So I was going over this basic problem in my head; what are the odds of getting a 2-hit kill with two headshots with a fresh magazine(EXCLUDING THE MINUSCLE CHANCE OF A OHK)? Spoiler alert: it's 39%. To get that answer, good way to think of it is in terms of the first and second bullet individually. Each time you reload, the 1st shot runs a 1/13 chance of being buffed THREE TIMES, as the buffs can stack. Therefore, the odds of it not being buffed is (12/13)^3 (to the third power), so the odds of the first bullet NOT being buffed AT LEAST ONCE is 1728/2197, or 79%, meaning the odds of the first bullet being buffed is 21%. After the first bullet is fired, whether or not it has been buffed has been revealed, and if it was buffed, you get a two-shot kill, end of story. Therefore the odds of getting a 2-hit kill are AT LEAST 21%, but for the other 79% of the time, you could also get a 2-hit kill if the second bullet is buffed. IF the 1st Bullet wasn't buffed, you can look at the enigma of the second bullet by using the same method, this time (11/12)^3, because one possible buff location has been eliminated. Therefore, the odds of the SECOND bullet NOT being buffed is 1331/1728, or 77%, so the odds of it being buffed is 23%. However, you cannot simply add 21% and 23% together, as the second percentage assumes a buff hasn't been placed on the first bullet. We can work around this though, by multiplying the the 23% chance of the second bullet being buffed assuming the first bullet is not BY THE ODDS OF THE FIRST BULLET NOT BEING BUFFED. So that would be 23%*79%, or 18%. Add that to the odds of the first bullet being buffed and you get a final probability of 39%. That means, directly after a reload, if you land two headshots then there is a 39% the guy is dead. End of story, perfect mathematics. BTW, the only reason I posted this was because of all the errors in mathematics I saw on the forum.