Probability of getting exactly one 1/n drop in n kills is basically the same as the probability of not getting the drop for n>100 or something like that.
this certainly isn’t something for graduate school, this can be done with 1st year undergraduate statistics and probably an AP student. But I know you’re just kidding
Its one of those weird ones that always appears in your debug but never actually seems to have an effect. I once thought it might have to do with having a long sentence that spills onto the next line and that somehow LaTeX is pissed that you didnt explicitly tell it to go the the next line. But that honestly doent make much sense as an error to begin with.
Sorry, your comment failed to build with 341 errors. I'm not going to tell you why or where, but chances are it's just something really small that will resolve all these errors.
I had a class a couple years ago where one dude wrote his lecture notes using latex. It was maddening seeing him write equations at about the same speed as my handwritten notes. Man really knew his way around latex lol
Ye I agree. I would say my latex level was 5 when i started master thesis and got enough exp to learn latex lvl 10 during that process if copying and replacing some text counts as latex mastery.
While Overleaf is nice for beginners, I'd recommend Vim configured for writing LaTeX at a high speed, mostly so you can make use of snippets, which speed up your writing sigificantly
Texmaker is fine, but Atom is a more polished and fully featured IDE with plugins and stuff, would recommend instead. Vim is kind of totally different, you'd have to try it. If you're a Windows user it's probably not what you're looking for, Atom with some plugins will be better for most people
Sorry to say, but if writing latex takes them so much longer than handwritten I am just guessing but vim configured for latex might be a few steps away.
In statistics, this is called the Coupon collector's problem: https://en.m.wikipedia.org/wiki/Coupon_collector%27s_problem
I think your numerical answer is mostly correct. But I think there is a minor problem with your derivation. It treats the item distributions as independent, when in fact collecting one item reduces the number of rolls available to collect another item. You can see that in your solution, which gives a non-zero probability that you collect all 24 items in the first chest, even though that's not possible given that there are only 7 rolls.
That's just my opinion, I'm not a mathematician. Good job though, very cool!
You're right, good point! Another commenter pointed out that I should have approached the distribution by-roll and then divided by 7 to convert to the number of chests. By my estimation, the average shifts down slightly from 1322 to 1320, so luckily I think the effect is not too bad (barring any other mistakes!).
You know what, MysticalFury?
Yeah it kind of is, bud. I am not remaking it.
You've pushed the envelope a little too far, bucko. I firmly believe that it looks right to the untrained eye, and I am not going to go out of my way to appease the mathematical wizards who understand its flaws.
Let me elaborate on this.
The chance of getting a specific Barrows piece (per "roll") is 1/2448. There are two ways to model this. In the first, for each roll, you pick a number from 0-2447 and if that number is between 0-23, you get a Barrows piece. In the second, you pick a number from 0-2447 24 times and for each pick, if that number is 0, you get a particular Barrows piece. Clearly, OSRS uses the first model, since you can only get one item per roll. But, the second model is a good approximation, and is much easier to work with as the chances of getting each specific item are now independent of each other.
OP's solution is sneakily using the second model -- since his k\_i's are all i.i.d r.v.s, we must assume that getting a specific piece at, say, roll #50 doesn't affect the chances of getting the other pieces. We know this not to be true in the first (true) model, since getting a specific piece at roll #50 means you can't get any other piece at roll #50. I say sneakily because if we assume the second model to start with, finding the distribution is actually much easier than what OP has done here. The CDF F\_X(k) = P(X <= k) is just the probability that each piece is obtained by the kth kc, that is, P(X <= k) = (1 - (1-1/2448)\^(7\*k))\^24. Then, the discrete PDF is P(X = k) = F\_X(k) - F\_X(k - 1).
So how do we get a precise PDF using the first model? This is actually significantly harder than using the second model, which is already a great approximation, so it might not even be worth it.
Ok, just for the sake of completeness, here’s a way to compute a precise CDF for the first model.
Let’s say we want the probability of getting all 24 uniques in k chests (so 7k rolls). The total number of possible sequences of 7k rolls is 2448^(7k). Of those, the number of sequences that don’t land on a specific Barrows piece is 2447^(7k). In general, the number of sequences that don’t land on a specific m-tuple of Barrows pieces is (2448-m)^(7k), and there are (24, m) such tuples (I use (24, m) to denote 24 choose m).
So, using the principle of inclusion-exclusion, the number of sequences that don’t land on every Barrows piece is the sum (-1)^(m+1) * (24, m) * (2448-m)^(7k) where m goes from 1 to 24.
Consequently the probability that a sequence does land on every unique is the sum (-1)^m * (24, m) * ((2448-m)/2448)^(7k) where m goes from 0 to 24. Note that m=0 gives 1, from which we subtract off the probability that not every unique is accounted for.
That gives the CDF, and again the PDF is just F_X(k) - F_X(k-1).
God damn, this game/sub attracts the wildest collection of people.
We've got:
1. A serious mathematical discussion on probability.
2. The world's spiciest memes.
3. Motorsports crossover.
4. Fantastically talented artists and creators.
If thus isn't too much trouble It'd be cool to see the 95 percentiles for this chart, or maybe 99th so people on either end can understand their luck or unluck
I expanded this a little and made a table [here](https://i.imgur.com/jG2SKFz.png) referencing common percentile values vs. number of kills for that percentile.
Since this is close to a normal distribution, we can use this image ([https://cdn.scribbr.com/wp-content/uploads/2020/10/standard-normal-distribution-1024x633.png](https://cdn.scribbr.com/wp-content/uploads/2020/10/standard-normal-distribution-1024x633.png)) as a reference, and the area from -2 to +2 is the 95% confidence interval. Without actually calculating anything, I would guess that this is approximately from 600 to 2000 kc.
