Droprates...

[evilpaul]

I think people are talking about two different things... if the NM has a 25% drop rate, then the odds of getting the drop on any given kill will be 1:4 regardless of how many times you've killed it. But, it becomes less likely that a person will kill the thing multiple times with no drop. So, Argus may only drop his PCA 25% of the time, but it's very unlikely that you could kill Argus 50 times in a row with no drops.

[Popsiclestick]

I think people are talking about two different things... if the NM has a 25% drop rate, then the odds of getting the drop on any given kill will be 1:4 regardless of how many times you've killed it. But, it becomes less likely that a person will kill the thing multiple times with no drop. So, Argus may only drop his PCA 25% of the time, but it's very unlikely that you could kill Argus 50 times in a row with no drops.

This is true. Statistically speaking, each roll of a die is independent of each other. Each kill or each die roll is independent from the previous, unless NM drops are based like ground king spawn conditions, where there is a limited number of chances of an occurrence until it becomes 100%, you could technically go on forever rolling 1's or getting vials of mercury

[Galois]

You spoke of flipping a coin four times and having heads come up each of the four times, then stating that the smarter bet would be to bet on tails being the next face to come up. For now, let's assume that the coin is a fair coin (even that it has been shown by Diaconis that a coin has a slightly higher probability to land on the side that it started on; about 0.51).

If you flip a (fair) coin 4 times and have the heads side come up four times, the probability of another heads coming up is still 1/2. People see a long streak and think to themselves "it is now time for a tails side to come up". This is gambler's fallacy. Remember that a coin has no memory; each of the flips are independent, so no matter what, the probability of obtaining a heads (or tails for that matter) is 1/2.

Coming back to Argus, if the drop rate is in fact 1/4 (and we make the assumption that each drop is independent-can me make this assumption?), then you would expect to see, in the long run, that 1 out of every 4 Argus kills drops the amulet. You may see steaks of many non-drops (or even a streak of drops), but that is just (pseudo) random chance. Sorry if I repeated anyone

[NynJa]

This is true. Statistically speaking, each roll of a die is independent of each other. Each kill or each die roll is independent from the previous, unless NM drops are based like ground king spawn conditions, where there is a limited number of chances of an occurrence until it becomes 100%, you could technically go on forever rolling 1's or getting vials of mercury

I doubt it works like that. Friend of mine killed Xolotl for 52 straight pops before they got the Harpe to drop.

[Seyrr]

I think people are talking about two different things... if the NM has a 25% drop rate, then the odds of getting the drop on any given kill will be 1:4 regardless of how many times you've killed it. But, it becomes less likely that a person will kill the thing multiple times with no drop. So, Argus may only drop his PCA 25% of the time, but it's very unlikely that you could kill Argus 50 times in a row with no drops.

I want to add one clarification here, because it's a really important (and kind of subtle) point that people miss a lot.

Assuming that Argus does have a 1/4 drop rate, BEFORE YOU START KILLING HIM it is extremely unlikely that you'd kill him 50 times in a row with no drops. If you've just wrapped up attempt 49 and you have no drop, those 49 attempts are in the past. You can't say "Oh, this next Argus is practically guaranteed to drop", because as was mentioned, it's a memoryless function (that is, future results are not affected by past results.)

It's only meaningful to talk about "it's very unlikely that you could kill Argus 50 times in a row with no drops" if you say that BEFORE you start your chain of 50.

[Yutuyu]

This is basically conditional probability, and what he's confusing is independence of events.

In probability P(x) where x is an event, is the probability of the occurance of event x.

P(x|y) is the probability of x knowing that y occured.

P(x|y) = P(x and y)/P(y) (Law of Conditional probability)

Say you flip a fair coin, one side is marked O one side is marked X. The two sides are equally weighted, and you use a machine to flip it, so that the chance of the coin landing on either side is perfectly 50%.

You flip it once, and you get X.

You want to know the chances of flipping it again and getting O.

P(O|X) = P(X and O)/P(X)

P(X and O) is the chance of getting X on the first roll, and O on the second. Lets see what the possibilities are.

