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For 2x2x2, I don't get how many moves the minimum scramble is in competition. In the WCA regulations, it says "random position" which I don't get. On the scramble program, it gives me http://www.worldcubeassociation.org/regulations/scrambles/scramble_cube_222.html which has 8 moves automatically, is this correct? And is the actual scrambler they use? Thanks.

If you look in the page source for the url you gave, you'll find this:
// Alternative mixing function, based on generating a random-state (by Conrad Rider)
function mix2(){
It's generating a random state for the puzzle, then finding a solution for that state. The number of moves doesn't matter. I'm not sure why you're saying "has 8 moves automatically"; the number of moves varies from scramble to scramble, and the default setting for minimum scramble length is zero, so it doesn't look like that's where you got it. It's the scrambler I use when generating scrambles for an official competition. I also used this code (copied separately and incorporated into a script) to generate the scrambles for the weekly competitions here.
 
If you look in the page source for the url you gave, you'll find this:

It's generating a random state for the puzzle, then finding a solution for that state. The number of moves doesn't matter. I'm not sure why you're saying "has 8 moves automatically"; the number of moves varies from scramble to scramble, and the default setting for minimum scramble length is zero, so it doesn't look like that's where you got it. It's the scrambler I use when generating scrambles for an official competition. I also used this code (copied separately and incorporated into a script) to generate the scrambles for the weekly competitions here.

Thank you very much for the explanation. I only counted the moves for the first scramble I got, and the others, just by a quick glance, looked the same length. So my bad on that, sorry.
 
I am unsure what you mean by this. First of all, the cross can sometimes require up to 8 moves. Second, where are you getting the 7's for each F2L pair?

I took a random F2L cheat sheet and counted:
4 + 4 + 3 + 3 + 8 + 7 + 8 + 7 + 8 + 8 + 8 + 11 + 8 + 8 + 8 + 8 + 7 + 7 + 7 + 7 + 5 + 5 + 11 + 9 + 8 + 8 + 7 + 7 + 7 + 7 + 7 + 11 + 8 + 8 + 8 + 8 + 0 + 11 + 11 + 11 + 11 + 12 / 42 = 7,5952380952380952380952380952381
So it should prob. be 8+7+8+7+8=38 instead
 
I took a random F2L cheat sheet and counted:
4 + 4 + 3 + 3 + 8 + 7 + 8 + 7 + 8 + 8 + 8 + 11 + 8 + 8 + 8 + 8 + 7 + 7 + 7 + 7 + 5 + 5 + 11 + 9 + 8 + 8 + 7 + 7 + 7 + 7 + 7 + 11 + 8 + 8 + 8 + 8 + 0 + 11 + 11 + 11 + 11 + 12 / 42 = 7,5952380952380952380952380952381
So it should prob. be 8+7+8+7+8=38 instead

I see. That probably isn't the best way to figure out F2L movecount, considering how people solve differently and don't always use what could be considered the most common 'alg', and I'm pretty sure many of those aren't optimal. I don't have any tips on a better way to count, though....
 
It is more of a guideline instead of a bottom line. You could get unlucky and have a solve with 11+11+11+12 F2L but as you get more XP you would avoid those if it wasn't the last unsolved slot, or you would learn advanced F2L tricks. Sub 40 MC for C+F2L is a nice goal in every solve.
 
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