text:
Lavinia, Azorius Renegade ruling
So my opponent is playing Jodah, Archmage Eternal, and they attempt to cast Omniscience for WUBRG. In response, I flash in Lavinia, Azorius Renegade.
What happens to the Omniscience since it's already on the stack and Lavinia says "each opponent can't cast noncreature spells with converted Mana cost greater than the number of lands that player controls"

While the answers cover what happens to Omniscience, keep in mind that Lavinia is still a great play in response to Omniscience, because that means it will resolve before the opponent gets to cast spells off of it without paying their mana cost

Nothing. Lavinia, Azorius Renegade has no effect on it, so it resolves normally.

To cast a spell is to put it on the stack, to make some choices (mode, targets, how the cost is payed, etc) and to pay its cost.

601.2. To cast a spell is to take it from where it is (usually the hand), put it on the stack, and pay its costs, so that it will eventually resolve and have its effect. Casting a spell includes proposal of the spell (rules 601.2a–d) and determination and payment of costs (rules 601.2f–h). [...]

Specifically, resolving a spell is not part of casting it. To respond to a spell is to do something after it has been cast but before it resolves, so Omniscience has already been cast by the time you get to respond to it.
Lavinia's static ability only affects spell that are cast while Lavinia is on the battlefield. It doesn't affect spells that have already been cast. What's been done is done. Nothing has a retroactive effect in Magic.[1]
Similarly, Lavinia's triggered ability only triggers for spells that are cast while Lavinia is on the battlefield. It doesn't trigger for spells that have already been cast.

Two exceptions:

In the extremely rare situation in which a spell becomes uncastable in the middle of the process of casting it, the game reverses to when the spell started being cast. (This isn't the case here because the player has finished casting Omniscience.)
When an error is made, the judge may undo certain actions in certain circumstances.

Lavinia's other ability is a triggered ability rather than a static one, but the result is similar. Once the opponent's spell has been cast, it's too late to flash in Lavinia.

@Arcanist Lupus, What I said applies equally to both abilities, yes. I've clarified.

@ArcanistLupus The triggered ability will never affect spells cast with Jodah's ability, because you will spend mana to cast it - just not the original amount.

@TheThirdMan, True, but noone said otherwise.

@ikegami The second ability was never addressed in the question, so bringing it up without clarifying might make someone think there may be an interaction. I'm not sure why you decided to include it in your answer, but I feel the comment is helpful.

Both spells resolve. Omniscience is put into the battlefield.
You need to cast Lavinia, Azorius Renegade before your opponent casts Omniscience if you want to prevent your opponent from casting it.


text:
Unknown Dice Game
Does anyone know what this game is? My wife picked up 2 of these shakers and I cannot identify the game.
Thanks in advance!

Are they all the tokens?  Only 8 numbered as show with the word 'nine'  or do others exist?

Looks like you're probably missing 3, 4, 5, six, 7, 8, 10,13 and possibly 17+. The only reason I can think of to write "nine" when the others are digits is to differentiate it from 6/six.

Looks like Kelly Pool, missing a few pieces (should have "peas" 1-16, per L. Scott Johnson's comment). Kelly pool's page doesn't mention a die, but you can probably just add the variant mentioned on the Bottle Pool page.

I can see a bottle in that link but I cant find any reference to the numbered tokens.  What bits do you think are missing?

Fixed the answer, set it to bottle pool originally because it mentioned using a die, guess I just assumed it used the red bits.

Kelly pea balls come 16 to  a set, numbered 1,2,3,4,5,six,7,8,nine,10,11,12,13,14,15,16. cf: https://www.dkgameroomoutlet.com/deluxe-leather-tally-bottle-wooden-kelly-pea-ball-set

Kelly Pool looks to be the most accurate answer for this one. It is true, my 4 year old daughter did a pretty good job of losing most of the numbered tokens. I'm still wondering what the 6 sided die is used for though.

From bottle pool link: "One further variation places a 6-sided die (although other sided die could also be used) on top of the bottle. If the bottle is knocked over and the die knocked onto the table, the player is awarded points equal to whatever side of the die lies facing upward when the die comes to a stop. This adds a random effect to hitting the bottle and in practice often results in players only striking the bottle in combination with other point-scoring shots."

