TDS Ultra rares in herocis is a myth (at least in this thulia HA)

Discussion in 'The Veterans' Lounge' started by rune00, Jul 15, 2015.

  1. Brohg Augur

    What part of grinding in Brother's Island for purity augs sounds like "The Thuliasaur HA" to you?
  2. Bigstomp Augur

    I've seen drops from Thalisaur HA - not myself but guildies.
    I've also seen random drops in as simple a place as katta (yay me!).

    They are just very very rare, not getting another is not suprising, and I've probably easily killed 50k tds mobs.
  3. Gatash Lorekeeper


    Not doubting the problem of a gambler's trap, but it is your understanding of probability that is wrong. I'll do the exercise in maths not to lecture, but to illustrate how 50K kills with no drops is extremely unlikely.

    We'll start with a 50% drop rate since it's easiest to demonstrate how the formula works. We'll also approach it from the angle of being wrong, which is analogous to killing a mob with a 50% drop rate and the item not dropping. For every kill, the probability of being wrong is 0.5, or 50%.

    The probability of killing one mob and not getting a drop = 0.5 = 50%
    The probability of killing two mobs and not getting a drop = 0.5 x 0.5 = 0.25 = 25%
    The probability of killing three mobs and not getting a drop = 0.5 x 0.5 x 0.5 = 0.125 = 12.5%

    And so on. The probability of each individual mob dropping your item doesn't change, but the overall probability that a person could experience the exact chain of events, as a whole picture, does change. This is logical; how likely do you think it is that you can flip a fair coin even 10 times and get heads each time? Try it. On average it would take you 1024 goes.

    If you wanted to know the probability of the outcome from any number of mob kills, you can get the answer with:

    Probability = (probability of failure)^(number of kills). So for 3 kills = 0.5^3 = 0.125, as above.

    Now for the actual drop rate. The probability of failure is 4999/5000 = 0.9998. 99.98% chance that each kill will NOT give my drop.

    The probability that I could kill 5k mobs and not get a drop = 0.9998^5000 = 0.367. So I have a 63.3% chance of getting the drop.

    The probability that I could kill 50k mobs and not get a drop = 0.9998^50000 = 0.000045

    That means that if 10,000 players each killed 50,000 mobs, not a single person (only 45% of a person) would be expected to not get a drop.
    Xianzu_Monk_Tunare, rune00 and Iila like this.
  4. porky Augur

    Just adding this in for statistics. I have personally seen 3 ultra-rares drop in TDS. One in Katta, one in brother's island, and one in Combine dredge. All three were in the static zones, not instanced.
  5. Endless Augur

    Please redo this with the actual drop rate of 1 in 5000. Doesn't really make sense, but maybe I am doing the math wrong, was never really good in math class and it's been a while, but your numbers make little sense. Cause with the actual numbers, you're saying by killing two mobs, there is a .000004% chance of not getting a drop. But maybe I am looking at it wrong, please correct me. The rest didn't seem right either. I added the in bold stuff.
  6. Gatash Lorekeeper


    Please take the time to read more than the first few lines of the post, where I already explicitly do exactly what you are requesting.

    EDIT: Ok, I understand now. You've done it the other way around to my example. It's far easier to follow the maths using the chance of failure to carry this forward. The result is still that with a 1/5000 chance of looting success, the chance of getting one drop in 50,000 kills is still 99.9955%
    Xianzu_Monk_Tunare likes this.
  7. Endless Augur

    Actually I plugged in the % for 1 in 5000.

    I'm not asking you to redo it using a made up 50% chance, I am asking you to redo it using the actual chance of 1 in 5000.
  8. Slasher Augur

    Don't bother dbcooper will never admit his gamblers anon class was wrong :) He fails to comprehend simple math.

    It's pretty simple for most of us to understand except him. We've tried to explain to him in simple terms. With each kill you have the same drop rate 1/5000 but when you take 50,000 kills as a whole the probability that X did not happen is dam near 0%.
  9. Gatash Lorekeeper

    It's a terribly sad evening for me that I have spent some time trying to boil this down into another approach; I've put more effort into this than a Stack Overflow answer for "machine learning"! There is no intelligible way I can find to describe the problem other than the way I did it for my original post, except for a very simple revision: the probability of both outcomes (a binomial distribution) must add to 1. You calculate the probability of the most likely event occurring over multiple instances (either success or failure), and subtract from 1 if that event is the opposite of what you wanted. What you attempted to do is opening a whole Pandora's box in terms of maths. Computers don't do well in calculating a factorial of numbers in the 10's of thousands.

