23 posts found  

I'll explain what I've done to make the results here clearer. Upon starting the game, I wanted to get a good gauge on just how "pay to win" the game is. I did the tutorial and then beat the practice NPCs once each to unlock all of the classes, then went for ranked mode. I made a point of using exclusively basic cards until I had all of themeven declining to open packs that I had earned. I've only played two ranked matches with a class that had all of its basic cards, and even then, only to do classspecific quests. With this approach, I quickly got to rank 20, then subsequently bounced around between ranks 20 and 16. I probably lost a little more than I won, but not that much moreand never fell to rank 20 with zero stars, the minimum possible after reaching rank 20. Remember, this is with exclusively basic cardsnot even common rarity expert cards. After getting all heroes to level 10 to unlock the basic cards, I did a single arena run, going 43 without the card disadvantage. At this point, I had 3 card packs and 975 gold, enough to buy 9 additional card packs. Satisfied that the game wasn't egregiously paytowin, I then spent $50 to buy another 40 packs of cards. Then I opened all 52 packs of cards consecutively. I meticulously recorded the results, which seems easy but is something that most people are apparently incapable of doing. And I'll share summaries of the results with you here. Quick summary: Total cards: 260 White (common) plain cards: 184 (70.77%) Blue (rare) plain cards: 58 (22.31%) Purple (epic) plain cards: 8 (3.08%) Orange (legendary) plain cards: 1 (0.38%) White (common) golden cards: 5 (1.92%) Blue (rare) golden cards: 3 (1.15%) Purple (epic) golden cards: 1 (0.38%) Orange (legendary) golden cards: 0 (0.00%) I also got the legendary Old MurkEye as a reward for getting all of the murloc cards in the process.  There are two other things that I'm interested in beyond the simple totals. One is how the cards within a pack are chosen. The other is which particular cards of a given rarity are chosen. Let's consider pack contents first, again by plain/golden and by rarity. 32 of the packs had 1 plain blue card and 4 plain white cards. That's the minimum guaranteed and the worst possible outcome. 6 of the packs had 2 plain blue cards and 3 plain white cards. Basically, two rares instead of one. 4 packs had 1 plain purple, 1 plain blue, and 3 plain white cards. Basically, an epic instead if one of the commons. 3 packs had 1 plain blue, 1 golden white, and 3 plain white cards. Basically, the minimum except that one of the commons was golden rather than plain. That leaves seven packs, and each of those seven packs had a unique combination. The combinations were: 1 plain purple, 4 plain white 1 plain purple, 1 plain blue, 1 golden white, 2 plain white 1 golden blue, 1 plain blue, 3 plain white 1 golden purple, 1 golden blue, 3 plain white 1 golden blue, 2 plain blue, 2 plain white 1 plain purple, 2 plain blue, 2 plain white 1 plain orange, 1 plain purple, 1 plain blue, 1 golden white, 1 plain white That last deck was something of a jackpot. The legendary, incidentally, was Tirion Fordring. So why do I bring this up? I'm interested in how the cards within a pack are chosen. One peculiar statistic is that of the 9 packs that had a purple, 8 (88.9%) also had a blue. That would be extremely unlikely if the cards were chosen by rejection samplingmeaning, pick 5 cards, and if they're all white, scrap it and pick another 5. So it's unlikely that rejection sampling is the mechanic. Another interesting statistic is that, while 73.3% of my cards are white, only 55.6% of the golden cards are white. Rather, my golden cards tend to have higher rarities. This is a small enough sample size that this could have reasonably happened by random chance. It's not possible to determine without the source code whether the game gives different rarities of cards different probabilities and chooses a particular card, or whether it chooses a rarity first and then picks a card within that rarity. You'll get the same distribution either way. But the latter is easier to analyze, so let's assume that it picks a rarity first and then the particular card of that rarity later. My best guess at a card choice mechanic is that it starts by picking 5 rarities from some fixed distribution along the lines of 1% orange, 4% purple, 5% blue, 90% white. After this, it picks one card and, if that card is white, changes it to blue. If the card is already blue or higher, it gets left alone. That way, you can ensure at least one blue or better in every pack. And it gives you probabilities within range of what I found above.  