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Post by JRandall0308 on Aug 16, 2017 15:13:13 GMT
This is the kind of question that would normally go on Mathematics Stack Exchange, but I am too lazy to make an account there. Plus I can't figure out how to make the fancy math sub- and super-scripts.What is the probability of getting at least (k) Ultra-Rares in (n) tries?I [ found] what I think is the correct formula. Where (k) = the number of URs we are trying to get (n) = the number of packs we're opening ('tries') (s) = 0.25, the assumed / derived chance to get a UR in a pack [note: ignoring chances for > 1 UR per pack because this would make the math super complex] Plugging numbers into this formula in Excel gives me the following results for "what is the probability of getting at least 3 URs in (1, 2, 3, ... 10) packs", which seem... way too high. In particular the row for n=2 doesn't make any sense. (There cannot be a positive probability of getting at least 3 URs in only 2 packs.) k= | 3 | n = 1 | -0.083333333 | n = 2 | 0.1875 | n = 3 | 0.390625 | n = 4 | 0.54296875 | n = 5 | 0.657226563 | n = 6 | 0.742919922 | n = 7 | 0.807189941 | n = 8 | 0.855392456 | n = 9 | 0.891544342 | n = 10 | 0.918658257 |
I also found [ an online calculator] (bottom of page) that gives the following probabilities when you plug in values. p = 0.25 [chance of success; was (s) in the formula above] N = 10 starting k = 1, 2, 3, ... [minimum number of URs we want the probability for, in 10 tries] ending k = 10 [maximum possible number of URs we can get in 10 tries] Pr(X >= k) or Pr( a <= X <= for k = 1 --> 0.9436... which matches the (relatively easy) brute force math I can do for k = 2 --> 0.7559... which seems way too high... your chances of at least 2 URs in 10 packs is 75%+??? for k = 3 --> 0.4744... which also seems pretty high for k = 4 --> 0.2241... still seems too high etc. Is anyone good enough at math to show me where I've gone wrong?
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Post by N7Mith on Aug 16, 2017 16:13:14 GMT
Is this some inside bff information again?
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Post by JuggerNuggss on Aug 16, 2017 16:35:12 GMT
Me looking at this like
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Post by JRandall0308 on Aug 16, 2017 18:15:39 GMT
A guy on reddit explained where I went wrong. The corrected table: n | Prob(k≥3) | 1 | 0 | 2 | 0 | 3 | 0.016 | 4 | 0.051 | 5 | 0.104 | 6 | 0.169 | 7 | 0.244 | 8 | 0.321 | 9 | 0.399 | 10 | 0.474 | 11 | 0.545 |
What this is saying is that your chances to get at least 3 URs in 10 packs is 47.4% (with the simplifying assumptions of never > 1 UR per pack, and no pity timer). The reason this is relevant this week is that with APs on sale on Xbox, you could buy enough APs for 3 Apex Elite Packs or 10 P/A/R packs. The math above shows that a little less than half the time, the 10 P/A/R packs are at least as good as 3 Apex Elite Packs looking only at URs (and obviously far better if you care about Rares and other stuff). TL;DR don't buy Apex Elite Packs.
