Fractals End Chest RNG?
Also, for reference, Wiki says there are 34 different ascended rings that can drop from FOTM end chests.
If there are 34 different ascended rings that can drop, and you can get either an infused or uninfused version, that’s really 78 different rings. The problem with doing these stats is I don’t know what tier I earned each of the rings on.
But suppose I earned them all on tier 50 and suppose the odds of getting an uninfused version is 0, then the odds of getting 2 identical rings in a row would be less than (1/34)*(1/34). This means the odds of getting 2 identical rings in a row would be less than .0865% if the drops were truly RNG.
I think they’re not.
Would love for John Smith/someone from the dev team/someone with better stats knowledge to comment.
Just RNG and humane reaction to try to find a pattern in everything.
Just RNG and humane reaction to try to find a pattern in everything.
You know, that excuse carries only so far. Human capacity for pattern recognition isn’t exactly a flaw on the whole, and some of us are extremely good at it. Even the worst of us is far, far better able to look at a picture of a child holding a hammer and standing over a broken vase and, with no further information, be very likely capable of figuring out what happened with accuracy than anything of A.I’s and algorithms are yet.
I’m somewhat tired of seeing human cognition offhandedly dismissed irrespective of evidence in accompaniment simply because some people’s faith in algorithms borders on deranged.
RNG doesn’t exist. There is code performing functions that simulate randomness, and sometimes, they are in fact flawed. Sometimes, they do wind up being essentially broken.
And they’re significantly easier to check for flaws than the human mind is.
So let’s not dismiss a thoroughly valid question out of hand. RNG is not always perfect. And sometimes, we much maligned pattern-seekers are very good at spotting the patterns that indicate something that’s supposed to be random being not so random.
Just RNG and humane reaction to try to find a pattern in everything.
That’s a lot of events with .0865% chance happening in a span of less than 500 full runs. Suppose we round that percent to .1%. That means every 1,000 runs we should see about 1 event. Yet we have many events over the course of far fewer runs. You’re in [KING]. For an odds comparison, each double event is about the same as getting a fractal tonic. Do you know anyone who has so many fractal tonics in so few runs?
Doubt it. Again, my stats may be shoddy, so feel free to correct.
An interesting observation, I will try to pay more attention to it. I do remember getting Raider’s chest and then 2 Red Deaths from 3 tiers of fractals in one day.
An interesting observation, I will try to pay more attention to it. I do remember getting Raider’s chest and then 2 Red Deaths from 3 tiers of fractals in one day.
Yeah. As you can see it doesn’t happen always, but it happens statistically more than it should. I wonder if there’s something weird with patches/daily resets? Really don’t know WHAT causes it but as you can see in my last 4 runs I’ve had 2 pairs. Again, if it was pure RNG according to wiki the odds would be about 1 in 1000 to get 2 of the same in a row. The last pair I know was done on tier 50 and was done the day before the last LS episode and the day after the patch.
“As you can see it doesn’t happen always, but it happens statistically more than it should”
Have you run an appropriate statistical test to determine this?
If so, please name the test, and provide your results along with any other pertinent information.
If not, please do not make this claim.
“As you can see it doesn’t happen always, but it happens statistically more than it should”
Have you run an appropriate statistical test to determine this?
If so, please name the test, and provide your results along with any other pertinent information.
If not, please do not make this claim.
Ok.
Again, not a stats person so tell me if this is wrong. Just ran a binomial stats test. Link here:
http://stattrek.com/online-calculator/binomial.aspx
A binomial experiment has the following characteristics:
-The experiment involves repeated trials.
-Each trial has only two possible outcomes – a success or a failure.
-The probability that a particular outcome will occur on any given trial is constant.
-All of the trials in the experiment are independent.
I defined a “success” as getting the same ring as before. There were 84 rings, so 84 trials. I assumed RNG as we’ve been told, and I assumed independence, IE one day is not linked to another day.
Again, as noted above, wiki says there are 34 fractal rings that can drop. So the probability of success for a single trial is .000865. There were 84 trials. Number of successes is 15, but I counted the triple ring as a single success so 14.
The calculator told me the probability of getting less than 14 doubles assuming a sample size of 84 and the odds listed is 0.999999999999999. That’s a 99.9999999999999 (13 9s) percent chance that I’d have fewer than 14 doubles.
(edited by Nevets Crimsonwing.5271)
It’s not just you OP. I know I get Ossa Family Signet Ring three fold any others. In my guild we know with shocking accuracy if someone is going to get something good who it’ll be. Hint* it’s the same person every time. Well after over two years it’s not really shocking anymore just expected.
