I'm creating this topic for players to post their numbers. To keep it simple, lets stick with a format like:
Number of items crafted:
Number of 1-UV items:
Number of 2-UV items:
Number of 3-UV items:
Lets figure out the chances for 2- and 3-UV items from crafting!
Are you sure that's how it is though Replicant? If so I'd like some links where devs state that it'll be this way. And also if this is true then I'm sorry.
Could it not be conceivable that they have chances which are independent of each other such that when you craft an item you get ChanceA to get 1-UV, ChanceB to get 2-UV, and ChanceC to get 3-UV instead of ChanceA to get 1-UV, ChanceA*ChanceA to get 2-UV, and ChanceA*ChanceA*ChanceA to get 3-UV?
It's a 1 out of 10 chance to get a single UV therefore, using that same ratio, any additional UVs would multiply the odds by 1 and 10 (i.e. 1/100, 1/1000). This is the most logical formula unless otherwise stated.
Could it not be conceivable that they have chances which are independent of each other such that when you craft an item you get ChanceA to get 1-UV, ChanceB to get 2-UV, and ChanceC to get 3-UV instead of ChanceA to get 1-UV, ChanceA*ChanceA to get 2-UV, and ChanceA*ChanceA*ChanceA to get 3-UV?
That's the same exact thing. I get what you're saying, however.
Just to make it clear, I'm suggesting that when an item is crafted it is decided right then and there whether it'll be 0- 1- 2- or 3- UVs based on some independent chances.
What your method implies is that there are 'UV slots' and each slot is rolled with the same chance, namely 1/10.
While I do agree that this is the most logical method to do this. Perhaps the devs thought a 1/1000 chance to get 3-UVs was too low. A guild mate of mine just got a 3-UV item after crafting 20 items. Though that's a single sample so only time will tell.
I'll update this every time I craft:
20/7: 24 Haze bombs. 1x Low Construct. 1x Med Fiend. 1x High Construct.
23/7: 21 Haze bombs. 0 Uvs.
25/7: 26 Haze bombs, 1x Low Slimes. 1x Low Beasts.
26/7: 24 Haze bombs, 1x Low CTR. 1x Undead Low.
27/7: 20 Haze bombs. 1x Fiend Low. 1x Undead Low. 1x Construct Low. 1x Construct Low. 1x Slime Medium.
28/7: 25 Haze bombs. 1x Undead VHigh. 1x Low CTR. 1x Slime Medium. 1x Low Beast. 1x Double UV: Low CTR & Low Slime.
29/7: 25 Haze bombs. 1x Low fiends. 1x Low gremlins. 1x Low Undead.
30/7: 30 Haze bombs. 1x Low Slimes. 1x Medium Construct. 1x Medium Slimes.
31/7: 30 Haze bombs. 1x Low Fiends. 1x Low Gremlins.
01/8: 30 Haze bombs. 1x Low Undead. 1x Low Beast. 1x Low CTR. 1x Low Undead. 1x Low Slimes.
Total: 245 Bombs. 29 Simple UVs. 1 Double UV.
Out of about 40 Haze Bombs I managed to get one with two UVs.
30 Haze bomb, no 2 u.v., 4 * 1 u.v. and needless to say zero 3 u.v.
/me backs up slowly
You're going to need a bigger sample size.
But just from this it looks like Replicant is might be right, each successive roll has 1/10 of hitting the next UV since out of almost 100 listed above there is only one with two UVs.
You're going to need a massive sample size to get any kind of accurate data on this.
It's probably just a 1/10 chance for each UV, so 1/1000 for a triple.
Using your math, this would give a 111/1000 chance of AT LEAST one UV. Not so sure that's right
Heck, I'm not totally convinced that the UV rate has ever been 10%. Any official posts about it?
Using your math, this would give a 111/1000 chance of AT LEAST one UV
1 x 1 x 1 is not 111.
Heck, I'm not totally convinced that the UV rate has ever been 10%
It's always been 1/10.
If each "Uv Slot" has a 1/10 chance of getting a UV and each of them are run regardless of the other slots, then the chance of getting 1 UV is 3/10.