    I'll provide anecdotal evidence for this rather than pure maths. The probability of rolling a 1 on a dice is 1/6. If I roll the dice 6 times, logic tells us that I should "probably" get a 1 in any one of those rolls. However, to approach it your way:

    Prob. of getting 1 in 6 rolls: (1/6)x (1/6) x(1/6) x(1/6) x(1/6) x(1/6) = 0.0000214

    So now it's saying that I have 0.00214% chance of getting a 1 in any of the 6 rolls. Nonsense.

    Let's look at the probability of not getting a 1:

    Prob. of NOT getting 1 in 6 rolls: (5/6)x (5/6) x(5/6) x(5/6) x(5/6) x(5/6) = 0.334

    Since the only other outcome is getting a 1, the probability of the above is 1-0.334 = 0.666 = 66.6%. This makes more sense. Originally the probability was 0.167 for a single roll, but over 6 rolls I have a 2/3 chance to get a 1.

    Now look back at my original answer. I suggest that for a 1/5000 probability of a drop, there is still only a 63.3% chance of getting a drop in 5000 kills. Does this seem unreasonable enough for you to question the outcome? I did it by the second approach presented above

    I don't know anyone who works in pure statistics that would approach this problem your way, because it just doesn't work. But now I have a fun time on my plate to understand exactly why it doesn't work in a purely mathematical sense. Thanks :p I stand 100% by my original answer, it is correct.
    Xianzu_Monk_Tunare likes this.
  10. Devildawg Elder

    *sigh* No. Just no.
  11. Brohg Augur

    Yes, just yes. If that's unclear maybe ask some kind of question, Devildawg? That's correct statistical analysis
  12. Belchere Lorekeeper

    To Gatash,

    You stated:

    Prob. of getting 1 in 6 rolls: (1/6)x (1/6) x(1/6) x(1/6) x(1/6) x(1/6) = 0.0000214

    Actually, that is the probability of getting a 1 in ALL 6 rolls, and its a correct value.

    You were trying to express the probability of getting a 1 in six rolls of a dice, that is the sum of the chance of getting a 1 multiplied by the number of rolls - so 1/6 * 6 or rather surprisingly 1 (100%).

    That's the answer, because if I told you to write down all the possibilities, you'd list 1, 2, 3, 4, 5 and 6, and none are more likely than the others, so in 6 rolls, the expected value would be 1 of every possible number. It's not likely to turn out that way all the time, but over very large numbers of rolls, it will indeed.

    Finally, what you were trying to do is come up with some sort of number that indicates how many times you would roll a dice to get your first one. That is where the backwards calculations come in - the probability of not rolling a 1 in the first roll is 5/6. The probability of not rolling a 1 in two rolls is (5/6)^2. The general formula for not rolling a 1 in N rolls is (5/6)^n, and the probability of a 1 coming up is then 1-(5/6)^n. If you put 6 in for N, you get 1-(5/6)^6 or 1 - 0.335 or 66.5%
  13. Belchere Lorekeeper

    Oh, and yes, then the chance of not getting a 1 in 5000 drop in 5000 tries would be:

    (4999/5000)^5000 or about 36.8%

    In 25000 tries, it becomes only 0.67% or about half a percent
  14. Iila Augur

    63.3% chance of getting a drop after 5,000 kills means a 36.7% chance of not getting a drop after 5000 kills.

    36.8% chance of not getting a drop after 5000 kills means a 63.2% chance of getting a drop in 5000 kills.

    Is this honestly something that's being argued? Or it just being repeated?
    (I'm not talking about the rounding.)
    Gatash likes this.
  15. DBCooper Elder

    No, I'm just not a completely math illiterate.

