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== Calculating Gains == | == Calculating Gains == | ||
{{: | Let's say you have 120 bars, 80% resource preservation, and 0% doubling chance. If a dagger takes one bar to create, how many daggers would you create on average out of 120 bars? | ||
The answer is <math>\frac{120}{1 - 80/100} = \frac{120}{0.2} = 600.</math> | |||
The main idea behind resource preservation is that you can reuse the preserved resources to make more items, but those resources when used will themselves be preserved also. | |||
Using the same example with 120 bars, you take 120 actions producing 120 daggers, but with 80% preservation you are left with <math>120 \times 0.8 = 96</math> bars. | |||
Now you can use those 96 bars to do the same thing, and you will be left with <math>96 \times 0.8 = 76.8</math> bars on average. | |||
Repeat that to infinity, and you will get the average number of daggers that you will produce from 120 bars. | |||
But how do we calculate this number? | |||
So the number of daggers produced is <math>120 + (120 \times 0.8) + (120 \times 0.8 \times 0.8) +\cdots + (120 \times 0.8^n)</math> where <math>n</math> goes to infinity. | |||
If we generalize and represent the number of starting resources (<math>120</math>) with <math>a</math> and the probability (<math>0.8</math>) with <math>r</math>, the number of actions performed with <math>a</math> action's worth of starting resources and a resource preservation of <math>r</math> (where <math>0 \le r \lt 1</math>) can be expressed as follows: | |||
<math>a + (a \times r) + (a \times r \times r) + \cdots + (a \times r^n) = \sum\limits^{\infty}_{n=0} ar^n</math> | |||
This expression is an [https://en.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series infinite geometric series], which can simply be expressed as <math>\displaystyle \lim_{n \to \infty}\tfrac{a(1 - r^n)}{1 - r} = \frac{a}{1 - r}</math>. | |||
So, coming back to our example where <math>a = 120</math> and <math>r = 0.8</math>, we get <math>\frac{120}{1 - 0.8} = 600</math>. | |||
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