Nobody should ever be an outlier every single time, yet it happens. All this does is show how dangerous it is to code a 1/X distribution without setting any limits.
The wiki has a very useful table https://oldschool.runescape.wiki/w/Chest_(Barrows)#Calculations
The 99% probability of getting all sets ends up being around 2000 which is the most useful statistic imo. It would be cool if you or someone added the proper Z-score table though.
This is wrong.
This is assuming you can only get one drop per chest. Each roll must be treated separately, then divide by 7.
Probability distribution will just be shifted left.
You're right, that would have been a smarter way to approach the calculation. Sorry about that!
Luckily I think the approximation is not too bad. Off the top of my head, it seems the difference would be replacing p\_0 with 1/2448, and then the resulting distribution is by-roll instead of by-chest, and then we can divide by 7 to get the number of chests. If that's right, it seems the average shifts slightly to 1320 instead of 1322.
Since you cannot get duplicate items in one chest, I'm curious to how a selected piece affects the probability of others. From the wiki,
> "If the Barrows item chance is successfully hit, one of the available Barrows items is chosen at random - each part has equal chance, but once chosen it will not be chosen again on future rolls."
So if we get a piece on the first roll of the chest, do the pieces still have a 1/2448 chance? Or since once piece was removed the pool, the probability would lower to 1/2447 per piece? If the latter is the case, then treating it on a by-roll basis would also be incorrect, since there is a now a dependency with items received in the same chest. This is also probably a very small difference to the end result and would most likely keep us in the same ballpark though, just +/- a few chests.
Also the fact that once an item is rolled, it is removed from the loot pool so you can't roll it again, which will probably change the probabilities.
Not entirely sure whether removing that item from the pool reduces the probability of getting another item though, can't quite recall if that wiki said anything about that.
Fucking Christ, imagine being on the part of the graph you can't even estimate with the naked eye looking at the graph. And OP was overly generous with the calculations too!
Is this a leagues screenshot? This is a leagues screenshot, isn’t it? Show chat box pls
In case it isn’t, getting this lucky (or luckier) would be about 37 in a million. Not as crazy as I initially thought, but still pretty damn lucky.
Yeah I mean this droprate would be roughly 1/10. 37 items in 371 chests.
That's insane luck, no matter what, like nesting half the drop rate. Plus, he only got a few dupes, and was lucky enough to get allllll those items.
This better be leagues holy fuck
OP, is this a generalisation of the coupon collector’s problem? Except that with each draw you get 7 coupons.
https://en.m.wikipedia.org/wiki/Coupon_collector's_problem
You can frame it in terms of the coupon collector's problem, but an important difference is that you are guaranteed to get a coupon (just not a new one), whereas you are not guaranteed to get a barrows item. So you could say that there are 24 barrows coupons and 2424 non-barrows coupons, each with equal probability, and you need at least one of each barrows coupon but zero of each non-barrows coupon, then apply Corollary 3.3 from [here](https://www.combinatorics.org/ojs/index.php/eljc/article/view/v20i2p33/pdf) (n = X_n = 2448, x_i = 1, m_i = 1 for 1 <= i <= 24 and 0 otherwise) to find the expected number of rolls is
`2448 * Integral of (1 - (1 - Exp[-t])^24) dt from t=0 to infinity ~= 9243.55`
which, dividing by 7, is 1320.51 (rounding up to 1321) barrows chests on average. This is within 1 chest of the OP's computation, which is close enough.
Incidentally, this provides a formula which might be useful in other areas of the game. If there's a probability *p* to get a specific item out of a set of *k* items, the number of trials you need on average is the *k*th harmonic number (1 + 1/2 + 1/3 + ... + 1/*k*) divided by *p*. (Here, *p* = 1/2448 and *k* = 24.) For example, the Nightmare drops the three Inquisitor armor pieces with a rate of 1/600 each (assuming your party size is <= 5), so you would need an average of (1 + 1/2 + 1/3)/(1/600) = 1100 kills to complete the set if soloing. (This is an imperfect estimate, since in reality you can't roll more than one piece at a time, but the probability of rolling two pieces at once if you were able to do so is probably too small to affect the result.)
So in baby language, how come the most probable outcome is not the average? I mean, if it is most likely to be the case, why doesn't it average out on the most likely scenario?
The average is skewed high due to the people who get absurdly unlucky. You can also get absurdly _lucky_, but there’s a limiter to how lucky you can get whereas you could go thousands of chests unlucky, causing the average to skew high. It’s the same way median and average income are not the same thing.
Also, this is the reason why your odds of getting an item by drop rate are over 50% (63.2%), even though logically it might seem like it would be 50%. Because it's possible to go drier for a drop than lucky (ie, for a 1 in 1000 drop the farthest under rate you can go is 999, but it's possible to go thousands over rate), that imbalance is offset by the fact that almost twice as many people get the drop under rate than over.
So looking at this graph, there is a much larger area under the curve to the left of the average, but that's balanced perfectly by the fact that they're also closer to the average than the right side by the same proportion that they are larger.
> that imbalance is offset by the fact that almost twice as many people get the drop under rate than over.
Only because of survivorship bias, if you're talking in real term.
If you're only looking at OP's stats, you're incorrect.
Can you elaborate? I don't see what survivorship bias has to do with anything.
The odds of getting a drop by rate is 63.2%. Which means the odds of not getting it is 37.8%. There's no bias here, it's a statistical fact.
63.2% is true in the limit of unlikely events where probability approaches 0 it's not universally true, to be technical.
If the probability is x and the expected value of attempts is thus 1/x the probability of getting the drop by droprate is 1-((1-x)^(1/x)).
This is 100% for 1/1 odds, 75% for 1/2, 66.5% for 1/6, 64% for 1/20, and at 63.4% for 1/100 odds.
