A possible explanation about Zeta drop rates (maths heavy)
TLDR - if zetas have a long term average drop rate that is similar to other gear/shards then we might expect them to have a higher chance of a 0 drop and a higher variance in the drops because of the possibility that we can draw 2 as well as 1 each time.
Many people are complaining about the low zeta drop rate they are experiencing. I just wanted to demonstrate using some basic probability theory why we might actually expect this given the fact we can draw not just 0 or 1 zeta each time, but we also have the possibility of drawing 2 in a single attempt.
Note that the examples below are purely hypothetical and are used just to illustrate the point of what 'might' be happening. I have no data on drop rates but the basic theory should be sound.
Example 1: Suppose we are drawing (e.g.) a character shard and the underlying drop rate is 1/3. We would expect to achieve a long term average drop rate of 1/3 (obviously).
We can write this as a discrete random variable, X. (X is the random variable, x is the value it can take):
x: 0 1
P(X=x): 2/3 1/3
The expectation of this random variable is then calculated as
E(X) = (0 x 2/3) + (1 x 1/3) = 1/3.
We can also calculate the variance:
E(X^2) = (0^2 x 2/3) + (1^2 x 1/3) = 1/3.
Var(X) = E(X^2) - (E(X)^2) = 1/3 - (1/3)^2 = 1/3 - 1/9 = 2/9.
Example 2: Suppose we are now drawing a zeta. We don't know what the actual drop rate is of course but we can still think about the underlying maths.
Our discrete random variable now has the form:
x: 0 1 2
P(X=x): P(X=0) P(X=1) P(X=2)
The expectation of this random variable is then calculated as
E(X) = (0 x P(X=0)) + (1 x P(X=1)) + (2 x P(X=2)).
Now, let's make a couple of big assumptions here to illustrate what might be happening. Firstly, let's suppose that the probability of 2 zetas dropping is half as much as the probability of a single zeta dropping (this is purely hypothetical but would make sense). Secondly, let's suppose that the devs want the long term average drop rate to be E(X) = 1/3 (again purely hypothetical but I'll use this value to directly compare to my Example 1. Assuming a different value here changes the numbers but not the key principle).
Then we have
E(X) = 1/3 = 0 + P(X=1) + (2 x P(X=2)) = 0 + P(X=1) + (2 x 1/2 x P(X=1)) = 2 x P(X=1)
From this we can find that P(X=1) = 1/6 and hence P(X=2) = 1/12 under my hypothetical assumptions.
Consequently, P(X=0) = 1 - P(X=1) - P(X=2) = 1 - 1/6 - 1/12 = 6/12 = 3/4
Note that this means that even though the long term average drop rate in both example 1 and example 2 is equal to 1/3, we are more likely to get a zero drop in example 2 where P(X=0) = 3/4 compared to P(X=0) = 2/3 in example 1.
We can also find the variance in example 2 using the values previously calculated for these assumptions:
E(X^2) = (0^2 x 2/3) + (1^2 x 1/6) + (2^2 x 1/12) = 6/12 = 1/2
Var(X) = E(X^2) - (E(X)^2) = 1/2 - (1/3)^2 = 1/2 - 1/9 = 7/18.
And hence the variance is higher in example 2 than in example 1 as well.
(If anyone spots any errors with the maths please let me know!)
Many people are complaining about the low zeta drop rate they are experiencing. I just wanted to demonstrate using some basic probability theory why we might actually expect this given the fact we can draw not just 0 or 1 zeta each time, but we also have the possibility of drawing 2 in a single attempt.
Note that the examples below are purely hypothetical and are used just to illustrate the point of what 'might' be happening. I have no data on drop rates but the basic theory should be sound.
Example 1: Suppose we are drawing (e.g.) a character shard and the underlying drop rate is 1/3. We would expect to achieve a long term average drop rate of 1/3 (obviously).
We can write this as a discrete random variable, X. (X is the random variable, x is the value it can take):
x: 0 1
P(X=x): 2/3 1/3
The expectation of this random variable is then calculated as
E(X) = (0 x 2/3) + (1 x 1/3) = 1/3.
We can also calculate the variance:
E(X^2) = (0^2 x 2/3) + (1^2 x 1/3) = 1/3.
Var(X) = E(X^2) - (E(X)^2) = 1/3 - (1/3)^2 = 1/3 - 1/9 = 2/9.
Example 2: Suppose we are now drawing a zeta. We don't know what the actual drop rate is of course but we can still think about the underlying maths.
