A critique of Hearthstone’s resource progression

I know this is going to be a controversial opinion, but I don’t think Hearthstone’s perfect. What I’d specifically like to address currently is the question, “is Hearthstone pay-to-win?”

Common Counterarguments

On some level, it’s pretty obvious that it isn’t pay-to-win in the strictest possible sense; TotalBiscuit’s Lord of the Legendaries series has illustrated that simply throwing every Legendary card in the game into a single deck only barely competes even at the lowest levels of play. Trump has also posted F2P videos which illustrate that a player can, with a bit of know-how, decent luck, and by exclusive dedication to a single class, achieve success in Hearthstone within a free-to-play context. I do not, however, think that either example actually proves that Hearthstone has a healthy progression model.

TotalBiscuit’s Lord of the Legendaries videos provide an entertaining series of flops with what may be the most expensive deck possible in Hearthstone. It’s not hard to see why he’s not ranking Legendary with it, either: the deck lacks any and all staple cards. The fact that these staples are accessible to new players (sometimes even as part of the free set) really doesn’t prove much, though. Yes, there are essential cards that you receive for free as a new player. Even at the professional levels of a generally P2W game [1] like Magic, as many as 24 of the 57 non-basic-land cards in a $230 professional-level Standard deck (with sideboard) [2] cost less than a quarter. That doesn’t mean that a cheapened version lacking the $30 Mutavaults, $20 Thoughtseizes, or even the $5 Desecration Demons will compete with the original.

Trump’s video series is more convincing; however, several obstacles prevent me from taking it as evidence for a healthy progression model:

  • Trump is a skilled player who knows what he’s doing, and most importantly knows what cards are most worth spending the precious little Dust one receives near the beginning of the game.
  • Trump received lucky pulls in his first few packs for both of his F2P attempts. For his mage, he received a Defender of Argus and a Mana Wurm in his first pack, and a golden Holy Wrath and a Doomhammer in his second pack — effectively giving him his choice of 5 commons, or two rares. His Shaman received a Feral Spirit in its first pack, and a Lightning Bolt and an epic to disenchant for his second. (I know I’d take that over the two Ancient Watchers I received in my first two packs.)
  • Trump was able to dedicate himself to a single class for the sake of his playthroughs, and he utilized this with strategic disenchanting to optimize the decks available to him, in a way which (though accessible to all players) would detract from the variety of the game. Furthermore, he only has gone through two of the classes, both of which are (arguably) stronger than the average.

All things considered, Trump’s videos only provide a conditional “no, Hearthstone is not entirely pay-to-win some of the time with some of the classes.”

Perhaps the only experiment which would really be fatal to the entire pay-to-win argument would be if a player were able to reach high rank with a deck composed solely of free cards, and repeat this feat for each of the nine classes in the game. I wouldn’t be surprised if someone’s already tried this, but I’m not aware of anyone who has succeeded in such an endeavor — and I have difficulty believing it’s even possible with some of the classes.

The Issue

Overall, the primary issue I have with Hearthstone’s resource progression is that, if a player doesn’t get lucky within their first few easy packs, or messes up with disenchanting (as newbies are prone to), it can be incredibly difficult to compete even at the lowly rank 20; when you can hardly even afford access to common staple cards from the measly 40 dust an unlucky pack will contain, much less the rares, or epics, or legendaries for an actual win condition, how are you supposed to win the games you need to actually unlock more cards?

Usually, this wouldn’t be an issue, because Hearthstone has a ranking system which should place you into competitive matchups; a player whose collection allows him to build a rank 15 deck will, if they fully exploit the resources available to them, hover around rank 15, winning approximately half of their games. A player with a rank 20 collection should hover around rank 20, and so on and so forth.

But, let’s consider the worst possible scenario — what happens when a player, by a stroke of bad luck, receives only terrible cards from their first few packs, and that collection only permits them to build a deck that would be competitive at rank 24?

For the sake of discussion, let P_{R}(A) denote the probability that a deck which should be at rank A will win a match at rank R — which is to say, P_A(A) = 0.5.

So, our unlucky player will start at rank 25, where they should win P_{25}(24) > 0.5 of their games, and they’ll rise to rank 24 as one would expect. At rank 24, they’ll win ~50% of their games on average — but since losses at ranks less than 20 do not lose stars, their rank will rise above the quality of the decks their collection enables. Similarly, at rank 23 they’ll have some P_{23}(24) < 0.5 probability of victory, but nonetheless they will eventually rise in the ranks to 22; to 21; and after a rather painful series of losses, they’ll be forced into rank 20. Depending on how steep the ladder is, P_{20}(24) may be incredibly low; and if a player is stuck in a state where their collection can only build decks to win one in five games [3], and they need a minimum of 5 wins to earn a pack (2.5 of the 2-win, 40 gold quests), then receiving packs becomes excessively rare [4]:

  • The probability that our unlucky protagonist will earn a pack in 5 games is 0.00032%. Instead, a properly ranked player has a probability of 3.13%.
  • …in 9 games, 2.00% instead of 50.0%
  • …in 12 games, 7.25% instead of 80.6%
  • …in 15 games, 16.4% instead of 94.0%
  • …in 20 games, 37.0% instead of 99.4%
  • …in 23 games, 49.9% instead of 99.9%
  • …in 30 games, 74.5% instead of ~100%
Pack receiving probabilities

A table describing the probability of earning at least one pack in \{10,15,20,25,30\} games if you win \{\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}\} of your games. (“100%” is rounded up, of course.)

How can a player forced through such a tedious grind, required to play over twice the intended amount, to receive the few packs they can get for free, and who has received little more in any pack than garbage amounting to a single common card’s worth of dust, ever see Hearthstone as anything other than blatantly pay-to-win? Even if the vast majority of Hearthstone is well-balanced, is it really fair to make a broad, sweeping statement that Hearthstone is not pay-to-win, when just under the bottom rung of the ladder there are players who are thrown into a situation which is just as tedious and as hopelessly unbalanced as any pay-to-win game on the market?

Footnotes and References

[1] I assume we’re in consensus on this; if MTG doesn’t meet a standard for “pay-to-win” I don’t know any game that would.

[2] http://www.wizards.com/Magic/Magazine/Article.aspx?x=mtg/daily/td/293

[3] I consider a 1:5 win-loss ratio generous, considering my W/L ratio isn’t much higher than that at rank 20 even after months of grinding through losses, slowly collecting the nonfree staple commons and a few rare cards. Believe me, it’s night and day between compromising with free cards and actually having a Defias Ringleader for board presence, or a Shadow Step to exploit battlecrys, or an Eviscerate for affordable early-game removal, or an Azure Drake to amplify Fan of Knives. Mind you, it might just be the Rogue class that’s unusable in free-to-play — particularly since almost all of its low mana-cost cards are nonfree and low ranks seem to be very aggro based (most bombs tend to be rare or higher, I guess?).

[4] Calculations done using \sum_{i=5}^{n} \left(\frac{1}{5} \right)^i \left(\frac{4}{5}\right)^{n-i} \binom{n}{i}. The term \left( \frac{1}{5} \right)^i \left(\frac{4}{5}\right)^{n-i} is the probability that a particular order-dependent sequence of i wins and n-i losses will occur. \binom{n}{i} is the number of distinct win/loss sequences containing i wins, so \left(\frac{1}{5} \right)^i \left(\frac{4}{5}\right)^{n-i} \binom{n}{i} is the probability of exactly i wins in a sequence of n games. When summed from i=5 \text{ to } n, they represent the probability of achieving at least 5 wins in n games.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s