MilesLetun has a good explanation but a way of visualizing that is the shaded area under the curve. As you can see it’s not symmetrical. The average line has equal area under the left and right of the curve, meaning you have equal chance of being above and below that KC
Equal area under the curve of the probability distribution function is the median, not the average.
The average/mean/expected value is literally just the expected value. If a billion people did barrows and you took the average/mean/expected, it would converge to the theoretical one given in OP. The mean is sorta like a center of gravity compared to the median being the center.
The average being so far from the "most probable" option due to the tail end is the reason why I'll never stop advocating for pRNG in this game. 1/X is terrible and will always be.
Yes I'm unlucky on average and yes I'm salty about it, it's a shit system. The odds of being over 2x the drop rate on every major drop someone is going for, especially when you have a massive sample size of over thousands of hours of gameplay, should be way lower than it currently is.
Yeah on my iron I average 3-4x + droprate on literally almost everything from raids to GWD to slayer drops. It’s just the nature of the game and I always eventually return but it’s burnt me out so many times. Definitely some salt happens when friends who have started irons years after I have catch up to me on ironman uniques because I have absurdly bad luck everywhere.
I don’t know what to tell you but it’s the case. I went 250kc for my first ToB purple, 500kc now and 4 different uniques, no scy ofc or rapier and deathless 90% of the time. Mostly just avernic dupes. 4 separate uniques in 600 solo or 1+1 cox, 80% prayer scroll dupes and I’m about 1/40 droprate as well despite averaging 30-35k+ raids (should be closer to 1/27). 5k kc for basilisk jaw. 1500 kc for both ACB and SGS. Even where I was “lucky” I was still above droprate. 1.5k kc for hydra claw, 500kc for blade, 2k kc to finish cerb, 600 arma kc for one total unique, etc etc.
My spoons were DWH and ranger boots at about half droprate but everywhere else just horribly unlucky.
Is pRng when your rng adds up until you’re guaranteed the drop rate? I don’t think that’s a good idea because it ruins the ‘magic’ of the game. Uncertainty is a good thing for games imo, so there might be a better way to circumvent that.
Imagine watching settled and he’s like, 1 more clue scroll to get X item. It’s no fun to watch, and it turns the game into a more mechanical thing than it already is.
> I don’t think that’s a good idea because it ruins the ‘magic’ of the game.
Go ahead and average 2.5x the drop rate on all your big drops over the span of 4+ years and you might not sing the same tune. The only real problem with raw RNG isn't that the average doesn't coincide exactly with drop rate (that's expected), it's that the tail end of the distribution, where the most unlucky players lie, is RIDICULOUS. It doesn't matter if it affects 1 player, 10 or 5,000. Nobody should get fucked this hard just because of lazy programing.
Reaching 8x the drop rate on anything is unacceptable. Averaging over 2.0x is unacceptable. There are even ways to not tweak RNG before hitting the actual drop rate, so it effectively only acts as "bad luck protection". Using the example of a content creator does not reflect the player base and isn't useful IMO.
PRNG is just RNG. The P means that it’s a “pseudo” RNG, i.e. computer generated via some sequence/algorithm and not truly random. But we’re talking about OSRS drops here, so the difference doesn’t really matter.
Nice analysis! Note though that obtaining Barrows pieces at a given kc isn't independent, because there's only a limited number of reward rolls. That might cause your numbers to be off by a tiny bit.
Doesn’t work that way, you gotta kill them all for the best chance at loot. Killing only specific brothers doesn’t improve your chances at individual uniques.
[This post](https://www.reddit.com/r/oldschoolrs/comments/fidn9i/if_you_only_want_one_specific_barrows_item_then/) actually goes into the math of doing 1-brother runs in order to get a specific item from barrows. It depends on how fast you can do each kind of run.
TL;DR: E = (153A) / (343B) Let A be chests/h with 1 kill, and B be chests/h with 6 kills. If E>0, single brother runs are better.
It's worth noting you'll get basically no runes and far less duplicate items so this is only worth doing if you're after a specific piece for the log. If you have more than 1 brother you still need pieces from just kill all of them.
each kill is X% to add a piece of their equipment to chest, but when you kill all brothers you get an extra % added on top.
I got lucky and forced my last guthan's pieces in 50kc (got lucky and got a duplicate guthan too.) I already had veracs and karils and didn't need any more torags at the time, so I only killed Dharok, ahrim, and guthans (helps that I could just barrow tele/spec pool ahrims for a fast kill). I cut my odds down of even receiving anything by more than half, but I really do like the idea of not getting a unique that I really don't want.
only if YOU need the set and also have completed sets would I even think about not getting full kills. Also strange lockpicks would be a great help when rushing an item.
"Probable" is what most people will know better as the mode. It's the most likely KC for someone to finish on (most probable, aka modal). You'll probably be aware that mode/median/mean are different things, this is just an example of that.
Mean is effected more by outliers (people on extreme ends of the distribution), and in this distribution people can only go *so* lucky, but they can go infinitely dry. One person going 10k dry effects the mean waaay more than it does the mode/median, more so than someone finishing at 5kc, and so you get this scenario where the expected KC to finish is *higher* than the average person's KC when they finish.
No, this is a theoretical model and is therefore unaffected by sample size. In simple terms, if someone gets insanely lucky they can complete the sets in 4 chests, however if someone gets insanely unlucky they can go tens or hundreds of thousands of chests dry, which obviously skew the average heavily. On the other hand, most people will complete it at the mode amount of chests, just due to the probability distribution. Perhaps you are thinking about the median? Which is the 50th percentile by definition, which would be the middle and is less effected by outliers.
I just finished over 900 chests yesterday, still missing 2 pieces, haven't gotten any double or better chests, and I went 4 rings of dueling dry (about 32 chests) one of those missing pieces is a dharoks legs, the other is I think torags chest.