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Post by The Biotic Trebuchet on Aug 16, 2017 19:39:00 GMT
This gave me flashbacks from my time in the University
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Post by JRandall0308 on Aug 16, 2017 19:51:36 GMT
Even more fun with math courtesy of another guy on reddit: note: ignoring chances for > 1 UR per pack because this would make the math super complex] It actually doesn't make it all too complex, you can just used your assumed net probability for the pack and extrapolate the probability each individual card in the pack is an UR by finding the fifth root of the inverse of the total probability and then subtracting that from 1. The reason this works is because the probability that you get at least 1 UR in a pack is can be defined the 1 - the probability that you get no UR cards in a pack. i.e inverse of total probability = 1 -0.25 = 0.75 (the probability you get 0 UR in a pack) fifth root of previous = 5 √0.75 = 0.944 (the probability each card has of not being an UR) inverse of that solution = 1 - 0.944 = 0.056 (the probability each card has of being an UR [what we care about] ) That is to say, the probability any individual card will be an UR is roughly 5.6% [based on your numbers] . Then you treat each pack as 5 individual tries and you can find the probability that you get ≥ 3 UR in 10 packs or 50 cards. To do this we apply the formula you mentioned, which plugged into wolfram alpha gives us: 0.536. In other words accounting for multiple UR in a single pack, the probability of getting at least 3 UR in 10 packs in is actually 53.6%
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Post by Kenny Bania on Aug 16, 2017 20:32:11 GMT
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Post by JuggerNuggss on Aug 16, 2017 20:59:10 GMT
I genuinely laughed out loud
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Post by ddrguy300 on Aug 16, 2017 22:41:33 GMT
While I certainly give you respect points for the, what I can only assume is real, complicated and sophisticated math...
What does this help exactly? We all know the drop rates...
Was this just a nerd-edong topic? Because you win all the convention credits.
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Post by pitka17 on Aug 16, 2017 22:53:38 GMT
Your estimate for Premium Packs seems quite reasonable, considering ~415-425k per UR via premiums. At 1M spent you are sitting at nearly 2.5 UR's on average. So roughly half the time you get 3+, and half you get 2 or less.
Overall though, shouldn't you apply the same logic to APEX packs. Suppose you buy 3 APEX packs, what is the likely-hood of getting 1 or more EXTRA UR? Based on the limited data you presented in the "What did you get from your elite pack" thread, lets say there is a 33% chance of getting at least one bonus in any one pack... Purchasing 3 APEX Elite's Odds of getting 0 bonuses (3 total UR): 28% (0.66^3) Odds of getting at least one bonus (4 or more total): 72% (1-[zero bonus pack value])
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Post by JRandall0308 on Aug 17, 2017 1:01:06 GMT
Well the point of the thread was a thought exercise to see if 10 Premium packs or 3 Apex Elite packs is a better use of APs. But as pitka17 points out, we would need to factor in the chances of getting *more* than 3 URs in your 3 Apex Elite packs. And even though I posted the 'data', I'm not confident enough in it to use that to make a decision. Basically I am too cheap to buy the Apex Elite pack especially *knowing* that it is likely priced too high because of Fourfold Pattern / Prospect Theory reasons. I have just enough risk tolerance that I would rather roll the dice with Premium packs over and over.
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Post by akots on Aug 17, 2017 1:28:48 GMT
Just keep in mind that this equation only really works well if you have normal random distribution of random numbers. Assuming that this is some regular computer generated pseudo-random number, there might be substantial deviations in the system. Moreover, it is not know whether "hits" are generated for each card or for only single pre-assigned card and if "hits" are generated sequentially, and whether they are independent events or not. So, strictly speaking, you cannot use neither 0.25 nor 0.75 in this exact equation. However, you can use these numbers for general approximation of probability with unknown margin of error.
My main gripe is that things are not completely independent but are rather determined by certain seed numbers used for various types of pRNG. So, for example, to reach conclusion about 0.25 hit rate, at 99% confidence, you might need either 1000 tests, or 100 tests, or 10000 tests.
tldr, mileage varies.
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Post by ddrguy300 on Aug 17, 2017 1:40:12 GMT
Well the point of the thread was a thought exercise to see if 10 Premium packs or 3 Apex Elite packs is a better use of APs. But as pitka17 points out, we would need to factor in the chances of getting *more* than 3 URs in your 3 Apex Elite packs. And even though I posted the 'data', I'm not confident enough in it to use that to make a decision. Basically I am too cheap to buy the Apex Elite pack especially *knowing* that it is likely priced too high because of Fourfold Pattern / Prospect Theory reasons. I have just enough risk tolerance that I would rather roll the dice with Premium packs over and over. I just go ahead and buy them. Last time it was up I thought, what the heck and bought it. Then realized it's just the one you can buy. So then it's extra "why not" because it's not a traditional "pack" being there's only one you can buy every once in a while, not deciding what string of packs to buy is the best. Crusader Siphon II for me in this one.