Cause you have 3 chance of getting a UV.
Which doesn't seem like it is the case here.
I'm making a few assumptions here:
1) The odds of obtaining at least one UV are exactly 1 in 10.
2) When you craft an item, each of 3 UV "slots" acts independently of each other and has the same chance of containing a UV.
Using this info, I have come up with the following numbers:
Odds of one particular "slot" to contain a UV: 1 in 28.977 (Probability: 0.034511)
Odds of no UVs: 1 in 1.111 (Probability: 0.900000) - based on my assumption
Odds of exactly 1 UV: 1 in 10.362 (Probability: 0.096509)
Odds of exactly 2 UVs: 1 in 289.885 (Probability: 0.003450)
Odds of exactly 3 UVs: 1 in 24329.988 (Probability: 0.000041)
I checked these results by adding up these four probabilities (the last four, obviously not the one for a particular slot), and it does come out to 1. If my initial assumptions are correct, then these are the odds of crafting each of the four possibilities.
I think it's probably 1/10 chance for a UV and then after a UV is created, they check again for a 2nd UV, so 1/10 for 1 UV, 1/100 for 2 UV, and 1/1000 for 3 UVs.
If they are independent slots with a 10% chance of having a UV, then the percentage of time you get no UV is 90% * 90% * 90% or 72.9%. I guess it could be a lower rate for each slot like in ThePiGuy's calculations, but that seems much more complicated. Once we get more data, it should be pretty clear whether the 3 UV items are 1/1000 or 1/24,000 (which would make the 3 UV tickets a huge bargain!)
I guess it could be a lower rate for each slot like in ThePiGuy's calculations, but that seems much more complicated.
It wouldn't be complicated at all. The code could basically say, "Do this 3 times: Roll for a 3.5% chance, and if that succeeds, add a new random UV."
Really, any consistent set of probabilities would be easy enough to implement. Guessing at what the probabilities are is just guessing. If you want to know, you need either a statement from the devs, or hard data. The devs probably won't tell us, since they usually avoid telling us such things, so that leaves data. And, as with past attempts at gathering statistical data, you need to be careful about how you collect it.
Thing to think about:
1 UV costs 15,000 Crowns
2 UVs cost 15,000 x 5 = 75,000 Crowns
3 UVs cost 75,000 x 5 = 375,000 Crowns
Each additional UV makes the operation cost 5 times more.
You could argue that this matches the relative probabilities of each result via crafting, or you could argue that it's a "discount" thereof. Make of this what you will, I don't have enough information to say either way.
Crafted 20 Magic Hoods today.
The Bad: 1 Magic Hood with UVs!!!!!! -_-;
The Good: That Magic Hood had 2 UVs: Max Shadow + Low Pierce
So ... ya :/
1 x 1 x 1 is not 111
Uh, do you know how to add? Your numbers were 1/10, 1/100, and 1/1000. Add those up and you get 111/1000 (probability of at least one UV).
It's always been 1/10.
How are you so sure? Has a developer revealed the numbers? Have enough people tested this number, without selection bias, that we can safely assume (with at least 95% confidence) that your number is correct? As far as I am aware, no. Just because most people tend to throw around this number does not make it true.
There are some threads very similar to this one under the old crafting rules that accumulated a fair few hundreds of testimonials that that 10% number came from. Individuals with large personal crafting records have reported numbers within 2% of that number, so it was near if not exactly that number, and 10.74% or something would be an odd threshold for a developer to choose.
This thread is unlikely to gain any noticeably more reliable data than those attempts, but I agree it's worth checking into because the crafting rules -did- just change dramatically.
Also, I have 4 crafts and no UVs so far.
Mind moving this thread to wikieditors? It will be easier for me to find there and update my post.
I started tracking my Crystal Bomb crafting on July 11th. Till the patch I received 16/81 with UVs, a 19.75% chance.
