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Probability Thread

What is the change of getting ( in comp ) on 4x4 on OLL OLL parity and the rest solved like 1 edge flipped?
And what is the change of getting edge skip on 4x4 REDUCTION
 
TeddyKGB said:
I just did a 4x4 solve and got a LL skip with the exception of OLL parity and PLL parity, are the odds of this the same as a regular LL skip?
Click to expand...

If you must have both OLL and PLL parity, and assuming the edge groups are all paired up, then this is:
\( \frac{16 \left( \begin{array}{ccc} 4 \\ 2 \end{array} \right)}{(4!)^2*2^4*3^3}=\frac{1}{2592} \)

Antonie faz fan said:
What is the change of getting ( in comp ) on 4x4 on OLL OLL parity and the rest solved like 1 edge flipped?
Click to expand...

Assuming you mean just 1 edge group flipped, and the rest of the LL is otherwise solved this would be:
\( \frac{16}{(4!)^2*2^4*3^3}=\frac{1}{15552} \)

Coincidentally, this is the exact same probability of a LL skip on 3x3x3, assuming no partial edge control, corner orientation control, etc. (!)

Antonie faz fan said:
And what is the change of getting edge skip on 4x4 REDUCTION
Click to expand...

Assuming you mean that after solving centers (not using Yau, but simply solving the centers), that every edge group is already paired up for you then this would be:

\( \frac{12!*2^{12}}{24!}=\frac{1}{316234143225} \)

yoshinator said:
Somebody at Vancouver open got a LL skip (either forte or turbo)
Click to expand...

-You're very nearly 300 times more likely to get at least one LL skip on 3x3x3 in an average of 5 than to get a LL skip on one solve of megaminx
-You're very nearly 12 times more likely to get a LL skip on one solve of 3x3x3 than you are to get at least one megaminx LL skip in an average of 5
-You're about 26 times more likely to get a megaminx LL skip on one solve than you are to get at least two 3x3x3 LL skips in an average of 5

That's very exciting!
 
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\( \sum_{k=0}^{12} \sum_{j=0}^{12-k} \frac{k}{24!}\left[ \left( \begin{array}{ccc} 12 \\ k \\ \end{array} \right)*\frac{12!}{(12-k)!}*2^k\right] \) \( \left[\left( \begin{array}{ccc} 12-k \\ j \\ \end{array} \right)*\frac{(12-k)!}{(12-k-j)!}*2^j*(24-2k-2j)!*(-1)^j\right]=\frac{12}{23} \)

When solving the 4x4x4 using reduction (not Yau method), the expected number of solved edge groups at the beginning of the edge pairing step is \( \frac{12}{23} \) edge group, or about 0.52 edge groups. Basically, about half the time no edge groups will already be solved, and about half the time 1 edge group will already be solved (expected value).

Mathematica is AWEsome! :D
 
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\( \sum_{k=0}^{12} \sum_{j=0}^{12-k} \frac{k}{24!*12!*2^{11}} \left( \begin{array}{ccc} 12 \\ k \\ \end{array} \right)*\frac{12!}{(12-k)!}*2^k* \) \( \left( \begin{array}{ccc} 12-k \\ j \\ \end{array} \right)*\frac{(12-k)!}{(12-k-j)!}*2^j*(24-2k-2j)!*(12-k-j)!*2^{(12-k)-1-j}*(-1)^j \)

This simplifies to

\( \sum_{k=0}^{12} \sum_{j=0}^{12-k} \frac{k}{24!}\left( \begin{array}{ccc} 12 \\ k \\ \end{array} \right)\left( \begin{array}{ccc} 12-k \\ j \\ \end{array} \right)(-1)^j(24-2k-2j)!=\frac{1}{46} \)

When solving the 5x5x5 using standard reduction, the expected number of solved tredge groups at the beginning of the edge pairing step is \( \frac{1}{46} \) tredge group, or about 0.0217 tredge groups. Basically, about one in every 46 of your solves will have one tredge group already solved at the start of the edges step (expected value).

Stefan said:
The complete opposite of tapatalk. Which replaces each of your formulas by a friggin smiley as if that's funny! Argh!
Click to expand...

That does sound like a pain :( I don't have a smart phone yet, so it'll be a little longer before I have to put up with that.
 