But what matters is not just which rarities of cards you get, but which particular cards. After all, if my 61 blue cards were 61 copies of a single card, that wouldn't be terribly useful. So let's have a look at how my cards were distributed within a rarity. Let's ignore the plain/golden distinction, as that only matters if you're looking to disenchant cards for dust and we have a small sample size on golden cards. There are 94 white (common) expert cards in the game. I drew 189 of them, which means I average about 2 copies of each card. Ideally, I'd get exactly two of each and be done with common cards, but that particular distribution isn't very likely. Rather, I'd expect to get three or four copies of some cards and none at all of others. A simple model is to assume that all cards of a given rarity are equally likely. With this model, we can compare how many cards I got exactly n of versus the expected number of cards that I'd get n of and see if they're similar. Let's look at white cards first. For small values of n: n me expected 0 10 12.45 1 28 25.31 2 27 25.58 3 16 17.14 4 07 08.57 5 05 03.41 6 01 01.12 7 00 00.32 8 00 00.08 In other words, there were 10 white expert cards that I didn't get at all. By random chance, there would on average be about 12.45 white expert cards that I didn't get. The numbers are pretty close, and surely within what one would expect from random variation. I actually got reasonably lucky, with at least one of more cards than expected, and fewer redundant cards than expected. Next, let's look at blue (rare) cards. There are 81 blue cards in the game, and I got 61 blues. For small values of n: n me expected 0 40 37.97 1 25 28.95 2 13 10.86 3 02 02.67 4 01 00.48 5 00 00.07 Again, the numbers are pretty close, and certainly within what one would expect from random variation. Here, I got a little unlucky, with more cards completely missing and more cards that I have multiple copies of than expected. Though looking at which particular blue cards I got, I'd say I did pretty well, as I got at least one of most of the cards that I particularly wanted. Let's have a look at purple (epic) cards. There are 37 purple expert cards in the game, and I got 9. For small values of n: n me expected 0 30 28.91 1 05 07.23 2 02 00.80 3 00 00.05 Here, I got unlucky, with fewer distinct purples than expected. But at least none of them were redundant, though it's unlikely to get 3 of the same card with only 9 cards in total. Looking at the particular cards I got, I fared worse, as among other things, I'm really not that excited about two pit lords. With only one legendary, doing the statistics above on the distribution of legendaries is pointless. The above statistics are all consistent with the assumption that, given a rarity, all expert cards of that rarity are equally likely. So that's the model that I think Blizzard went with, or at least something within rounding of it. As for how strong my hand will be, I'd contend that having all of the blue and lower cards available but no purple or better is substantially closer to having everything available than it is to having only the basic cards. Many blue cards are clearly stronger than white cards. But epics and especially legendaries are often rather specialized, not general purpose cards that can do everything. Legendaries may well be awesome cards for their particular purpose, but a deck consisting of 28 legendaries (the max possible) and 2 epics probably wouldn't steamroll everything else and, given its dearth of spells and lowmana cards, might not even be a good deck at all. That's probably intentional on Blizzard's part. 