Our discrete random variable now has the form:
x: 0 1 2
P(X=x): P(X=0) P(X=1) P(X=2)
The expectation of this random variable is then calculated as
E(X) = (0 x P(X=0)) + (1 x P(X=1)) + (2 x P(X=2)).
Now, let's make a couple of big assumptions here to illustrate what might be happening. Firstly, let's suppose that the probability of 2 zetas dropping is half as much as the probability of a single zeta dropping (this is purely hypothetical but would make sense). Secondly, let's suppose that the devs want the long term average drop rate to be E(X) = 1/3 (again purely hypothetical but I'll use this value to directly compare to my Example 1. Assuming a different value here changes the numbers but not the key principle).
Then we have
E(X) = 1/3 = 0 + P(X=1) + (2 x P(X=2)) = 0 + P(X=1) + (2 x 1/2 x P(X=1)) = 2 x P(X=1)
From this we can find that P(X=1) = 1/6 and hence P(X=2) = 1/12 under my hypothetical assumptions.
Consequently, P(X=0) = 1 - P(X=1) - P(X=2) = 1 - 1/6 - 1/12 = 6/12 = 3/4
Note that this means that even though the long term average drop rate in both example 1 and example 2 is equal to 1/3, we are more likely to get a zero drop in example 2 where P(X=0) = 3/4 compared to P(X=0) = 2/3 in example 1.
We can also find the variance in example 2 using the values previously calculated for these assumptions:
E(X^2) = (0^2 x 2/3) + (1^2 x 1/6) + (2^2 x 1/12) = 6/12 = 1/2
Var(X) = E(X^2) - (E(X)^2) = 1/2 - (1/3)^2 = 1/2 - 1/9 = 7/18.
And hence the variance is higher in example 2 than in example 1 as well.
(If anyone spots any errors with the maths please let me know!)
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Replies
But if Jane has 3 apples, and Jonny has 7 shiny nickels, how long would it take to find Waldo if the sun sets at 2.3 degrees right ascension of Vernal Equinox at the peak of the Summer Solstice?
But you shouldn't, probability doesn't work that way.
You flip 100 coins, the 48 of the first 50 are heads. Should you expect a really long 0 run? No because the probability is that you will get 25 heads, it doesn't matter that you got an outlier earlier.
Common core.
Your math is fine. The underlined part is where I disagree .
The fact that we can't buy more Zeta challenge attempts with crystals makes me believe the expectation of 1/3 is too low. I think it's more like 3/4, but we shall see over time.
In the example you gave the probability of 48 heads in a row is extremely low. You should do millions or milliards flips till you see that streak. While in pseudo rngs you can face them quite often as they usually guarantee the average constant probability on really large sample size. And the lower the probability (drop rate) the bigger sample size you will need. In other words, if you need to farm just 3-5 small droid callers or stuncuffs you can get **** up by rng very easy and you will want to go and complain about the drop rate on forum. And in this case I would advice to buy/ask for donation of such gear. But if you need 2-3 sets of cuffs - farm them from normal nods and be sure - when you will reach 100+ drops the drop rate will be close to 20% (or 25) I don't remember, but there was post from sneekypants recently, plus I have done small statistics by myself
I thought perhaps you meant to find the variance of the expected proportion amongst n independent trials, but then you would need to include information about the number of trials to avoid confusing readers. Your example two, though, is a little confusing. Again you mention that this is a variable with discrete outcomes, but then proceed to treat it as a continuous variable in your example. Your formula for the expected value would yield an expected value on the continuous interval [0,2]. So for example, if the expected drop rate for a single zeta were one-third, or 3/9, the drop rate for zero zeta materials was 5/9, and the drop rate for two zeta materials were 1/9, then your example yields an expected value of 0*(5/9)+1*(3/9)+2*(1/9), which would be 0.55556, or just over one-half zeta material per challenge. Again, if you wanted to provide an example of the long term proportion, say if you were to do 100 challenges, you would expect 55-56 zetas to drop, then I think I follow your reasoning. However, I feel you need to be more clear in your wording to show that this is the case. Is this what you mean in your examples?
