Drop Rates

So anyone who has ever played an MMO or RPG will be familiar with these, the probability that the monster you just killed drops a valuable. This can be modelled by the binomial distribution. This is a distribution with two outcomes termed success and failure, denoted as p = success and q = failure, failure is not succeeding so q = 1 - success (in statistics 1 denotes 100% probability). So lets assume that the monster in question has a 1% probability of dropping a particular valuable item. This is: \[ \begin{array}{rlc} p = \frac{1}{100} & q = \frac{99}{100} \end{array} \] The probability of succeeding is the same as not failing which can be denoted as: \[ 1 - (1-p)^n\] In this context: \[1 - \Big(1 - \frac{1}{100} \Big)^{100} \approx 0.6339 \approx 63.4 \% \] This can be generalised for any drop rate, if the drop rate is $\frac{1}{n}$ and you kill $n$ monsters the probability of receiving a drop is always around 63%, providing that $n$ is sufficiently large.*

So what if we wanted a 90% to be the minimum probability of getting a drop? Well consider the above equation for a probability of 0.90 \[ 1 - \Big( \frac{99}{100} \Big)^{n} > 0.9 \] Solving this expression for n yields: \[ n = \frac{\ln \frac{1}{10}}{\ln \frac{99}{100}} = 229.1 \approx 230 \] So if you were to kill 230 monsters you'd have a 90% probability of receiving a drop.

It is very important to note that each trial is independent of all other trials. That is to say that if you have killed 10 or 10,000 monsters the probability that the next kill will yield a drop is still $p$.

A practical example is tossing a coin. If I flip a coin and get 10 heads in a row, what is the probability that the next toss is heads? It might seem that the "law of averages" would make the next toss tails. However there is no such force, this is another independent trial, so the probability is still $\frac{1}{2}$ regardless of how many heads were tossed beforehand.

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*there is a reason that as $n$ increases the probability tends to 63%. This is because: \[ \lim_{n \to \infty} \Big(1 - \frac{1}{n} \Big)^n = \frac{1}{e}\] I proved in my last post that: \[ \lim_{n \to \infty} \Big(1 + \frac{1}{n} \Big)^n = e \] This is a specific case of the formula: \[ \lim_{n \to \infty} \Big(1 + \frac{x}{n} \Big)^{n} = e^x \] In the case where $x=-1$: \[ \lim_{n \to \infty} \Big(1 - \frac{1}{n} \Big)^{n} = \frac{1}{e} \] So as $n \to \infty$: \[ 1 - \lim_{n \to \infty} \Big(1 - \frac{1}{n} \Big)^{n} = 1 - \frac{1}{e} \approx 0.6321 \approx 63.21 \text{%} \]