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| Author | Flooders Tenthouse |
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| Alms gatherer. | https://youtu.be/wyydWaVyJB8
Hehehe | | is the move out button on MG quests free? too scared to click on it :P | for Pankaj_Kalra21:
0.1 diamond. | | You receive 69 gold, Moonstone | 2017-06-27 22:56:00: HailHydra [16] has discovered Servant of Darkness cuirass [100/100]
2017-06-27 22:54:35: HailHydra [16] has discovered Dwarf warrior boots [100/100]
That is some luck :P | for Pankaj_Kalra21:
I have seen that many time though. Many people probably open a lot of chests back to back... that person could have opened 50 or so chests by that time. If we assume getting a chest to be 1/100 chance, then the chance for getting one more chest in 50 attempts is about 40%. If we assume 1/1000 chance, it is about 5%. And it is typically believed that the chance for a rare loot is between that range.
So the chance is 5% to 40%... which is not that bad... you are likely to find a few of them if you go through the pages of rare art loots. | The probability of getting a certain item is largely different from other possible items. For instance, windflowers are far more common than a rare art.. Furthermore, i also believe that there is exists a descrepency between probabilities of rare arts ie. Unruly barb arts are pulled few and far between items like dwarf/amphibian etc..
Tbh, i dont know the odds of getting a rare art but i know the chances are very slim.. However, i can imagine the sheer number of diamonds that are donated to the game each day, maybe several thousands of dollars worth and with that comes an inconceivable number of chests.. The rare arts page has 30 entries per page.. So as we speak only 30 rare arts have been pulled in the period 15-28 july.. I hope u understand now that the frequency is very-very low..
Maybe you could've phrased your post a bit better as i cant comprehend a small and vital part of it.. But probabilty of pulling rare arts and clauses like 40% dont belong in the same sentence
Anyway, pulling 2 rare arts in the span of 2 minutes is really lucky at least by my standards | Anyway, pulling 2 rare arts in the span of 2 minutes is really lucky at least by my standards
Well who knows how many chests he opened simultaneously in 2 minutes ;) ? it takes 2 sec anyways to open 1 box :P | | This year they had also changed the frequency of drops to more often, don't remember the numbers, it is somewhere here in the forums if you care to find it. | for Pankaj_Kalra21:
Ah I think I didn't explain things clearly there.
I know well that the probability of looting a rare art is very low. So I assumed the range is between 1/100 and 1/1000. So between 1% and 0.1%
I also assumed in about 1-2 minutes, you can open 50 chests fairly easily with good internet and fast reflexes. So you have 50 attempts for the second one after you have already opened the first one.
Next, mathematically speaking, we therefore want the probability for opening a second chest, with 1/1000 to 1/100 probability in 50 attempts.
For 1/100, the probability will be 1-(99/100)^50 = 40%
And for 1/1000 it will be 1-(999/1000)^50 = 5% approx.
Hence the probability of getting a second one within 1-2 minutes, assuming you keep opening 50 chests back to back, is 5%-40%. And the rough lower bound of 5% isn't as incredibly low as you might intuitively think... getting a straight up win in roulette is twice as unlikely.
As for some minute evidence that this logic is correct, even from the current page of loots you can see 3 times that the same person got two rare arts back to back in 1-2 minutes.
I hope it makes more sense now ^^ | Probability has never been my strong point in mathematics.. I lean towards planar math, calculus etc as these ties in with my physics..
Let's just assume 0.1% for everyone's sake.. 5% is the probability of pulling 1 rare art in 50 attempts..
Assuming what windrider said and supposing an infinite number of chests were opened, the probability of pulling 2 in any arbitrary sample size of 50 must be much less than 5% (intuitively speaking).. I need to think how to exactly formulate that %
Or maybe ur just right and i just cant wrap my head around the 5% figure (cuz i realize how much bigger it is than the initial possibility of 0.1) | | 0.1/per box, i meant | | Obviously conditional probability has to be used.. Ill get to it when i dont feel as lazy as right now :p | for Pankaj_Kalra21:
Hmm yeah I think calculus is interesting too, but probability is one of my favourites ^^ Though to be honest I know very little on statistical theory.
