scramble has junk like R L R' L' in it. I'm guessing by a factor of a hundred.
So, it *could* happen, given millions of cubers doing hundreds of thousands of solves in a lifetime. And I want to see the dude's face when it does. He'd be all like "Holy crap!"
After solving a 2x2x3 block in Petrus, how many combinations are there left?
Yes.Is there any possibility that with a random scramble the cube will solve itself? how about being 1 move away? or a u2?
I'm not 100% sure what you mean by relative, but see http://www.cubezone.be/crossstudy.html.What the probability of a relative cross solved on any colour?
6*5*4 * 3^5 * 7! / 2 = 73,483,200How many combinatons are there when scambling using <R,U>?
6! * 3^5 * 7! * 2^6 / 2 = 28,217,548,800After solving a 2x2x3 block in Petrus, how many combinations are there left?
After the OLL is finished, what is the probability that the PLL is in the right place (no U, U', U2, y, y', y2 needed to apply the alg) ??
1/4: there are 4 ways to AUF, and 1 is right.
Why? Unless I've misunderstood the question, just divide the number of 2-move QTM scrambles (I think it's 27) by the number of combinations?Probably only Ryan Heise knows. E-mail him here: webmaster at hi-games.net
I'm not sure I get this - sorry if it's just me. Anyway: By stickers do you mean edge + corner stickers as well? Why is x between 0 and 10? What do you mean ignoring twisted corners - we leave them there twisted? And what about flipped edges?
What is the probability in a BLD solve that, ignoring twisted corners in place and using a fixed buffer, you will have to shoot to x stickers? (0 <= x <= 10)
I mean only corners. And yes, completely ignoring twisted corners as if they are solved.
x is 10 or less because that is the maximum number of stickers you would have to shoot to in order to solve the corners. It's intuitively pretty clear. Assuming you start with the buffer piece solved, There can at most be 3 cycles you have to break into. Two of these will be of length 2 and the other of length 3, and you will have to shoot to 3 stickers to solve a 2-cycle and 4 to solve a 3-cycle. 3+3+4=10
Even Permutations
{ 7 } 5760
{ 6, 2 } 3360
{ 5 } 1344
{ 5, 3 } 2688
{ 4, 4 } 1260
{ 4, 2 } 2520
{ 3, 3 } 1120
{ 3, 2, 2 } 1680
{ 3 } 112
{ 2, 2, 2, 2 } 105
{ 2, 2 } 210
{ } 1
Odd Permutations
{ 8 } 5040
{ 6 } 3360
{ 5, 2 } 4032
{ 4, 3 } 3360
{ 4, 2, 2 } 1260
{ 4 } 420
{ 3, 3, 2 } 1120
{ 3, 2 } 1120
{ 2, 2, 2 } 420
{ 2 } 28
So, if my maths are correct, 16065/40320 or 39.8% of corner solves do *not* require breaking into a new cycle.