|
|
| (9 intermediate revisions by 6 users not shown) |
| Line 1: |
Line 1: |
| Let's say you have 120 bars and 80% resource preservation and 0% doubling chance. | | {{V|1.1.1}} |
| | <onlyinclude>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? |
|
| |
|
| If a dagger takes one bar to create, how many daggers would you create on average out of 100 bars?
| | The answer is <math>\frac{120}{1 - 80/100} = \frac{120}{0.2} = 600.</math> |
|
| |
|
| 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. |
| | |
| | |
| The main idea behind the preservation is that you can reuse what you preserved to make more stuff. | |
|
| |
|
| 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. | | 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. |
| Line 17: |
Line 15: |
|
| |
|
|
| |
|
| So the number of daggers produced is <math>120 + (120 \times 0.8) + (120 \times 0.8 \times 0.8) + ... + (120 \times 0.8^n)</math> where <math>n</math> is a number as huge as you can imagine. | | 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 replace the number <math>120</math> with any other number <math>a</math> and the probability 0.8 with any other probability <math>r</math>, the number of produced daggers would look like this:
| |
| | |
| <math>a + (a \times r) + (a \times r \times r) + ... + (a \times r^n)</math>
| |
| | |
| or in short
| |
|
| |
|
| <math>\sum\limits_{n=0} ar^n</math>. | | 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: |
|
| |
|
| This thing is called a geometric progression, and it [https://en.wikipedia.org/wiki/Geometric_progression#Derivation turns out] that its value is equal to <math>\frac{a(1 - r^n)}{1 - r}</math>.
| | <math>a + (a \times r) + (a \times r \times r) + \cdots + (a \times r^n) = \sum\limits^{\infty}_{n=0} ar^n</math> |
|
| |
|
| And because <math>r < 1</math> and <math>n</math> is very big, it can be simplified to <math>\frac{a(1 - 0)}{1 - r} = \frac{a}{1 - r}</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 | | 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>.</onlyinclude> |
|
| |
|
| <math>\frac{120}{1 - 0.8} = 600</math>.
| | {{Menu}}[[Category:Math]] |
| This page was last updated for (v1.1.1).
|
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]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ 120 + (120 \times 0.8) + (120 \times 0.8 \times 0.8) +\cdots + (120 \times 0.8^n) }[/math] where [math]\displaystyle{ n }[/math] goes to infinity.
If we generalize and represent the number of starting resources ([math]\displaystyle{ 120 }[/math]) with [math]\displaystyle{ a }[/math] and the probability ([math]\displaystyle{ 0.8 }[/math]) with [math]\displaystyle{ r }[/math], the number of actions performed with [math]\displaystyle{ a }[/math] action's worth of starting resources and a resource preservation of [math]\displaystyle{ r }[/math] (where [math]\displaystyle{ 0 \le r \lt 1 }[/math]) can be expressed as follows:
[math]\displaystyle{ 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 infinite geometric series, which can simply be expressed as [math]\displaystyle{ \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]\displaystyle{ a = 120 }[/math] and [math]\displaystyle{ r = 0.8 }[/math], we get [math]\displaystyle{ \frac{120}{1 - 0.8} = 600 }[/math].
Combat:
