Hello, today I finish my soj farming from Nightmare Andariel. I decide to collect some data about ring drop rate and quality of the rings. Every day I made 50 runs with 428 mf sorc. Here is results (M - magic, R - rare, S - set, U - unique)
Set rings drops more frequently then unique. I hope extra 100 - 200 mf will be enough to make chances equal.
When I found 40 unique rings (38 from andariel + 2 extra rings from mobs) I decide to stop running. Before I identify them I calculate chances to get 0, 1, 2 … 6 soj from 40 unique rings (I supposed that every unique ring has chance 1/33 to become the soj)
Finally I bought 2 identify toms and quite unexpectedly found three SOJ! A year ago I identified 20 rings and I have not found a single soj :(. Today I was more lucky…
Nice analysis. Helps put things in perspective. I was planning on trying to calculate the optimal number of runes necessary to have a decent chance for Andy to drop a soj. Cheers!
You would actually have to reduce your mf for that. Diminishing returns have a bigger weight factor on uniques than on sets; and more on sets than on rares.
So as you increase mf you will:
greatly increase the chance for rares
moderately increase the chance for sets
slowly increase the chance for uniques
With more MF you will have even a bigger number of sets compared to uniques.
With more MF you will have even a bigger number of sets compared to uniques
You’re right. I forget about it. But as far as I know, the game first applies the chance for a unique item then set item then rate and magic. It means that unique items number will increase with more MF.
I agree. Indeed…if it doesn’t roll unique…with each extra mf point you have, you get increased chances to roll set vs rare compared to the effect each mf point has in increasing the probability to roll unique when you look at the graphs.
Hi Icanfly, thanks for that! Could you put up a quick example formula for lets say the “0” (29.2) and “1” (36.5) case? Would be of practical interest for many of us!
Hi Icanfly, thanks for that! Could you put up a quick example formula for lets say the “0” (29.2) and “1” (36.5) case? Would be of practical interest for many of us!
I’m not expert in probability theory but 0 and 1 soj is simple cases for calculating. Every unique ring from Nightmare Andariel has chance 1/33 to became soj. Chance to get other ring (Nager or Mandal) is 32/33. Probability to get Nager or Mandal 40 times in row is just:
p_0 = (32/33) ^ 40 = 0.2920
When you get 1 soj it means 39 times an event occurred with probability 32/33 and 1 event occurred with probability 1/33, BUT it is necessary to take into account how many ways you can get 1 soj. In this exmple 40 ways (first ring can became soj, or second … or last 40). So formula for 1 soj:
p_1 = (32/33) ^ 39 * (1/33) ^ 1 * 40 = 0.3650
General formula using Number of k-combinations C(N, k) to calculate last term:
p_i = (32/33) ^ (40 - i) * (1/33) ^ i * C(40, i)
If you have matlab or octave you can run this script:
% Number of unique rings
N = 40;
% Probability for single unique ring became soj
p = 1/33;
x = zeros(N + 1, 1);
for i = 0:N
% Calc probability for get exactly i soj
x(i + 1) = p ^ i * (1 - p) ^ (N - i) * nchoosek(N, i);
% Print result
fprintf('%d: chance %2.1f%%\n', i, x(i + 1) * 100);
end