Assuming, then, that this is what you were trying to show, I would say yes, obviously the variance is larger in the second example because you are dealing with numbers on a larger scale. Now, whether this is the reason people are perceiving drop rates to be lower than drop rates where a single piece of gear is being dropped, I think depends on the amount of skew in this distribution. How heavily is the distribution weighted towards zero? I do feel obliged to point out, however, that a better comparison would be the other challenges, where between 3 and 6 mats are expected to drop. Now, perhaps people feel that they are getting fewer zetas than in these challenges, but it would be because in the other challenges, the EXPECTED VALUE is larger, not the variance. I would think that the reason for the perceived lower drop rate is the OPPOSITE of the reason you supply; that is, not because the distribution includes 2 (a higher value), but because it includes 0 (a lower value). Although maybe people indeed are comparing the drop rate of zetas to those of say, stun cuffs, but I think in that case we can't leave it rest at the fact that the variance is larger. We would need to calculate the third and fourth moments, the skewness and kurtosis, to determine if the probability of a 0 for zeta mats is greater than the probability of a zero for stun cuffs. The fact of the matter is, we don't know the drop rates, so all this rests on assumptions.
I think another assumption here is what is determining the drop rates. Does a chance of two zeta materials mean there were two independent draws from a binomial distribution with a single drop rate, or a single draw from a multinomial distribution, perhaps with varying drop rates for each outcome. It seems that you are assuming the latter, and I feel inclined to agree, just because that would seem to make more sense. But I could be wrong. Knowing the drop rate parameters for each outcome, we could then calculate the joint PDF for any particular outcome over n trials. So we could find the probability that people would receive a string of n zeroes. Still, I don't know that we need to look to the variance to explain the perceived drop rates, but rather the weightings...as this density function will be weighted towards the value with the larger probability (zero). This then, reflects the drop rates.
Still another assumption we are making is that the people complaining about the drop rates on these forums are representative of everyone simming zeta materials. It seems likely that the people who experience a long string of zeroes are more likely to come onto the forums and complain than the people who receive a string of ones. And perhaps another issue here is that the probability of AT LEAST 1 material is equal to 0.20 or 0.33, such that the probabilities of 1 and 2 individual at up to that drop rate. This would make the drop rate for either of those values lower than the rate for gear in battles, and if people are somehow attending to that, it might seem less. Still, I think if people are comparing to other challenges where there is a MINIMUM drop of 3 and you always get something, having a challenge where 0 is a potential result could lead to a lot of frustration.
Thanks for the brain exercise and kudos for the effort!
I agree w Raphs on both the math. I'm not so sure about the assumptions you made in the middle.
To put it more simply, if you have an average of less than 0.5 with only whole number outcomes, you will always have to have more 0 outcomes. The more possible outcomes farther from the average, the variance has to go up unless you continually counter weight the outcomes closer to the average.
From casual observation, it's probably close to two rolls of 1/3 for an event EV of 0.66. 0 = 4/9, 1 = 4/9, 2 = 1/9 per sim.
We don't know how the drop rates are rolled though. The game could use conditional probability, where it rolls 0 or 1 zeta, and only if you roll 1 zeta the game rolls again for a second. This theory would explain why so many people complain of the low drop rate.
The game could also do independent rolls, where it rolls for the second chance at a zeta regardless of whether you get one the first time.
About your other point though (and the point made by @OnEMesSduPKiD ) - I completely agree. It is very much an assumption I am making that all draws are independent and that you draw 0,1, or 2 in one go. We just don't know exactly how it works and we can't really start to extrapolate much further until someone collects a load of data on zeta drops.
Thank you. Working through my response, I started to gather that this is what you mean, and I think I may have been confusing myself a little because my default approach to dealing with a discrete variable is to hit the exact probabilities head on. You are taking a large sample approximation kind of approach.
The term "pseudo-random number generator" is used to describe such generators in computing, because of the way seeds are generated sometimes results in the repeat of strings of values, and sometimes other anomalies with the values that are a little outside the expectations of chance. My understanding is that there are methods that are better about this now. Your argument is that this is simply "not random," and this is strictly true. However, the term pseudo-random is used to imply that randomness is reasonably approximated for purposes of the exercise at hand.
This was the only part of this thread I understood. Thank you for this.
We also don't know if getting 2 depends on getting the first one. Are the drops independent of eachother?
Like is the chance to get zeta A 20% and and the chance to get zeta B 10%? Or do you have to get the first one to drop before it even gives you a chance to get a second one as well? Or is it all different where you have a chance 20% chance to get 1 zeta and a 5% chance to get 2 and a 75% chance to get none? Like is getting 0, 1, or 2 zetas each a different set drop from the challenge, or is there just a set chance to get a zeta material and is that drop chance run twice? Like whats the mechanic behind this.
This is the worst RNG bad luck string I have been on since playing games like this.
Hopefully, Our Lord RNGesus bestows a full week of two zetas each chance upon me, to even out as She tends to do.
The dumbest thing you can actually do:-) unless you want your opponents progress faster