Droprates...

[Demosthenes11]

Zam!ish question, but I did a bit of thinking today about droprates. For my example of my calculation, I will use Argus.

Argus has about a droprate of 1/4 total. This is true for all kills, each individual kill should be 1/4. But at the same time, for every 4 total argus pops killed by all people, it SHOULD drop once. This is also true for your own kills. For every 4 you personally kill, it should drop once.

So here is when calculations start. Say you have seen 2 argus pops in 3 days (the other spawn being leech king) and you have no idea about the previous spawns. In those 2 spawns, you have gotten claim once. Neither of these Argus' dropped. 2/4 of the Argus needed to be spawned have, and 1/4 of your own personal kills have occurred. But at the same time, the droprate is technically static at 1/4. Add these together (1/2 from total Argus spawns + 1/3 droprate from personal kills + 1/4 static droprate) yields 13/12, and average these out for a total of 13/36, or about 36%.

So in effect, I believe the next personal kill will drop 36% of the time. I doubt I explained that well, but I think it makes pretty good sense (because lets be honest, the chance you get drop increases the more times you kill). Also, with this model, the highest the probability drop could be is 1+1+1/4 / 3 or 75%, meaning it could never approach 100%.
I think there is a way this could also be calculated with crafting, though not sure exactly since my first few calculations made no sense.

Is this at all valid? or have I just gotten too little sleep?

[Kaylia]

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

[Demosthenes11]

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

[Kinto]

Bleh, 0/8. Wish I could believe it's a 25% drop rate. But I'd have a hard time believing that over time the probability of an item dropping increases. I would think every time that monster depops, it just goes back into it's original state.

[Demosthenes11]

Think of a coin. If you flip 4 heads in a row, more times than not the next flip will be tails (and a smart bet). The only difference is, this coin is a dice with 4 sides, with only 1 tails. Also, you may not be the one rolling the dice. As you roll more and more heads, your chances of rolling tails will increase purely do to probability.

[Kaisha]

It's an individual 1/4 chance for every kill, so killing it twice warrents a 2/8th chance at a drop overall, not 2/4.

[Kaylia]

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

Gonna keep it simple and give you an example of what would happen after 10 "no drop" days.

The chance of not getting any drop for 10 day in a row is calculated like this: ¾*¾*¾*¾*¾*¾*¾*¾*¾*¾= 59049/1048576 (~6%). However, that number won't effect the next drop in any way, the world isn't trying to preserve the balance in the universe and "increase your chance" to get it. Basically, the die that determine the drop on next day is the same, and there is still ¾chances it land on nothing, and ¼ it land on drop.

¾*¾*¾*¾*¾*¾*¾*¾*¾*¾*¼ = 59049/4194304 = ~1.4% to get that particular streak with a drop after 10 nothing
¾*¾*¾*¾*¾*¾*¾*¾*¾*¾*¾ = 177147/4194304 = ~4.2% to get a 11 days without drop streak

You can see the ratio is still 3 chance to get nothing for 1 (25% drop rate) chance to a drop.

[Weeks]

If you flip 4 heads in a row, more times than not the next flip will be tails (and a smart bet).

Is this a joke, or do you actually believe this?

It's actually slightly true, since coins will fall tails more than 50% of the time, but do you honestly believe that probability works this way?

[Demosthenes11]

If you flip 4 heads in a row, more times than not the next flip will be tails (and a smart bet).

Is this a joke, or do you actually believe this?

It's actually slightly true, since coins will fall tails more than 50% of the time, but do you honestly believe that probability works this way?

you just contradicted yourself by saying it is "somehow" more than a 50% chance. That "somehow" is what im trying to account for

[Kaylia]

Think of a coin. If you flip 4 heads in a row, more times than not the next flip will be tails (and a smart bet). The only difference is, this coin is a dice with 4 sides, with only 1 tails. Also, you may not be the one rolling the dice. As you roll more and more heads, your chances of rolling tails will increase purely do to probability.

That's not a correct way to think. Getting same result 4 times in a row is a rare (not really) occurance with 1/16 chances to happen. Getting same results 5 times has 1/32 chances to happens. However, it's the same for any series of heads and tails.

