Rules Tip of the Day: Blocking creatures are declared at the very beginning of the declaration of blockers step. In order for a creature to be declared as a blocker, it must be in play when this step begins. This means that the last chance you have to either get a creature into play in order to block is in the declaration of attackers step. Keep in mind that if you animate a permanent in play, or play a creature as an instant, your opponent can still respond before the declaration of blockers step begins and destroy this creature so it cannot block.

Q: If I have two Stinkdrinker Bandits in play, do they stack?

A: Yes. Each of their abilities will trigger independently and separately. An unblocked attacking Rogue you control will end up getting +4/+0.

Q: Arbiter of Knollridge is played with Rain of Gore in play. I believe that the only person that isn't going to take damage is the person with the highest life, and any player who is below half the life total of that person is going to lose. Am I right?

A: No, only the controller of the Arbiter of Knollridge will take damage, and only then if his life total is not the highest. This is because Rain of Gore only affects what happens when a spell or ability causes the controller of the spell or ability to gain life. It does not change what happens when an spell or ability causes other players to gain life.

Q: Can I use Trickbind to counter the ability of Doran, the Siege Tower, and if so, for how long?

A: No, that does not work for any length of time. Doran has a static ability, not an activated or triggered ability. That ability, like all static abilities, does not use the stack and its effect is always applied, as long as the ability exists.

Q: I Lignify a creature with a +1/+1 counter on it. Does it lose the counter?

A: No. Removing abilities from a creature does not remove counters from it. Also keep in mind that, because the effect from counters is applied after the power and toughness setting effect from Lignify, this creature will end up being a 1/5 creature.

Q: Can creatures with summoning sickness tap because of Drowner of Secrets or Summon the School?

A: Yes, as they are not tapping to activate one of their own abilities that has the tap symbol as part of the cost. This also means that the Drowner can tap to activate its own ability on the turn that it enters play, as that ability does not have the tap symbol.

Q: I have an indestructible 1/6 Doran, the Siege Tower, thanks to my Timber Protector. My opponent blocks Doran with Gaddock Teeg, then plays a Sudden Death on it. Does my Doran die?

A: No. It will end up as a -3/2 creature with two points of damage on it. Normally it would be destroyed due to lethal damage, but it is indestructible, due to the Timber Protector.

Q: I have a Magus of the Scroll in play, and my opponent has Telepathy, so I play the game with my hand revealed. I have three different cards in hand. If I use the activated ability of the Magus and the ability resolves, what happens after I name a card? The card I named would have been one of the cards in my hand.

A: You still reveal a card at random from your hand; it does not matter that they are known and currently begin revealed. You can do this be rolling a die or some other mutually agreed method to randomly choose a card. If the randomly chosen card matches the one that was named, then the targeted creature or player will receive two points of damage.

Q: I have Aether Storm in play; can my opponent pay a suspend cost for a creature?

A: Yes, suspending a card is a special action and is not the same as playing it. But if the Aether Storm is still in play when this card unsuspends, your opponent will not be able to play it, and it will remain removed from the game.

Q: If I have Swirl the Mists in play, naming red, and a White Knight is in play. My opponent plays Mind Bend, targeting the White Knight and choosing blue when Mind Bend resolves. What happens?

A: The timestamp from the Mind Bend will be later than the timestamp from Swirl the Mists, so this creature will have protection from blue.

Q: How do Rock Hydra and Balduvian Hydra interact with Ancient Ooze? I think you would need to keep track of what the original X was, regardless of how many "heads" are currently on the Hydra, since the "total converted mana cost" was X when the Hydra was put on the stack. Is this correct?

A: No. When in play, the value of X in the converted mana cost of a permanent is zero, regardless of what was chosen and paid to play these permanents. The converted mana cost of a Rock Hydra or a Balduvian Hydra in play is two. Ancient Ooze's power and toughness will not change as the number of counters on these permanent change; it does not use any other value other than zero.