# [Feedback] Concerns about drop chance of some MI legendaries

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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It’s ok to love the “Z” - just remember that he operates a 0 fux given ship…

5 Likes

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 1 Like

I may just be an Acolyte but I stand proud amongst you 3 Likes

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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@Ericberic did research on new dungeon drops, I will put the link to the topic, since I have no idea about calculations.

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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: 5 Likes

50 more runs, it really much faster and more enjoyable to run only to mages rather than full dungeon. Respect to this dev that placed all this vases and bowls in dungeons which player can target with SS/blitz Hope one time in this game or gd2 we will have option to reforge one MI legendaries to another randome one (after long quest chain, which forces you to run each dungeon to complete it and open possibility to exchange, scam rates like 2 to 1 are welcome). So you will be rewarded for completing all content what devs created but not feel frustrated when you can’t beat RNG wall and legendaries you wishing about are not dropping

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For me reforging one set item into another is already a stretch.

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