It’s not just you OP. I know I get Ossa Family Signet Ring three fold any others. In my guild we know with shocking accuracy if someone is going to get something good who it’ll be. Hint* it’s the same person every time. Well after over two years it’s not really shocking anymore just expected.
Haha same here. I casually mentioned in a run, “I wonder if I’ll ever get a fractal weapon.” Lo and behold, that special someone in my group gets the fractal axe that very run…
For the OP, I’ve never gotten the same ring back to back. I did get two ascended armor chests a few days apart, and none since.
It’s not just you OP. I know I get Ossa Family Signet Ring three fold any others. In my guild we know with shocking accuracy if someone is going to get something good who it’ll be. Hint* it’s the same person every time. Well after over two years it’s not really shocking anymore just expected.
I wish someone with a better stats background or an ANET employee would comment. But if I did the stats right (no guarantees) there’s a 99.9999999999999% chance I’d have less than 14 doubles, implying that something is sort of weird that makes drops “couple” to one another. I’ve also gotten multiple of the SAME chest – like same stats AND same piece. I’m not going to bother trying to combine the data though. Just wanted to point this out. If my stats are correct, then it should take roughly 10 trillion players before we find one with the same kind of doubles that I experienced. Again, this is assuming that I did stats right and drops really are RNG which I don’t think they are.
You can’t do what you did. For one, you absolutely must have independent trials, and defining success based on a previous trial means that this condition does not apply.
Humans are very good at spotting patterns, even where none exist. If you give a human a string which is “0123456789”, they’ll see a pattern and probably think it’s not random. If you give them the string “2579825731” you can probably convince them it was randomly generated. However, the probability of randomly generating either string is exactly the same, and is the same as “1111111111”.
You can’t do what you did. For one, you absolutely must have independent trials, and defining success based on a previous trial means that this condition does not apply.
I don’t understand. The trials (each fractal run that yielded a ring) were independent. A success occurred when the ring was the same as the previous (independent) trial.
Anyway, you seem to have some knowledge of stats, can you help me out here? If what I did isn’t appropriate, what is?
You can’t do what you did. For one, you absolutely must have independent trials, and defining success based on a previous trial means that this condition does not apply.
Humans are very good at spotting patterns, even where none exist. If you give a human a string which is “0123456789”, they’ll see a pattern and probably think it’s not random. If you give them the string “2579825731” you can probably convince them it was randomly generated. However, the probability of randomly generating either string is exactly the same, and is the same as “1111111111”.
I guess you missed the bolded part
Ok.
Again, not a stats person so tell me if this is wrong. Just ran a binomial stats test. Link here:
http://stattrek.com/online-calculator/binomial.aspx
A binomial experiment has the following characteristics:
-The experiment involves repeated trials.
-Each trial has only two possible outcomes – a success or a failure.
-The probability that a particular outcome will occur on any given trial is constant.
-All of the trials in the experiment are independent.I defined a “success” as getting the same ring as before. There were 84 rings, so 84 trials. I assumed RNG as we’ve been told, and I assumed independence, IE one day is not linked to another day.
Again, as noted above, wiki says there are 34 fractal rings that can drop. So the probability of success for a single trial is .000865. There were 84 trials. Number of successes is 15, but I counted the triple ring as a single success so 14.
The calculator told me the probability of getting less than 14 doubles assuming a sample size of 84 and the odds listed is 0.999999999999999. That’s a 99.9999999999999 (13 9s) percent chance that I’d have fewer than 14 doubles.
You can’t do what you did. For one, you absolutely must have independent trials, and defining success based on a previous trial means that this condition does not apply.
Humans are very good at spotting patterns, even where none exist. If you give a human a string which is “0123456789”, they’ll see a pattern and probably think it’s not random. If you give them the string “2579825731” you can probably convince them it was randomly generated. However, the probability of randomly generating either string is exactly the same, and is the same as “1111111111”.
I guess you missed the bolded part
Ok.
Again, not a stats person so tell me if this is wrong. Just ran a binomial stats test. Link here:
http://stattrek.com/online-calculator/binomial.aspx
A binomial experiment has the following characteristics:
-The experiment involves repeated trials.
-Each trial has only two possible outcomes – a success or a failure.
-The probability that a particular outcome will occur on any given trial is constant.
-All of the trials in the experiment are independent.I defined a “success” as getting the same ring as before. There were 84 rings, so 84 trials. I assumed RNG as we’ve been told, and I assumed independence, IE one day is not linked to another day.