Assuming what windrider said and supposing an infinite number of chests were opened, the probability of pulling 2 in any arbitrary sample size of 50 must be much less than 5% (intuitively speaking).. I need to think how to exactly formulate that %
That is very different. Pulling out at least TWO in a sample size of 50 is much less likely. To find that, you have to subtract from 1, the probability of either exactly 0 arts being pulled in 50 attempts, or exactly one(so you add the probabilities, and the subtract the sum from 1). And to find the probability of exactly k boxes being pulled, you will need to use P(k) = (50 choose k)(1/1000)(999/1000)^(50-k).
However as you correctly mentioned, conditional probability should be used, since we want the probability of 2 boxes being pulled, such that the second box is 50 attempts within the first box, NOT both of them being pulled in 50 attempts.
Therefore the condition in this case is that the first box has already been pulled. Which is a practical assumption, because we don't care(and we don't know) how many attempts it took the person to get the first box, could be 1000 attempts. Therefore the calculation boils down to the chance of a 1/1000 event occurring in 50 attempts(in which the result of a previous attempt does not affect the probability of the next one), hence, assuming 0.1% chance for 1 box, it is 1-(999/1000)^50 = 5% approx.
i just cant wrap my head around the 5% figure (cuz i realize how much bigger it is than the initial possibility of 0.1)
Yes probability is often counter intuitive, such as the well known Monty Hall problem ^^ In this case an intuitive way of thinking is, something has 1 in 1000 chance, and you have 50 tries, so it's about 50 in 1000 chance, which is 5%. Obviously mathematically this is incorrect, but since the number of attempts(50) is far less than the reciprocal of the chance(so 1000 in this case), it is a close approximation. Also, I'm too lazy for this, but using calculus you can take the limit of their ratio approaching zero, then the probabilities calculated mathematically, and "intuitively" should end up approaching the same value.
Sorry for blabbering on so much haha, that's when you do when you have got little to do in life lol... | | Laborers' guild: 7 (6666) +1334 :P | Therefore the calculation boils down to the chance of a 1/1000 event occurring in 50 attempts(in which the result of a previous attempt does not affect the probability of the next one), hence, assuming 0.1% chance for 1 box, it is 1-(999/1000)^50 = 5% approx
Exactly, glad we're on the same page now..but the odds of pulling the second time clearly DEPEND upon the previous successful outcome..that is what conditional probability is exactly for..
now, assuming 0.1% chance for 1 box, it is 1-(999/1000)^50 = 5% approx. is still the standalone probability of pulling one and nothing to do with the previous (successful), which we have no use for..
my arguments maybe correct but so far ive not provided any mathematics to prove them..tbh, ive forgotten several methods in this topic..im sure it can be formulated by conditional probability or even bernolli's principle..next time i talk about this i wanna talk numbers | but the odds of pulling the second time clearly DEPEND upon the previous successful outcome
No not at all, that's why I ignored it. There is no "dynamic balancing" in these loots. Whether you got a fern flower last chest, or a thief invitation, the chance to get a thief invitation on the next loot is the same.
I'm slightly confused, do you mean you agree with what I'm saying or not? :P
PS: What do you mean by Bernoulli's principle? Isn't that a physics thing? xD Maybe you mean Bayes' theorem or something? That has nothing to do with this though. | depend upon each other not directly..but the entire event of 2 arts can only take place when it's initiated..that is the basis of conditional probabilities..of course pulling one is independent in itself
https://en.wikipedia.org/wiki/Bernoulli_trial | And for 1/1000 it will be 1-(999/1000)^50 = 5%
that is the odds of occurring ONCE. what we're dealing with is a poisson distribution. The Poisson distribution describes the probability of an event happening k times during a "memoryless" process with an average rate of X. A memoryless process is one where events happen at a certain average rate but the probability of an event happening at any given time is independent of the past. |
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