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Help identify this game with four divided concentric circles
I inherited the a wooden game from my parents. Probably date from 1960-1975. No box, no instruction. Can you help me identify it and locate rules for it?
It's a 9.5" square with 4 indented concentric circles. The outer ring is divided into 12 sections, the next into 8 sections, the 3rd into 4 and the center - approx 1" diam. is not divided.White and black dots are in seemingly random sections. Inside the black leather pouch are 6 brown wooden marbles and two chrome cylinders. One cylinder has 3 raised concentric circles on one end, the other has 2 concentric circles, the innermost indented. "Skjode Skjern Danmark" is etched on the bottom.

Could you post images of both the games, as separate questions?

Removed one of the two games. Please post it as a separate question (with a picture). Your description can be found here.

The first game has been identified as "Patience" a form of Solitaire. Thanks for your interest.

It is a Morra board game.
You will find details here on Board Game Geek :
https://boardgamegeek.com/boardgame/41679/morra-board
Here is their explanation of it:

From the instructions: Morra-board builds on an ancient game - renewing and varying it by linking it to a board.
  The old game on which Morra-board is based is in Danish called 'Klunse', in English 'Odds and Evens'. Each of the two players presents a clenched fist containing a few small stones, coins or matches - both players taking turns at guessing if the sum of the objects in the two hands is even or uneven.
Players pieces move around the track based on how many total marbles the two players reveal. If the inner track has a marking on it, the piece advances. The goal is to be the first to advance all the way to the center.

it could be Byzantine Chess , search using the pinterest search by image. you may want to include other pieces included with that board

There does not appear to be nearly enough spaces on the board for that

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Which rule(s) govern the difference between {T} activated abilities and activated abilities that require tapping?
There seems to be a difference in Magic between activated abilities with the tap symbol, and activated abilities with a cost that requires you to tap permanents, and I'm unsure which rules specifically detail this difference.
For example: Assume I have four vampires in play and I have just used Sorin, Vampire Lord's -8 ability stating:

Until end of turn, each Vampire you control gains "{T}: Gain control of target creature."

I play a Captivating Vampire. Thanks to Sorin he has "{T}: Gain control of target creature." in addition to his other abilities, but I can't activate it because he has summoning sickness. For some reason, however, I can activate Captivating Vampire's second ability by tapping himself and my four other vampires. Why is this not similarly prohibited by his summoning sickness?

Note: I'm aware that I can do this. The Gatherer rulings call it out specifically. I'm just wondering why. Is the only reason really because the activated ability lacks the Tap symbol?

Related, maybe duplicate? https://boardgames.stackexchange.com/questions/25234/what-is-summoning-sickness-and-how-does-it-affect-my-creatures?rq=1

302.6. A creature’s activated ability with the tap symbol or the untap symbol in its activation cost can’t be activated unless the creature has been under its controller’s control continuously since their most recent turn began.

So it's the tap symbol vs. the word 'tap'.
I'm not sure why this is ruled differently; one reason could be that the tap symbol is always an ability of the creature itself, while the word 'tap' indicates an external ability. But there are (of course) exceptions: there's an Elf which allows you to tap three Elves for three green mana. It can tap itself the turn it enters the battlefield.
But I guess one could question the origin of most Magic rules this way; for instance, why do creatures suffer from this but lands do not? (To which the answer probably is: lands are played and creatures are summoned. It does make some sense, flavour-wise.)

There is actually nothing in the rules that prevents a creature with summoning sickness from being tapped.  It's only attacking and abilities with the tap or untap symbol that can't be used.
Rule 602.5a is the rule that prevents creatures with summoning sickness from using tap or untap abilities.

602.5a A creature’s activated ability with the tap symbol ({T}) or the untap symbol ({Q}) in its activation cost can’t be activated unless the creature has been under its controller’s control since the start of their most recent turn. Ignore this rule for creatures with haste (see rule 702.10).