Again, as noted above, wiki says there are 34 fractal rings that can drop. So the probability of success for a single trial is .000865. There were 84 trials. Number of successes is 15, but I counted the triple ring as a single success so 14.
The calculator told me the probability of getting less than 14 doubles assuming a sample size of 84 and the odds listed is 0.999999999999999. That’s a 99.9999999999999 (13 9s) percent chance that I’d have fewer than 14 doubles.
Essence, he’s saying the way I defined “success” broke the assumptions necessary for that test to be valid. I’m not sure he’s correct. But if he is, surely he can help out by showing me the proper way to test the odds here.
I think I might know what to do. The “event” has to be a pair of ring yielding runs. So two ring yielding runs makes 1 “event” and a “success” occurs when the rings are the same. That way the trials are independent. I’ll recalculate in an hour or so after my class.
Ok Zui and everyone:
I defined an event as the results of 2 ring yielding fractal runs. I had 84 rings, so there were 42 “events.” The outcome of each event can either be a yield of 2 of the same ring or 2 of a different ring. The odds of getting the same ring for a given event are still .000865 or (1/34)^2. I had to go through and “pair” each of the rings with its subsequent ring. Thus if one of my yellow “doubles” wasn’t in a “paired” group, I didn’t count it as a success. For example, in the image posted, the first row of rings had 1 success because the pairings matched up with the couplings, but rows 4 and 5 had 0 successes. The total number of successes was 8. If you need a visual for this, please let me know.
The odds of receiving fewer than 8 doubles in a sample size of 42 pairs, assuming RNG is:
1. Yes, that’s right, it’s even higher than my previous calculation. The odds of this happening are so high the calculator didn’t bother with decimal places.
for me i have 0.001 % chance to get a useful ring … so …
can i trade my 14 Lost Seal of Usoku and 8 Healing Signet for one zerker ring … seems fair to me ?
(edited by Farming Flats.5370)
There are 26 different Rings that can drop from the Fractalchest(in your bank are 27 different ones, because the Circle of Light is discontinued and no longer obtainable). Namely any that you can also buy from the Vendor.
Secondly the chance to get doublerings is, under evenly distribution of the rings, 1/26. Because whatever ring drops first, only the second ring decides if its a doublering or not(and the chance for getting a specific ring is 1/26). Chance of a Tripplering is 1/26² or 1/676
You had done 84 runs.
1 Tripplering (~1/8 would be expected)
12 Doublerings (~3,23 would be expected)
So yes, your amount of doublerings is a bit high, but nothing too unusual.
(edited by Artemi.1437)
There are 26 different Rings that can drop from the Fractalchest(in your bank are 27 different ones, because the Circle of Light is discontinued and no longer obtainable). Namely any that you can also buy from the Vendor.
Secondly the chance to get doublerings is, under evenly distribution of the rings, 1/26. Because whatever ring drops first, only the second ring decides if its a doublering or not(and the chance for getting a specific ring is 1/26). Chance of a Tripplering is 1/26² or 1/676You had done 84 runs.
1 Tripplering (~1/8 would be expected)
12 Doublerings (~3,23 would be expected)So yes, your amount of doublerings is a bit high, but nothing too unusual.
What’s the odds of getting two heads in a row in two flips of a coin?
It’s 1/4, not 1/2.
Wait. You’re right, because in two flips of a coin you can also get two tails.
Ok, redid math.
99.9825172916403% chance that I’d get fewer than 8 doubles out of 42 trials. Definitely a lot more plausible that I just got unlucky, but with others seeing doubles too, with yellows dropping in pairs from world bosses etc. etc., I still think their RNG does a poor job giving a normal range of outputs on the INDIVIDUAL level.
Perhaps on the scale of the whole game their RNG system outputs a normal distribution of drops, but individuals tend to see duplicates. On the other thread people were talking about bad seeds and other things like this, which is quite beyond my level of knowledge, but there it is anyway.
Perhaps on the scale of the whole game their RNG system outputs a normal distribution of drops, but individuals tend to see duplicates. On the other thread people were talking about bad seeds and other things like this, which is quite beyond my level of knowledge, but there it is anyway.
Don’t worry, most of the people talking about “bad seeds” etc have no idea either.
Mind you, after observing a lot of fractal drop data, and doing calculations of my own, i’m currently of the mind that the drop chance is weighted, and some stat sets (zerker being most visible) do have lowered chance of appearing. Which actually increases chances of getting doubles (from those sets with above average drop chances).