X O
X X
O O
O X

So, one of these four possibilites is the desired outcome. Therefore the probability of getting X then O is 1/4.

Thus P(O|X) = P(X and O)/P(X) = (1/4)/(1/2) = 1/2.

Thus you see that knowing X occured the first time has absolutely no effect on O occuring or not occuring.

That is because X and O are statistically independent events. Anytime a particular experiment's trials are indenpendant it's said that such an experiment has "no memory." Thus the comment about coins having no memory.



EDIT: To sum up, the drop rate of an item isn't an absolute law. It is an amount of time an item will drop, on average. If an item drops on average 50% of the time, then with a sample size (killing the mob and seeing if it drops or not) of 1000 trials, you will have a number somewhere around 500 for how many drops, but it it will not actually BE 500. It will be closer to 500 or not depending on pure chance and the kind of distribution (type of randomness) being employed.

[alt]

Assuming it's a 1/4 drop rate...
/snip

I don't mean to be rude, but get some sleep, ask your math teacher/prof for a good statistics book, and chant "coins have no memory" while you read it.

Or go back and read Zosi's post again.

Basically the OP is asking if an unlucky streak changes his chance at the next kill, which is no. In this case, past events don't affect future ones, and next argus pop still has a 1/4 chance of dropping (or whatever the droprate is supposed to be)

Correct. Each event is unique and unrelated to any previous coin flip. If you flipped a coin and got heads 100 times in a row, you would still have exactly a 50% chance of it landing heads or tails on the next toss. Irregardless of what side it landed on the last 100 times.

Mathematicians have done experiments where they had random people toss a coin in the air 100 times, and record the results. And then make up 100 coin tosses and record those as well. Virtually every single time, the mathematicians could pick out which set of results were made up, and which set was the result of real coin tosses.

This is because when people make up results, they get focused on the 50/50 chance of heads of tails, and thus their results become "too random".
However, when you flip an actual coin, each event is distinct and unrelated to any previous event, thus you often get strings of maybe 10 heads tosses in a row, followed by 1 tails, then 6 more heads, etc.

[Ambir]

[quote=Francisco II]

Assuming it's a 1/4 drop rate...
/snip

I don't mean to be rude, but get some sleep, ask your math teacher/prof for a good statistics book, and chant "coins have no memory" while you read it.

Or go back and read Zosi's post again.

Basically the OP is asking if an unlucky streak changes his chance at the next kill, which is no. In this case, past events don't affect future ones, and next argus pop still has a 1/4 chance of dropping (or whatever the droprate is supposed to be)

I know... I was actually trying (and failing) to remember the math to figure out the probability of a drop over a predetermined number of kills.

1/4th drop rate, one kill = 25% chance of drop

1/4th drop rate, two kills = ??? chance of drop[/quote:cb28d]

43.75%

(1-(0.75*0.75))*100


(seeing noone answered)


But yeah, theres never been any indication that drops are anything but independant events. LSs do camps of NMs all the time, and drop rates have never been seen to be dependant on anything but the mob that was being fought.

[NiggaFromHarlem]

/confused

[Zosi]

43.75%

(1-(0.75*0.75))*100

(seeing noone answered)

25% drop rate on TWO kills =
0 drops: 1 * (0.75 ^ 2) = 56.25% chance
1 drop: 2 * (0.25 * 0.75) = 37.5% chance
2 drops: 1 * (0.25 * 0.25) = 6.25% chance
At least one drop: 43.75% chance

I answered.

Guess that's the downside to posting a wall-o-text with lots of math, nobody reads it.

[Spekkio]

http://en.wikipedia.org/wiki/Memorylessness
that's about all that needs to be said. keep the pseudomathematics to indiana, where they tried to legally mandate pi to be equal to 3.

[Necronus]

Like said many times above, drop rates, coins, and dice have no memory. What you are calcuating is "What are the odds that I got to where I am now along the sequence of events I observed".

If Argus is truly coded to have a drop rate of 0.25, then EVERY kill will have a 25% chance, no more no less. Even if Argus has not dropped in is past 20,000 kills, the odds of a drop on the 20,001th kill is not 99.99999~, it is exactly 25%.