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Post by akots on Aug 17, 2017 2:00:15 GMT
Well the point of the thread was a thought exercise to see if 10 Premium packs or 3 Apex Elite packs is a better use of APs. ... From simplistic point of view provided that you have enough money, you can actually benefit in the long run by mixing them up. But it all depends on how pRNG works and how pity timer is implemented. For simplicity, it is possible to assume that premium and elite packs use same seed for pRNG (for elite it is forced to be 100% for one of the cards) and that pity timer is reset every time you get some UR from anywhere (which is not the case with premium packs apparently, but assuming otherwise will make life very complicated). In this case, if you have bought premium pack and got there an UR, you might consider buying elite pack right after that. Which is kind of intuitive but nobody does that for some reason. Because it is akin of quitting after big win in a casino. No, everyone just wants to try again and again until they lose everything. In general, it is similar to boy-girl paradox or other cases of dynamic odds. en.wikipedia.org/wiki/Randomness
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Post by JRandall0308 on Aug 17, 2017 12:35:57 GMT
Just keep in mind that this equation only really works well if you have normal random distribution of random numbers. Yes, I'm assuming things are normally distributed because to assume otherwise rapidly leads to insanely complex math that I can't even conceive, let alone execute. Not *known*, but do we have any reason to believe the events are *not* indepdent? crashsuit opened over 2,000 packs and that's good enough for me. From simplistic point of view provided that you have enough money, you can actually benefit in the long run by mixing them up. But it all depends on how pRNG works and how pity timer is implemented. As far as we know it's quite simple: The 10th packs in a P/A/R streak will contain a UR (guaranteed) if and only if the previous 9 packs (of that same type) did not. The pity timer persists across buying other non-P/A/R packs, resetting your console/PC, exiting the store, game patches, and apparently (for a single test I did) even across buying some packs in-game and some in the Apex HQ app. That being the case, I am not clear on why after I get a UR I should switch packs. If for no other reason than it makes my tracking more cumbersome. Eh? Yes it is. If by "reset" you mean "need to look for a new streak of 9 packs without a UR". I don't see why I would switch, or how this is like the boy-girl paradox. Hitting the pity timer on a P/A/R pack gives us no new information about Expert packs.
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Post by akots on Aug 17, 2017 13:04:37 GMT
... I don't see why I would switch, or how this is like the boy-girl paradox. Hitting the pity timer on a P/A/R pack gives us no new information about Expert packs. But it gives you new information about your next premium pack. Let's assume you buy one pack. There are two outcomes, -UR/+UR, probability of +UR = 0.25 Now, you buy two packs. There are more outcomes, you have -UR-UR/-UR+UR/+UR+UR/+UR-UR . Since your first pack has +UR, you just eliminated -UR-UR and have 3 events left instead of 4. This is exactly boy-girl paradox as you chance of getting another girl after the first girl is lower than the first girl. As you have predefined pseudo-random dice throw coming up, you might as well "waste" that "bad" throw with reduced chances onto something that can guarantee change of probability which is buying a different pack. You then can go back to whatever pRNG has in store for you assuming that at least to some extent, worsening of chances has been avoided. Question is whether different types of packs have their own different dice (pRNG sequences of numbers) or pRNG throws are generated for all packs from the same sequence? Hitting pity timer gives no new information about anything, I assume the UR hit was not due to pity timer as it is not the situation that is being considered. But are there any dice throws wasted on pity timer drop? Does it consume a slot of the pRNG sequence? That might make the whole sequences not independent.
