Hearthpwn

Quote from rolaran >>

I bought 5000 Coins last week for 35€. But when i want to buy the Preorder-Pack it only shows the 49,99 Payment Method and not the Coin-Method. On other Packs the Coins are available. I do not need the Cardback, so is it possible that the Coins are available with the launch?

Hi, I noticed that, since yesterday, Hearthstone listed under games that give coins back on amazon https://www.amazon.com/s/ref=sr_pg_1?rh=n:2350149011,n:!2478760011,n:!2492217011,n:10238536011&ie=UTF8&qid=1509813546&ajr=2. The amazon page doesn't give any information about the deal itself, only that it exists, and it is applied to one or more in-app purchases.

Also, on disguisedtoast webpage he mentions that the last deal started on 1Nov https://disguisedtoast.com/articles/27-how-to-get-cheaper-hearthstone-packs-with-amazon-coins. I didn't see any mention anywhere else though.

I tried to pre-purchase, but 1-The button is not there yet on the Amazon version of Hearthstone for android and 2-The purchase of coins never finishes to load, so I can't get the coins for the moment (apparently they will send me an email when it is processed). 

I'm on EU servers, so that would explain a certain delay. Has anyone on US been able to pre-purchase with amazon coins yet? Did you get coins back?

I don't understand why you want that information.

Yes, I have Nat Pagle, and with that system you would know that 10% of users have Nat Pagle. But I got him in a pack, randomly. The system wouldn't tell the difference if i crafted it because it was good for every deck (nope), or was a key card of a certain deck (nope) or simply got him.

The numbers you will probably see: the same % for every legendary. The amount of legendaries I crafted is very small vs the number I got in a pack. From 40 leg I have I crafted only 4. The best legendaries everybody crafts will have a veeeeery slightly greater %, but not enough to tell them apart from the random never-used legendaries obtained in packs.

(Sorry for my english)

Allows for board clears as soon as turn 2 if combined with Circle of healing. 

Quote from Hadesitotedapapaya >>

Well,Deathwing is already corrupted,so he cant be corrupted anymore,sooooo it might be a spell cataclysm themed.

Assuming the same rarity distribution as past expansions (37%-27%-20%-15%), and that you disenchant every golden card you get and every nongolden past 2 copies, 150 packs will get you on average:

• 100 % of the commons (of those,  99% doubled, 1% singled)
• 98.7 % of the rares (of those 93.3% doubled 5.4% singled)
• 68.4 % of the epics (of those, 32% doubled, 36.3 % singled)
• 27% of the legendaries
• 8500-8700 Dust (from disenchanting 433 commons, 87 rares, 4 epics, 1 legend, 9 golden commons, 11 golden rares, 3 golden epics)

I didn't consider the case when you get a golden card of which you didn't have the normal version, and decide to keep it instead of disenchanting it. Let me know if you want another simulation with different parameters, or want the formulas to try yourself.

Today I logged in to see the 18 deck slots. They were there, as they should, but my happines came from an unexpected source. A button to go to the Collection from Ranked play window!  and it works both ways!!! This is one of those times when you don't want something until you see it

Thanks, good to know. Entomb is a very good card as-is, allowing to choose the druid version would be too much. 

The fact that the Druid of the claw and Ancient of war work differently despite their texts being the same is widely known. If you return an Ancient to your hand, you can choose the buff again, but if you return a druid of the claw, the stats are "fixed". Does the same happen with entomb? (i.e. Can I choose the version of Druid of the claw if I entomb it and draw it later?)

I was playing vs a druid near fatigue and he played a druid of the claw in charge mode. If i could get a taunt next turn, i would win, but if it was fixed in charge mode i would lose. 

Couldn't find anything on the advanced rulebook. Any of you faced that situation in a game?

To the OP: I found the "get the cards that are competitive" concept a bit vague. If you provide the number of commons, rares, etc you want from each set, and how many of them you already have, I can do the math and tell you the average packs needed. Here's an example. If:

• You want every card in a set, at least one copy
• You start from scratch (you have no cards from that set at the beginning)
• Only packs from one expansion, with the size of GVG (123 cards)
• If you have 3 goldes, disenchant one
• If you have 3 normals, disenchant one.

Then the evolution of your collection (one type of pack only) goes like this (% means cards you have at least one of)

Quote from Abzeroth >>

Holy crap! 100 dust a pack?! You're probably the luckiest player I know. The most I ever get is between 40 (1 rare)  and 60 (2 rares). 

there are only so many commons and rares that you need to compete, that having 3140 dust would easily round out a deck with crafting.  

The time of the day you play is also a factor. In my country, playing from 19:00 to 2:00 is suicide. The best players usually play during those hours. Early in the morning you may find office workers playing one game during a bathroom break, instead of the college student who plays games until 6 a.m. and that is now asleep. Which one do you want as an opponent? If you want legend, there is no shame in avoiding opponents better than you and preying on weak ones. That's the only way you have to affect matchmaking, use it! (obviously consider your job/study obligations if you have them. I'll assume you're a sensible person)

PS:

Quote from iMPose >>

• Anything above 60% is a very good winrate.  You might be tempted to think that playing perfectly = 100% winrate, but the game has too much RNG for that.  If someone advertises their deck as "85% winrate to Legend!," know that they A) are lying, B) got very lucky, C) are very practiced with the deck, D) legitimately broke the meta, which will quickly self-correct, or E) some combination of those

I'd like to add F) They are ommiting the sample size. I can get a 75% winrate with my decks every day (if I only consider the last 4 games played, that is). Way more useful to me is that they said "Rank 20->legend in 400 games", you can't cherrypick that number as easily as percentages. 

