Hi,

A few days ago ChrisH asked me for some quesitons about the exact 
edge distributions for the 4x4. I'd like to report my findings here.

Due to the rather technical nature of how I obtained these number 
(quadruple nested summations and products)..., I omit the details. 
Besides I don't want to ruin it for ppl who'd like to solve this 
problem on their own.

For the 3x3, the cross *exact *probabilities on a random cube is:
0    84.43%           160489/190080
1    14.51%           6893/47520
2    1.03%            65/6336
3    0.046%           17/47520
4    (5.26*10^-4)%    1/190080

For the 4x4, the chances of getting a specific number of solved 
edges on a random scramble or after doing centers randomly is:
0    59.38%
1    30.92%
2    8.08%
3    1.41%
4    0.186%
5    0.0198%
6    etc.....
7
8
9
10
11
12

For the 5x5, .... well you get the idea...


Ok it's better if you just look at the following pages if you are 
intersted in this discrete distribution:
http://www-personal.umich.edu/~dlli/3x3_ExactCross.jpg
http://www-personal.umich.edu/~dlli/4x4_EdgePairs.jpg
http://www-personal.umich.edu/~dlli/5x5_EdgePairs.jpg

That is Mathematica code, btw (but those functions are certainly not 
built in, lol).

These took me several hours to figure out, the 5x5 probabilites 
where based on the other two and easily found after I figured out 
the cross stuff...

Everything adds to 1, so it checks out. I also matched the numbers 
to doing it by case count for some base cases so I'm confident it's 
correct.

In case you're interested there is a great deal of probability 
theory involved and what Chris showed me: Inclusion-Exculsion 
Principle. Also I had to resort to what I would call "the Theory of 
Highly-Generalized Partial-Derangments" (lots of Subfactorials 
involved and most math ppl don't even know what that is).

I can show you how I setup the summations if soneone's interested. 
Plus I can generalized it to NxNxN cubes.

Tell me if someone already did this, I'd like to know if I was the 
first to bother doing this.


-Doug





 
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