However, this rule doesn't matter when the ability is something like that on Gravespawn Sovereign.

Tap five untapped Zombies you control: Put target creature card from a graveyard onto the battlefield under your control.

This is not a tap ability because it doesn't have the tap or untap symbol in the cost.


text:
Can Ring of Ma'rûf call in cards outside of MtG?
I’m building a completely casual deck with friends, and the the restrictions on Ring of Ma'rûf aren’t entirely clear. Could I, for example, use Ring of Maruf to gradually bring in a YuGiOh card such as Exodia The Forbidden One. Could the Ring allow that?

In a casual game, you can do whatever you want... but... what are you going to do with Exordia once you have it in your hand?

No, you can only get a Magic: the Gathering card.  The meaning of the word "card" is given explicitly in the rules:

108.2. When a rule or text on a card refers to a “card,” it means only a Magic card or an object represented by a Magic card.

Moreover, you can only get traditional Magic cards, so no Archenemy schemes, Planechase planes, etc:

108.2a Most Magic games use only traditional Magic cards, which measure approximately 2.5 inches (6.3 cm) by 3.5 inches (8.8 cm). Certain formats also use nontraditional Magic cards, oversized cards that may have different backs.
108.5. Nontraditional Magic cards can’t start the game in any zone other than the command zone (see rule 408). If an effect would bring a nontraditional Magic card into the game from outside the game, it doesn’t; that card remains outside the game.

So while you could choose your copy of All in Good Time with Ring of Ma'rûf, it would not be able to enter the game due to 108.5, and you effectively just waste your draw and exile the Ring for nothing.

In most competitive situation, they limit these pulls to cards in your sideboard and not your complete collection.   An example of this is Karn, the great creator which is  from War of the spark.


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Catan - Discarding fish to draw more from the supply
In The Fishermen of Catan, if a player has 7 fish tokens and a roll would have them receive more, can they discard the ones they don't want to draw the new ones?

Yes.

In the original scenario named The Fishermen of Catan, there was no limit to how many fish tokens could be kept. A revised version was published in Traders & Barbarians. This version included the following rule:

7 fish tokens at most: You are not allowed to have more than 7 fish tokens at any one time. If you already have 7 fish tokens and would obtain another 1 or 2 fish tokens for a settlement or city, you may exchange one of your fish tokens with a fish token from the supply (hoping for a token with more fish on it).

So if you couldn't draw a fish tile because you already have 7, you may first discard one of your fish tokens.
However, it's not clear exactly when and how often this can be done.

What if you have 6 fish tokens and would obtain 2 from a city?
What if you have 6 fish tokens and would obtain 1 each of two settlements?
What if you have 7 fish tokens and would obtain 2 from a city?
What if you have 7 fish tokens and would obtain 1 each of two settlements?
What if you have 7 fish tokens and would obtain 1 from a settlement and 2 from a city?


text:
MTG - Priority and timing between activated abilities and instants
If I have got a creature with "inflatable abilities" - just like Frozen Shade, or Granite Gargoyle - I want to know what happens when activating it. Since inflatable abilities are activated abilities, if I activate three times Granite Gargoyle's ability ({R}: +0/+1 until end of turn) in response to each other, they go on the stack one after the other, and my opponent may respond to this with a Shock targeting the Gargoyle. So, in such a situation, the Gargoyle dies, and goes to the graveyard  before his ability comes to rescue. Is this correct?

Please tell Ikigami it is this a kind of question slightly different among others I already asked for.1°one concerned about the correct timing between enchantments/instants;2°one was about the timing between instants and other instants too;this one –3°– just involved special abilities and instants.The 3 questions ARE NOT exactly the same,and may serve well in order to understand whyLIFOrule is basically the correct way to recall any order of resolution.In this way, the 3 simple questions are summarized,and fast-recovering,for all beginners/old-time players returning playing after many years.

I'm joking, Ikigami, you're welcome...obviously, this is not the kind of questions you really like...since you're an expert one and stuff like this is simply boring.

The question is identical. They're all "how does the stack work". The very fact that you use "in respond" in all questions makes then identical.