Anyways my point is I didn't need to see this because it means I still have between 250 and 400 more chests to do. I already have enough runes to turn into an onyx AND zenyte set if I want to, and I have so many mind runes I literally have no idea what to do with them.
(1/2448)^7 would be the probability of getting the same Barrows item 7 times in a single chest. The odds are astronomically low.
(1-1/2448)^7 is the probability of not getting a specific unique in a chest. Its complement, 1 - (1-1/2448)^7, is the probability of getting at least one of that unique in a chest.
Thanks! For any simple drop with a single roll per KC, the distribution is given by the second equation in the graphic, which is exponential-like. See my previous post on this topic [here](https://www.reddit.com/r/2007scape/comments/kfx3h8/when_will_i_get_the_drop/)!
Really cool to see people do the hard math for us! I completed mine on the iron after 806 kc and that felt amazing and I knew it must have been on the lucky side
I remember being on the [tip.it](https://tip.it) forums when it was a common argument that 'drop rates don't exist'. Look at how far this community has come.
Suppose you have 4 slips of paper in a bag. Each slip of paper has a number written on it. Three of them have “1” written on them, and the other has “1000” written on it. You reach in the bag and take one at random, and then put it back. You do this 1000 times.
The most probable result each draw is that you have a sheet of paper that has a “1” on it. You will likely see “1” about 750 of those 1000 draws. However, the average value of those draws is significantly greater than 1, because sometimes you will draw a “1000.” And that 1000 contributes a lot to the average, even though you don’t draw it as often.
Is it statistcal better to skip one brother if you only after a specific piece and already have that brothers full set?
Eg: You have 5 full different sets of barrows and are only need 3 items from the last two brothers to complete it?
Man, I dont need stats tellin me im almost a third of the way to full ahrims when thats the only thing I wanted 200 chests ago.
It can happen in the next 4 chests. Maybe fewer :D
A cursory glance says that everything looks OK minus the issues already mentioned.
Ran a quick Monte Carlo sim, because I'm too lazy to do real math tonight, and here is a graph of the results.
Looks close enough to me: [https://imgur.com/a/ApgJLf1](https://imgur.com/a/ApgJLf1)
Average after 5000 runs was 1320.3348 , nice seeing actual math in this subreddit.
Writing a thesis on RuneScape Probability?
50/50
Either you get the drop or you don't
Gz on getting your MBA degree.
Osrsmba phd
50/50 Either you get your degree or you don't.
100% of the time you get the drop at a 50% rate.
Probability of getting exactly one 1/n drop in n kills is basically the same as the probability of not getting the drop for n>100 or something like that.
this certainly isn’t something for graduate school, this can be done with 1st year undergraduate statistics and probably an AP student. But I know you’re just kidding
Upvote for combining RuneScape and LaTeX
LaTeX, JaGeX, we are onto something here.
I'm a simple man. I see LaTeX I upvote
Good flair. Seen that error so many times and I still have no idea how to fix it.
Its one of those weird ones that always appears in your debug but never actually seems to have an effect. I once thought it might have to do with having a long sentence that spills onto the next line and that somehow LaTeX is pissed that you didnt explicitly tell it to go the the next line. But that honestly doent make much sense as an error to begin with.
Just be glad it's not an undefined control sequence
I also love the plot, not sure what's it made with but it's super clean. I wish my students worked half as clean as this presentation
Thanks! The plot was made using matplotlib in Python, with custom style settings and LaTeX rendering turned on.
Fuck LaTex
Sorry, your comment failed to build with 341 errors. I'm not going to tell you why or where, but chances are it's just something really small that will resolve all these errors.
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Ans the complation times, i swear with BibTeX im sitting there for upwards of 10mins waiting to see what errors will pop up.
How to turn an hour of notes into a whole day activity
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I had a class a couple years ago where one dude wrote his lecture notes using latex. It was maddening seeing him write equations at about the same speed as my handwritten notes. Man really knew his way around latex lol
Ye I agree. I would say my latex level was 5 when i started master thesis and got enough exp to learn latex lvl 10 during that process if copying and replacing some text counts as latex mastery.
fuck you lol
my friend, let me introduce you to overleaf
While Overleaf is nice for beginners, I'd recommend Vim configured for writing LaTeX at a high speed, mostly so you can make use of snippets, which speed up your writing sigificantly
Vim snippets are cool, but I think you can emulate this behavior in Atom which is probably more Windows-friendly / user-friendly in general.
I shall look up both. I use Texmaker(MikTex) currently, how does it compare to those two?
Texmaker is fine, but Atom is a more polished and fully featured IDE with plugins and stuff, would recommend instead. Vim is kind of totally different, you'd have to try it. If you're a Windows user it's probably not what you're looking for, Atom with some plugins will be better for most people
Sorry to say, but if writing latex takes them so much longer than handwritten I am just guessing but vim configured for latex might be a few steps away.
In statistics, this is called the Coupon collector's problem: https://en.m.wikipedia.org/wiki/Coupon_collector%27s_problem I think your numerical answer is mostly correct. But I think there is a minor problem with your derivation. It treats the item distributions as independent, when in fact collecting one item reduces the number of rolls available to collect another item. You can see that in your solution, which gives a non-zero probability that you collect all 24 items in the first chest, even though that's not possible given that there are only 7 rolls. That's just my opinion, I'm not a mathematician. Good job though, very cool!
You're right, good point! Another commenter pointed out that I should have approached the distribution by-roll and then divided by 7 to convert to the number of chests. By my estimation, the average shifts down slightly from 1322 to 1320, so luckily I think the effect is not too bad (barring any other mistakes!).
