ATMA 5.04 and new DropCalc

[hakai_no_tenshi]

I have not looked at this point (do not have the time to try to check), but did you take into account that there is a maximum of 6 items that can be dropped from a single initial call to the first TC?  In other words there is a 6 item cap on the drops of the act end bosses and the countess (the only ones that I recall off the top of my head that end up with more than 6 picks for items).

I am currently working on doing this properly ... it's amazing what a pain in the royal behind her TC is but I believe the others are as accurate as they come. Anyway, Thrugg just sent me email outlining what would have to be done:

t's easyish if you know the right numbers at the right time.  It might
be trickier depending on what you are storing at the time you need
to do this calculation.

This will be easier with an example, so I might as well use a topical
example - Countess Rune.  This has 3 picks, at players 1 it has 5/20
no-drop and 15/20 going on to Runes 12.  And let's say the chance of
getting a Tir assuming we go to Runes 12 is 1/5 (I know it isn't
but to make the numbers easier).  With complete freedom (ie unlimited
drops allowed), the chance of getting a Tir from the whole TC is
(1 - (1 - (15/20 * 1/5))^3) = 0.385875
However, this number is counting the possibility that you get two
non-Tir drops and then a Tir with the third drop.  Suppose you
had a max of 2 drops to come, you have to discount that possibility.
The chance of that is ((15/20)*(4/5))^2 * ((15/20)*(1/5)), or
more neatly (15/20)^3 (4/5)^2 (1/5).  This is very similar to binomial,
the difference is you only want to count the exact non-Tir non-Tir Tir
combination (not "Tir non non" or "non Tir non") so there is no
(3 choose 1) factor at the start.

Pr (Tir | 3 drops remaining)
= (1 - (1 - (Pr(drop) * Pr(Tir | drop)))^3)
= 0.385875

Pr (Tir | 2 drops remaining)
= Pr (Tir | 3 drops remaining)
  - Pr (non non Tir)
= above
  - (Pr(drop)^3 * (1-Pr(Tir | drop))^2 * Pr(Tir | drop))
= 0.331875

Pr (Tir | 1 drop remaining)
= Pr (Tir | 2 drops remaining)
  - Pr (non Tir anything)
  - Pr (nodrop non Tir)
  - Pr (non nodrop Tir)
= above
  - (Pr(drop)^2 * (1-Pr(Tir | drop)) * Pr(Tir | drop))
  - (Pr(nodrop) * Pr(drop)^2 * (1-Pr(Tir | drop)) * Pr(Tir | drop))
  - (Pr(nodrop) * Pr(drop)^2 * (1-Pr(Tir | drop)) * Pr(Tir | drop))
= above - ((1 + 2 Pr(nodrop)) * Pr(drop)^2 * (1-Pr(Tir | drop)) * Pr(Tir | drop))
= 0.217125
("anything" includes no-drop, non and Tir)

Pr (Tir | 0 drops remaining)
= Pr (Tir | 1 drop remaining)
  - Pr (Tir anything anything)
  - Pr (nodrop Tir anything)
  - Pr (nodrop nodrop Tir)
= above
  - (Pr(drop) * Pr(Tir | drop))
  - (Pr(nodrop) * Pr(drop) * Pr(Tir | drop))
  - (Pr(nodrop)^2 * Pr(drop) * Pr(Tir | drop))
= above - ((1 + Pr(nodrop) + Pr(nodrop)^2) * Pr(drop) * Pr(Tir | drop))
= 0 (as you would expect).

You could go in the opposite direction instead of course, start with
0 and add combinations on.  When I do such a thing by hand I start
from both ends, because it gets more complex as you get towards the middle,
but a computer can't do that as easily.

--T