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Post by akots on Aug 17, 2017 13:16:18 GMT
Not *known*, but do we have any reason to believe the events are *not* indepdent? ... IDK, we can try to do some test for randomness on the pack data, but pity timer will skewer it as we will never know if 10th drop was pity timer trigger or just regular normal "hit". You can do some simple test like Kolmogorov-Smirnov's test online. www.physics.csbsju.edu/stats/KS-test.n.plot_form.html There is no need to compare anything, just use it to check normality of the array. I have Sigmaplot, I can run some simple scripted non-parametric tests also. There are a bunch of scripts for R (R is open source statistical software www.r-project.org/ ), these might be possible to use as well. In general, any independently generated array should produce results close enough to normal provided that there are sufficient data. Opening large number of packs like Crash did only gives you better approximation of the probability bracket used to determine the outcome (0.25). It has other valuable data that can be extracted as well somehow, just need to ask the right questions.
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Post by JRandall0308 on Aug 17, 2017 13:36:10 GMT
The K-S online tester only wants numbers (not words like 'Weapon_Ultra-Rare' or 'Character_Ultra-Rare')... can I just arbitrarily assign numbers like '1' and '2'? Does it matter what numbers I assign?
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Post by akots on Aug 17, 2017 13:56:14 GMT
The K-S online tester only wants numbers (not words like 'Weapon_Ultra-Rare' or 'Character_Ultra-Rare')... can I just arbitrarily assign numbers like '1' and '2'? Does it matter what numbers I assign? There are multiple possible input formats. I presume the purpose is to check randomness of gaps between UR drops. The test itself is simple, so it won't tell you clearly whether distribution is normal or log-normal. So, you can either input the outcome for UR drops (0/1/2) or input number of empty slots with outcome 0, IDK what works better. Typically, people do it both ways and P value for normality should be similar. If not, then it points out some irregularities. Also, it would be nice to know how skewered limited set of data is. For example, if input is large enough, that would more likely to be random with higher probability than some shorter sequence. The minimal length of the shorter sequence that still gives good P (like maybe 95%) can be a measure how streaky is pRNG. Although, it is all relative and there is not much to compare to. It is always possible to use something more advanced like www.cacert.at/random/
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Post by TheThirdRace on Aug 17, 2017 14:03:30 GMT
akotsThe problem with your affirmations is they don't necessarily apply to MEA. First, it really doesn't matter if it's boy-girl or girl-boy. As long as you get an UR, the pack sequence is moot. Second, you assume buying an different pack will break the probability of bad draw. It's already been proven buying different packs doesn't affect the pity timer so why would we assume it could have an impact on your draw luck? If the packs are isolated systems, which community tests seem to agree with this theory, the "breaking bad draw luck" principle wouldn't apply in this particular case. I think you would be right in most cases, just not for MEA given the system seems different than what you described.
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Post by JRandall0308 on Aug 17, 2017 14:04:47 GMT
Let's assume you buy one pack. There are two outcomes, -UR/+UR, probability of +UR = 0.25 Now, you buy two packs. There are more outcomes, you have -UR-UR/-UR+UR/+UR+UR/+UR-UR . Since your first pack has +UR, you just eliminated -UR-UR and have 3 events left instead of 4. This is exactly boy-girl paradox as you chance of getting another girl after the first girl is lower than the first girl. As you have predefined pseudo-random dice throw coming up, you might as well "waste" that "bad" throw with reduced chances onto something that can guarantee change of probability which is buying a different pack. You then can go back to whatever pRNG has in store for you assuming that at least to some extent, worsening of chances has been avoided. Question is whether different types of packs have their own different dice (pRNG sequences of numbers) or pRNG throws are generated for all packs from the same sequence? And circling back to follow up on this. We don't know how the RNG store works. Basically, does it work like the game XCOM? In that game, the 'random' (I know, pseudo-random, work with me here) numbers are all pre-determined for each mission. Whenever you do something in a mission the game will pick the next number in its list and use that for the result. This is why if you take a shot and miss, then you can reload and do something else to use up that random number (say, shoot a lower priority enemy, or try to Hack something you don't care about) which will force the game to pick the next number in its list. We don't know if the MEA store works like that. For example: you have just bought 5 Premium packs in a row with 0 URs. Unbeknownst to you, the next two 'random' numbers are going to be 0.74 (let's say low is bad so this is a 'miss' and you won't get a UR) and 0.99 (high is good; this will definitely be a UR). Now, can you force the store to 'use up' the 0.74 by purchasing some pack you don't care about, like an Expert pack? The 0.74 will roll and you'll get 'not-a-Rare'. Then on your next pack you go back to Premiums, the 0.99 rolls, and you get a UR. The problem is that we not only don't know how the store works, we don't know what 'random' numbers are being rolled.