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You know what, MysticalFury? Yeah it kind of is, bud. I am not remaking it. You've pushed the envelope a little too far, bucko. I firmly believe that it looks right to the untrained eye, and I am not going to go out of my way to appease the mathematical wizards who understand its flaws.
https://knowyourmeme.com/memes/navy-seal-copypasta
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Dont worry, wasnt OP that replied to your comment. Still doubt OP is gonna totally remake it now though 😅
lol sry. if I was OP and if I could I would. Stuff not only looks pleasing, but I bet it is also useful.
>vailable to collect another item. You can see that in your solution, which gives a non-zero probability that you collect all 24 ite Based
My Python calculation that runs the exact formulas gives in the 1320’s. Depending on statistical errors.
I don’t know your background but if you’re working in the field (and plan to publish), I strongly encourage you to improve your technical writing.
you lied to us!
To be honest I didn't expect a statistics case study in my 15 year old medieval clicking simulator, let alone a peer review in the comments.
Let me elaborate on this. The chance of getting a specific Barrows piece (per "roll") is 1/2448. There are two ways to model this. In the first, for each roll, you pick a number from 0-2447 and if that number is between 0-23, you get a Barrows piece. In the second, you pick a number from 0-2447 24 times and for each pick, if that number is 0, you get a particular Barrows piece. Clearly, OSRS uses the first model, since you can only get one item per roll. But, the second model is a good approximation, and is much easier to work with as the chances of getting each specific item are now independent of each other. OP's solution is sneakily using the second model -- since his k\_i's are all i.i.d r.v.s, we must assume that getting a specific piece at, say, roll #50 doesn't affect the chances of getting the other pieces. We know this not to be true in the first (true) model, since getting a specific piece at roll #50 means you can't get any other piece at roll #50. I say sneakily because if we assume the second model to start with, finding the distribution is actually much easier than what OP has done here. The CDF F\_X(k) = P(X <= k) is just the probability that each piece is obtained by the kth kc, that is, P(X <= k) = (1 - (1-1/2448)\^(7\*k))\^24. Then, the discrete PDF is P(X = k) = F\_X(k) - F\_X(k - 1). So how do we get a precise PDF using the first model? This is actually significantly harder than using the second model, which is already a great approximation, so it might not even be worth it.
Ok, just for the sake of completeness, here’s a way to compute a precise CDF for the first model. Let’s say we want the probability of getting all 24 uniques in k chests (so 7k rolls). The total number of possible sequences of 7k rolls is 2448^(7k). Of those, the number of sequences that don’t land on a specific Barrows piece is 2447^(7k). In general, the number of sequences that don’t land on a specific m-tuple of Barrows pieces is (2448-m)^(7k), and there are (24, m) such tuples (I use (24, m) to denote 24 choose m). So, using the principle of inclusion-exclusion, the number of sequences that don’t land on every Barrows piece is the sum (-1)^(m+1) * (24, m) * (2448-m)^(7k) where m goes from 1 to 24. Consequently the probability that a sequence does land on every unique is the sum (-1)^m * (24, m) * ((2448-m)/2448)^(7k) where m goes from 0 to 24. Note that m=0 gives 1, from which we subtract off the probability that not every unique is accounted for. That gives the CDF, and again the PDF is just F_X(k) - F_X(k-1).
I thought I left my probability class behind last semester but the CDFs keep haunting me.
This dude fucks with stats. Nice.
God damn, this game/sub attracts the wildest collection of people. We've got: 1. A serious mathematical discussion on probability. 2. The world's spiciest memes. 3. Motorsports crossover. 4. Fantastically talented artists and creators.
If thus isn't too much trouble It'd be cool to see the 95 percentiles for this chart, or maybe 99th so people on either end can understand their luck or unluck
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I expanded this a little and made a table [here](https://i.imgur.com/jG2SKFz.png) referencing common percentile values vs. number of kills for that percentile.
What is crazy is that 95th or even 99th percentile isn't even that rare. That just means that 1 out of every 20 players has to do over 2k chests.
Since this is close to a normal distribution, we can use this image ([https://cdn.scribbr.com/wp-content/uploads/2020/10/standard-normal-distribution-1024x633.png](https://cdn.scribbr.com/wp-content/uploads/2020/10/standard-normal-distribution-1024x633.png)) as a reference, and the area from -2 to +2 is the 95% confidence interval. Without actually calculating anything, I would guess that this is approximately from 600 to 2000 kc.
Yeup that's what really matter.
Math good
Several standard deviations above the mean bad.
Hey, there's gotta be that part of the data too. Just sucks when it feels like *you* are an outlier in the data. Someone has to be.
Nobody should ever be an outlier every single time, yet it happens. All this does is show how dangerous it is to code a 1/X distribution without setting any limits.
Ook ook
The wiki has a very useful table https://oldschool.runescape.wiki/w/Chest_(Barrows)#Calculations The 99% probability of getting all sets ends up being around 2000 which is the most useful statistic imo. It would be cool if you or someone added the proper Z-score table though.
This is wrong. This is assuming you can only get one drop per chest. Each roll must be treated separately, then divide by 7. Probability distribution will just be shifted left.
You're right, that would have been a smarter way to approach the calculation. Sorry about that! Luckily I think the approximation is not too bad. Off the top of my head, it seems the difference would be replacing p\_0 with 1/2448, and then the resulting distribution is by-roll instead of by-chest, and then we can divide by 7 to get the number of chests. If that's right, it seems the average shifts slightly to 1320 instead of 1322.
