Can’t thank you enough for all the sweet things you’ve done for us. 
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+50% additive or multiplicative? Must you strain our literal minds with this vagueness? /s
Pretty sure it’s multiplicative or we’ll see alkamos rings used in place of gems on bedazzled armor.
Rough math for people expecting to get a set of alkamos rings. 50th percentile yadda.
1.02/.82 drop rates (off grimtools) yields average 135 runs to see a functional pair, seeing about 2.484 rings throughout.
1.53/1.23 (+50%) not at all surprisingly cuts this to 90 runs.
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OMG, I 
You. Thanks so much Z!
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I’ve got 165.64 expected runs. The general formula I’ve got is
1/p + 1/q - 1/(p+q)
I needed to figure out a formula for 1 + 2x^1 + 3x^2 + 4x^3 + … = (1-x)^(-2) to calculate it
(calculation or logic mistake possible though because the first time I tried it I got complete bullshit)
How did you solve it?
Wait, is it possible you’ve just made a typo and wrote 135 instead of 165? 
If so, can you get the answer without complicated calculations? Is the formula obvious?
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Haha. Because he is the boss, doesn’t that mean he can’t or shouldn’t!?
It’s okay. As long as he knows that this decision was greatly appreciated 
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I may just be an Acolyte but I stand proud amongst you 
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First I broke it down into a simple “did a ring drop?” with 1.84/98.16 split
R = ring chance
X = number of runs
N = number of rings
chance of occurrence = R^N * (1-R)^(X-N) * ( X choose N )
This merely gives us the spread of ring quantities, we are looking at getting a proper pair however. For a given N rings there are only two combinations of ring drops that do not yield a complete set. The ratio of ring occurrences give us a 55.43% / 44.57% split. Let’s call them A and B. So the chance of a usable pair among N rings is 1 - A^N -B^N.
This leads us to the final percentile chance of expecting at least one pair to drop being…
Summation from N = 2 to X
R^N * (1-R)^(X-N) * (X choose N) * (1-A^N -B^N)
Haha, we’re calculating different things 
I haven’t 100% checked your reasoning but seems correct and imo you’re calculating this
What is the probability of getting a pair of rings in X runs
Then you’re finding X big enough for appropriate percentile
For example you found that in 135 runs the probability of getting a pair of rings is 90% [or whatever]
whereas I calculated
average number of runs [Expected value] to get a pair of ring
For example if P(K) = probability of getting a pair of rings in exactly K runs
and I calculated the following infinite series
weighted average mean of number K of needed runs where weights are probabilities of getting a pair in K runs
Summation from K=2 to Infinity
P(K) * K
P(K) is simple if you try to figure it out (without binomial symbols) but quite long and unreadable so I skip it.
so you didn’t calculated expected aka average number of runs needed as is incorrectly assumed (I was misled by a word “expecting”) but numbers of runs needed to be i.e. 95% sure of getting a pair
nothing wrong with that since I think both our values are useful information
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Nery the herald will play some warming music with his fanfare trumpet. What an awful choice for instrument!
Btw with next patch I will rerun SoT. I kinda gave up to have legit Alkamos ring after 300+ runs.
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LMFAO! 
Thats just too good!
A simple miscommunication in these otherwise tumultuous times. I worked off your definition and concur with a mean 165.64 runs to yield a pair. 50% adjusted droprate yielding the expected 110.428
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Again I was wrong to calculate anything here (this infinite series and stuff). Here’s without calculation.
p - chance for 1st ring
q - chance for 2nd ring
Expected runs to get 1st ring - 1/p
(this is kinda basic so no proof here, for anyone reading - think 100 tosses of a coin needed for heads if heads has 1/100 chance)
Expected runs to get 2nd ring - 1/q
So
1/p = Sum to infinity (K * P(K)) [K is number of runs, P(K) is probability of getting 1st ring in K runs]
1/q = Sum to infinity (K * Q(K)) [similar to above]
Now let’s add it together and see what in means
1/p + 1/q = Sum to infinity (K * [P(K) + Q(K)])
P(K) + Q(K) is the probability of the following event
EVENT 1
- 1st ring from K-th run
AND
no 1st ring in any of the 1, 2, … K-1 runs
OR
- 2nd ring from the K-th run
AND
no 2nd ring in any of the 1, 2, …, K-1 runs
But what we would like to calculate?