3/27/14 6:56:42 PM#2
Hardcore.


4/05/14 6:14:31 PM#3
IMO Legends are 1% so 2/240 = 40 pack purchase Epics are around 24%. Rare are exactly 1/6 so 16.5%? The disenchanting system is about on par with that so it makes sense. I am convinced there is a hard cap within the system that does not totally run on a RNG but is more a rng within the cap based on YOUR total purchases. The Rare are 100% all the time 1 per pack.I do not believe the Epics and Legendary are totally RNG random but have a base hard cap system than monitors each individual.This again would make sense otherwise you might NEVER get a Legend while some lucky guy might get your legend that was in the last 240 cards purchased. Example is if Blizzard releases say a Legend at the 45...185 interval,if you bought cards at 50184 each time you may never get a legend EVER.This is not the case,i guarantee you will get 13 Legends EVERY 40 packs..Point being the system monitors each person and is not totally rng.That keeps everyone happy knowing they will not go $200 in purchases and not get a single Legend. http://www.youtube.com/user/Napolianboo#p/u/15/rCYLLQCNc1w 

Originally posted by Wizardry While I don't have enough data to pin down the legendary or epic rates very well, you're wildly wrong on rares. For starters, there are 5 cards per pack; you seem to have the impression that there are 6. As explained above, it is not the case that you're guaranteed exactly one rare per pack. One pack had no raresbut it did have an epic, so it fit Blizzard's promise of one rare or better. Another pack had 3 rares, one of which was golden. Overall, I had 61 rares (3 of which were golden) in 52 packs, so more than 23% of the cards I got were exactly rare. I'd regard it as unlikely but not impossible that Blizzard would rig the legendary system so that you don't get too few or too many. While that's doable, it's harder to code and debugand that means greater risk that you botch the system badly. The only game I've seen do anything along those lines while claiming it was random is Wizard 101 in their "fetch 10 of item X" type quests. More likely, they let the various laws of large numbers handle it. If each card independently has a 1% chance of being legendary (I want to say, "if the card rarities are iid", but most people probably wouldn't follow that) and you get 1000 cards, it's pretty trivial that your expected number of legendaries is 10. The probability that you get between 7 and 13 legendaries (inclusive) is about 73.7%. The probability that you get between 5 and 15 (inclusive) is over 92.3%. The probability that you get no legendaries at all is less than 1 in 23163exceedingly rare. Ultimately, looking for outliers is impractical to do empirically, as you'd need too many people to record exactly what they got on every single pack without doing it wrongly. The basic scientist's skill of being able to record exactly what happened on every single trial without any bias of being more or less likely to record it depending on the outcome sounds pretty simple, but my experience with A Tale in the Desert is that most people are completely incapable of doing so. 

4/05/14 9:36:28 PM#5
I got 7 legendary from 80 packs.


4/05/14 9:43:56 PM#6
Thats nice work and some good info on the card rarities , but i dont believe your "just how pay to win " Hearthstone is works for me .. TCG at there root are all about buying decks and collecting cards all of them are , always have been and always will.. so imo the ptw doesnt apply to TCG its a natural mechanic of that genre ... 

4/07/14 9:54:51 AM#7
This is a brilliant thread because it has real content.
The only problem is that I think playing arena is better value overall. This is because:
1 pack = about £1. 1 arena = £1.48
After each arena you get 1 pack. Therefore, for the extra 48p you get to play 39 games (good experience) + whatever you win.
What you win is mostly stuff to craft the cards you need (disenchanting cards to make other cards is very inefficient, so getting the mana dust is very good) + possibly gold.
If you make it to 7 fights, you get enough gold so that you don't have to pay real money for the next arena match.
However, I concede that very few people are good enough to make it freetoplay  maybe only children and students have that sort of time on their hands to get that good. 

4/07/14 3:37:36 PM#8
I decided to buy 40 packs after all, because I have a specific deck in mind to build and I wanted to crack on with it.
I made a spreadsheet of the cards obtained. The results are amazingly similar to the audit above.
The biggest difference is that I got 2 legendary cards  one of them was a golden one!
However, I was not lucky enough to get all the murlocks  so I guess Old Murk Eye will have to wait for another day! :)
I still think that if you want to win games with a specific setup in mind, arena is the way forward because you have the advantage of getting crafting materials.
I doubt that I will be using half the cards I obtained by buying packs alone, but we will see! 

Whether you get your card packs as Arena rewards or from buying them more directly (regardless of whether with gold or dollars), the probable contents of the card packs still matter. Being so good at Arena that you can live off your winnings there forever is very, very hard to do. If you have a 50% chance of winning each Arena match, your odds of winning at least 7 matches in one run are a little under 9%. A pretty good player who has a 60% chance of winning a match has only about a 23% chance of getting to 7. Even an exceptionally good player with a 2/3 chance of winning each match has under a 38% chance. And one should note that this is made harder by the matchmaking tending to put you against harder opponents if you're winning most of your matches. Nor can you count on getting a ton of spare gold from the matches where you do get there to subsidize those where you don't. My arena run with 8 wins got me 150 gold exactlyrefund the entrance fee, yes, but no spare to cover runs where I did more poorly. Even my 121 arena run (Lay On Hands + 3x Consecration helps a lot) only got me 250 goldthough it also gave me an epic card, a golden blue, and the customary card pack, so it was plenty rewarding, but not that helpful in terms of staying in Arena forever. My best guess would be that the win percentage needed to break even in gold alone (excluding other rewards) is in the ballpark of 75%80%and that's exceptionally hard to do when you're mostly going to get matched against high quality opponents and high quality decks because you're winning so much. Now, there are other rewards, of courseincluding an automatic card pack for every arena run. Even so, by playing in arena, you're forgoing the rewards that you'd have gotten from ranked play, and that needs to be accounted for, too. I concluded that for an average player who wins 50% of his arena matches and who does enough ranked play to complete all daily quests, arena would tend to be slightly more rewarding than ranked playbut only slightly. Ultimately, if you're not an exceptionally good or exceptionally bad player, I'd argue that your choice to play arena or not should depend mainly on whether you think it's fun. 