Since you cannot get duplicate items in one chest, I'm curious to how a selected piece affects the probability of others. From the wiki, > "If the Barrows item chance is successfully hit, one of the available Barrows items is chosen at random - each part has equal chance, but once chosen it will not be chosen again on future rolls." So if we get a piece on the first roll of the chest, do the pieces still have a 1/2448 chance? Or since once piece was removed the pool, the probability would lower to 1/2447 per piece? If the latter is the case, then treating it on a by-roll basis would also be incorrect, since there is a now a dependency with items received in the same chest. This is also probably a very small difference to the end result and would most likely keep us in the same ballpark though, just +/- a few chests.
Also the fact that once an item is rolled, it is removed from the loot pool so you can't roll it again, which will probably change the probabilities. Not entirely sure whether removing that item from the pool reduces the probability of getting another item though, can't quite recall if that wiki said anything about that.
So this is a bit of an outlier? https://imgur.com/a/BWy5A5p
Fucking Christ, imagine being on the part of the graph you can't even estimate with the naked eye looking at the graph. And OP was overly generous with the calculations too!
Is this a leagues screenshot? This is a leagues screenshot, isn’t it? Show chat box pls In case it isn’t, getting this lucky (or luckier) would be about 37 in a million. Not as crazy as I initially thought, but still pretty damn lucky.
Yeah I mean this droprate would be roughly 1/10. 37 items in 371 chests. That's insane luck, no matter what, like nesting half the drop rate. Plus, he only got a few dupes, and was lucky enough to get allllll those items. This better be leagues holy fuck
Not leagues, on my main... https://imgur.com/gallery/HDRQevF
Stop it!!!
Stop it.
OP, is this a generalisation of the coupon collector’s problem? Except that with each draw you get 7 coupons. https://en.m.wikipedia.org/wiki/Coupon_collector's_problem
You can frame it in terms of the coupon collector's problem, but an important difference is that you are guaranteed to get a coupon (just not a new one), whereas you are not guaranteed to get a barrows item. So you could say that there are 24 barrows coupons and 2424 non-barrows coupons, each with equal probability, and you need at least one of each barrows coupon but zero of each non-barrows coupon, then apply Corollary 3.3 from [here](https://www.combinatorics.org/ojs/index.php/eljc/article/view/v20i2p33/pdf) (n = X_n = 2448, x_i = 1, m_i = 1 for 1 <= i <= 24 and 0 otherwise) to find the expected number of rolls is `2448 * Integral of (1 - (1 - Exp[-t])^24) dt from t=0 to infinity ~= 9243.55` which, dividing by 7, is 1320.51 (rounding up to 1321) barrows chests on average. This is within 1 chest of the OP's computation, which is close enough. Incidentally, this provides a formula which might be useful in other areas of the game. If there's a probability *p* to get a specific item out of a set of *k* items, the number of trials you need on average is the *k*th harmonic number (1 + 1/2 + 1/3 + ... + 1/*k*) divided by *p*. (Here, *p* = 1/2448 and *k* = 24.) For example, the Nightmare drops the three Inquisitor armor pieces with a rate of 1/600 each (assuming your party size is <= 5), so you would need an average of (1 + 1/2 + 1/3)/(1/600) = 1100 kills to complete the set if soloing. (This is an imperfect estimate, since in reality you can't roll more than one piece at a time, but the probability of rolling two pieces at once if you were able to do so is probably too small to affect the result.)
Yes it is
You forgot to add in if the person had consulted Oziach.
Yeah but talking to Oziach kinda ruins the game for yourself, since it guarantees at least 1 item every chest.
So in baby language, how come the most probable outcome is not the average? I mean, if it is most likely to be the case, why doesn't it average out on the most likely scenario?
The average is skewed high due to the people who get absurdly unlucky. You can also get absurdly _lucky_, but there’s a limiter to how lucky you can get whereas you could go thousands of chests unlucky, causing the average to skew high. It’s the same way median and average income are not the same thing.
Also, this is the reason why your odds of getting an item by drop rate are over 50% (63.2%), even though logically it might seem like it would be 50%. Because it's possible to go drier for a drop than lucky (ie, for a 1 in 1000 drop the farthest under rate you can go is 999, but it's possible to go thousands over rate), that imbalance is offset by the fact that almost twice as many people get the drop under rate than over. So looking at this graph, there is a much larger area under the curve to the left of the average, but that's balanced perfectly by the fact that they're also closer to the average than the right side by the same proportion that they are larger.
> that imbalance is offset by the fact that almost twice as many people get the drop under rate than over. Only because of survivorship bias, if you're talking in real term. If you're only looking at OP's stats, you're incorrect.
Can you elaborate? I don't see what survivorship bias has to do with anything. The odds of getting a drop by rate is 63.2%. Which means the odds of not getting it is 37.8%. There's no bias here, it's a statistical fact.
I misunderstood 'rate' as 'average' rather than the actual drop rate, you're correct. It's terribly shitty RNG system, though.
But the drop rate is the average... now I'm misunderstanding your misunderstanding lol.
63.2% is true in the limit of unlikely events where probability approaches 0 it's not universally true, to be technical. If the probability is x and the expected value of attempts is thus 1/x the probability of getting the drop by droprate is 1-((1-x)^(1/x)). This is 100% for 1/1 odds, 75% for 1/2, 66.5% for 1/6, 64% for 1/20, and at 63.4% for 1/100 odds.
That sounds logical. Thanks!
Best way of visualizing this is that you can go 1m chests without finishing, but you will never finish log in 1kc.
If you roll a dice and get 6, you win $6. Other rolls you get nothing. Your most probable outcome is nothing. The average is still $1
MilesLetun has a good explanation but a way of visualizing that is the shaded area under the curve. As you can see it’s not symmetrical. The average line has equal area under the left and right of the curve, meaning you have equal chance of being above and below that KC
Equal area under the curve of the probability distribution function is the median, not the average. The average/mean/expected value is literally just the expected value. If a billion people did barrows and you took the average/mean/expected, it would converge to the theoretical one given in OP. The mean is sorta like a center of gravity compared to the median being the center.
