# 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

$1-\left(\frac{14999}{15000}\right)^{15000}$

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]:

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

Where:

• $p$ - 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:

$C_{lowerbound}=p-z_{1-\frac{\alpha}{2}}\sqrt{\frac{p(1-p)}{n}}=0.02-1.96\sqrt{\frac{0.02(1-0.02)}{500}}=0.00772846\approx\frac{1}{129}$

And...

$C_{upperbound}=p+z_{1-\frac{\alpha}{2}}\sqrt{\frac{p(1-p)}{n}}=0.02-1.96\sqrt{\frac{0.02(1-0.02)}{500}}=0.0322715\approx\frac{1}{31}$

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 $e$:

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

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 $r$.