Originally posted by Mors.Magne I'd be curious to see whatever data you're willing to post. 

4/07/14 5:43:00 PM#11
Originally posted by Quizzical
Yes, you could be right.
In addition, one advantage to buying packs that I have just found is that you can replicate the same setup that you have chosen for your arena deck (with the exception of the legendary card, in my case). This must help in terms of practicing with that deck before you enter into arena combat. 

4/07/14 5:44:55 PM#12
Originally posted by Quizzical
I would be willing to post it, but it would be a bit tedious to do  so I'm not sure if I will find the time.
The results look truly unremarkable! They look like the same odds that have been posted on other forums before. 

For what it's worth, I'm currently 4728 in arenaabout 63%. That's substantially above average. If one looks only at gold and ignores quests completed in arena, the net result of some pretty good runs is to have lost about 400500 goldeven though one of those arena runs was free because it was my first. And that's even ignoring the 150160 gold that those wins would have been worth in ranked play. (Of course, if I had played all of those games in ranked, I'd probably have won around 35, as I'd face opponents with stronger decks.) Of course, if one looks at other earnings, I've come out way ahead, with 10 card packs, 85 arcane dust, and 8 cardsincluding 3 golden, 2 rare, and 1 epic. (The golden rare that I got counts as both a golden and a rare there.) 

Originally posted by Mors.Magne Actually, I think that would be of limited use. In ranked play, you'll tend to play opponents with much stronger decks. In Arena, if you don't have much in the way of board clears or card draw, you're going to be really hurting. In Ranked, it's fairly trivial to put quite a bit of each in your deck. And even that is completely ignoring that in Ranked mode, you'll often face decks with several rares and/or legendaries. On net, even a very strong Arena deck would struggle to get past the high teens in Ranked mode. For example, I just had an 83 run with a priest, and most of my wins weren't close. My deck was based around 3x Northshire Cleric, 2x Holy Nova, a Stampeding Kodo, and a Cabal Shadow Priest. Most of the key cards that my deck relied on were free, basic cards. One could readily build a stronger deck than that using just basic cards. One major factor is that in Ranked mode, you can get both extensive synergies among your cards, as you can pick one card while guaranteeing that you'll get others to pair with it. In Arena, if you pick card A hoping to pair it with card B, card B may never come. In Ranked, you can grab two of each and not just have both of the cards, but make it likely that you'll be able to play both together, especially with the basic cards. Another factor is that you can make sure that every card is useful. In Arena, you're going to get a number of draws where you pick between three cards and don't particularly want any of them. In Ranked, if you don't like some cards, you can ignore them and pick something else that will be useful. 

4/08/14 4:06:01 AM#15
Well at the moment, I'm having quite a lot of fun with a murloc deck in ranked play.
When things get really bad, I whip out my Warlock's Lord Jaraxxus, so whatever happens it's rather fun. :o) 

4/09/14 2:27:06 AM#16
Impressive thread OP. Very nice info. Hearthstone is up in my Legendary kind of games status right now :) What I mean is that I go to some extreme personal goal in a game I really like. In WoW that was going for Justicar solo, in D3 going HC and down Diablo after launch. In Heavy Rain it was going to see all endings. While Hearthstone is a perfect casual game, I am going to get Gold capped. 20.000 Gold just for the fun. While I initially paid 80 Euros in packs I no longer do that. Just grinding out Gold in daily and Arenas. I reckon that by Oct I will have reached that point. Still lacking 19 Legendaries and 2 Epics to have a full collection, but Hearthstone is played at 2 hours a day right now. Mostly on the iPad since I downloaded it on the Canadian shop. I reckon that I'll complete the full collection along the way to Oct. 

4/09/14 11:22:38 AM#17
I played last sunday and got 12 wins in an arena run. Got 415 gold and 2 card packs. "Tiny clown, he got wet. I was talking to a psychic and I can't sleep in the ozone. There are too many different peanuts, looking sad. 

