Complete Doom Solution

[Terrorcorp]

oh yeah that s the same...

because u say the prob to get >=1 of successes .... cuz u say the prob not to drop is (1-p)^n so when u do 1-(1-p)^n is automatically as saying the prob the get 1 or more drops.

Since ur formula gave(on my mistake) another prob i was trying to convince myself about the error

[DaniOvertures]

dude , 1-0.9975 it is wrong as probability because it considers only the fact to process n-1 fails on every event and the last must be success so it considers only this event:

0 0 0 0 0 1 on n=6 but if want to give a probability on the number of successes on n tries u have to consider 0 0 1 0 0 0 aswell...



So that formula gives the probability of n consecutive fails and then u r saying that after the Nth toss u have 1-0.9975^n prob to drop.... so it comes out that on n=100 u have a 52% chance to drop on the next try... u can't calculate probability on rolling basis (in that case prob of drop remains 0.0025)

My formula says on 100 tries there's a 22% prob to drop (and that's not considering luck etc only by mobs) wich is by far more realistic

...the formula NOT says the prob that you have on the next cycle, it only says how much prob you have to drop 1 art in n cycle =)
btw if you try to calculate what you have written is:
1 - 0,9975^100 = 0,22% not 0,52 as you said
the probably of drop every cycle remains the same cause (for now without fix) client has not memory so you can be at your first cycle or at you 100th cycle and you have still the same prob to drop, but the prob to drop 1 art in 100cycle is not the same to drop 1 art in 1cycle and it's approx give by the theory

[DaniOvertures]

okay... i want to say that i studdied theory of probability, math statistics, math analys ) and i know exact formulas ) i dont need to know probobility of 1 drop. for week it will be 99.99% and this numeric shows nothing, becuase for a week i can drop 5 arts or nothing. if u calculate probobility of one side of coing for 100 turns, it will be 99.999% and what for do u need this value? the value u re interested is 50 times in average.
i caclualted that for 2 hours i drop 0.23 of art )) or in average 1 art for 8 hours of playing. i dont care that it's wrong probobility calcualtion or whatever. actualy i calculated 30 min. in fact it's 20 most of the time ) and it doesnt matter, because i dont care about probobility of my drops. the aim of previuos post was to show that a member of banzai-horde drops ~15 times less, than me ) and actualy that's all

yeah, if you've read i said that ur post was nice, and i haven't say anything about your math preparation, it's only to give a correct way to read the % formula for everyone

[Terrorcorp]

dude , 1-0.9975 it is wrong as probability because it considers only the fact to process n-1 fails on every event and the last must be success so it considers only this event:

0 0 0 0 0 1 on n=6 but if want to give a probability on the number of successes on n tries u have to consider 0 0 1 0 0 0 aswell...



So that formula gives the probability of n consecutive fails and then u r saying that after the Nth toss u have 1-0.9975^n prob to drop.... so it comes out that on n=100 u have a 52% chance to drop on the next try... u can't calculate probability on rolling basis (in that case prob of drop remains 0.0025)

My formula says on 100 tries there's a 22% prob to drop 1 or more times (and that's not considering luck etc only by mobs) wich is by far more realistic

[Terrorcorp]

btw , theorically for probability in this case is calculated as a binomial...

so in the case of the n=5 mobs the probability for the monster to drop(no your probability to get the drop) assuming that P=0.0075 is the prob of success k=#successes it gets:

P(k=1) = n!/(n-k)!*p^k*(1-p)^(n-k)

for P(k>1) i think the prob is so low that u can easily forget about it


i prefer fnurov's empirical prob

[Terrorcorp]

okay... i want to say that i studdied theory of probability, math statistics, math analys ) and i know exact formulas ) i dont need to know probobility of 1 drop. for week it will be 99.99% and this numeric shows nothing, becuase for a week i can drop 5 arts or nothing. if u calculate probobility of one side of coing for 100 turns, it will be 99.999% and what for do u need this value? the value u re interested is 50 times in average.
i caclualted that for 2 hours i drop 0.23 of art )) or in average 1 art for 8 hours of playing. i dont care that it's wrong probobility calcualtion or whatever. actualy i calculated 30 min. in fact it's 20 most of the time ) and it doesnt matter, because i dont care about probobility of my drops. the aim of previuos post was to show that a member of banzai-horde drops ~15 times less, than me ) and actualy that's all

fnurov is right... there's no need for theorical probability when you have the luxury to see probability with an empirical approach.

Empirical in this case is better because we r not sure if the data we have is perfectly applicable to the one of the server... (for example a lot of skill training tables in sticky are wrong...)

