i dont understand yours logic cause all 3 bears made 64 attacks but only 49 hits me. and who told you critical hits are counted from properly attacks and not from miss hits? we dont know mechanics and we only can guess or ask
It's possible that critical hits bypass both evasion and block, or maybe just one of them. And indeed, we can only ask. This would, of course, make your case even more plausible; for 64 attacks at 1% critical, 3 or better would have a chance of 2.65%, or one-in-forty
and still dont understand yours counting. 3/49*100%=~6%
That's right, it's close to 6% - but what are the chances of that? Here's the breakdown.
Each attack has a 99% chance of
not being a critical hit. For 0 criticals, all attacks need to be in that category. Thus, the chance for 0 criticals is 0.99^49=61.11%.
For 1 critical, we expect a chance of 0.01*0.99^48 - 1 critical and 48 non-crits - but since the one critical can be any of the 49, there are 49 equal options, which means the chance for one critical in the entire battle is 49*0.01*0.99^48=30.25%.
For 2 criticals, the same logic applies. 0.01^2*0.99^47, this time with 49*48/2 possible arrangements for the criticals (one can be any of 49, then it's one of the remaining 48, then divide by 2 to nullify the fact that we counted each possible combination twice). 49*48/2*0.01^2*0.99^47=7.33%.
So the chance for 2 criticals or less in a battle with 49 hits (assuming criticals are determined after the hit) would be 61.11%+30.25%+7.33%=98.69%. Thus, the chance for 3 or more criticals - all other outcomes - is 100%-98.69%=1.31%. Q.E.D.
This is basic combinatorial mathematics, I must have been taught this stuff at least 4 times before and I keep forgetting it...