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General drop formula


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#1 Drak231

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Posted 02 November 2010 - 06:31 PM

Here's what I use when I want to drop a card bad :

With a time component, the actual formula is actually (this is not including the drop change with your level => http://rode-r.doddle...om/viewpage/d/3 ) :

1-((1-probability)^occurrence) = chances of having the item

For example, if you kill approximately 10 andre/minute, and the drop chance of the card is 0,01% (0,0001)

1-((1-0,0001)^10)) = 0.000999550119979002519790012 chances of having the item after 1 minute-> 0,1%

Now after an hour : you killed about 600 andres :

1- ((1-0,0001)^600) = 0.058238291893480576241755707966 chances of having the card after 60 minutes -> 5,82%

More explanation with a poker example :

You have 9 and J of hearts. On the flop there is a K of spade, a 5 and a 7 of heart.

If you want to chase the flush and don't count the cards in the hand of the other player, you have 1/4 of obtaining it on the turn and 1/4 of obtaining it on the river.

NOW, the total chance is calculated by the formula...

1-((1-probability)^occurrence) = 1-((1-0,25)^2) = 0,4375 = 43,75% of obtaining your flush.

NOTE THAT THIS IS IF YOU DON'T COUNT THE CARDS OF THE OTHER PLAYERS!!!


Hope it helps somehow!

Edited by Drak231, 02 November 2010 - 06:32 PM.

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#2 Daray

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Posted 09 November 2010 - 01:11 AM

The example about poker doesn't hold up, because poker is the type of game where cards aren't put back before pulling and so it has a different mathematical formula for chance.

Also while I understand what you mean, there is no way to get to the 100% point, so whatever calculations you make, you could get the card at the first drop, or you could never get it - because both have chance <=1.
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#3 Drak231

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Posted 09 November 2010 - 06:40 PM

The example about poker doesn't hold up, because poker is the type of game where cards aren't put back before pulling and so it has a different mathematical formula for chance.


Not true. I know what you want to suggest is to consider that every time you draw, the actual chance change, but in the example I gave, the only thing that would be needed to be considered is the hands of the other players. In fact, the exact probability wouldn't be 1/4, but 9-(number of heart in other players hands)/(52-(2cards*number of players)+3(flop)). The 1/4 is an approximate result to give an example.

Also, it only applies in Texas Hold'em

Also while I understand what you mean, there is no way to get to the 100% point, so whatever calculations you make, you could get the card at the first drop, or you could never get it - because both have chance <=1.


I know, that's not the point. The only reason the result would give 100% is because the probability of obtaining the item is 100%. Try and put any probability (other than 1) and occurence and you will never get 100%.

I only use the formula to motivate myself when I absolutly want to drop a card.
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