Stefan
Member
Hmm, I'd say that depends on your personal F2L then. A beginner might take more moves for F2L and thus have more chances to build such a block and thus have a higher probability.
If I do a first face (not a layer), how many positions are possible then?
How many cases are there to solve the cube in 1 look if one has built a tripod already? Assume that cases that are simply a cube rotated version of another case can be considered the same.
How many cases are there to solve the cube in 1 look if one has built a tripod already? Assume that cases that are simply a cube rotated version of another case can be considered the same.
So the total number of cases, excluding the solved case which is of course symmetrical, are: 7749/3+26=2609 cases.
How probable is this to happen?
https://www.youtube.com/watch?v=iaTUkd948JM
My number is off by almost exactly a factor of 2 from yours. Can you help me figure out where I went wrong? I can explain my calculation if you need. It's been so long since I've used the (not) Burnside's lemma and I'd like to make sure I'm not misusing it.
I really am unsure the correct search terms to use to find the answer so that's why I am creating a new thread. What are the odds of skipping one of the last 4 edges after realigning the centers on 7x7? This happened to me in a solve so I was curious.
You can also use the four middle edges as reference.
http://www.wolframalpha.com/input/?i=sum+(4+choose+i)*(-1)^(i+1)+*+(16-4*i)!+/+16!+for+i+from+1+to+4
Edit: Hmm, just saw your edit and I don't understand your changed answer.
You can also use the four middle edges as reference.
I was treating the 16 wings as 1 orbit of 16 pieces
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