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Post by akots on Aug 17, 2017 14:14:40 GMT
The problem is that we not only don't know how the store works, we don't know what 'random' numbers are being rolled. We do know if the number is less than 0.75 or higher than 0.75, that might be sufficient for statistical purposes to check randomness.
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Post by JRandall0308 on Aug 17, 2017 14:20:00 GMT
Results from the K-S tool. I took a list of the # of URs in my P/A/R packs and pasted it in. (So the list looked like 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, ... etc.) No idea what these mean but it says "it's unlikely this data is these data are normally distributed". Huh! Data Set 1:
278 data points were entered
Mean = 0.2590
95% confidence interval for actual Mean: 0.2016 thru 0.3163
Standard Deviation = 0.486
High = 2.00 Low = 0.00
Third Quartile = 0.00 First Quartile = 0.00
Median = 0.000
Average Absolute Deviation from Median = 0.259
KS says it's unlikely this data is normally distributed: P= 0.00 where the normal distribution has mean= 0.6543 and sdev= 0.4768
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Post by akots on Aug 17, 2017 14:21:33 GMT
akots The problem with your affirmations is they don't necessarily apply to MEA. First, it really doesn't matter if it's boy-girl or girl-boy. As long as you get an UR, the pack sequence is moot. Second, you assume buying an different pack will break the probability of bad draw. It's already been proven buying different packs doesn't affect the pity timer so why would we assume it could have an impact on your draw luck? If the packs are isolated systems, which community tests seem to agree with this theory, the "breaking bad draw luck" principle wouldn't apply in this particular case. I think you would be right in most cases, just not for MEA given the system seems different than what you described. I would presume like with any pRNG, that triggering pity timer does not reset the seed number for pRNG. For example, you have predefined sequence of 10 numbers. Pity drop may eliminate a certain number in that sequence or may not, and there is no way of knowing it unless Bioware tells us how it works. So, for simplicity, it might be easier just forget about pity counter and rely only on probability of regular drops. For this, it'll be nice to know whether pity timer drops can be just a minor nuissance or do they seriously contribute to number of URs you can get? Drop rate 0.25 is not that low, I would presume that very few UR drops are actually due to pity counter, maybe something in the range of 10% of all UR. Which makes it possible to throw them out of the equation? So, I don't see how this system is so much different from any regular casino gambling theory. Ideally, assuming that -UR drop is 0.75, having 9 -UR drops in a row to trigger pity counter has a chance of 0.75^9 = 0.075. That means that only 7.5% of all URs you get are due to pity counter, and the number is probably less, something in between 0.075 and 0.056 (0.75^10).
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Post by akots on Aug 17, 2017 14:26:07 GMT
Results from the K-S tool. I took a list of the # of URs in my P/A/R packs and pasted it in. (So the list looked like 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, ... etc.) No idea what these mean but it says "it's unlikely this data is these data are normally distributed". Huh! For this input, you need a binary calculator. This online calculator is not binary, apparently, it needs gap numbers. Can you please link the spreadsheet, I can try to run it through a Sigmaplot script for binary randomness.
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