thanks, now i even more frustrated after watching this
Fixed incorrect drop chance in my post, apparently right number is 3% (at least now i know how to properly use monster db)
The magi rings are the only “issue”, if you think there is one. However they are comparable in drop rate to other Rogue Dungeons, with Alkamos rings being the jewellery comparison. I think jewellery has always had significantly lower drop rates.
Personally, I just ignore these items because Rogue Dungeons are not my favoured endgame activity and I find that much farming/travelling to be unpleasant.
I would love for those Rogue jewellery pieces to be transmutable for a large fee, simply to reduce the burden of having to farm specific content over and over. If that happened I would venture to Alkamos more often, and a specific Magi ring (let us be real, most people will want a specific ring) would become more realistic to acquire.
While i really really wish i had many items in this game, i don’t have a problem with the drop rates because its another item i havent gotten yet which is another reason to keep playing.
My feelings are its kinda like in a free-to-play CCG, such as Hearthstone or MTG Arena, in that farming for MIs is RNG on the level of getting a pack of cards. You get a free pack every day or so, and get more cards. Hopefully anticipating you get that card you really want in the pack. When theres still cards you want that you dont have yet, it draws you in and makes you want to keep playing. But once you have all the cards, it loses a lot of its shine, unless youre a die hard top tier player playing ranked matches, in which case youre probably buying the entire set of cards (using grimtools).
No, this game wouldnt become stale once I have all the items, because the build diversity and depth of customization is so great id be able to make TONS of builds, but the dopamine rush from actually USING those builds would be a little less, knowing theres no items to get that i havent gotten.
For me, the sheer rarity of these items, and the struggle to acquire them, increases their value and the special feeling I will have when I finally get them. Same with the legendaries that drop from Totems- ive still only gotten 1 Phasebreaker. Nothing else. But thats part of the excitement, I think. Right?
Wish I can disenchant all of hundreds of “final stops”, “guillotines” and “silverbolts” to craft few of these rings. HC mode is really great way for me to counter inflation of items, cause in any shard or crucible run i can lose them.
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?
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
more purples than blues
thanks, now i even more frustrated after watching this
thanks
You. Thanks so much Z!
