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Probability Thread

Stefan

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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.
 

cmhardw

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If I do a first face (not a layer), how many positions are possible then?

Permute the solid color layer:
(4!)^2

Permute and orient the rest of the cube:
(4!)*3^3 corners
(8!/2)*2^7 edges

For edge permutations you divide by 2 since the edge permutation parity must match the corner permutation parity.

Altogether:
(4!)^2*(4!)*3^3*(8!/2)*2^7 = 963158 999040
 

guysensei1

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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.
 

TMOY

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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.

First we count the total number of cases (considering cases which are identical up to rotation as different); we obtain 3^3 corner orientations, 4! corner permutations, 2^2 edge orientations and 3! edge permutations, dividing by 2 because of parity, we obtain 3^3*4!*2^2*3!/2=7776 positions.

Next we count the number of symmetrical positions. For a position to be symmetrical, the corner on top of the tripod must be in place (not necessarily oriented), and the other three must all have orientations compatible with the orientation of the top corner (only 1 possibility), and their permutation must be a 3-cycle. The edges have only one possible orientation, and their permutation must also be a 3-cycle. Hence 3*3*3=27 symmetrical positions, and 7776-27=7749 asymmetrical positions..

So the total number of cases, excluding the solved case which is of course symmetrical, are: 7749/3+26=2609 cases.
 
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cmhardw

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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.

That's a difficult question. In terms of number of combinations in total there are:
4!*3^3*3!/2*2^2=

I tried using the lemma that is not Burnside's (and I may have used it incorrectly) and got:
1/6 * (7776 + 3*(4(2)) + 1(3(2*2))) = 1302 unique cases, which is more than total LL combinations.

Since the tripod is essentially the last layer with 1 solved edge, I would think it would be less than the 1211 algs required for 1LLL, but maybe since the tripod has fewer symmetries (I count 8 for a layer and 6 for a tripod, but that may be wrong) then there would be more tripod cases?

Can someone with a better math background comment on this?

--edit--
Ninja'd by TMOY
 
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guysensei1

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Thanks for the calculation! What if we used 2 look tripod? Which set will have less cases: solve corners in 1 look, then edges, or edges first then corners?
 

cmhardw

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So the total number of cases, excluding the solved case which is of course symmetrical, are: 7749/3+26=2609 cases.

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.
 

TMOY

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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.

The difference between our two calculations is simply that you're counting two cases which are mirror images of each other as only one while I count them as two.
 

SirWaffle

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7x7 Last 4 edge odds?

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.
 

cmhardw

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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.

The probability that at least 1 edge group is solved is approximately 0.127%

Clicky

--edit--
I fixed a conceptual math error about 5 minutes after sending this post.
 
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cmhardw

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cmhardw

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You can also use the four middle edges as reference.

Wow, I understand what you did there using the edges as reference. I also just realized that in my expression those terms would cancel out, effectively using the edges as reference.

I would not have thought to use inclusion-exclusion like you did where you start by counting the number of states with 1 (really at least 1) edge group solved. That makes sense now that I look at it. For some reason I always want to use it to find derangements, when it's obviously a more powerful method with further uses.

Thanks for the math tips! As always, Stefan, you've got skills!
 
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