ok, so we can say if you have three 100% chances to "win", then you have a 300% chance overall, because you will win 3 times. however, if you have three 50% chances, you do not have a 150% chance to win. you still have a 50% chance to win, just 3 times. what if you had sixty 50% chances, your overall chance has to be more than 50%.
i'm trying to figure out how i can determine, or at least get as close a possible, your chances at getting an item from a chest. usually a chest will execute 2 or 3 different blocks multiple times. lets use smoking blackwood chest for example ... it executes: random 9060:random 9059:random 9059:random 9059:random 9059:random 9061. now we'll just use a few items for example.
from 9059 we could get a sharktooth trident 9/100 times, a beaded belt 9/100 times, trollskin boots 8/100 times, and an ogre bone bracelet 2/100 times.
now block 9059 is executed 4 times, but we don't have a 36% chance at getting a beaded belt. however, wouldn't the overall chance be greater than 9%?
what if you had a 2/100 chance and a 10/100 chance, would it just be your highest % chance of 10%? or would the only way to show these right we be to say something like "beaded belt (9% x4)" and then even further "diamond (5% x3, 20% x2, 10% x1)" etc?
Calling all Math Geeks (Chest Drop %)
Re: Calling all Math Geeks (Chest Drop %)
ok, so we can say if you have three 100% chances to "win", then you have a 300% chance overall, because you will win 3 times. however, if you have three 50% chances, you do not have a 150% chance to win. you still have a 50% chance to win, just 3 times. what if you had sixty 50% chances, your overall chance has to be more than 50%.
i think you're confusing % of success with chance to "win". for the first case, you'd have a 100% chance now matter how many tries. for a chance of "win" you're looking for what i believe is statistically called the probability density (?). so you have 1.00 (for 100%) x 3 tries = 3 items. 0.50 x 3 tries = 1.5 items on average that you would "win".
i'm trying to figure out how i can determine, or at least get as close a possible, your chances at getting an item from a chest. usually a chest will execute 2 or 3 different blocks multiple times. lets use smoking blackwood chest for example ... it executes: random 9060:random 9059:random 9059:random 9059:random 9059:random 9061. now we'll just use a few items for example.
from 9059 we could get a sharktooth trident 9/100 times, a beaded belt 9/100 times, trollskin boots 8/100 times, and an ogre bone bracelet 2/100 times.
now block 9059 is executed 4 times, but we don't have a 36% chance at getting a beaded belt. however, wouldn't the overall chance be greater than 9%?
technically i believe you should have a 36% chance of getting a beaded belt. this is true because 0.09 x 4 tries = 0.36 items "won", AKA 36% chance of getting it
what if you had a 2/100 chance and a 10/100 chance, would it just be your highest % chance of 10%? or would the only way to show these right we be to say something like "beaded belt (9% x4)" and then even further "diamond (5% x3, 20% x2, 10% x1)" etc?
i think you're saying "what if you had a 2/100 chance and a 10/100 chance of getting the same item". if that is true, you would have a 12% chance because 0.02 x 1 try + 0.10 x 1 try = 0.12 items "won"
i think you're confusing % of success with chance to "win". for the first case, you'd have a 100% chance now matter how many tries. for a chance of "win" you're looking for what i believe is statistically called the probability density (?). so you have 1.00 (for 100%) x 3 tries = 3 items. 0.50 x 3 tries = 1.5 items on average that you would "win".
i'm trying to figure out how i can determine, or at least get as close a possible, your chances at getting an item from a chest. usually a chest will execute 2 or 3 different blocks multiple times. lets use smoking blackwood chest for example ... it executes: random 9060:random 9059:random 9059:random 9059:random 9059:random 9061. now we'll just use a few items for example.
from 9059 we could get a sharktooth trident 9/100 times, a beaded belt 9/100 times, trollskin boots 8/100 times, and an ogre bone bracelet 2/100 times.
now block 9059 is executed 4 times, but we don't have a 36% chance at getting a beaded belt. however, wouldn't the overall chance be greater than 9%?
technically i believe you should have a 36% chance of getting a beaded belt. this is true because 0.09 x 4 tries = 0.36 items "won", AKA 36% chance of getting it
what if you had a 2/100 chance and a 10/100 chance, would it just be your highest % chance of 10%? or would the only way to show these right we be to say something like "beaded belt (9% x4)" and then even further "diamond (5% x3, 20% x2, 10% x1)" etc?
i think you're saying "what if you had a 2/100 chance and a 10/100 chance of getting the same item". if that is true, you would have a 12% chance because 0.02 x 1 try + 0.10 x 1 try = 0.12 items "won"
Re: Calling all Math Geeks (Chest Drop %)
i think you're confusing % of success with chance to "win". for the first case, you'd have a 100% chance now matter how many tries. for a chance of "win" you're looking for what i believe is statistically called the probability density (?). so you have 1.00 (for 100%) x 3 tries = 3 items. 0.50 x 3 tries = 1.5 items on average that you would "win".
i was just using "win" as an example of "winning the item" or "winning the lottery." the first example was assuming there was only 1 item to be won/gotten. the problem with saying "0.5 * 3 = 1.5 items won" is because i'm not interested in how many items i win, i want to know the chance of winning those items.
technically i believe you should have a 36% chance of getting a beaded belt. this is true because 0.09 x 4 tries = 0.36 items "won", AKA 36% chance of getting it
this isn't true because if it were just two items at 50%, then .5 * 2 = 1.0 (or 100% chance). that's false.
i think you're saying "what if you had a 2/100 chance and a 10/100 chance of getting the same item". if that is true, you would have a 12% chance because 0.02 x 1 try + 0.10 x 1 try = 0.12 items "won"
again, what if it were a 90% chance and a 10% chance. (.9 * 1) + (.1 * 1) = 1.0. you don't have a 100% chance.