Remember, remember, 15th of November
99.9825172916403% chance that I’d get fewer than 8 doubles out of 42 trials.
You did 83 trials, because you did 84 runs.See the following sequence:
1-2-3-4-5-6-7
How much chances for doubles are in this?
6!
1-2, 2-3, 3-4, kitten -6, 6-7
You can’t just half the number of attempts because doubles obviuosly take up two attempts to reach^^
I have no idea what on earth makes the number combination a swear word. Can someone enlighten me of there harmfull nature?
99.9825172916403% chance that I’d get fewer than 8 doubles out of 42 trials.
You did 83 trials, because you did 84 runs.See the following sequence:
1-2-3-4-5-6-7How much chances for doubles are in this?
6!
1-2, 2-3, 3-4, kitten -6, 6-7
You can’t just half the number of attempts because doubles obviuosly take up two attempts to reach^^I have no idea what on earth makes the number combination a swear word. Can someone enlighten me of there harmfull nature?
Why can’t we model 2 drops as a “trial” where “successes” means 2 of the same ring, and “failure” means 2 different rings?
We’d have 42 trials, we know the odds of a success is 1/26, odds of failure is 25/26, and can calculate the odds using the above linked binomial probability calculator to see the odds of getting 8 or more successes out of 42 trials.
Also, 4 can be construed as an A, 5 can be construed as an S, so kitten could be a donkey.
The real thing that needs to happen is for ArenaNet to revisit the rewards in dungeons and Fractals. If they aren’t going to add new dungeons or Fractals, the least they can do is make the rewards up to par with Silverwastes. I’ll be specific as well because somehow ArenaNet finds a way to muddle up what the player base really wants.
- Fractal Weapon Boxes. You get one of these as a drop and you *get to pick the Fractal Weapon you want.
- Fractal Armor Sets. Ascended grade Fractal Armor sets that we can earn a currency in Fractals and then use it to buy the armor. It also has a small chance to drop a piece of armor(box) in the Fractal itself.
- Same for dungeons. Allow us to play for ascended armor now. You ahve your new gating. It’s called Masteries
I’m somewhat tired of seeing human cognition offhandedly dismissed irrespective of evidence in accompaniment simply because some people’s faith in algorithms borders on deranged.
I’d put that in my signature if it wasn’t too long. Beautifully well written post. Thank you.
Why can’t we model 2 drops as a “trial” where “successes” means 2 of the same ring, and “failure” means 2 different rings?
Thats what we do, but you are forgetting nearly half the trials you have with this method by only counting to 42.
To ellaborate: The first ring is the start of our first trial. The second ring is the second ring of our first trial but also the first ring of our second trial! The third ring that drops is the second ring to our second trial and the first for our third. And so on.
Because if you get a double ring you don’t dismiss it with the notion: Oh its the second and third ring, those don’t count.
Why can’t we model 2 drops as a “trial” where “successes” means 2 of the same ring, and “failure” means 2 different rings?
Thats what we do, but you are forgetting nearly half the trials you have with this method by only counting to 42.
To ellaborate: The first ring is the start of our first trial. The second ring is the second ring of our first trial but also the first ring of our second trial! The third ring that drops is the second ring to our second trial and the first for our third. And so on.Because if you get a double ring you don’t dismiss it with the notion: Oh its the second and third ring, those don’t count.
But if you do the counting that way don’t you violate the independence assumption for the trials?
Why can’t we model 2 drops as a “trial” where “successes” means 2 of the same ring, and “failure” means 2 different rings?
Thats what we do, but you are forgetting nearly half the trials you have with this method by only counting to 42.
To ellaborate: The first ring is the start of our first trial. The second ring is the second ring of our first trial but also the first ring of our second trial! The third ring that drops is the second ring to our second trial and the first for our third. And so on.Because if you get a double ring you don’t dismiss it with the notion: Oh its the second and third ring, those don’t count.
But if you do the counting that way don’t you violate the independence assumption for the trials?
If you do the counting your way, you lose 6 of your doubles that you highlighted in your OP.
Ok Zui and everyone:
I defined an event as the results of 2 ring yielding fractal runs. I had 84 rings, so there were 42 “events.” The outcome of each event can either be a yield of 2 of the same ring or 2 of a different ring. The odds of getting the same ring for a given event are still .000865 or (1/34)^2. I had to go through and “pair” each of the rings with its subsequent ring. Thus if one of my yellow “doubles” wasn’t in a “paired” group, I didn’t count it as a success. For example, in the image posted, the first row of rings had 1 success because the pairings matched up with the couplings, but rows 4 and 5 had 0 successes. The total number of successes was 8. If you need a visual for this, please let me know.
