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:

• Bones have a 100% drop rate from chickens.
• Feathers have approximately a 75% drop rate from chickens.

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. An exception to this is if there is a bad luck mitigation mechanic listed on certain monsters and events; this will increase the chance of receiving a drop after a dry streak. 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

See also: Calculator:Dry calc

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.

$\lim _{x\to \infty }{\left(1-\left({\frac {x-1}{x}}\right)^{x}\right)}=1-{\frac {1}{e}}\approx 0.6321...$

In another example, the probability of receiving at least one draconic visage in a task of 234 Skeletal Wyverns can be received by changing the drop rate and kill count in the equation:

${\begin{aligned}&1-(1-0.0001)^{234}\\=&\ 1-0.9999^{234}\\\approx &\ 1-0.976871\\\approx &\ 0.023129\end{aligned}}$

Therefore, there is approximately a 2.3% chance of receiving at least one 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.

This is an instance of the famous coupon collector's problem, but with unequal probabilities. The problem was solved and the solution published in 1954 by H. von Schelling. For any given amount of drops, and with any probability for each drop, one can use the following formula:

$\int _{0}^{+\infty }\left(1-\prod _{i=1}^{N}\left(1-e^{-p_{i}x}\right)\right)dx$ where p_i are the probabilities.

An example of this formula in use can be found and interacted with here. As one can see, finishing bandos(including boots) will take about 867 kills.

See also

• Calculator:Dry calc
• Drops
• Drop table
• Rare drop table
• Items
• Kill count
• Bestiary