[fnurov]

nice!
this kind of things are approx 'cause if you launch 2 times a coin you haven't 2*(50%) = 100% of probability to take a specific side cause you've not to do 50% + 50% = 100% but 50% + 50%of50% = 50% + 25% = 75% and next 75% + 50%of25% = 87.5%

okay... i want to say that i studdied theory of probability, math statistics, math analys ) and i know exact formulas ) i dont need to know probobility of 1 drop. for week it will be 99.99% and this numeric shows nothing, becuase for a week i can drop 5 arts or nothing. if u calculate probobility of one side of coing for 100 turns, it will be 99.999% and what for do u need this value? the value u re interested is 50 times in average.
i caclualted that for 2 hours i drop 0.23 of art )) or in average 1 art for 8 hours of playing. i dont care that it's wrong probobility calcualtion or whatever. actualy i calculated 30 min. in fact it's 20 most of the time ) and it doesnt matter, because i dont care about probobility of my drops. the aim of previuos post was to show that a member of banzai-horde drops ~15 times less, than me ) and actualy that's all

[DaniOvertures]

drop chance per cycle = (1.25 (df) + 0.75 (mobs) * 5 + [0..0.9] (luck bonus))*number of mobs in each room

personal drop chance per cycle = drop chance per cycle / number of players*number of mobs in each room

personal drop chance per cycle (time based) = (time played / time to make cycle )*(drop chance per cycle / number of players)*number of mobs in each room

"me alone in doom for 2 hours"
(time played / time to make cycle )*(drop chance per cycle / number of players)= (2h/30m)*(5.8/1)=4*5.8=23.2%

"i with 2 good players for 2 hours"
(2h/12)*(5.8/3)=10*1.93=19.3%

i with "boomstick/soulseeker horde" 10 players for 2 hours
(2h/35m)*(5.2/10)*2=3.42*0.52*2=3.42%

"boomstick/soulseeker horde" without me (or another good doom player) 10 players for 2 hours
(2h/1h)*(5.1/10)*2=2*0.52*2=2.08%

hello to everybody who still think, that drops don't depand on damage, especially whan good players logout in face of "boomstick/soulseeker horde"

nice!
this kind of things are approx 'cause if you launch 2 times a coin you haven't 2*(50%) = 100% of probability to take a specific side cause you've not to do 50% + 50% = 100% but 50% + 50%of50% = 50% + 25% = 75% and next 75% + 50%of25% = 87.5%

so you have % to drop in N cycle
5,8% - 1
11,26% - 2
16,41% - 3
21,26% - 4 (this is your % of drop alone in 2hours - 4cycles)

the easiest way to calculate it is to do:
1 - 0,058 = 0,942 = 94,2% to NOT drop in 1 cycle (to adjust with personal luck ecc)
1 - 0,942^n % to drop in n consecutive cycle

but we've to remind that this is only theory and it is like real only with a big amount of cases

[smallangel]

hi fnurov, can u ask me a tanker with disco for doom ?

i want to do doom easy or in solo.

tank you!

[fnurov]

math for people, that don't drop

drop chance per cycle = (1.25 (df) + 0.75 (mobs) * 5 + [0..0.9] (luck bonus))*number of mobs in each room

personal drop chance per cycle = drop chance per cycle / number of players*number of mobs in each room

personal drop chance per cycle (time based) = (time played / time to make cycle )*(drop chance per cycle / number of players)*number of mobs in each room

"me alone in doom for 2 hours"
(time played / time to make cycle )*(drop chance per cycle / number of players)= (2h/30m)*(5.8/1)=4*5.8=23.2%

"i with 2 good players for 2 hours"
(2h/12)*(5.8/3)=10*1.93=19.3%

i with "boomstick/soulseeker horde" 10 players for 2 hours
(2h/35m)*(5.2/10)*2=3.42*0.52*2=3.42%

"boomstick/soulseeker horde" without me (or another good doom player) 10 players for 2 hours
(2h/1h)*(5.1/10)*2=2*0.52*2=2.08%

hello to everybody who still think, that drops don't depand on damage, especially whan good players logout in face of "boomstick/soulseeker horde"

[DaniOvertures]

i hate you f***** russian

(joking)
i have not drop anything yet ARGH

[cobravendicatore]

[fnurov]

[belinelli]

se avere fc4, fcr6, swing 1.25s, hml 40+ - si

con soulseeker - nò

p.s. io non parlo italiano

se uso spade slayer?

[fnurov]

ho visto che per doom ( in solitaria ) le skill meglio sono queste

Mace 120
Tactics 120
Anatomy 120
Bushido 120
Chiva 70
Res Spell 70
Necro 100

giusto??

se avere fc4, fcr6, swing 1.25s, hml 40+ - si

mi chiedo se al posto di mce ci metto sword è la stessa cosa?

con soulseeker - nò

p.s. io non parlo italiano