The odds of receiving fewer than 8 doubles in a sample size of 42 pairs, assuming RNG is:
1. Yes, that’s right, it’s even higher than my previous calculation. The odds of this happening are so high the calculator didn’t bother with decimal places.
I have a little problem with the bolded part because it doesnt matter, which ring you get in the first place. Only the second ring should matter and that is 1/34 and not (1/34)^2.
Bloin – Running around, tagging Keeps, getting whack on Scoobie Snacks.
Why can’t we model 2 drops as a “trial” where “successes” means 2 of the same ring, and “failure” means 2 different rings?
Thats what we do, but you are forgetting nearly half the trials you have with this method by only counting to 42.
To ellaborate: The first ring is the start of our first trial. The second ring is the second ring of our first trial but also the first ring of our second trial! The third ring that drops is the second ring to our second trial and the first for our third. And so on.Because if you get a double ring you don’t dismiss it with the notion: Oh its the second and third ring, those don’t count.
But if you do the counting that way don’t you violate the independence assumption for the trials?
If you do the counting your way, you lose 6 of your doubles that you highlighted in your OP.
Yeah, I know, but I don’t violate the assumptions of the stats test which I think is more impt.
@Wanze
I know, I fixed it and changed the data later on:
“Ok, redid math.
99.9825172916403% chance that I’d get fewer than 8 doubles out of 42 trials. Definitely a lot more plausible that I just got unlucky, but with others seeing doubles too, with yellows dropping in pairs from world bosses etc. etc., I still think their RNG does a poor job giving a normal range of outputs on the INDIVIDUAL level."
^that’s with changed odds to 1/26
Did Anet ever say that the drop chance for every ring is the same?
Maybe certain rings have a higher chance at certain fractal levels.
Bloin – Running around, tagging Keeps, getting whack on Scoobie Snacks.
Did Anet ever say that the drop chance for every ring is the same?
Maybe certain rings have a higher chance at certain fractal levels.
Definitely possible.
Did Anet ever say that the drop chance for every ring is the same?
Maybe certain rings have a higher chance at certain fractal levels.
Nope. I would also like to know if ascended armor boxes for chests and legs are more rare. I have so many feet/hands/head boxes… Seems more rare to me anyway but could just be RNG
But if you do the counting that way don’t you violate the independence assumption for the trials?
No, because the first ring of the pair doesn’t matter. Meaning that when the first and the second ring aren’t a pair, it’s still a 1/26 chance that the second and the third ring could be.
The same is true if the first and second ring are the same: still a 1/26 chance that the second and third are also a pair (making the whole a tripple at least).
It would be a different story if you would do this with tripples. Then 1-2-3 and 2-3-4 wouldn’t be independent, because if 1-2-3 would be a tripple the chances would be higher that 2-3-4 would also be a tripple(1/26), then if 1-2-3 weren’t already a tripple(1/676). Because the information that 1-2-3 is a tripple already makes sure that 2 and 3 are the same ring and we could use that information to decrease the possibilities for 2-3-4.
But if you do the counting that way don’t you violate the independence assumption for the trials?
No, because the first ring of the pair doesn’t matter. Meaning that when the first and the second ring aren’t a pair, it’s still a 1/26 chance that the second and the third ring could be.
The same is true if the first and second ring are the same: still a 1/26 chance that the second and third are also a pair (making the whole a tripple at least).It would be a different story if you would do this with tripples. Then 1-2-3 and 2-3-4 wouldn’t be independent, because if 1-2-3 would be a tripple the chances would be higher that 2-3-4 would also be a tripple(1/26), then if 1-2-3 weren’t already a tripple(1/676). Because the information that 1-2-3 is a tripple already makes sure that 2 and 3 are the same ring and we could use that information to decrease the possibilities for 2-3-4.
Oh, I get it!
Thanks for the explanation.
SO! Final (I hope!) maths:
http://stattrek.com/online-calculator/binomial.aspx
Probability of success on a single trial: .0385
Number of trials: 83
Number of successes: 15
Cumulative Probability: P(X >/= 15): 6.12868871985306E-07
Total Odds of Getting Greater Than or Equal to 15 Successes in my 83 Trials: 1 in 1,631,670
I’m pretty sure that’s what Tim got on the other thread.
So, is that possibly just from random chance? Sure, there are 3.5 million accounts purchased. That said, I doubt even 1% of those have done that many fractals, and from what people have been saying, I’m not the only one who has received patterns of doubles like this.*