I think people should stop complaining about drop rates. Its a video game, you're either going to play it or you're not. If the drop rate is 2% are you going to play the game more than if it was 1%? Just play the damn game and quit complaining. Play whatever torment level you're comfortable with or have fun on. The more gear you get the harder it will be to find upgrades, that's true for any game built around this concept.
Think you stumbled into the wrong thread my friend.
This post is built on a completely incorrect assumption. It was confirmed by Lylirra in that very same thread that it is, in fact, a 15% multiplicative per torment level buff (eg 1.15^N):http://us.battle.net/d3/en/forum/topic/12413271220?page=13#241
Taking this into account, the drop chance for 150 MF is effectively the same thing as farming a rift one torment level higher. That is to say, farming a rift on T3 with 150 MF is effectively the same thing as farming a rift on T4 with 0 MF. It is nowhere near the drop chance of T6 rifts, as shown in your standard extrapolation graph on the right.
Very disappointed to see such clearly incorrect and misleading information be prominently displayed on the front page of the site.
Does this take into account that MF towards legendaries is only 10% effective?
The derivation will be omitted, but the end result is the following expression (where N again is a torment difficulty value):
Can you please post the derivation because I don't see where that equation comes from. It seems unreasonable. Increasing the health of the mods increases the drop rate of legendaries, what? And (health_n+1/health_n)^1/3 will always be 1.6^1/3 ~= 1.17 since health increase is 60% per T.
Also y-axis from 0 would make for a more honest comparison.
He probably assumed that blizzard wanted some kind of roughly equivalent droprate based on the time it takes to kill enemies so he used health to calculate it. The math doesnt necessarily check out though, as you say. His formula has no more ground to stand on than the assumption of 1.15^N.
Really, you cant calculate the droprate bonuses when given only 2 points of data. There are an infinite number of equations that would satisfy those requirements. For now at least, the best guess might be the simplest (1.15^n)
EDIT: In light of the post above me, it really is 1.15^n
Really, you cant calculate the droprate bonuses when given only 2 points of data. There are an infinite number of equations that would satisfy those requirements. For now at least, the best guess might be the simplest (1.15^n)
EDIT: In light of the post above me, it really is 1.15^n