Originally Posted by

**Massacre**
Any skew on crit rate for normal attacks based on mastery would have been obvious when I was attacking with 2 swordsmanship. If the skill made any difference, I'd expect to be hitting roughly 0% crits with no ranks in it, yet I was hitting them about 20% of the time, same I was was with 12 ranks. My samples for the normal hits were about 100 each, tho I will admit my samples for the skills were smaller.

Again tho, if swordsmanship skewed crits on autoattacks, there would be an immediate and obvious near-absence of them while running no or next-to-no points in the attribute. This was not the case.

I'm sorry, Massacre, but your sample population simply wasn't large enough to see valid trends. To prove my point (and to double-check your observations/conclusions), I created a PvP warrior with a +15% when hexed axe and beat on AL 60 target dummies for 2000 hits at 5 different ranges of Axe Mastery. That's 400 auto-attack hits each at ranks 0, 1, 4, 9, and 13.

Here are the results, which clearly show a progression of crit rate that is consistent with SonOfRah's *fuzzy* formula. I've broken down each of the results into a total crit amount for the entire 400 hits, followed by a breakdown of the crits that occurred in each 100 hits to show how wide the variance can be per hundred.

Also note that 400 hits is still way too small of a sample size to show a conclusive percentage crit rate. But it *is* large enough to clearly show trend, which is what I was after.

Code:

0 ranks: 0 crits (0, 1, 0, 1)
1 ranks: 10 crits (1, 4, 2, 3)
4 ranks: 22 crits (7, 6, 4, 5)
9 ranks: 49 crits (6, 14, 17, 12)
13 ranks: 60 crits (11, 16, 20, 23)

How do these observations compare with what SonOfRah's formula would predict?

Critical hit chance = (1+[1.25*Attribute])*2^ ([CLVL - TLVL]/5)

Code:

0 ranks = 1.00% (1+[1.25* 0])*2^ ([20 - 20]/5)
1 ranks = 2.25% (1+[1.25* 1])*2^ ([20 - 20]/5)
4 ranks = 6.00% (1+[1.25* 4])*2^ ([20 - 20]/5)
9 ranks = 12.25% (1+[1.25* 9])*2^ ([20 - 20]/5)
13 ranks = 17.25% (1+[1.25*13])*2^ ([20 - 20]/5)

And here we go... My observed results, with a sample population of only 400, is still obviously very close to the predicted results. SonOfRah might not have derived the *exact* formula, but his fuzzy formula is obviously damn close to the actual formula.

Code:

0 ranks: 0.50% observed, 1.00% predicted
1 ranks: 2.50% observed, 2.25% predicted
4 ranks: 5.50% observed, 6.00% predicted
9 ranks: 12.25% observed, 12.25% predicted
13 ranks: 15.00% observed, 17.25% predicted