# Drop Rate Inspired.

• Sep 6th 2009, 07:35 PM
Frggr
Drop Rate Inspired.
OK so i was playing a RPG earlier today and was astonished at how many people couldn't understand that drop rates are not linear so I decided to create a simple formula as to how to determine %chance to drop after each kill.

I started with this:
(Note)
a: percent to drop
ptd: means percent to drop (the acctual words not the number)
k:kill
(End Note)

ptd 1k: a
ptd 2k: a+(1-a)a= 2a-a^2
ptd 3k: a+(1-a)a+[1-(a+(1-a)a]a
=2a-a^2+[1-2a+a^2)]a
=2a-a^2+a-2a^2+a^3
=3a-3a^2+a^3
ptd 4k:4a-6a^2+4a^3-a^4
ptd 5k:5a-10a^2+10a^3-5a^4+a5

Now i know there's a pattern in there and I'm sure it has something to do with geometric series but I just cant define the pattern.

(I'm sure there's an easier way to do this question than my method and I would love to know it so I could give them a formula real fast but I am also curious as to how to find the pattern in these series)

^^Sorry for bad grammer up there if you didnt understand plz ask me to explain I will be looking back at this every hour or so.
• Sep 6th 2009, 08:37 PM
mr fantastic
Quote:

Originally Posted by Frggr
OK so i was playing a RPG earlier today and was astonished at how many people couldn't understand that drop rates are not linear so I decided to create a simple formula as to how to determine %chance to drop after each kill.

I started with this:
(Note)
a: percent to drop
ptd: means percent to drop (the acctual words not the number)
k:kill
(End Note)

ptd 1k: a
ptd 2k: a+(1-a)a= 2a-a^2
ptd 3k: a+(1-a)a+[1-(a+(1-a)a]a
=2a-a^2+[1-2a+a^2)]a
=2a-a^2+a-2a^2+a^3
=3a-3a^2+a^3
ptd 4k:4a-6a^2+4a^3-a^4
ptd 5k:5a-10a^2+10a^3-5a^4+a5

Now i know there's a pattern in there and I'm sure it has something to do with geometric series but I just cant define the pattern.

(I'm sure there's an easier way to do this question than my method and I would love to know it so I could give them a formula real fast but I am also curious as to how to find the pattern in these series)

^^Sorry for bad grammer up there if you didnt understand plz ask me to explain I will be looking back at this every hour or so.

So in a nutshell, you want the rule for generating the following sequence:

$\displaystyle s_1 = a$

$\displaystyle s_2 = 2a - a^2$

$\displaystyle s_3 = 3a - 3a^2 + a^3$

$\displaystyle s_4 = 4a-6a^2+4a^3-a^4$

$\displaystyle s_5 = 5a-10a^2+10a^3-5a^4+a^5$

Correct?
• Sep 6th 2009, 08:39 PM
Frggr
Quote:

Originally Posted by mr fantastic
So in a nutshell, you want the rule for generating the following sequence:

$\displaystyle s_1 = a$

$\displaystyle s_2 = 2a - a^2$

$\displaystyle s_3 = 3a - 3a^2 + a^3$

$\displaystyle s_4 = 4a-6a^2+4a^3-a^4$

$\displaystyle s_5 = 5a-10a^2+10a^3-5a^4+a^5$

Correct?

Correct -- My bad making it more confusing than it should be :)
• Sep 7th 2009, 09:26 AM
awkward
Quote:

Originally Posted by mr fantastic
So in a nutshell, you want the rule for generating the following sequence:

$\displaystyle s_1 = a$

$\displaystyle s_2 = 2a - a^2$

$\displaystyle s_3 = 3a - 3a^2 + a^3$

$\displaystyle s_4 = 4a-6a^2+4a^3-a^4$

$\displaystyle s_5 = 5a-10a^2+10a^3-5a^4+a^5$

Correct?

Just eyeballing it, it appears that

$\displaystyle s_n = 1 - (1 - a)^n = \sum_{i=1}^n (-1)^{n+1} \binom{n}{i} a^i$

where

$\displaystyle \binom{n}{i} = \frac{n!}{i! \; (n-1)!}$

is a binomial coefficient.