Ahaha!! Ok, the MacDonald situation is really funny. Isn't it, Lause&Ikigami? Ok, so I'll assume that: the next time I'll go lunching at Mac, instead of paying money I will have a new doggy-dog suit  - with the word Shinigami on it - to protect my dog Ajani from the rain (and lightning bolts too),as a tribute immediately after buying a cheesburger!!!

You are an expert, Ikigami, not just like me...The question is identical for you. There's lots of people actually having  real hard and bad discussions in order to understand the rules properly.

@Massimiliano before you ask more questions on this topic I suggest reading https://boardgames.stackexchange.com/questions/12652/in-plain-english-how-does-casting-spells-and-using-creature-abilities-work-with. That may help you understand this topic more generally.

This is pretty much the same as the Shock vs. Titanic Growth example in the online rules, it's just that it's an ability here and not a spell, but then the rules also mention casting spells and activating abilities both use the stack ("When you cast a spell or activate an ability [...] it goes on the stack.") Of course, with an ability, it's easier to just reply to Shock by pumping again.

Re "in such a situation, the Gargoyle dies, and goes to the graveyard before his ability comes to rescue. Is this correct?", According to the original wording of the question, that is not correct. I fixed the question to be what the OP believes it is based on extensive comments by the OP's on a now-deleted answer. This is the third question by the OP with an identical answer, but rather than closing this one as a dup of an earlier one, I closed an earlier one as a dup of this one as it has a better answer.

What do you mean with OP's? What does it mean?

Ah, ok , it stands for "Original Poster", or "Overpowered", sometimes...

Yes, there's not much difference between activated abilities and instants in this particular case. If you look at the Comprehensive Rules, you'll see that they are treated similarly, e.g. in rule 405:

405. Stack
405.1. When a spell is cast, the physical card is put on the stack (see rule 601.2a). When an ability is activated or triggers, it goes on top of the stack without any card associated with it (see rules 602.2a and 603.3).
405.2. The stack keeps track of the order that spells and/or abilities were added to it. Each time an object is put on the stack, it’s put on top of all objects already there.
405.4. Each spell has all the characteristics of the card associated with it. Each activated or triggered ability that’s on the stack has the text of the ability that created it and no other characteristics.

In this case, you're better off activating the Gargoyle's ability once, and let it resolve before activating it again. If your opponent responds with Shock, you can respond to that with activating the ability again, saving the Gargoyle.

Exactly. Thank you Glorfindel. It is a simple kind of question, but it could be important, since some players may think at "inflatable abilities" just like they were static ones:that is wrong.

That's why it's best to wait for the Gargoyle's ability to resolve before activating it again. In fact, under tournament rules, that's what "I activate three times" means

Well, properly talking...returning back to the Gargoyle example, activating the inflatable ability just one time - instead of two - in response to Shock, it is enough to rescue her.

Yes, but that would require spending an extra {R} you didn't need to spend,

'In fact, under tournament rules, that's what "I activate three times" means' This was unknown to me. I thought the tournament rules were pretty darn pedantic about what you actually say, and what you actually say in that case is that you activate it three times without paying priority and therefore without letting them resolve. Do you have a reference?

@Arthur, Tournament rules under "Tournament Shortcuts". See my answer for the relevant passage.

@Arthur, If you were forced to do what you say, you couldn't say "I activate this two times." It presumably means "I activate this, and in response, I activate this again" or "I activate this, I wait for it to resolve, then I activate this again". But either way, you are proposing a shortcut you need to define. The TR just define a few shortcuts so you don't have to do so every game. It also defines a few shortcuts to handle forgotten things without having to invoke a judge. (e.g. Forgetting to scry is like opting not to look and not to change the position of cards, something scry allows.)

Where all my comments about Ikegami’s shortcuts have been end up?Please someone resume them,there were important things they describe.Or maybe it was just you, Ikegami, to hide them?Shortcuts are interesting,but it is important to warn your opponent that the next game is about to use them.Do not doing this means  trickery.Much more than this, it is not your interest judging if a question is identical to another one.Looking at your editing,and your comments about it,I'm making sure about the question I asked is really useful,expecially for beginners.This is not you,Ikegami,I already told you

I guess they were not trying to improve the post, so they were deleted. Please check the Help Center for when it is appropriate to comment and when not.