i was just using "win" as an example of "winning the item" or "winning the lottery." the first example was assuming there was only 1 item to be won/gotten. the problem with saying "0.5 * 3 = 1.5 items won" is because i'm not interested in how many items i win, i want to know the chance of winning those items.
technically i believe you should have a 36% chance of getting a beaded belt. this is true because 0.09 x 4 tries = 0.36 items "won", AKA 36% chance of getting it
this isn't true because if it were just two items at 50%, then .5 * 2 = 1.0 (or 100% chance). that's false.
i think you're saying "what if you had a 2/100 chance and a 10/100 chance of getting the same item". if that is true, you would have a 12% chance because 0.02 x 1 try + 0.10 x 1 try = 0.12 items "won"
again, what if it were a 90% chance and a 10% chance. (.9 * 1) + (.1 * 1) = 1.0. you don't have a 100% chance.
Re: Calling all Math Geeks (Chest Drop %)
yea but if you divide items "won", or how many items you get divided by the tries, you get the effective percentage of getting BOTH. but i guess you're looking for the probability that you get either of the items? if there are 2 items and each gives a 50% probability, then there is 25% probability you will get neither, 50% probability you will get one of them and 25% probability you will get both of them. i believe this can be calculated with a probability tree, with independent results. chuck is getting a master's in probability, how about a little input from him?
but i think you're saying that there is only one item, and there is a 50% chance applied twice that you will get it (for example). this would be dependent, which can still be calculated with a probability tree, but a different one
but i think you're saying that there is only one item, and there is a 50% chance applied twice that you will get it (for example). this would be dependent, which can still be calculated with a probability tree, but a different one
Re: Calling all Math Geeks (Chest Drop %)
yea but if you divide items "won", or how many items you get divided by the tries, you get the effective percentage.
using this formula with my example of 1 item at 50% with sixty tries, 0.5*60=30 items won on average... 30/60 tries = 0.5 effective percentage or 50%. the second calculation is the opposite of the first.
i believe this can be calculated with a probability tree, with independent results. chuck is getting a master's in probability, how about a little input from him?
lemme formulate a reply and i'll get it out
i agree, i just dont think it's as simple as what has been suggested.
using this formula with my example of 1 item at 50% with sixty tries, 0.5*60=30 items won on average... 30/60 tries = 0.5 effective percentage or 50%. the second calculation is the opposite of the first.
i believe this can be calculated with a probability tree, with independent results. chuck is getting a master's in probability, how about a little input from him?
lemme formulate a reply and i'll get it out
i agree, i just dont think it's as simple as what has been suggested.
Re: Calling all Math Geeks (Chest Drop %)
sorry, i edited it after that, you weren't supposed to read it yet :P
but then i contradicted myself...
:wstupid:
but then i contradicted myself...
:wstupid:
-
Scruff
Re: Calling all Math Geeks (Chest Drop %)
using this formula with my example of 1 item at 50% with sixty tries, 0.5*60=30 items won on average... 30/60 tries = 0.5 effective percentage or 50%. the second calculation is the opposite of the first.
The only problem with that conclusion is that percentages in mmud are not cumulative. On a 50% chance, with 60 tries, there is a possiblity to get the item 60 times or 0 times because EVERY try the % resets itself. Granted, on average the number of items will fall around 50% (usually +/- 15% or so with that small a sample size). Like Ray said, the best way is with a tree because probability is really what you're interested in.
edit: damned typos
The only problem with that conclusion is that percentages in mmud are not cumulative. On a 50% chance, with 60 tries, there is a possiblity to get the item 60 times or 0 times because EVERY try the % resets itself. Granted, on average the number of items will fall around 50% (usually +/- 15% or so with that small a sample size). Like Ray said, the best way is with a tree because probability is really what you're interested in.
edit: damned typos
Re: Calling all Math Geeks (Chest Drop %)
but a standard tree does not really account for a limited number of items...
if you get 60 shots at a lim1 item that drops 50%, the probability would be like .99995 that you would get it. i guess you could get that through a tree, but it'd be a damn ugly tree, more like a big diagonal
edit, probability would be 1-(0.50)^60 actually, which is 0.9999999999999999991326382620116
if you get 60 shots at a lim1 item that drops 50%, the probability would be like .99995 that you would get it. i guess you could get that through a tree, but it'd be a damn ugly tree, more like a big diagonal
edit, probability would be 1-(0.50)^60 actually, which is 0.9999999999999999991326382620116
Re: Calling all Math Geeks (Chest Drop %)
uh, this is statistics guys. They actually have formulas for this. I'll look in my notes and post it, and you guys can have fun figuring out wtf it all means. :)
Basically, you need to figure out how many "items" you want to win. If you want to know your chance of winning atleast 1 item, then you have a very high %. If its win ALL of them, then you have a very low percent. So, have to figure out what you want.
Basically, you need to figure out how many "items" you want to win. If you want to know your chance of winning atleast 1 item, then you have a very high %. If its win ALL of them, then you have a very low percent. So, have to figure out what you want.
Re: Calling all Math Geeks (Chest Drop %)
yeah, if you define very carefully what specific probability you're trying to find, this'll be a lot easier to answer :)
In answer to your first question in the initial post, the chance of getting a beaded belt is 1 - (.91 ^ 4)... which is the inverse prob of the chance of not getting a belt in 4 trials...
some prob/stats experience is helpful here, a lot of prob results are somewhat counter-intuitive.
In answer to your first question in the initial post, the chance of getting a beaded belt is 1 - (.91 ^ 4)... which is the inverse prob of the chance of not getting a belt in 4 trials...
some prob/stats experience is helpful here, a lot of prob results are somewhat counter-intuitive.