# CumulativeDropProbability

## Template page

*103,497*pages on

this wiki

Syntax: {{**CumulativeDropProbability**|*<p=0.001>*}}

- |p= is the probability of the item drop rate as a decimal number. This defaults to 0.001 = 0.1% = 1 / 1000.

Number of Kills | 100 | 288 | 401 | 693 | 1386 | 2302 | 2995 | 4603 | 9206 |
---|---|---|---|---|---|---|---|---|---|

Cumulative Probability | 9.5% | 25% | 33% | 50% | 75% | 90% | 95% | 99% | 99.99% |

Neither the drop rate nor cumulative probability increases based on the number of monsters you have already killed. |

## Computing the probabilities Edit

The table shows the cumulative probability of getting at least one desired item while killing a particular number of monsters that drop the desired item with a constant probability *p*.

Mathematically, the probability of *not* getting the desired drop as a direct consequence of any particular kill is (1-p). We assume that each mob drops the item independently of any previous mobs killed, so the probability of *not* getting a drop as a direct consequence of *any* of N kills is (1-p)^N. The opposite of this -- the probability of getting at least one drop as a consequence of all of N kills -- is 1 - (1-p)^N. This probability converges to 1 as N goes to infinity; one is **not** guaranteed to get a drop in a finite number of kills.

Note that as the mobs drop items independently of each other, the process is essentially memoryless: having killed (N-1) mobs without getting a drop, the probability of the item dropping on N'th kill is still p, as opposed to the cumulative probability for N kills.

The above describes cumulative probability P as a function of number of kills N; we can also describe N as a function of P:

- P = 1 - (1 - p)
^{N} - N = log(1 - P)/log(1 - p)
- N should be rounded up to an integer: can't loot half a monster.