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Probability Thread

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JPerm
last year while i was trying my friend's cube, the very first 2 solves i got were both non-AUF PLL skips.
1/(72*4)^2 = 1/82944.

based on these results, his cube doesn't need PLL. i should get his cube.
 
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Acts of God?

I was practising my algorithm s earlier and remembered an OLL case I remembered that I hadn't used in ages. I practised the algo until I remembered it again, and since that, it has appeared in 2 look OLL for the last 15-ish solves in a row. I used a scramble given by Speed Cube Timer for Android, and scrambled with yellow on top. I wanted to know what 'acts of God' could have happened, like a number of OLL/PLL/ LL skips in a row?

[moved here from being a separate thread by AvGalen]
 
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If you count mirrored PLLs as the same, then H is the rarest (1/72 chance). If you don't, the N perms also each have a 1/72 chance.
Thanks for the answer! But I would like to understand why some PLL's are rarer than others. When I started learning CFOP I thought every PLL have the same probability.
 
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it's to do with symmetry. The symetric (in terms of their diagrams) PLLs have less chance of occuring. For example U perms can happen in 4 ways due to the 4 AUFs. However Nperms can happen in 2 ways, since they are symetrical about an axis there are only actually 2 AUFs so they have 2 ways to occur instead of 1. hence occur less frequently. Now with Hperm it is symetrical about both axes and so has 1 AUF and 1 way of occuring. That is why hperm will occur less frequently.
Got it! Good explanation, thanks man!
 
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it's to do with symmetry. The symetric (in terms of their diagrams) PLLs have less chance of occuring. For example U perms can happen in 4 ways due to the 4 AUFs. However Nperms can happen in 2 ways, since they are symetrical about an axis there are only actually 2 AUFs so they have 2 ways to occur instead of 1. hence occur less frequently. Now with Hperm it is symetrical about both axes and so has 1 AUF and 1 way of occuring. That is why hperm will occur less frequently.
thats what i said

also nperms only have 1 auf
 
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Chance of a single correctly placed Edge/Corner

So I've been thinking about this for a while, but the only answer I can seem to get is 7/8 for corners and 11/12 for edges. Can someone help me out here, and tell me where I'm going wrong?
 
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So I've been thinking about this for a while, but the only answer I can seem to get is 7/8 for corners and 11/12 for edges.
I assume that:
[1] "Correctly placed" means that the corner/edge may be unoriented or oriented in its location.
[2] You're allowing any random configuration of the 3x3x3 cube (including configurations in which only a few pieces are unsolved).

Can someone help me out here, and tell me where I'm going wrong?
You have to take the cycle classes (cycle types) into account as you're doing a calculation like this. Despite all of the possible cycle types, for this type of problem, we just group all cycle types which involve the same number of pieces with each other. So we can use my formula (which I did research on and developed here...you can see that post to see all of the cycle types, of you're interested):

\( \sum\limits_{k=p-x}^{p}{\left( \begin{matrix}
p \\
k \\
\end{matrix} \right)!k} \)

Where:
p = the total number of the piece type. For wing edges, p = 24, for middle edges, p = 12, and for corners, p = 8.
No more than x of each piece type are solved. \( 0\le x\le p \).
and k is just the index of the summation.

Corners
As you probably know, the corners have 8! different permutations. The sum of all of the possible cycle types (and all versions of each cycle type) for 8 objects is equal to this amount. That is, if we let x = 8 (meaning that we sum all cycle types involving 0 objects to all 8 objects...all cycle types), we have:

\( \sum\limits_{k=8-8}^{8}{\left( \begin{matrix}
8 \\
k \\
\end{matrix} \right)!k}=\sum\limits_{k=0}^{8}{\left( \begin{matrix}
8 \\
k \\
\end{matrix} \right)!k}=8! \)

[Link]

But since we are just trying to count how many permutations involve exactly 7 pieces (because involving 7 pieces is equivalent to saying that one piece is left alone/solved), we just use the expression \( \left( \begin{matrix}
8 \\
k \\
\end{matrix} \right)!k \) and not the summation. That is, we substitute 7 for k in the expression, and we get 14, 832.

So we simply do (14832/8!)*100 = 36.8%.

Edges
We use the same reasoning as we used for the corners, and we also get 36.8%.

Corners and Edges
If we want to calculate the chance that only one corner and one edge are correctly placed, we simply multiply them together, but we must be careful to not forget to multiply by two because the parity of the edges and corners must match (cube law of permutations).

\( \left( \frac{\left( \begin{matrix}
8 \\
7 \\
\end{matrix} \right)!7}{8!} \right)\left( \frac{\left( \begin{matrix}
12 \\
11 \\
\end{matrix} \right)!11}{\frac{12!}{2}} \right)\times 100= \) 27%.
 
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So I was practicing yesterday and hit the same LL twice in a row with random scrambles. Completely different solves up to that point, but then identical LL, including AUF - both dot OLLs (OLL 2 from the wiki), followed by J perms with no AUF.

I was just thinking about this today, because I had the same OLL 2 solves in a row... which subsequently both turned into an N perm (but this time with different AUF).

So I guess my question is: what is the probability of getting the same LL (CFOP) 2 solves in a row. This is assuming no edge control, so all OLLs are possible. I'd be interested to know how much lower the probability of my first experience was (i.e. identical LL: same OLL, same PLL, same AUF) than the likelihood of just the same OLL and PLL would be (my second experience).

Answers on a postcard! TIA

edit: I realise as I'm rereading my own post that the probability would be different for more/less common OLL/PLL cases - just how to calculate would be very much appreciated.
 
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