Quote from klaassmeerkaas >>

Nice explanation, very clear. I must admit that I'm (no longer) skilled enough in statistics to check whether you've covered every win condition. Did you take into account that for instance (3,6,0) is also a win condition?

 

Yes, it is considered. To get to isntance (3,6,0) you can either come from (3,5,0) and hit minion B, or come from (2,6,0) and hit minion A. Both (3,5,0) and (2,6,0) are considered, and any instance of (3,6,0) would be included in them and would yield repeated cases. The reason I didn't analyze any higher win condition is because once you get to (3,5,0), anything that happens later results in victory. It doesn't mean that the game doesn't continue, only that further hits are irrelevant.

Quote from klaassmeerkaas >>

In any case I ran the simulation again, you'll find the source code in the spoiler below. I'm assuming that bouncing blades only stops when a minion dies and thus continues even when your opponent has 0 or negative health. I don't know for sure if this is the case but this is how it works with Coldlight Oracle (in fatigue): if the first drawn card is lethal, then the second card will still be drawn.

 

Quote from klaassmeerkaas >>

I can verify the top two cases analytically quite easily, they seem to match which gives me quite some confidence that simulation is accurate. Unfortunately there seems to be a mismatch between the calculated value (15.05%) and the simulated one (23.63%). Any ideas what's the root cause? feel free to check my code for mistakes, I make those every now and then :)  

Found the error.

Don't worry, I revised your code and found nothing odd (and learned some matlab tricks along the way). The error was mine: i forgot the winning  scenarios (1,7,0) with last hit on minion B, (4,3,0) with last hit on minion A, and all its versions with different damages on Taurisan. Those cases would kill one axe flinger, yes, but the 2 damage would happen anyway and win. I'll asume those 8 cases add up to the 8% chance missing. If I have a moment during the weekend I'll update my results. 

Challenge accepted!

Let me start by defining some word i will use.

A scenario is the final situation when the spell stops or you win, describing how much total damage each minion took. There are winning scenarios (3 dmg to flinger A, 5 dmg to flinger B, 0 to Taurisan), losing scenarios (0 to A, 0 to B, 4 to Taurisan) and impossible scenarios (99 to A, etc)

A sequence is a way of hitting the minions that leads to a scenario. I will say A when the hit is for the small flinger, B for the big flinger, and T for taurisan. Example: the sequence ABABABBB leads to scenario (3,5,0). There are other ways to arrive to (3,5,0) 

Chances of a given scenario:

The problem can be simplified knowing that all minions stay alive while the spell is running (Because if not, his death would stop the spell and you lose). So each minion has 1/3 chance of being hit, always. The chances of a scenario is:
(Number of different sequences that lead to it)/3^(number of bounces required)
For examples, the chances of getting to scenario (1,1,1) are 6/3^3=2/9 (6 sequences to arrive: ABT, ATB, BTA, BAT, TAB, TBA and 3 bounces long each)

Number of sequences to a given scenario:

Once you decide an scenario, the number of ways it can be reached is (! means factorial, and the formula comes from https://en.wikipedia.org/wiki/Permutation#Permutations_of_multisets):
#Sequences= (# total hits)! / (#hits to A) ! · (#hits to B) ! · (#hits to T) !

Sometimes a scenario requires that the last bounce falls into a specific minion (to avoid continuing counting after you already won). If that is the case, the last letter of the sequence is fixed, so is one less hit to randomize.

Winning scenarios, and their chances:

(3,5,0) = 8!/(3!·5!·3^8) = 56/6561
(3,5,1), last hit to A = 8! / (2!·5!·1!·3^9) = 56/6561
(3,5,1), last hit to B = 8! / (3!·4!·1!·3^9) = 280/19683
(3,5,2), last hit to A = 9! / (2!·5!·2!·3^10) = 28/2187
(3,5,2), last hit to B = 9! / (3!·4!·2!·3^10) = 140/6561
(3,5,3), last hit to A = 10! / (2!·5!·3!·3^11) = 280/19683
(3,5,3), last hit to B = 10! / (3!·4!·3!·3^11) = 1400/59049
(2,6,0) = 8! / (2!·6!·3^8) = 28/6561
(2,6,1), last hit to A = 8! / (1!·6!·1!·3^9) = 56/19683
(2,6,1), last hit to B = 8! / (2!·5!·1!·3^9) = 56/6561
(2,6,2), last hit to A = 9! / (1!·6!·2!·3^10) = 28/6561
(2,6,2), last hit to B = 9! / (2!·5!·2!·3^10) = 28/2187
(2,6,3), last hit to A = 10! / (1!·6!·3!·3^11) = 280/59049
(2,6,3), last hit to B = 10! / (2!·5!·3!·3^11) = 280/19683

TOTAL = 15.506%

Conclusion:

You are in a bad situation. It is more likely to get a legendary from a unstable portal!.

Small sample size, but I filtered the replays I had from the last 900 games or so, to see wich dream cards were more frequent. The number of "games where X dreamcard was generated by me" was:

Laughing sister 19
Ysera awakens 14
Emerald drake 27
Dream 20
Nightmare 30

I don't know how to count individual dream cards generated, it would require me to join thousands of XML files into one.  Also, I don't know what the OP means by "lately", last week?

Conclusion: I feel that laughing sister appears not more often, but in the worst moment possible. Unfortunately, my own data doesn't support that feeling. Gotta stick to the facts I guess