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Drop rate

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Drop Rate is the probability that a monster is expected to yield a certain item when killed once by a player. When calculating a drop rate, divide the number of times you have received the certain item, by the total number of that NPC that you have killed. For example:

A common misconception is that you are guaranteed that item when you kill the NPC x number of times, where \frac{1}{x} is the drop rate. You are never guaranteed anything, no matter how many times you kill that monster. The drop rate is simply the probability of getting a certain drop in one kill. The probability that a monster will drop the item at least once in x kills is 1 minus the probability that it will not drop that item in x kills, or 1 - \left(1 - \frac{1}{y}\right)^x, where x= number of kills, and y= drop rate.

For example, if dust devils are expected to drop a Dragon chainbody once out of 15000 kills, then the probability that a player will get at least one Dragon chainbody after 15000 kills is


Which is approximately 63.21%. Similarly, we can solve for the number of Dust Devils you need to kill to have a 90% probability of getting one when you kill them:

1-\left(\frac{14999}{15000}\right)^{x} > 0.9

\left(\frac{14999}{15000}\right)^{x} < 0.1

Which yields the answer x>34538. There is also an equation for computing the probability of a certain amount r of a particular drop after n amount of kills:

P(r,n)={}^{n}\textrm{C}_{r} p^{r}q^{n-r}

And if you take the sum of this equation from when r=1 until r=n you get the probability of at least 1 drop of a particular item after n kills:

\sum_{r=1}^{n}{}^{n}\textrm{C}_{r} p^{r}q^{n-r}=1-\left (1-\frac{1}{n} \right )^{n}

Confidence IntervalsEdit

It is given to us that the confidence interval for the success probability of a model X\sim B(n,p) may be expressed as the formula[1]:

C=p\pm z_{1-\frac{\alpha}{2}}\sqrt{\frac{p(1-p)}{n}}


  • p - the assumed probability of success given by the ratio of successes to sample size. To clarify: if one were to gain 2 Divine Sigils after 2000 Corp kills, the assumed probability of success would be \frac{2}{2000}=\frac{1}{1000}=0.001
  • z_{1-\frac{\alpha}{2}} - this is the critical standard score such that P(Z\leq z_{1-\frac{\alpha}{2}})\approx1-\frac{\alpha}{2} for Z\sim N(0,1). This z-value may be found by checking with this table. Information on how to read this table may be found here.
  • \alpha - the confidence error you wish your interval to represent. An example value may be 0.05 (this represents 95% confidence).
  • n - the amount of trials you've conducted. In the example used in the definition of 'p', this value would be 2000.

To save the reader time, a list of possible z-values is supplied:

\alpha Confidence level z_{1-\frac{\alpha}{2}}
0.2 80% 1.28
0.1 90% 1.64
0.05 95% 1.96
0.01 99% 2.57
Example of usage

Consider the following case: we have killed a combined total of 500 Black Dragons and have gained 10 Draconic Visages between us. This suggests that we take p=\frac{10}{500}=0.02 and n=500. Now let us say that we wish to create a 95% confidence interval for our p-value (this is to say that \alpha=0.05 and z_{1-\frac{\alpha}{2}}=1.96). Our confidence interval is constructed as follows:




What this means is that we can be about 95% sure that the drop rate of Draconic Visages (from Black Dragons) is somewhere between 1 in 31 and 1 in 129.

Notes on usage
  • This method of calculating confidence intervals relies on being able to approximate our binomial model as a normal distribution -- as such, most statisticians will not use this method unless np>5 and n(1-p)>5.[2]

Trivia Edit

If we let x be an arbitrary number and 1/x be the drop rate for a particular drop, the larger x gets (in other words, the rarer the drop is), the closer the probability of obtaining that item in x kills approaches 1 - \frac{1}{e}, or approximately 0.63212, where e is the exponential constant \approx{2.718281828459045}. We can express this limit as follows:

\lim_{x \to \infty} 1 - \left(1 - \frac 1x\right)^x = 1 - \frac 1e

This follows from the definition of e:

e = \lim_{n \to \infty} \left(1 + \frac 1n\right)^n<br>
 =\sum_{i=0}^{\infty} \frac{1}{i!}

This leads to the conclusion that, given a drop rate of \frac{1}{r}, the approximate chance of not receiving a drop after n kills is \left(\frac{1}{e}\right)^\frac{n}{r}. Note that this is only accurate for large values r.

Notes Edit

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