HHHH = ½*½*½*½ = 1/16
HHHT = ½*½*½*½ = 1/16
THTH = ½*½*½*½ = 1/16
TTTT= ½*½*½*½ = 1/16
...
In the end, there is 16 differents way to rolls a coin 4 times. If you add a 5th roll, you will get a total of 32 combinations, but since you are already rolled 4 (1/16), only the last 1/2 need to be rolled (1/16 to get a particular series *1/2 to get head or tails = 1/32).


Actually, if there is an random event in real life that give you biased result like this, you should always bet for the side that has the best success, since it might be fixed.

[Zosi]

Simple answer. The dice don't have memory. If you flip a coin 10 times and they're all heads, the next flip has the exact same chance of being heads as the previous 10, 50%. What you're confusing is individual events and sequences of events. Assuming coins are equally weighted (they aren't), each flip would have a 50/50 chance of landing heads/tails, and so it's easy to calculate the probability of getting 10 heads in a row, which is 0.50 ^ 10, which calculates out to about 0.1%. Here's a sequenced list of the chances, if it helps you understand.

1 Flip = 50% chance of heads.
2 Flips = 50% on first, 50% on second = 0.50 * 0.50 = 25% chance of both being heads.
3 Flips = 50% on each = 0.50 * 0.50 * 0.50 = 12.5% chance of all three being heads.
4 Flips = 50% on each = 0.50 * 0.50 * 0.50 * 0.50 = 6.25% chance of all four being heads.
5 Flips = 50% on each = 0.50 ^ 5 = 3.125% chance of all five being heads.
...
9 Flips = 50% on each = 0.50 ^ 9 = 0.1953125% chance of all nine being heads.
10 Flips = 50% on each = 0.50 ^ 10 = 0.09765625% chance of all ten being heads.

As you can see, the chance of each is simply half of the previous, because you're adding in another flip with an independent 50% probability of going either way. Each flip is an independent event, meaning that no other flips influence it, and it influences no other flips, so the results are the same whether all are flipped simultaneously or in sequence.

The same applies to your loot drop rates. If Argus has a 25% chance to drop, there's no cosmic rule saying he MUST drop exactly one charm in any 4 kill period, and his drop rate on any individual kill is exactly 25%, whether there hasn't been a drop for 30 kills previously or it had dropped on the last three kills.

Regarding the "slightly more than 50% chance" thing, that's because most coins don't have precisely equally distributed weight. Quarters (I think?) have something like a 52% chance to land tails because the designs on the sides result in an unequal weight distribution. I'd test it myself, but I kinda used all my quarters doing laundry earlier. In any case, it has nothing to do with drop rates, because it's just a slightly different heads/tales chance than you'd expect because coins aren't identical on both sides.

[Evelyn]

If you flip 4 heads in a row, more times than not the next flip will be tails (and a smart bet).

Is this a joke, or do you actually believe this?

It's actually slightly true, since coins will fall tails more than 50% of the time, but do you honestly believe that probability works this way?

you just contradicted yourself by saying it is "somehow" more than a 50% chance. That "somehow" is what im trying to account for

The "somehow" he's talking about has nothing to do with probability. Its got to do with imperfections in the coin, aerodynamics, how the coin is oriented when you flip, how you flip, etc.

[Shuvo]

If you flip 4 heads in a row, more times than not the next flip will be tails (and a smart bet).

Is this a joke, or do you actually believe this?

It's actually slightly true, since coins will fall tails more than 50% of the time, but do you honestly believe that probability works this way?

you just contradicted yourself by saying it is "somehow" more than a 50% chance. That "somehow" is what im trying to account for

Not at all. One side is slightly heavier then the other, hence the irregularity.

[Francisco II]

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

Kill #1  Kill#2   Kill#3  Kill#4
 0         0         0        0
 1         1         1        1
 2         2         2        2
 3         3         3        3

There are 256 possible outcomes here, where 0 = Drop and 1,2,3 = No Drop

That's just waaaay too much typing, and I forget the math to figure it out at the moment, so a simpler example.

Assuming it's a 1/2 drop rate...

Kill #1  Kill#2   Kill#3  Kill#4
 0         0         0        0
 1         1         1        1

0 = drop
1 = no drop

Possible Outcomes:

0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

After four kills, there's a 32/64 chance you get the drop, which is 50%... which doesn't make any sense logically.