It was not my fault, you know Glorfinder...in fact, what you say it is true: the last comments are going to describe Ikegami's personality much better than giving info about the question I have made...

Who knows how the stack really works?Beginners still don't know very well if a certain type of spell goes on the stack or not,and therefore they need to know if there are differences in interaction between enchantments/instants/abilities,with other instants.To show the validity of the question I may talk about the case of a creature unmorphing itself:in this case,since it is a special action, the unmorphing WILL NOT GO on the stack.Therefore, interaction with it would be different (since opponent cannnot respond to it).Only experienced players,like Ikegami&Murgatroid,know how the stack works!

This is why I explicitly proposed three similar situations, and to show their differences with the "special case" in which an opponent responds to a creature's unmorphing - a case I will perhaps show later, with another example - and that looks very different because opponent cannot respond to the "unmorphing" of a creature in itself (since it is a special action), but can respond to the triggered ability due to the unmorphing of the creature, if it has one.


text:
What is this 2 pack patience game?

Can anyone tell me what the name of this 2 pack solitaire game is called? I have played it since I was a child but would like to discover the full rules. It is much like standard one pack solitaire except the middle column can only be moved directly onto the ace piles. Other cards can just be moved into black and red sequences

This appears to be a variant of Russian Bank (which is known to sport quite a few variants, some of which seemingly more common in some countries than others). I don't know which particular variant this is, but maybe someone else does.

Thank you for the suggestion. The game I play is for one person only.


text:
Do you need to reveal which specific tunnel you’re attempting to claim?
In Ticket To Ride, if you want to claim a tunnel path, you go through the process of laying out your cards for the tunnel and drawing 3 cards to determine if you’re required to pay more cards. My question is: are you required to say which particular tunnel path you’re trying to claim, or do you only say the color and length of the tunnel, revealing which particular tunnel only once you’ve successfully claimed it (after drawing the 3 cards and paying up if needed)?
The rules don’t specifically mention this, but it seems like an important distinction to make. If the player fails to claim the tunnel, then his opponents either will or won’t have an additional advantage, depending on whether or not they know which tunnel it was (assuming there are multiple tunnels of the same length/color still available).

Yes, you need to indicate which tunnel you are attempting to claim.
This has been a somewhat contentious question among other players; because you are correct that the rulebook doesn't not directly address this. However, Alan Moon, the designer of the game, stated1 that you do reveal which route you are attempting.
Source:
https://boardgamegeek.com/thread/303859/how-claim-tunnel-definitive-answer/page/1
1 We only have the word of a BGG user that Alan Moon did indeed say this. But I did not see suggestions by the BGG population at large that it was a lie.

As further corroboration, the digital version by Asmodee does force you to pick before the test (and it reveals where you are attempting to build to all players).

This is another Yes, you need to indicate which tunnel you are attempting to claim. but with other sources.
Using the rules here look at the examples for claiming a route and claiming a route with a tunnel.   Both sets of example show a red arrow pointing at the section being claimed.
I think the classic rule of interpreting instructions applies here.  If something isn't explicitly stated then it probably isn't allowed.  If a player was allowed to lay out cards, see if tunnel was successful and THEN decide where to build it then that would surely be stated as its a huge difference from the regular claim a route rules.
Another source (which is harder to reference on here) is the Ticket to Ride App.  When I play that and attempt a build a tunnel route I have to drag cards I want to play to the selected route before I see if it successful or not.   There is no interface to say "I want to see if a 3 red tunnel is successful before saying where" interface which again suggests you are required to state the route you wish to claim.

I'm fairly certain you're correct and I think the app is useful evidence, but to really make a convincing case with the app, you have to talk about what it looks like to other players - can you confirm that it says to them e.g. "Player X tried to build tunnel Y and failed" rather than just "Player X tried to build a tunnel and failed"?