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Expected number of semi-pairs at the start of 5x5x5 edge pairing (reduction method)

\( \sum_{k=0}^{24} \sum_{j=0}^{24-k} \frac{k}{24!} \left( \begin{array}{ccc} 24 \\ k \end{array} \right) \left( \begin{array}{ccc} 24-k \\ j \end{array} \right) (24-k-j)! (-1)^j = 1 \)

When solving 5x5x5 using reduction, at the beginning of the edge pairing step you are expected to have 1 semi-paired edge group (a middle edge paired with exactly one wing edge) at the start of the step. This assumes that you do not use any techniques to either influence or preserve semi-pairs while solving centers.

This also assumes that the utility of a solved tredge is the same as that of two semi-pairs in the strict sense of the term "semi-pair" (basically if you have no preference whatsoever between getting two semi-pairs vs. a solved tredge).
 
cmhardw said:
\( \sum_{k=0}^{24} \sum_{j=0}^{24-k} \frac{k}{24!} \left( \begin{array}{ccc} 24 \\ k \end{array} \right) \left( \begin{array}{ccc} 24-k \\ j \end{array} \right) (24-k-j)! (-1)^j = 1 \)

When solving 5x5x5 using reduction, at the beginning of the edge pairing step you are expected to have 1 semi-paired edge group (a middle edge paired with exactly one wing edge) at the start of the step. This assumes that you do not use any techniques to either influence or preserve semi-pairs while solving centers.

This also assumes that the utility of a solved tredge is the same as that of two semi-pairs in the strict sense of the term "semi-pair" (basically if you have no preference whatsoever between getting two semi-pairs vs. a solved tredge).
Click to expand...

I don't quite understand that formula (haven't really tried, though) but here's how I think about it:

The middle edges provide reference, so we can just look at this whole thing as a permutation of 24 objects (the 24 wings) and ask how many we expect to be at the correct place. Now we could go over all 24! permutations and for each one, count how many of the 24 objects are at the correct place, add them all up, and divide by the number of permutations, 24!. But imagine the 24! permutations as a static huge table with 24! rows and 24 columns. What we've done is count "correct places" anywhere in the table, going over the table row by row. Rows have different numbers of correct places, from 0 (much of the time) to 24 (just once). Now instead of row by row, let's count the correct places anywhere in the table, but going over the table column by column. Due to symmetry, column 1 contains object 1 just as often as any other of the 24 objects, so 1/24th of the time, column 1 has a correct place. So unlike rows, all 24 columns have the same number of correct places, namely (1/24)*(24!). Summing them all up (i.e. multiplying with 24) and dividing by 24! of course gives you: 1. No Mathematica needed for this one :p

Btw: {n \choose k} and \binom{n}{k} are much nicer ways to get \( {n \choose k} \).
 
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Stefan said:
I don't quite understand that formula (haven't really tried, though) but here's how I think about it:

The middle edges provide reference, so we can just look at this whole thing as a permutation of 24 objects (the 24 wings) and ask how many we expect to be at the correct place. Now we could go over all 24! permutations and for each one, count how many of the 24 objects are at the correct place, add them all up, and divide by the number of permutations, 24!. But imagine the 24! permutations as a static huge table with 24! rows and 24 columns. What we've done is count "correct places" anywhere in the table, going over the table row by row. Rows have different numbers of correct places, from 0 (much of the time) to 24 (just once). Now instead of row by row, let's count the correct places anywhere in the table, but going over the table column by column. Due to symmetry, column 1 contains object 1 just as often as any other of the 24 objects, so 1/24th of the time, column 1 has a correct place. So unlike rows, all 24 columns have the same number of correct places, namely (1/24)*(24!). Summing them all up (i.e. multiplying with 24) and dividing by 24! of course gives you: 1. No Mathematica needed for this one :p

Btw: {n \choose k} and \binom{n}{k} are much nicer ways to get \( {n \choose k} \).
Click to expand...

Stefan, you never cease to amaze me. That is a very clear explanation of how to count the number of times a piece ends up at the correct location. That is a really elegant proof!

My calculation is an application of the inclusion/exclusion technique.

Sum the number of "correct" pieces, k, as k goes from 0 correct to 24 correct. For each k, calculate using index j the number of derangement permutations of the remaining (24-k) "unsolved" pieces. This gives me 24 terms over the k index. Multiply each of the 24 terms by k and divide by 24! to provide it's contribution to the expected value result.

Also, thank you for those ways to produce a binomial term more elegantly than with arrays!
 
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