# Drop rate

"Rng" redirects here. For the Combat skill, see Ranged.

The drop rate is the frequency at which a monster is expected to yield a certain item when killed by players. When calculating a drop rate, divide the number of times you have gotten the certain item, by the total number of that monster that you have killed. For example:

## Drop rate

All items have a chance of being dropped that is expressible as a number—their drop rate. In Old School RuneScape, drop rates are treated independently from one to the next; meaning no previous drops, decrease or increase the drop rates for future kills.

Drop rates such as "1 in 5" do not guarantee such a drop will be obtained after 5 kills. Instead, it can be thought that on average, a player would receive the drop once per 5 kills, with the chance to receive more or less. While a player killing such a creature 30 times and not obtaining such a drop is unlikely, the players' chance of their next kill obtaining the drop remains the same.

While somewhat counter-intuitive, probability itself cannot be simply multiplied per drop. For example if one were to calculate the chance of receiving a drop at least once after 100 kills as 10%, it would not become 20% after 200 kills. If it was additive, then the chance after 1100 kills would be 110%, which is impossible as probabilities are always equal to or between 0 and 1.

### Common Terms

A common list of terms is provided below:

• Average or Expectation is fairly intuitive, though not to be confused with the probability; if players seek a 1 in 5 drop and kill a monster 10 times, the average would be 2 drops, though a player could receive more or less. However the average doesn't refer to a specific outcome, for example, killing a monster with a 1 in 5 drop 3 times, would yield 0.6 of that drop on average, which is impossible as an outcome itself.
• Probability or Chance refers to the likelihood of a specific outcome occurring. This can never be below zero or above one.
• Independent/Dependent refers to two events influence each others outcome. Generally drops are treated independently.
• Probability Distribution is a mathematical function that defines the probabilities of certain occurrences in specific circumstances, such as the commonly used binomial distribution.

## Calculating the odds of receiving items

If the King Black Dragon is expected to drop a draconic visage once out of 5,000 kills, then the probability of getting a drop from one kill is as follows:

{\begin{aligned}&1-\left(1-{\frac {1}{5000}}\right)^{1}\\=&\ 1-\left({\frac {4999}{5000}}\right)^{1}\\=&\ 1-0.9998\\=&\ 0.0002\\\end{aligned}} That is 0.02%. To find the drop chance in 5,000 kills, the expression inside the parenthesis can be raised to the 5,000th power, which yields a meagre 63.2% chance of getting a draconic visage, i.e. $1-\left({\frac {4999}{5000}}\right)^{5000}\approx 0.632$ .

More generally, while not holding true for extremely common drops, the chance of obtaining at least one drop, of chance ${\frac {1}{n}}$ once you have $n$ kills trends towards $1-{\frac {1}{e}}\approx 0.6321...$ or $63.212\%$ . This estimation becomes accurate extremely quickly for even moderately rare drops. This value can be determined by determining following limit.

$\lim _{x\to \infty }{\left(1-{\frac {x-1}{x}}\right)}^{x}=1-{\frac {1}{e}}\approx 0.6321...$ In another example, the probability of receiving one draconic visage in a task of 234 Skeletal Wyverns can be received by changing the drop rate and kill count in the equation:

{\begin{aligned}&1-(1-0.0001)^{234}\\=&\ 1-0.9999^{234}\\\approx &\ 1-0.97687\\\approx &\ 0.023129\end{aligned}} Therefore, there is approximately a 2.3% chance of receiving a draconic visage during this task.

### Dry streaks and the odds of receiving a drop

It is crucial to understand that there is a difference between the odds of receiving a drop $P(X>0)$ and receiving exactly one drop $P(X=1)$ . Hence the probabilities are not the same. The example above calculates the odds of receiving any amount of visages from 1 to 5000, i.e. the odds of not going empty-handed. The term $\left({\frac {4999}{5000}}\right)^{5000}\approx 0.368$ seen in the expression above describes the probability of not receiving an item $P(X=0)$ or more commonly spoken of as a dry streak of length 5000.

### Binomial model

Given a known drop rate of ${\frac {1}{x}}$ , the chance of receiving such an item $k$ times in $n$ kills can be calculated using binomial distribution. All the examples above are special cases where the amount of drops, $k=0$ or $k>0$ . For finding the probability of obtaining an item at least once, rather than a specified number of times, the binomial coefficient can be simplified into this equation:

$1-(1-p)^{x}$ , where $(1-p)^{x}$ is calculating the probability of not receiving the item, and that is used to calculate the inverse.

The probability of receiving an item $k$ times in $n$ kills with a drop rate of ${\frac {1}{x}}=p$ follows:

${\binom {n}{k}}p^{k}(1-p)^{n-k}$ , where ${\binom {n}{k}}={\frac {n!}{k!(n-k)!}}$ As this formula can be rather tedious to work by hand, many prefer using a binomial probability calculator where you enter $p$ ,$n$ and $k$ .

### Reaching a percentile

Similarly, the number of King Black Dragons needed to kill to have a 90% probability of getting one when you kill them:

{\begin{aligned}1-\left(1-{\frac {1}{5000}}\right)^{x}&=0.90\\x&={\frac {\log(1-0.9)}{\log({\frac {4999}{5000}})}}\\x&=11511.774\ldots \approx 11512\end{aligned}} This yields the answer 11,512 meaning that you need to kill 11,512 for at least 90% chance of getting a draconic visage.

### Back to back drops

Again, using the King Black Dragon expected to drop a draconic visage once out of 5,000 kills, the probability of getting said drop twice from two kills or back-to-back is as follows:

$\left({\frac {1}{5000}}\right)^{2}=\ 0.00000004=0.000004\%$ For a back to back streak of any desired length, simply exchange the exponent. This is equivalent to using the binomial model and solving for the probability of receiving an item $n$ times in $n$ attempts.

### Finishing a boss

Players may wonder how many kills one should expect on average to "finish a boss", i.e. receiving at least one of all the desired drops. While an activity or boss with a high amount of items, such as Chambers of Xeric or Barrows, is easier to simulate than calculate, there is a short and exact one for receiving two items. This formula can be used to approximate more than two drops.

Given two known drop rates of $p={\frac {1}{x}}$ and of $q={\frac {1}{y}}$ , the following formula is the expected amount of kills to receive both items, in any order:

${\frac {1}{p}}+{\frac {1}{q}}-{\frac {1}{p+q}}$ derived from the sum seen here

For example, for players attempting to get both Bandos hilt (1 in 508) and Bandos tassets (1 in 381), the formula would be:

${\frac {1}{\frac {1}{508}}}+{\frac {1}{\frac {1}{381}}}-{\frac {1}{{\frac {1}{508}}+{\frac {1}{381}}}}=508+381-{\frac {508*381}{508+381}}\approx 671.3$ In other words, players have to do 672 kills to "hit the drop rate" of these two items combined.