IS is controlling The Actual five star rate at the background?

Now I don’t know nothing about that math there

But in the end it’s still RNG and it varies from person to person. Is there another person out there who got 7 Bram and 1 Camilla and 1 Robin?

Yeah probably. It’s just random.

And another thing is that the system is critiqued when it doesn’t go ‘well’ but never praised when it does.

If you got 7 Brammimonds, would this thread exist? No you most likely wouldn’t make a thread about how the drop rates for Bram are higher. That’s the sort of thing I mean. You’re only looking at a single case, your sample size is minuscule, you would need to look on a much larger scale to prove something like this. Or just datamine the games code or whatever, but again, I don’t know much about coding so :feh_lucyshrug:


On the L Chrom banner, I pulled Chrom, then 6 LAzura in a row before my second Chrom. Yes I was frustrated, cuz I wanted Chrom merges of course, but it’s just RNG. Sometimes I do amazing on banners and pull insane numbers of 5* heroes in seemingly little orbs, and sometimes my pity rate reaches obscene heights. It’s just RNG.

Example of RNG at work, with a different game. Right now in RuneScape, a new skill, Archeology just dropped. I’ve been camping at big game hunter since Saturday for the new Draggon mattock, and have managed to snag this rare tool 11 times in 300 Dinos hunted with a drop rate between 1/77 to 1/100. Those are lucky odds. While my friend has gotten 4 in 500 hunts. I’m making bank, and he’s frustrated. It’s just RNG. Though I feel bad for him, Jagex isn’t screwing him over. It’s RNG.


Math definitely doesn’t check out though, cause OP is color sniping and assuming the condition of 5 star pulls. So 8% chance is irrelevant.

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that moment when the math is less fuzzy than the but the english. :thinking:

intuitively, this figure is too small. term by term, i can see what you were doing. i don’t think conditional properly was accounted for properly among other things but still not sure what precisely that figure is supposed to be.

what you are probably looking for is a multinomial distribution with k=3 categories (robin, camilla, bram), all categories having a 0.3333… success rate, and category k1 with frequency 7, k2 with frequency 2, and k3 with frequency 1. this gives me 0.00610, so a bit over half a percent.


Lol, being nerdy here. What was your math? I got almost half of that, where P=0.00366.

Using your notation, I thought P=[(# of trials)!/[(Freq of k1)! x (Freq of k2)! x (Freq of k3)!]] x [(Prob of k1)^(Freq of k1) x (Prob of k2)^(Freq of k2) x (Prob of k3)^(Freq of k3)]

Or more clearly: P = (9!/(7!x1!x1!)] x (1/3)^7 x (1/3)^1 x (1/3)^1 = 0.00366

EDIT: Mistakes in what I wrote and nvm, I see the discrepancy.

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Glad to know this, just still feel bad for being home this many flying grima lol

Yep that’s how I freshly +10ed my L!Robin ;) (got three of them within 10 pulls) within 473 orbs I don’t know it’s my super luck or … lol :joy: omg I cant stop laughing

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The reason why 0.08 is relevant here is that I considered all Orbs I spent in this pool so 0.08 dropping rate measures all possibilities that includes 3stars and four stars. Because not every single pull drops five stars :joy: after all. The logic here is: Colourless five stars —> 7 robins —> only one Bramimond —> out of nine pulls :open_mouth::thinking:

Thank for the thinking’s here mate
Yes, I considered the whole perspective for the reality Math of Event model calculation. This event is: the pulling result 7 As, 1 B and 1C out of nine trials ( I ignored other colours because I was sniping colourless even tho many circles didn’t even have the colourless orb, but I treated it as ignorable in terms of this mathematical even if considering that might even make the possibility even smaller).
Firstly: given the dropping rate of each A, B and C are considered as 0.08. And each A, B & C share 1/3 in the group of three colourless heroes. 7 As dropping rate is (1/3)^7 (power of seven), B and C both appear any order and any times in this sequence, so that gives the calculation: [(1/3)^2]x2
Secondly: B and C could appear as BC or CB: (1/3)^2+(1/3)^2=2x(1/3)^2

(The significant part of this calculation is to put the order into consideration. for example: a: AAAABAAAC, b: ACAAABAAA …etc.)