Since the change on July 21st for Crystal Bombs I'm currently at 1/24 or 4.17%. (It was low fiend for those that want to know...) Hope my luck picks up again soon. In the long run it should. :)
So another 84 Magic Hoods
8 UVs.
:/
Edit: one of the 8 has 2 UVs on it
Uh, do you know how to add? Your numbers were 1/10, 1/100, and 1/1000. Add those up and you get 111/1000 (probability of at least one UV).
Oh... Oh wow. That's not how fractions and probabilities work at ALL. That's just... Wow.
How are you so sure? Has a developer revealed the numbers? Have enough people tested this number, without selection bias, that we can safely assume (with at least 95% confidence) that your number is correct?
Yes.
1/10 = 100/1000
1/100 = 10/1000
1/1000 = 1/1000
100/1000 + 10/1000 + 1/1000 = 111/1000
P(One UV) + P(Two UVs) + P(Three UVs) = P(At least one UV), or P(One, Two, or Three UVs)
You're misunderstanding how UVs work; it doesn't roll them together, it rolls them individually. That makes the probability of getting at least a single UV 10%, not 11.1% [based on previous calculations].
Oh... Oh wow. That's not how fractions and probabilities work at ALL. That's just... Wow.
Actually, it is. Perhaps you're thinking that you proposed 1/10 is the probability of 1 or more UVs, 1/100 is the probability of two or more UVs, and 1/1000 is the probability of 3 UVs. But it's not. Maybe it's what you intended, but you expressed it as three disjoint outcomes of probability 1/10, 1/100, and 1/1000. In the case of disjoint outcomes, the probability of any one of them happening is indeed the sum of their individual probabilities.
Actually, it is.
You are correct and I admit my error that I made in haste.
Perhaps you're thinking that you proposed 1/10 is the probability of 1 or more UVs, 1/100 is the probability of two or more UVs, and 1/1000 is the probability of 3 UVs. But it's not.
You're right, it's not. The odds of getting a single UV is 1/10, the odds of getting 2 is 1/100, and the odds of getting 3 is 1/1000. It's like this because they are all rolled individually rather than together.
Maybe it's what you intended, but you expressed it as three disjoint outcomes of probability 1/10, 1/100, and 1/1000. In the case of disjoint outcomes, the probability of any one of them happening is indeed the sum of their individual probabilities.
But it's still wrong. As I said above.
It's like this because they are all rolled individually rather than together.
This is pure speculation. There isn't nearly enough data to tell the difference between 1/10 and 111/1000 as an overall probability of one or more UVs. It could easily be more like 1/10, 1/50, 1/250 than 1/10, 1/100, 1/1000. UV ticket price scaling suggests it *could* have roughly those proportions.
This is pure speculation. There isn't nearly enough data to tell the difference between 1/10 and 111/1000 as an overall probability of one or more UVs.
True. There's absolutely no evidence to suggest whether UVs are rolled together rather than apart. Even if it's one or the other, the odds are so close together it's irrelevant.
This is an absolute pointless discussion.
Unless you do know the actual source code or have the data of a really really large statistical ensemble, u can not say if the probability is 1/10 or 111/1000 or 10*(e/2)/(h*c/MeVfm) (http://www.wolframalpha.com/input/?i=10*%28e%2F2%29%2F%28%28%28Planck+co...) for one UV.
Even if you know the probability for one UV is somewhere near 1/10,
if OOO roll for the second UV only if the first was successful (and so on) and each roll would have a probability of P=1/10 you would get P(1UV)=1/10, P(2UVs)=1/100, P(3UVs)=1/1000.
If they roll 3 times with P=1/10, P=1/100 and P=1/1000 and add the UVs, its 111/1000, 111/100000, 1/1000000.
Maybe the probability is higher for new users and drops if you mass-craft.
-> We dont know the probability distribution
200 haze bombs crafted
construct low
slime low
undead low
slime low
beast low
gremlin medium
undead low
gremlin low
beast low
beast low ctr low
construct low beast med
slime high construct low
med gremlin
slime low
fiend low
beast low
construct low
undead low
construct low
so I ran a chi-square test on the data (I ignored the ones which were not specific enough).