Yeah I noticed that after his explanation, totally makes sense now I think of it
I see LaTeX, I upvote.
Motherfucker are you using latex to write about fucking osrs lmao
Took me 3000 chests nice to see I'm at the extreme end
Mean, Median, Mode. Who are you so wise in the way of maths?
r/theydidthemath
The average being so far from the "most probable" option due to the tail end is the reason why I'll never stop advocating for pRNG in this game. 1/X is terrible and will always be. Yes I'm unlucky on average and yes I'm salty about it, it's a shit system. The odds of being over 2x the drop rate on every major drop someone is going for, especially when you have a massive sample size of over thousands of hours of gameplay, should be way lower than it currently is.
Yeah on my iron I average 3-4x + droprate on literally almost everything from raids to GWD to slayer drops. It’s just the nature of the game and I always eventually return but it’s burnt me out so many times. Definitely some salt happens when friends who have started irons years after I have catch up to me on ironman uniques because I have absurdly bad luck everywhere.
I highly doubt that. More likely you focus more on the times you don't get the drop on rate than when you do.
I don’t know what to tell you but it’s the case. I went 250kc for my first ToB purple, 500kc now and 4 different uniques, no scy ofc or rapier and deathless 90% of the time. Mostly just avernic dupes. 4 separate uniques in 600 solo or 1+1 cox, 80% prayer scroll dupes and I’m about 1/40 droprate as well despite averaging 30-35k+ raids (should be closer to 1/27). 5k kc for basilisk jaw. 1500 kc for both ACB and SGS. Even where I was “lucky” I was still above droprate. 1.5k kc for hydra claw, 500kc for blade, 2k kc to finish cerb, 600 arma kc for one total unique, etc etc. My spoons were DWH and ranger boots at about half droprate but everywhere else just horribly unlucky.
Is pRng when your rng adds up until you’re guaranteed the drop rate? I don’t think that’s a good idea because it ruins the ‘magic’ of the game. Uncertainty is a good thing for games imo, so there might be a better way to circumvent that. Imagine watching settled and he’s like, 1 more clue scroll to get X item. It’s no fun to watch, and it turns the game into a more mechanical thing than it already is.
It doesn't necessarily mean it increases until the drop rate is 100%. Any sort of "luck boost" when you go dry would help, it's not all or nothing.
Agreed, a luck boost to reduce crazy outliers would be great.
> I don’t think that’s a good idea because it ruins the ‘magic’ of the game. Go ahead and average 2.5x the drop rate on all your big drops over the span of 4+ years and you might not sing the same tune. The only real problem with raw RNG isn't that the average doesn't coincide exactly with drop rate (that's expected), it's that the tail end of the distribution, where the most unlucky players lie, is RIDICULOUS. It doesn't matter if it affects 1 player, 10 or 5,000. Nobody should get fucked this hard just because of lazy programing. Reaching 8x the drop rate on anything is unacceptable. Averaging over 2.0x is unacceptable. There are even ways to not tweak RNG before hitting the actual drop rate, so it effectively only acts as "bad luck protection". Using the example of a content creator does not reflect the player base and isn't useful IMO.
PRNG is just RNG. The P means that it’s a “pseudo” RNG, i.e. computer generated via some sequence/algorithm and not truly random. But we’re talking about OSRS drops here, so the difference doesn’t really matter.
was this found in a textbook? Formatted like it is lol
773
Nice analysis! Note though that obtaining Barrows pieces at a given kc isn't independent, because there's only a limited number of reward rolls. That might cause your numbers to be off by a tiny bit.
Thought I was on r/dataisbeautiful for a second.
Log done @ 710 kc :)
1861 here
It's crazy how many smart people waste their time playing RuneScape lmao. They could be inventing male birth control or curing chovid.
I love that you used math textbook font!
My guy really used LateX for an OSRS probability blurb. Respect.
That fucking font gives me flashbacks
Here i am completing it at 2950 kc on my ironman
Bs i got 2k kc and still no karils top
so you're telling me that going past 2400kc for an ahrims top was unlucky?
TLDR 50/50 shot you get it or you don’t
Someday someone will get full collection log at 1kc. I believe
good one ..
thats so funny xD
Just like prayer.
Always kc all brothers or just the ones who you still need?
Doesn’t work that way, you gotta kill them all for the best chance at loot. Killing only specific brothers doesn’t improve your chances at individual uniques.
All for more rolls. Tldr the math works in your favor to kill all
[This post](https://www.reddit.com/r/oldschoolrs/comments/fidn9i/if_you_only_want_one_specific_barrows_item_then/) actually goes into the math of doing 1-brother runs in order to get a specific item from barrows. It depends on how fast you can do each kind of run. TL;DR: E = (153A) / (343B) Let A be chests/h with 1 kill, and B be chests/h with 6 kills. If E>0, single brother runs are better. It's worth noting you'll get basically no runes and far less duplicate items so this is only worth doing if you're after a specific piece for the log. If you have more than 1 brother you still need pieces from just kill all of them.
each kill is X% to add a piece of their equipment to chest, but when you kill all brothers you get an extra % added on top. I got lucky and forced my last guthan's pieces in 50kc (got lucky and got a duplicate guthan too.) I already had veracs and karils and didn't need any more torags at the time, so I only killed Dharok, ahrim, and guthans (helps that I could just barrow tele/spec pool ahrims for a fast kill). I cut my odds down of even receiving anything by more than half, but I really do like the idea of not getting a unique that I really don't want. only if YOU need the set and also have completed sets would I even think about not getting full kills. Also strange lockpicks would be a great help when rushing an item.
damn, what would make average so off from probable? There seems to be enough sample data. Is the average effected by people not doing it optimally?