@BenjaminCosman In the Web version, you could clearly see the cards attempting to go to the tunnel and bouncing back.


text:
Hardened Scales + Champion of Lambholt + Token Generator

Champion of Lambholt: Gets a +1/+1 counter whenever a creature enters the battlefield.
Hardened Scales: Whenever 1 or more counters are put on a creature, put one more of those counters on that creature.
Gather the Townsfolk: Put two 1/1 human tokens on the battlefield.

If I play Gather the Townsfolk, does Champion get 3 counters or 4?

"Whenever" indicates a triggered ability, but HS has a static ability that creates a continuous effect that's a replacement effect. Avoid using "when" or "whenever" for these.

Four.
When the two tokens enter the battlefield, Champion of Lambholt's triggered ability triggers twice. It will be placed on the stack twice. When the topmost instance resolves, it would add a +1/+1 counter to Champion of Lambholt. Instead, it adds one extra, so two. When the other instance resolves, it will also add two, so a total of 4 +1/+1 counters are added to Champion of Lambholt.

Is this an issue of timing  (one token gets summoned, then another, in near instant time), or are the tokens spawning simultaneously but each token is a valid trigger for Champion (causing each trigger to be an isolated incident, and both valid for Scales)?

It depends on the wording of the trigger condition. If it says "Whenever one or more creature enters the battlefield", then the ability would only trigger once when GotT resolves. But here we have the far more common "Whenever a creature enters the battlefield". It doesn't matter if it enters at the same time as another creature. 603.2c An ability triggers only once each time its trigger event occurs. However, it can trigger repeatedly if one event contains multiple occurrences


text:
How does Bolas's Citadel work?
If I play the card Bolas's Citadel, do I have to pay mana costs for Planeswalkers and Creatures or do I play them for free?

Can you clarify what your uncertainty is? Are you asking about casting a card from your hand normally while Bloas's Citadel is out? Or are you asking about when you play the top card of your library? In the latter case, the wording on the card seems straight-forward; so not really sure what you're asking.

Or are you asking if playing Planeswalkers and Creatures count as playing "spells"?

Yes actually.i wasnt sure if they counted as spells or not.from the way the text was worded, it seemed as if only sorceries costed life if you were to play them from the top of my library.

Bolas's Citadel doesn't allow you to cast anything for free.
To play a Planeswalker card or a Creature card is to cast it as a spell. You will need to pay life instead of mana if you play one from the top of you deck.

A spell is a card on the stack.[1] It doesn't matter if it's an instant, a sorcery, a creature, a planeswalker or something else. Cards of any type can be spells.[2] All that matters is that it's on the stack.
Bolas's Citadel imposes an alternative cost to spells when they are cast from the top of your library.[CR 118.9] Alternative costs replace the mana cost of spells.[CR 601.2f]
total cost to cast
= mana cost or alternative cost
+ additional costs and cost increases
- cost reductions
In this case, to cast a card from the top of your library, the cost would be calculated as follows:
total cost to cast
= mana cost or [life equal to the card's converted mana cost]
+ additional costs and cost increases
- cost reductions
For example, if you wanted to cast Llanowar Elves from the top of your library, you would have to pay 1 life (but no mana).
Spells cast from elsewhere (e.g. from your hand) are cast normally (because they aren't "spells cast this way").

It could also be a copy of a card on the stack (e.g. see Isochron Scepter) or a copy of spell (e.g. see Bonus Round).[CR 112.1, 112.1a, 112.1b]
Lands are never cast, so they never go on the stack, so there can never be a land spell.[CR 305.9]
Bolas's Citadel does allow you to play a land from the top of your deck, but it's played normally (because it's not "a spell cast this way"). It even counts towards the limit on the number of lands per turn.[CR 305.2]

