What? Doing a Rubik's cube Behind ur back?

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lets say you make 8 random moves ( L L' stuff does not count ) there is no chance at all that you could have solved one of ALL possible states since the cube could be scrambled in a way that AL LEAST more than 8 moves are necessary.
I think you mean ANY, not ALL.

so I'm sure the chance of solving a random scrambled cube with random moves is les than 1 in 43 quintillion if you are using less than 20 moves.

Also, if you are using a random number of moves the chance of solving it is less than 1 in 43 quintillion…
False. If you have "random scrambled cube" (with "random moves" or whatever), then doing 1 blind move is as likely to put it into a solved state as any other number of blind moves, even a "random" number.
 
I agree with Lucas. If you don't have any knowledge about your starting position, and you make eight moves, you still have no knowledge about your ending position.

A random position is highly unlikely to be within 8 moves of being solved. But if you make the assumption that your are certain that it is not within 8 moves of the solved position, then you know something about that position, and the assumption of it being a completely random position is false.

Without that assumption, then you have that there is a small probability that it is within 8 moves of the solved state, and there's a small chance that some chosen 8 moves will solve it. The end result is still a 1 in 43 quintillion chance (approximately, of course) of the ending position being a solved cube.

The Cayley graph for Rubik's (for face turns, let's say) looks exactly the same from any node. Because of that symmetry, you can't pick a set of moves that will tend to favor one position with respect to others.
 
Superflip: U R2 F B R B2 R U2 L B2 R U' D' R2 F R' L B2 U2 F2

Too long and hard to remember. I just set it up into two six-flips.

You mean do (R U R' F)*5 for one six-flip then set up for the 2nd six-flip, then undo? Wouldn't that be 40 twists plus setup moves? Would that be hard to track blindfolded? (I got that six-flip from Macky's BLD page.)

I prefer ((M' U)*4 x y')*3. No setups, easy to perform. It leaves the cube rotated correctly when finished. (I discovered this super flipper 20+ years ago. I don't remember if I found it myself or if I read it in David Singmaster's book.)
 
He didn't look at the cube before? How should that be possible.. the same thing as when I say: I'm think of a number between 0 - 99999999999999999999. Gues it..

The answer has to be 42 because obviously we are all hunting for the answer and make this the great question.

@topic I think he just set it up with a scramble he knew so that it looks really scrambled or it was an easy blind scramble for him which he still knew, so that you if you cube don't see the cube and think, haha it's nearly finished..
 
lets say you make 8 random moves ( L L' stuff does not count ) there is no chance at all that you could have solved one of ALL possible states since the cube could be scrambled in a way that AL LEAST more than 8 moves are necessary.

so I'm sure the chance of solving a random scrambled cube with random moves is les than 1 in 43 quintillion if you are using less than 20 moves.

Also, if you are using a random number of moves the chance of solving it is less than 1 in 43 quintillion…

so I think Swoncen is right

Remi

Like Lucas said, I am right...
 
Superflip: U R2 F B R B2 R U2 L B2 R U' D' R2 F R' L B2 U2 F2

Too long and hard to remember. I just set it up into two six-flips.

You mean do (R U R' F)*5 for one six-flip then set up for the 2nd six-flip, then undo? Wouldn't that be 40 twists plus setup moves? Would that be hard to track blindfolded? (I got that six-flip from Macky's BLD page.)

I prefer ((M' U)*4 x y')*3. No setups, easy to perform. It leaves the cube rotated correctly when finished. (I discovered this super flipper 20+ years ago. I don't remember if I found it myself or if I read it in David Singmaster's book.)

It's easier to execute than the optimal one. Plus, superflip isn't really need in a blindfold solve.
 
Superflip: U R2 F B R B2 R U2 L B2 R U' D' R2 F R' L B2 U2 F2

Too long and hard to remember. I just set it up into two six-flips.

You mean do (R U R' F)*5 for one six-flip then set up for the 2nd six-flip, then undo? Wouldn't that be 40 twists plus setup moves? Would that be hard to track blindfolded? (I got that six-flip from Macky's BLD page.)

I prefer ((M' U)*4 x y')*3. No setups, easy to perform. It leaves the cube rotated correctly when finished. (I discovered this super flipper 20+ years ago. I don't remember if I found it myself or if I read it in David Singmaster's book.)

It's easier to execute than the optimal one. Plus, superflip isn't really need in a blindfold solve.

If you do 3OP then it may come in handy when there are 10-12 edges misoriented.
 
Superflip: U R2 F B R B2 R U2 L B2 R U' D' R2 F R' L B2 U2 F2

Too long and hard to remember. I just set it up into two six-flips.

You mean do (R U R' F)*5 for one six-flip then set up for the 2nd six-flip, then undo? Wouldn't that be 40 twists plus setup moves? Would that be hard to track blindfolded? (I got that six-flip from Macky's BLD page.)

I prefer ((M' U)*4 x y')*3. No setups, easy to perform. It leaves the cube rotated correctly when finished. (I discovered this super flipper 20+ years ago. I don't remember if I found it myself or if I read it in David Singmaster's book.)

It's easier to execute than the optimal one. Plus, superflip isn't really need in a blindfold solve.

If you do 3OP then it may come in handy when there are 10-12 edges misoriented.

Well, if you use a method that doesn't orient first...
 
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