You can stop here, but if you want a real world example, read below:


Go look for a Roulette wheel in a casino where they keep a electronic "history" board next to the table. Note that they usually don't just list the past 20 or so numbers landed on. If the casino is being especially clever, the board will put all of the red numbers on one side of the board in red lights and the black numbers on the other side of the board in white lights. This is to lull betters into a false sense of security in 2 possible ways.

First: Consider a table with 40 betters. Had all 40 walked up to a table without a history board without seeing any spins, they would likely make average sized bets spread more evenly across the table. But if the last 16/20 spins were red, you can be sure most people will not only cluster on black, but will put down much larger bets, thinking its a "sure thing" when in fact their odds are still 50/50. Since a casino shaves off the top anyway, if it does land on black they take a shortterm loss, but if it lands on red they make more net gain then they would have lost on black. But in this case since people were lulled into MUCH larger bets, over time players lose money faster, potentially making them bring more money into the casino sooner.

Second: Same situation as above, but the color sorting also makes it easy to see if a specific number was repeated 2-3 time in the past X spins. People will avoid these numbers even though the odds of that number being landed on a 3rd or 4th time is the same as the first.

The crux of the casino example is that the "memory" of a odds calculation of a random number is an illusion. If in a day, all betters at a table bet $1,000,000 and the casino is skewing payouts for a ~5% profit, then they will make ~$50,000 profit ... BUT if you could go back in time and add a memory board at the start of a day, those same people potentially might have bet a sum total of $1,500,000, due to this illusion of security, meaning the casing made ~$75,000.

[Septimus]

I remember in one of my college courses my professor tried to get across the point of memorylessness for coin flips and asked the class that if he flipped a coin 10 times and it came up heads every time which they would bet on for the next flip. I was the only person who said "heads". When he asked me why, I responded, "because it's probably a double-headed coin". He really appreciated that answer.

[Doc]

Good example with the casino. Also of course for people who are camping Argus the likelihood of it not dropping 20000 times in a row is very, very low. :>

[Demosthenes11]

meh, none of this matters when shit like this happens:
http://i2.photobucket.com/albums/y25...813_120450.png

That search before argus appears in the log was 15 minutes before it spawned, and I fell asleep (had been camping for 9 hours).

GG luck 0/4 LK, 0/1 in one week of camping

[divisortheory]

You lost me. >_> I don't understand why you add the 3 stats (½pop * ¼ drop rate = 1/8chance to get an amulet)

The probability to get the drop on your next kill will ALWAYS be 25%, no matter what you do (excluding TH obviously), because everything that happened before has no impact on the next event.

that's what we are conditioned to think though. If you look at it from the outside though, the probability that you kill 4 Argus' and don't get drop is pretty low on a 1/4 droprate. Probability is what is increasing, not droprate I guess is what im trying to say

No, that's the way probability works.

[riddaraan]

If you flip a perfectly balanced coin a dozen times and it lands on heads every time, what's the probibility that it will land on tails?

[Cadena]

If you flip a perfectly balanced coin a dozen times and it lands on heads every time, what's the probability that it will land on tails?

4095/4096

[Yutuyu]

If you flip a perfectly balanced coin a dozen times and it lands on heads every time, what's the probibility that it will land on tails?

And the only part that matters:

[...] you flip a perfectly balanced coin [...] what's the probibility that it will land on tails?

[levish]

Like said many times above, drop rates, coins, and dice have no memory. What you are calcuating is "What are the odds that I got to where I am now along the sequence of events I observed".

If Argus is truly coded to have a drop rate of 0.25, then EVERY kill will have a 25% chance, no more no less. Even if Argus has not dropped in is past 20,000 kills, the odds of a drop on the 20,001th kill is not 99.99999~, it is exactly 25%.

Still the likelyhood of walking away with a P. Charm in (for example) 4 kills is much higher than the single instance kill drop probability of 25%. it is not still 25% chance because you have more opportunities (and that has a statistical value given earlier).