112.1. A spell is a card on the stack. As the first step of being cast (see rule 601, “Casting Spells”), the card becomes a spell and is moved to the top of the stack from the zone it was in, which is usually its owner’s hand. (See rule 405, “Stack.”) A spell remains on the stack as a spell until it resolves (see rule 608, “Resolving Spells and Abilities”), is countered (see rule 701.5), or otherwise leaves the stack. For more information, see section 6, “Spells, Abilities, and Effects.”
112.1a A copy of a spell is also a spell, even if it has no card associated with it. See rule 706.10.
112.1b Some effects allow a player to cast a copy of a card; if the player does, that copy is a spell as well. See rule 706.12.
118.9. Some spells have alternative costs. An alternative cost is a cost listed in a spell’s text, or applied to it from another effect, that its controller may pay rather than paying the spell’s mana cost. Alternative costs are usually phrased, “You may [action] rather than pay [this object’s] mana cost,” or “You may cast [this object] without paying its mana cost.” Note that some alternative costs are listed in keywords; see rule 702.
305.2. A player can normally play one land during their turn; however, continuous effects may increase this number.
305.9. If an object is both a land and another card type, it can be played only as a land. It can’t be cast as a spell.
601.2f The player determines the total cost of the spell. Usually this is just the mana cost. Some spells have additional or alternative costs. Some effects may increase or reduce the cost to pay, or may provide other alternative costs. Costs may include paying mana, tapping permanents, sacrificing permanents, discarding cards, and so on. The total cost is the mana cost or alternative cost (as determined in rule 601.2b), plus all additional costs and cost increases, and minus all cost reductions. If multiple cost reductions apply, the player may apply them in any order. If the mana component of the total cost is reduced to nothing by cost reduction effects, it is considered to be {0}. It can’t be reduced to less than {0}. Once the total cost is determined, any effects that directly affect the total cost are applied. Then the resulting total cost becomes “locked in.” If effects would change the total cost after this time, they have no effect.

A well phrased, well reasoned, and well researched, in-depth answer to what most could dismiss as a simple, unnecessary question. Definitely a +1 from me!

You pay for their cost with your life points instead of spending mana, assuming the card you want to play is on-top of your library. You can not use Bolas's Citadel to play cards from your hand.
Below is some of the text from Bolas's Citadel.

You may look at the top card of your library any time.
You may play the top card of your library. If you cast a spell this way, pay life equal to its converted mana cost rather than pay its mana cost.

As you can see, you pay life equal to its converted mana cost instead of paying its mana cost if the card you are trying to play is the top card of your library.


text:
How likely am I to have a key card in my starting hand in Brawl?
I'm playing the Brawl format (a 60-card singleton format), and I really want one specific card in my starting hand.
I'm willing to take two mulligans (leaving me with a starting hand of 5 cards after putting 2 on the bottom of my library).
What are the odds of drawing the card in my opening hand within 2 mulligans?

Hers already a hint how to calculate this

You currently have a few questions peppered in there. For the avoidance of doubt, it would be good to remove all but the questions in the last paragraph.

The question Aetherfox refers to is very different from this. In my opinion, this is not a duplicate.It is true that they are similar; however, this is a different probability problem, as this question asks about the probability of obtaining a specific card even AFTER making two mulligans - and therefore it is about "probability with re-entry in the initial sample".Furthermore, in the other question a very different example is made, in which the probability of having a particular card in the initial hand is requested, but of which there are 4 copies in the deck.I don't think that's the case...

@Aetherfox Re "Does this answer your question?", It does not. The linked question doesn't cover mulligans whatsoever. (It also about a different deck size and a different number of target cards.)

Doh! I didn't want to unilaterally reopen the question

The first mulligan is free in Brawl[CR 903.11g], so it's 3 mulligans leaves you with 5 cards. I shall reply accordingly.
After removing the Commander, 59 cards remain in the deck. 58 of those aren't the card.
The probability that the 1st card drawn isn't the card is 58/59.
The probability that the 2nd card drawn also isn't the card is 57/58.
The probability that the 3rd card drawn also isn't the card is 56/57.
etc.
So the probability that you don't have the card in your opening hand
= (58/59)*(57/58)*(56/57)*(55/56)*(54/55)*(53/54)*(52/53)
= (58*57*56*55*54*53*52)/(59*58*57*56*55*54*53)
= 52/59
Since you draw seven card from a freshly shuffled deck when taking a mulligan, the probably of not getting the card in a mulligan is the same as for the initial hand.
So the probability that you don't have the card after up to three mulligans
= (52/59)^4
= 60%
The probability that you do have the the card after up to three mulligans
= 1 - 60%
= 40%

