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m (stray >)
(Adjust note, force calculation to use 33% as opposed to 1/3.)
 
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! style="text-align: right" | Number of Kills
 
! style="text-align: right" | Number of Kills
! 100 || {{#expr: ceil(ln(1 - 0.25)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.33333333)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.50)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.75)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.90)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.95)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.99)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.9999)/ln(1 - {{{p|0.001}}}))}}
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! 100 || {{#expr: ceil(ln(1 - 0.25)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.33)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.50)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.75)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.90)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.95)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.99)/ln(1 - {{{p|0.001}}}))}} || {{#expr: ceil(ln(1 - 0.9999)/ln(1 - {{{p|0.001}}}))}}
 
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|-
 
! style="text-align: right" | Cumulative Probability
 
! style="text-align: right" | Cumulative Probability
 
| {{#expr: floor(1000 * (1 - (1 - {{{p|0.001}}})^100))/10}}% || 25% || 33% || 50% || 75% || 90% || 95% || 99% || 99.99%
 
| {{#expr: floor(1000 * (1 - (1 - {{{p|0.001}}})^100))/10}}% || 25% || 33% || 50% || 75% || 90% || 95% || 99% || 99.99%
 
|-
 
|-
| colspan="10" | <small>The probabilities shown here are cumulative. The drop rate does not increase as you kill additional monsters.</small>
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| colspan="10" | <small>Neither the drop rate nor cumulative probability increases based on the number of monsters [[Template:CumulativeDropProbability#Computing the probabilities|you have already killed]].</small>
 
|}</onlyinclude>
 
|}</onlyinclude>
   
----
 
   
This 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''.
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== Computing the probabilities ==
  +
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.
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  +
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.
   
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.
 
   
 
The above describes cumulative probability P as a function of number of kills N; we can also describe N as a function of P:
 
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)<sup>N</sup>
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: P = 1 - (1 - p)<sup>N</sup>
 
: N = log(1 - P)/log(1 - p)
 
: N = log(1 - P)/log(1 - p)
: N should be rounded up: can't loot half a dead monster.
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: N should be rounded up to an integer: can't loot half a monster.

Latest revision as of 18:12, December 22, 2009

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.
Probability of at least one drop in X kills
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.

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