Drop rate

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:


 * Bones have a 100% drop rate from Chickens
 * Feathers have a 75% drop rate from Chickens

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 Intervals
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 :

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

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:

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:
 * Example of usage:

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

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 $$n(1-p)>5$$.

Trivia
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 $$r$$.