We want to get both 1st ring and 2nd ring in K runs which means the following event
EVENT 2
- 1st ring from the K-th run
AND
no 1st ring in any of the 1, 2, … K-1 runs
AND
2nd ring in any of the 1, 2, …, K-1 runs
OR
- 2nd ring from the K-th run
AND
no 2nd ring in any of the 1, 2, … K-1 runs
AND
1st ring in any of the 1, 2, …, K-1 runs
It just so happens that these events can be easily substracted because
- the 1st subevent of EVENT 1 contains 1st subevent of EVENT 2
- the 2nd subevent of EVENT 1 contains 2nd subevent of EVENT 2
The result of this substraction is (see NO added)
EVENT 3
- 1st ring from the K-th run
AND - no 1st ring in any of the 1, 2, … K-1 runs
AND - NO 2nd ring in any of the 1, 2, …, K-1 runs
OR
- 2nd ring from the K-th run
AND - no 2nd ring in any of the 1, 2, … K-1 runs
AND - NO 1st ring in any of the 1, 2, …, K-1 runs
But this EVENT 3 is basically and event of getting 1st ring OR 2nd ring in K runs which has expected number of runs of 1/(p+q). We can think of it as one item of probability p + q
Let E be expected number of runs. We want expected number of runs of EVENT 2. We showed that
EVENT 1 = EVENT 2 + EVENT 3
and that
EVENT 2 and EVENT 3 are exclusive
Thus
E[EVENT 2] = E[EVENT 1 - EVENT 3] = E[EVENT 1] - E[EVENT 3] = (1/p + 1/q) - 1/(p+q) ≈ 165.64
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Love this thread.
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@Ericberic did research on new dungeon drops, I will put the link to the topic, since I have no idea about calculations.
https://www.reddit.com/r/Grimdawn/comments/ghjcrb/100_runs_of_tomb_of_the_heretic_magi_only_rings/
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Very nice of you!
I am running another 100 or more sets. Why? Because after 232 runs (32/100 of my 3rd set) I am completely missing 2 Legendary Rings. And swimming in Tawrots. I wish I had the discipline to write down the exact spawns of each Magus. But at this point my fingertips are bleeding and I just want to get it over with and dull the pain.
Curently 232 runs. Missing 2 rings. Plenty of Khonsar or Tawrot. Z E R O Sethris or Orissia. RNG will RNG.
@grey-maybe counted 0.71% for specific ring so no Sethris in 232 runs is like 19% so pretty unlucky but reasonable. But then if we include that you got neither Sethris nor Orissia in these 232 it drops to a few % (don’t know exactly) which is unfortunate but still in the realms of possibility.
By the way the expected runs for getting a pair of legendary Alkamos rings was 1/p + 1/q - 1/(p+q)
I calculated the formula for 3 items and it’s
1/p + 1/q + 1/r - 1/(p+q) - 1/(q+r) - 1/(r+p) + 1/(p+q+r)
It’s most probably the same for more items so let’s use it to find expected runs to get the full Lokkar set
Lokarr’s Boots (Level 1)27.78%
Lokarr’s Coat (Level 1)22.22%
Lokarr’s Gaze (Level 1)27.78%
Lokarr’s Mantle (Level 1)22.22%
1/p + 1/q + 1/r + 1/s - 1/(p+q) - 1/(q+r) - 1/(r+p) - 1/(p+s) - 1/(q+s) - 1/(r+s) + 1/(p+q+r) + 1/(p+q+s) + 1/(p+r+s) + 1/(q+r+s) - 1/(p+q+r+s)
1/0.2778 + 1/0.2222 + 1/0.2778 + 1/0.2222 - 4 * 1/0.5 - 1/0.4444 - 1/0.5556 + 2 * 1/0.7778 + 2 * 1/0.7222 - 1 ≈ 8.5
The expected number of Lokkar Kills to get the full set is 8.5
It’s actually higher, Ricadio gave us wrong info, chance is not 2.5%, but 3%, so the chance is 0.857%
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50 more runs, this time only for magic mans:
Finally something, hope I can drop one more sethris and anubar
And one thicc helmet:
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