Thirdly: THAT treatment pulls this math model even closer to its natural behaviour. All the orders possibilities: 7!/9!
three parts of logic together give us : 0.08 x [(1/3)^7] x [2x(1/3)^2] x (7!/9!) = 1/8857350 = 0.0000001129

[Five Star Rate] x [Seven L!Robins (with any appearance order)] x [One B!Camila and One Bramimond (with any appearance order)] x [The probability of any three colourless heroes with any orders within Nine trials] = Result.

That should be clearer than before :)

Just some advise there as the entire trial probabilities calculated as 7!/9! instead of another way around.
and secondly, there are two orders as well for B and C which are BC & CB, and that give us 1/3 x 1/3 + 1/3 x 1/3 = 2x(1/3)^2 as I listed above.
so that leads the completed equation (if ignoring five star dropping rate 0.08): (7!/9!) x (1/3)^7 x [2x(1/3)^2] = 0.000001411
Hope that making it easier to understand lol sorry pardon my English, its not my first language :joy:

Hold up. Now I ain’t challenging anyone in math here because I’m a big dumb.

But can’t you simplify that whole thing, like a lot?

Isn’t your argument in colours new summons are less likely?

Therefore you just have 9 5* summons and 7 A, 1 B and 1 C

Can’t you just get those results, what does the 8% have anything to do with it?

Because you aren’t arguing about the 8* you’re a arguing about stuff within the 8*

Or is my math wack af

And honestly I won’t likely understand the answer you give anyways :feh_lucyshrug:

Hi there
0.08 means the very first condition of what we are talking about here but like the first skin of onion that 0.08 can be abandoned due to the situation description that
a: Given detail overall orbs during summoning --> include 0.08
b: Given detail --> without 0.08
The thinking of 0.08 is that the REAL situation is the actually five-star dropping rate IS 0.08 which we cannot ignore all of those 3-4 stars in the middle of the summoning session right haha but I also got the result without 0.08 consideration is 0.000001411 which … you know still :rofl:

I have a degree in mathematics. I think I know a bit about probability.

This equation makes no sense on the context of this discussion. Where did you get it?

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alrite, i think i know what he was trying to calculate and where the approaches diverges, ill write it up later when i have time. he/she was actually pretty close to the right answer (and there is a right answer, it’s just a question of why that figure doesn’t mesh with the multinomial figure).


This game is all RNG and you cant do anything about it. I do get lucky sometimes but sometimes I summon I get a single five star at 300 orbs + on a legendary banner… RNG


thanks for explaining.

ignore the 0.08 term for now, it doesn’t explain the whole discrepancy in the approaches regardless.

this ended up being pretty mundane as far as the why - you flipped the numerator and denominator in the combination term and made a small mistake with the bram and camilla terms with the factor of 2. it seems to me that the real expression for the given event as per your logic should be (1/3)^7*(1/3)*(1/3)*9!/7! = 0.00365797896. look familiar? yep, it’s just the multinomial case as above.

now - the 0.08 term. 0.08 is the naive probability for ONE SUMMON being a focus unit on a legendary banner, not the probability of the “dropping rate of each A, B and C”. we don’t know how many summons we made to get 9 5* units, and we shouldn’t care. if you want to generalize this and factor in orbs/summons as a limiting factor, this becomes a poisson distribution with lambda = ~55-60 orbs/5* and is much messier.

side note: we managed to take the fun out of complaining about IS, and that is saying something. you fookin nerds ruin everything


This is why it’s good to have a crapton of orbs


this guy knows his stuff, pretty much took the words right out of my mouth



Anon already wrote a good writeup, but since you responded directly to me, I’ll just add this:

That’s not true though. Because you’ve only calculated for nine trials (n=9). But you yourself said you spent hundreds of orbs getting those 9 five star pulls. Which means overall, you have more than nine pulls. You can’t say you’re considering all orbs while only calculating for the odds of pulling only 9 five star units. You need to include the number of pulls it took to also get the other 3-4 star units.

Stated another way, by including the overall odds of a five star pull but only calculating for the nine pulls, you’re effectively saying you pulled nine five star units in exactly nine pulls.

EDIT: Realizing you’re not a native English speaker, I reworded some things for clarity.

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I had the same experience with Roy.
Its like… not what I was looking for, but I’ll take it.