8 years ago
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!)