There are three main hypotheses:
Hypothesis 1 (fixed probabilities):
1 UV exactly= 0.1 (probability of course)
2 UV exactly= 0.01
3 UV exactly= 0.001
Hypothesis 2 ("Roll once; if you get a UV; roll again to see if you get another; otherwise stop rolling"):
1 UV exactly= 0.09
2 UV exactly= 0.009
3 UV exactly= 0.001
Hypothesis 3 ((UV Slot theory) "Roll three dice at once; count the number of UV's"):
1 UV exactly= 0.243
2 UV exactly= 0.027
3 UV exactly= 0.001
The last two assume that on a single roll the chance of UV is 1/10.
Running the Chi-square test on the sample size of 597, we get that if hypothesis 1 is true, we have about 50% chance that the deviation from predicted is due to randomness, for hypothesis 2 it is 70%, and hypothesis 3 is just waaaay off. The standard is that if the probability is <5%, the hypothesis is considered falsified, so the most we can say is that at the moment hypothesis 2 is doing better than 1. (In my opinion 2 is the more accurate though, but more data is necessary.)
Feel free to object if there is something wrong and please keep contributing data :3
Thanks for the Chi-squared analysis.
See ThePiGuy's post for a more reasonable version of hypothesis 3, a single roll succeeding with a probability of 0.034511.
For a sample size of 597, I take it you're including Dirt's report? I found it ambiguous as to whether there were any single UVs, and "Out of about 40" is imprecise. (Could you clarify, Dirt?) Leaving that out, I count 557 trials, 50 single UVs, 6 double UVs, 0 triple UVs. Did I miss any?
Yeah, hypothesis 2 looks like the front runner so far.
uh no. i've crafted over 50 items and have never gotten a uv on 1.
uh no. i've crafted over 50 items and have never gotten 1 uv.
sorry for some reason it posted 2 comments instead of one
Actually as it turns out, Pi Guy's probability distribution (which I call hypothesis 3') is still very off (although much better than the original hypothesis 3), with a chi-square value of 12.89 which with 3 degrees of freedom gives a probability of less than 1% that the deviations are due to random error alone, thus we should probably reject this hypothesis.
(sorry for people who don't like math >_<)
70 haze Bombs
0 triple UVs
1 double UV
9 single UVs
60 vanilla bombs.
Seems as though I have had a decent run thus far. Ill post up more as I get them
38 haze bombs
0 triple UV's
1 double UV
6 single UV's
31 vanilla bombs.
20 haze bombs
0 triple UV's
0 double UV
3 single UV's
17 vanilla bombs.
a few days ago
180 haze bombs
172 vanilla
8 single UV
0 double UV
0 triple UV
as you can see, i have terrible luck
With the current sample size of 921 trials, and no triple UVs. The upper bound on the probability of getting a triple UV is 0.0032 with 95% confidence.
the possibilities here are endless guys, i believe that there is not enough data to accurately assume any of the things that have been assumed thusfar. for example, because of the fact that haze bombs are/were easy to make, most of our data stems directly from haze bombs. Have we stopped to consider that this is rather presumptuous of us? That haze bombs are exactly like every other piece of equipment? For example, i craft mainly 3 star swords to sell on the ah for profit. in the last two days i have made exactly 8 Shockburst Brandish, a Tempered Calibur, and a Grintovec, resulting in the tempered calibur receiving 2 uvs, beast medium gremlin low, a shockburst with fiend low, and a shockburst with construct low, gremlin medium. Granted, this is a horribly small test group, is it not equally as possible that larger tiered weapons have a larger chance of receiving uv? and thus double uv? Again, this is an assumption, exactly like the assumptions we have all made thusfar. until we break out of the haze bomb only crafting group, however, we will never fully understand any of the mechanics of uv generation except haze bombs.
It is much more likely the the system is something simple, rather than something complicated. The data collected previously for other items, although not a big sample, was at least consistent with a 10% chance across the board.
1 UV = 1/10.
2 UV = 1/100.
3 UV = 1/1000.
Yay basic math.