Editing Item drops

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The overall chance of obtaining items by killing several monsters or destroying several boxes can also be calculated. For multiple rolls, the overall chance to receive a drop naturally increases. However, this increase is not linear. The increase is determined by following the binomial distribution formula:
The overall chance of obtaining items by killing several monsters or destroying several boxes can also be calculated. For multiple rolls, the overall chance to receive a drop naturally increases. However, this increase is not linear. The increase is determined by following the binomial distribution formula:


<code>''chance of at least'' 1 ''drop'' = 1 - (1 - ''drop rate'') ^ (''number of events'')</code>
<code>''chance of at least 1 drop'' = 1 - (1 - ''drop rate'') ^ (''number of events'')</code>


Using the above example, the overall chance of receiving at least 1 weapon box from a Claw by killing it '''10''' times is:
Using the above example, the overall chance of receiving at least 1 weapon box from a Claw by killing it '''10''' times is:


<code>''chance of at least'' 1 ''drop'' = 1 - (1 - ''1/10.89'') ^ (''10'') = '''1/1.62 (61.83%)'''</code>
<code>''chance of at least 1 drop'' = 1 - (1 - ''1/10.89'') ^ (''10'') = '''1/1.62 (61.83%)'''</code>


=====Average drop chance=====
=====Average drop chance=====
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<code>1/2 + 1/4 - 1/2 * 1/4 = '''1/1.6 (62.5%)'''</code>
<code>1/2 + 1/4 - 1/2 * 1/4 = '''1/1.6 (62.5%)'''</code>
====Multiple drops====
The numbers above show the process of obtaining at least 1 item, which is sufficient for most players who hunt for a certain drop once. The chance to obtain multiple items can be calculated too. For example, if a monster drops an item with a drop rate of <code>1/200</code> and is killed 100 times, the chance of it dropping <code>3</code> items or more can be calculated:
<code>''chance of at least'' 3 ''drops'' = 1 - P(0) - P(1) - P(2)</code>, where <code>P(k)</code> is the chance of ''exactly'' <code>k</code> item drops.
So, in this example, the chance of receiving exactly 0, 1 or 2 drops has to be calculated. The general formula for <code>P(k)</code> is:
<code>P(k) = n_C_k * p^k * (1-p)^(n-k)</code>, where:
* n: number of trials, in this example '''100''' monster kills
* k: number of successes, which for P(0) is '''0''', P(1) '''1''', etc.
* p: probability of success, which is the drop rate of '''1/200'''
* n_C_k: the binomial coefficient, which is defined as '''n!/(k!(n-k)!)''' with "!" referring to the factorial
The process of evaluating each <code>P(k)</code> value is the same, it is shown here for <code>P(2)</code>:
* n = '''100'''
* k = '''2'''
* p = '''1/200'''
* n_C_k = 100!/(2!(100-2)!) = '''4950'''
<code>P(2) = 4950 * (1/200)^2 * (1-1/200)^(100-2) = '''1/13.21'''</code>
Repeating the process for <code>P(0)</code> and <code>P(1)</code> we arrive at:
<code>''chance of at least'' 3 ''drops'' = 1 - 1/1.65 - 1/3.29 - 1/13.21 = '''1/70.91 (1.41%)'''</code>


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