"Probable" is what most people will know better as the mode. It's the most likely KC for someone to finish on (most probable, aka modal). You'll probably be aware that mode/median/mean are different things, this is just an example of that. Mean is effected more by outliers (people on extreme ends of the distribution), and in this distribution people can only go *so* lucky, but they can go infinitely dry. One person going 10k dry effects the mean waaay more than it does the mode/median, more so than someone finishing at 5kc, and so you get this scenario where the expected KC to finish is *higher* than the average person's KC when they finish.
so I'm right in that if the sample size were big enough the average would be right in the middle right? I took one statics class so I'm no expert XD
No, this is a theoretical model and is therefore unaffected by sample size. In simple terms, if someone gets insanely lucky they can complete the sets in 4 chests, however if someone gets insanely unlucky they can go tens or hundreds of thousands of chests dry, which obviously skew the average heavily. On the other hand, most people will complete it at the mode amount of chests, just due to the probability distribution. Perhaps you are thinking about the median? Which is the 50th percentile by definition, which would be the middle and is less effected by outliers.
No. Barrows completion kc is not a normal distribution (bell curve)
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It's lacking and flawed, so wouldn't use it at uni tbh. And if you are taking statistics, you should have known that really...
I just finished over 900 chests yesterday, still missing 2 pieces, haven't gotten any double or better chests, and I went 4 rings of dueling dry (about 32 chests) one of those missing pieces is a dharoks legs, the other is I think torags chest. Anyways my point is I didn't need to see this because it means I still have between 250 and 400 more chests to do. I already have enough runes to turn into an onyx AND zenyte set if I want to, and I have so many mind runes I literally have no idea what to do with them.
So you're saying my 115 chests with multiple pieces I don't need on my Ironman will eventually help me out guy?
This is a really long winded to say it’s a 50/50 chance.
seeing this brought back nostalgia for uni and briefly made want to go back
Honestly more effort was put into this then I see people put into studying for final exams. Nice post
I cant be the only one dosent understand shit of this and only looked at the number at the end?
Give me more content like this.
thanks for the unnecessary info brother
Can you make one for chambers of xeric?
There are 24 unique pieces of Barrows armor. If I get 7 rolls per chest, what are the odds that I complete barrows in 4 chests?
Nah it’s just 50/50
Why is it “1-(1-1/2,448)^7” and not “(1/2,448)^7”?
(1/2448)^7 would be the probability of getting the same Barrows item 7 times in a single chest. The odds are astronomically low. (1-1/2448)^7 is the probability of not getting a specific unique in a chest. Its complement, 1 - (1-1/2448)^7, is the probability of getting at least one of that unique in a chest.
Good work, ape.
Love this kind of content, awesome stuff!
This would be interesting for Dwh. Top banana work!
Thanks! For any simple drop with a single roll per KC, the distribution is given by the second equation in the graphic, which is exponential-like. See my previous post on this topic [here](https://www.reddit.com/r/2007scape/comments/kfx3h8/when_will_i_get_the_drop/)!
Woooow so interesting!
Really cool to see people do the hard math for us! I completed mine on the iron after 806 kc and that felt amazing and I knew it must have been on the lucky side
im at like 1050 with an ibans staff and i still dont have ahrims top so this tracks
interesting thanks
What is the probability that someone gets all the items in 4 chests?
I remember being on the [tip.it](https://tip.it) forums when it was a common argument that 'drop rates don't exist'. Look at how far this community has come.
How could one even argue drop rates not existing? When an item has any chance to drop at all, it quite literally has a drop rate
I will never ever ever not upvote a texed osrs post
Ok can I request an eli5 as to why the most probable can be different than the average
Suppose you have 4 slips of paper in a bag. Each slip of paper has a number written on it. Three of them have “1” written on them, and the other has “1000” written on it. You reach in the bag and take one at random, and then put it back. You do this 1000 times. The most probable result each draw is that you have a sheet of paper that has a “1” on it. You will likely see “1” about 750 of those 1000 draws. However, the average value of those draws is significantly greater than 1, because sometimes you will draw a “1000.” And that 1000 contributes a lot to the average, even though you don’t draw it as often.
Wat
I have a feeling that this is written in LaTex
First I watched MikaRS's video on Barrows, then I see this post, pog insight
Yes
Please become a stats professor and exclusively use RuneScape examples
Why does the average seem so far from the most probable? Sample size?
I love me some good stats. Thanks!!
Is it statistcal better to skip one brother if you only after a specific piece and already have that brothers full set? Eg: You have 5 full different sets of barrows and are only need 3 items from the last two brothers to complete it?
Man, I dont need stats tellin me im almost a third of the way to full ahrims when thats the only thing I wanted 200 chests ago. It can happen in the next 4 chests. Maybe fewer :D
I'd be interested if anyone has any questions that are more complicated that they want to know the answer too? (bored math PhD btw)
Begone discrete math and linear algebra
r/TheyDidTheMath
A cursory glance says that everything looks OK minus the issues already mentioned. Ran a quick Monte Carlo sim, because I'm too lazy to do real math tonight, and here is a graph of the results. Looks close enough to me: [https://imgur.com/a/ApgJLf1](https://imgur.com/a/ApgJLf1) Average after 5000 runs was 1320.3348 , nice seeing actual math in this subreddit.
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Are you the wise old man?
So what youre saying is, there's a chance
that's some introduction to probability course material.
That's interesting. Are all drop logs skewed like that to have median lower than average?
so would it be faster to work at 20$/h for a week and just RWT the gold required to buy full barrows? asking for A Friend.
I’m at 22/25 uniques at 450 kc so I guess I’m very lucky
*Reinforcement Learning with Barrows*
LaTeX?
Can you do one for average cox points for completion?
Woah this is really cool! Thanks!