Edit: the computations in this post are mistaken. I'll explain why in a subsequent post. Upon opening more packs, I've stumbled across something that surely tells us something about the distribution of cards. At this point, I've obtained 398 plain white expert cards (nongolden common)including cards that I have since disenchanted. There are 94 distinct white expert cards. I've received at least two of each of the white expert cards without crafting any. (The only card I've crafted at this point is Ragnaros.) You might think that's fortunate, as you want 2 of each card rather than an eighth copy of something white. But most people wouldn't recognize offhand how vanishingly unlikely such a card draw is. If each plain card is chosen uniformly at random and independent of all others, the probability of getting at least 2 of each in the first 398 draws is exceedingly unlikely: about 4.76 x 10^(23). If the distribution is not uniform, but the card draws are iid (independent and identically distributed), the probability would be even lower. So I can confidently say that if the game is going to give you a plain white expert card, it does not pick the card uniformly at random from among all possible. If you want to know where that number came from, there are {n + k  1 \choose k1} ways to put n indistinguishable balls (since I don't care about the order in which I obtained cards) into k distinguishable boxes (since I do care which particular card I got): if you have n balls and k1 box dividers in a line for n + k  1 total objects, that's the number of ways to pick which objects are the balls and which are the box dividers. Meanwhile, if we want to have at least c balls in each box, there are {n + k  ck  1 \choose k  1} ways to do it: you start by putting c balls into each box, then have n  ck balls left, so it's the number of ways to put n  ck balls into k boxes. This gives us {491 \choose 93} / {303 \choose 93}, which evaluates to about 4.76 x 10^(23), as stated above. Because those computations will cause overflow errors on many devices (e.g., {491 \choose 93} > 10^(102), and if you take 491! as part of the computation, that's over 10^(1000)), one simplification that is guaranteed to avoid such overflow errors is to compute it as: (303 / 491) * (302 / 490) * ... * (211 / 399) This has the added advantage that, as a product of 93 terms that are each between 1/2 and 2/3, you can get that the number is between (1/2)^(93) and (2/3)^(93) without the need for a calculator. That, I think, makes it clear without the need for a calculator that such an event is vanishingly unlikely to happen by random chance, and so the cards you get are clearly not iid, even if you restrict to white common cards. Of course, demonstrating that the cards are clearly not random in the obvious manner raises the issue that if you think the game's card draw mechanism is biased against you, it might be. This bias above is in players' favor, but if there is a bias, it could also work against you. 

5/04/14 1:48:14 AM#19
The obfuscated pricing of cards is very good for business but very bad for the consumer, which is why I choose to enjoy an otherwise highly enjoyable game like this without paying them a single cent. If this was a €20 where you easily got all the cards, I would have bought it. They'd be missing out on bleeding lots of other cystomers dry, though, so it would be a bad move for them in terms of profit. For many people, the game "appears" cheap, but despite being a low budget title with a tiny dev team allocated, is a lot more expensive than many AAA box games. "Tiny clown, he got wet. I was talking to a psychic and I can't sleep in the ozone. There are too many different peanuts, looking sad. 

Apparently my computations above are wrong. The balls need to be distinguishable, as without it, we're implicitly assuming that we're just as likely to get 2 of card A and 2 of card B as 4 of card A and 0 of card B. The former should be six times as likely as the latter. A quick and dirty approximation is to assume that the number I get of each card is a Poisson distribution and independent of the number I get of each other card. On average, I get 398/94 ~ 4.234 of each card. For a Poisson distribution with expected value 4.234, the probability that I get at least 2 copies of a card is approximately 0.9242. The probability that this happens for all 94 cards is this to the 94th power, or about 0.0006. I'm pretty sure that this overestimates the probability that I'll get at least two of each card, though it might not be by all that much. This obviously is not exactly correct, as it means that the total number of cards that I get is a Poisson distribution with expected value 398, rather than the known value of exactly 398. I can do the exact computation, but it's a bit of a pain. In particular, the numbers involved will overflow any standard 64bit floating point data type, which rules out the use of a spreadsheet. It's an inclusionexclusion argument, so tricks to make the numbers more manageable like two posts above would be awkward or impractical. I'm going to try using Java's BigInteger data type. 