Ok so this quest... i cant seem to figure out how does it exactly work. So you play your first minion and it immediately counts it as 1 of the 4 minions played having the same name which doesnt make sense cuz its your first minion... and then when you play your first minion it also seems to lock on that minion and only starts counting number of times that minion is played even though you play two same minions different than the first one you played..whether it was played on the same turn or on another turn.. seriously this is so confusing I've lost so many games due to this weird mechanic can someone explain to me exactly how it works... thank you.
It's simple. It counts your first minion to quest. If played 2 of them it still there, but if you played another minion 3 times then this minion played 3 times does count. It just counts the biggest amount of minions played.
The quest shows the name and count of the minion you've played the most copies of, if I recall. The first count when you first play a minion is just for tracking purposes and has no effect on the conditions of the quest other than notifying you that three more of that minion triggers it. As soon as you exceed the number of the current minion, the name should change. There's no locking in or anything like that. Any 4 of a minion is all you need.
It goes by whichever of the same minion gets played 4 times. For example you have to play 4 Swashburglars to activate the quest. Doesn't matter which minion gets played or when, as long as it is played 4 times.
The quest shows the name and count of the minion you've played the most copies of, if I recall.The first count when you first play a minion is just for tracking purposes and has no effect on the conditions of the quest other than notifying you that three more of that minion triggers it. As soon as you exceed the number of the current minion, the name should change. There's no locking in or anything like that. Any 4 of a minion is all you need.
Whichever minion you can play 4 of, the quest will work.
Quest only shows one name if there are several minions tied for the most played. Not that hard to understand..
First minion to be played is the first to get counted
Whenever a minion gets played more times than the minion that the quest is currently showing, that one shows instead
Although the quest only shows the one that has been played the ''most'' (or atleast shares the #1 spot) it counts all
So if you for instance play firefly firefly on t2, it shows 2 fireflies. But if you play Swashburglar Swashburglar on t3, it still shows the fireflies until you play a 3rd swashburglar, then the quest gets overwritten.
Ok so this quest... i cant seem to figure out how does it exactly work. So you play your first minion and it immediately counts it as 1 of the 4 minions played having the same name which doesnt make sense cuz its your first minion... and then when you play your first minion it also seems to lock on that minion and only starts counting number of times that minion is played even though you play two same minions different than the first one you played..whether it was played on the same turn or on another turn.. seriously this is so confusing I've lost so many games due to this weird mechanic can someone explain to me exactly how it works... thank you.
It's simple. It counts your first minion to quest. If played 2 of them it still there, but if you played another minion 3 times then this minion played 3 times does count. It just counts the biggest amount of minions played.
The quest shows the name and count of the minion you've played the most copies of, if I recall. The first count when you first play a minion is just for tracking purposes and has no effect on the conditions of the quest other than notifying you that three more of that minion triggers it. As soon as you exceed the number of the current minion, the name should change. There's no locking in or anything like that. Any 4 of a minion is all you need.
So everytime i want to play a different minion i have to play it +1 times more than the last one?
It goes by whichever of the same minion gets played 4 times. For example you have to play 4 Swashburglars to activate the quest. Doesn't matter which minion gets played or when, as long as it is played 4 times.
Easy N'zoth Quest Priest
It's pretty easy:
So if you for instance play firefly firefly on t2, it shows 2 fireflies. But if you play Swashburglar Swashburglar on t3, it still shows the fireflies until you play a 3rd swashburglar, then the quest gets overwritten.
Fuck cubelock
http://www.hearthpwn.com/forums/class-discussion/rogue/190512-is-rogue-quest-bugged-why-does-playing-1?comment=3
http://www.hearthpwn.com/forums/hearthstone-general/general-discussion/190420-rogue-quest-not-counting-same-name-minions