    Try actually READING and COMPREHENDING. I quite clearly said that if you killed 25k mobs and didn't get one, yes, that was bad luck, but it had no bearing on your chance of getting one in the next 25k kills. Those large numbers were clearly to make a point, and the point was that - contrary to the poster I quoted and therefore was contextually responding to - every kill was an independent chance.

    I never once made ANY statistical interpretation of the validity of the null hypothesis that the number of kills being talked about without a drop was reasonable - to whatever degree you find significant - as being part of the same population of all kills where the mean frequency was .0002. Heck, I never even offered an OPINION on it.

    The simple fact is that the person I quoted was clearly stating the gambler's fallacy (he flat out said every kill you didn't get one increased the chance you would get one later), which is wrong.

    But you want to talk about comprehending simple math, then maybe you should go take some basic statistics class, because doing a mean frequency calculation based off the number of kills and then eyeballing it to claim "the chance is dam* near zero", or the other guy you were replying to that did something similar, has absolutely ZERO relevance or factual support for your statement. You might as well go "big number is bigger than small number, so can't be right", because that's about all you did.

    Here's a hint: 0.0002 is the mean frequency for the putative population. Assume a normal distribution. So if it really is 1 in 5000 (incidence of one 1 and 4999 zeroes) you can get a standard deviation (population so N not N-1) and then run a statistical test of a sample of 25k or 50k with 0 drops.

    Now, I wouldn't be at all surprised that statistically something is off, particularly if someone is running the same mission every time, but no one here in this thread has proven it.

    Oh, and the entire presumption you make in what I quote above clearly shows that YOU do not understand probability and statistics. You make the same mistake that someone that wins the lottery does when they go "its a miracle!", or when someone says something like, "it can't be chance that we have life on this planet when the chance of it happening is so small", or the intelligent design wackos that talk about the probability of an eye evolving "by chance" (putting aside the factual issue, just talking the numbers). And if you don't understand what I mean, then go look up the "prosecutor's fallacy" which describes a major error in the common understanding of the probability of a DNA sample matching someone.
  16. Matari Augur

    People need get there head out of ground and look at the big picture. The OP is 100% correct- he should have gotten an item. If rares were intended to be this rare then they some explaining to do.
  17. Slasher Augur

    No where in this thread did the OP say the 1/5000 ratio goes up with each kill. Anyone understands each kill in itself is 1/5000. It is entirely possible someone can kill 5000 mobs and never see the item it happens frequently. However anyone can see what the OP was talking about here you just assume he meant the odds go up with each kill when he never said that.

    If I /ran 1 5000 50k times is it possible I will not hit every number ? Yes Is it likely ? no.

    Now onto the topic. A dev response would be nice, but I am not holding my breath.
    Xianzu_Monk_Tunare and Gatash like this.
  18. Gatash Lorekeeper


    This is correct. The problem was that Endless was not satisfied with my initial analysis and was attempting to do his/her own analysis that involved getting the drop on every single kill. My original post was correct, but I sought another route to illustrate the same effect. Sadly it was a thankless and painful journey to arrive at the answer that it could not be done any other way, as the dissenters' ranks grow.
  19. Xianzu_Monk_Tunare Augur

    No, that is the gambler's fallacy.

    Incorrect again, there are 6^6 = 46,656 possibilities; because you have to look at the whole set, not just each individual roll; and unless you are making order important then some outcomes are more likely than others.

    Since 46,656 posibilities are just way too many to write out, I'm going to instead use a magic coin which has an equal chance of landing on Heads, Tails or Side; which we flip 3 times; this means there are 3^3 = 27 possibilities:

    HHH, THH, SHH
    HHT, THT, SHT
    HHS, THS, SHS
    HTH, TTH, STH
    HTT, TTT, STT
    HTS, TTS, STS
    HSH, TSH, SSH
    HST, TST, SST
    HSS, TSS, SSS

    Now let's look here and see what the probability of it never landing on its side would be; (2/3)^3 = 8/27 which matches what we find from the list above. So the probability of the magic coin landing on its side at least once (which includes landing on its side twice or all three times) would be 1 - (2/3)^3 = 19/27; this is the probability regardless of when the side occurs.
    Gatash likes this.
  20. Goth Augur

    Hey i just finally got a CotF ultra rare... Though i dont think these have much value on FV...