Which means I don't know what I'm talking about, and someone else will have to explain it, because I did it wrong.

[Evelyn]

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)

[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

[Zosi]

No, you're just interpreting the results wrong.

Four kills, 50% drop rate means that there's a 0.50 ^ 4 chance (6.25%) of none at all dropping. In other words, there's a 93.75% chance of at least one drop over 4 kills. 6.25% is also the chance of any SINGLE combination in your list happening, a fact we'll use for the rest. An easy-to-prove side note is that you have the exact same chance (6.25%) of getting 4 drops in a row as getting zero drops in 4 kills, which is only true for a 50% drop rate.

To figure out the chance of exactly one drop happening in four kills, you're calculating the odds of getting one of the following combinations:

1000
0100
0010
0001

Which can be calculated as the sum of the chances of each happening individually, which is 4 * 0.0625 = 25% chance of exactly ONE drop.

Our two-drop situations are the following combinations:

0011
0101
0110
1001
1010
1100

Calculated the same way, 6 combinations at 6.25% each = 37.5% chance of getting exactly TWO drops.

Our three-drop situations are thankfully shorter:

0111
1011
1110
1101

Calculated the same as above, 4 combinations at 6.25% each = 25% chance of exactly THREE drops.

Summarized:
0 drops: 6.25% chance
1 drop: 25% chance
2 drops: 37.5% chance
3 drops: 25% chance
4 drops: 6.25% chance

Which adds up to 100%, meaning that you've accounted for all possibilities.

Double edit because I'm both bored and unable to sleep. Coefficients are derived from the counting method above, the actual potential results are the same no matter what the drop rate on individual kills is.

25% drop rate on ONE kill = 25% drop, 75% no drop.

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

25% drop rate on THREE kills =
0 drops: 1 * (0.75 ^ 3) = 42.1875% chance
1 drop: 3 * (0.25 * 0.75 * 0.75) = 42.1875% chance
2 drops: 3 * (0.25 * 0.25 * 0.75) = 14.0625% chance
3 drops: 1 * (0.25 ^ 3) = 1.5625% chance
At least one drop: 57.8125% chance

25% drop rate on FOUR kills =
0 drops: 1 * (0.75 ^ 4) = 31.640625% chance
1 drop: 4 * (0.25 * 0.75 * 0.75 * 0.75) = 42.1875% chance
2 drops: 6 * (0.25 * 0.25 * 0.75 * 0.75) = 21.09375% chance
3 drops: 4 * (0.25 * 0.25 * 0.25 * 0.75) = 4.6875% chance
4 drops: 1 * (0.25 ^ 4) = 0.390625% chance
At least one drop: 68.359375% chance

[Kaylia]

Which means I don't know what I'm talking about, and someone else will have to explain it, because I did it wrong.

Not every outcome have the same chance to happen .
1111 = 1/256 ( ¼*¼*¼*¼)
0000 = 81/256 (¾*¾*¾*¾)


The total chance to get a drop is the addition of the probability of every different outcome that have a drop

Basically, the probability of 1111+ 1110+ 1101+ 1011 +0111+ 1100+ 1001+0011 + 0101 + 1010 + 0110 +0001+0010+0100+1000.
But in this case, it's faster to simply substract from the total the probability you get nothing (only 1 case)
100% - probability(0000) = 256- 81 = 175/256 = 68% chance to get a drop

[Priran]

http://img150.imageshack.us/img150/7554/mathqr6.gif

[NynJa]

Think of a coin. If you flip 4 heads in a row, more times than not the next flip will be tails (and a smart bet). The only difference is, this coin is a dice with 4 sides, with only 1 tails. Also, you may not be the one rolling the dice. As you roll more and more heads, your chances of rolling tails will increase purely do to probability.

No its not...
the effects of the past have no impact on the effects of the future.

If I flip heads 4 times in a row, the 5th time is still 50% heads, 50% tails. Probability does not go up in favor of tails because of the 4 heads I flipped before. It is just made to seem that way as the probability of landing 5 heads in a row is, I believe, 3.125%. That 3.125% stat means nothing on the 5th flip though: 50% heads and 50% tails, thats all that matters.