'Summon' inconsistencies [Unseal the Vault]
Submitted 4 years, 11 months ago by
Nuagoo
So I just played Quest Hunter and encountered a very peculiar (in my opinion) inconsistency:
- Leeroy Jenkins accounts for 1 summon, though stating 'summon 2 minions for your opponent'
- Any cards your opponent plays, that summon minions for you, count towards your Quest
My question is WHY?
Apparently 'summon' is NOT dependent on the actual player summoning minions (otherwise the opponent summoning minoins shouldn't count) but 'players' (?) summoning minions?
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So I just played Quest Hunter and encountered a very peculiar (in my opinion) inconsistency:
My question is WHY?
Apparently 'summon' is NOT dependent on the actual player summoning minions (otherwise the opponent summoning minoins shouldn't count) but 'players' (?) summoning minions?
I noticed that as well, but... hey... Blizzard isn't exactly known of consistency.
My guess would be that the game is only looking for how many minions are added to your side of the game board, not how many minions the player summons in general.That is also why [Hearthstone Card (Faceless Corrupter) Not Found] only counts for one.
Gosh dang it, cards bad.
Totally agree. While the priest quest counts anduin healing the ennemy properly, the hunter quest and side quest mechanics seems sloppy as hell and full of inconsistencies
ARES summons Erymanthian Huffer.
ARES declares attack with Huffer .
Adonis' hp reached 0. ARES wins!
Pretty likely so. Still the In-Game description is very much misleading...
Well at least it's implemented in the most fair way.. imagine playing leeroy and advancing your opponent's quest, also leeroy is good enough in quest hunter as well so buffing that interaction is also totally not needed..
Faceless Corrupter don't summon a new one, but he transforms an existing one. It would be weird, if you get a second count you already got for the usual 1/1 you had on the board. This is fine and consistent.
Works as intended, you summon the minions for your opponent, not for you. That's why you don't get the quest count. Why should the quest condition consider the summoning for your opponent? I don't understand and it has nothing to do with consistency.
Because you get minions on your board that's why it counts. Why shouldn't it? I don't understand your problem.
Pretty sure you DO advance your opponents quest by playing Leeroy...
This can be pretty much summarized by 'who summons a minion' and hence who should benefit by a summon. There are several minions summoning tokens for you and/or your opponent. So which of these summons should count towards your quest? That's the question at hand.
It does? I missed your second point probably :/ (read too fast) well it's a rare scenario when was the last time you saw your opponent play leeroy and you didn't die?
guess it just counts your side of the board in summon and corrupter is not a summon but a transform and that doesn't count.
It doesn't... Just tried it -_-
I saw someone playing paladin and was shocked when brazen zealot got a buff when a friendly minion was magnetised, but not the case for Hunter quest
Honestly the main reason for issues like this is simply keeping card text clean. Blizzard have always held a design philosophy with Hearthstone that certain things simply won't be stated in any depth in the client simply because it would lead to UI cluttering.
What's actually happening when you play something like Leeroy, rather than you summoning the whelps, is you force your opponent to summon whelps. But that would lead to multiple lines of additional game text to explain what's happening, and the cases in which that actually matters are absolutely tiny. So, they leave it unexplained because to even encounter it you're going to need a specific, niche set of legendaries - and when it happens once, you then know the interaction going forward anyway.
I support that philosophy from a game design perspective, personally. Though the card game purist in me hates it, because I'm used to physical games where the card text has to be read in fine print to understand perfectly.
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