Probabilities are easy to get wrong, so I like to verify through experimentation.
#!/usr/bin/perl

use strict;
use warnings;

use List::Util qw( shuffle );

use constant NUM_TRIALS => 10_000;

my @deck = ( 1, (0)x58 );
my $successes = 0;
for (1..NUM_TRIALS) {
   for (7,7,6,5) {
      if ( grep { $_ } ( shuffle(@deck) )[0..6] ) {
         ++$successes;
         last;
      }
   }
}

printf("%.0f%%\n", $successes/NUM_TRIALS * 100);

Output:
$ ./a
40%

$ ./a
40%

$ ./a
39%

The current rules for mulligans:

103.4. Each player draws a number of cards equal to their starting hand size, which is normally seven. (Some effects can modify a player’s starting hand size.) A player who is dissatisfied with their initial hand may take a mulligan. First, the starting player declares whether they will take a mulligan. Then each other player in turn order does the same. Once each player has made a declaration, all players who decided to take mulligans do so at the same time. To take a mulligan, a player shuffles the cards in their hand back into their library, draws a new hand of cards equal to their starting hand size, then puts a number of those cards equal to the number of times that player has taken a mulligan on the bottom of their library in any order. Once a player chooses not to take a mulligan, the remaining cards become that player’s opening hand, and that player may not take any further mulligans. This process is then repeated until no player takes a mulligan. A player can take mulligans until their opening hand would be zero cards, after which they may not take further mulligans.
903.11g In any Brawl game, the first mulligan a player takes doesn’t count toward the number of cards that player will put on the bottom of their library or the number of mulligans that player may take. Subsequent mulligans are counted toward these numbers as normal.

(Emphasis mine.)

This answer appears to use a mulligan rule different than given on the Wizards website:  https://magic.wizards.com/en/game-info/gameplay/formats/brawl  Card draws are 7, 7, 6, 5 as I read the rules.  Not 7, 7, 7, 7 as this answer states.

But with the London mulligan rule you  draw 7 and discard down to 6 or 5. So the odd can be calculated on 7 if you are only looking for one card.

The question is specified in the sense that the player starts with 7 cards in hand; and then the first mulligan is made, drawing only 6;then, the second mulligan consists, based on what is requested by the application, of a third drawing of only 5 cards. The answers must be based on this data.

Thank you @Styxsksu, I wasn't aware of the London Mulligan rule.

@John, You should follow your own link :) That is not how mulligans work in MTG (anymore). Please read 103.4 (and 903.11g), now conveniently quoted in my answer.

@ManoFromBerlin, Re "The question is specified in the sense that the player starts with 7 cards in hand; and then the first mulligan is made, drawing only 6", No, it's not. That is not how mulligans work in MTG (anymore), the OP specifically indicated they know how it works (since they referenced putting cards on the bottom of the deck as part of doing a mulligan).

If so, you are right Ikegami. Even if this really leaves me perplexed ... I can't believe it: can a player see all seven cards even in the three mulligans after the first draw, and choose the 5 or 6 ones to hold in his hand? That's why this game has lost charm ... in this way it is practically always worth taking Mulligan ... –

@ManoFromBerlin, It is more favorable than the old method. (Answer to the question using the old way: 32% Answer to the question using the new way: 40%). But not by much. That means that every 25 games, you'll get the card in your starting hand twice more.

Odds for non-Brawl using the pre-Core 2020 mulligan rules (according to Ikegami's editing)
The probability of having a desired card in the hand,at the first draw of seven cards, is, by removing the commander, (7/59) * 100 = 11.8644%
If, at the first draw, the desired card has not arrived, then, according to what is stated in the question, you will draw again using the mulligan; and then the probability of reaching this point, which is obviously greater than 7/59, is as follows -

first of all, this probability is due to the probability of not drawing the desired card to the first set of seven cards, which is: (52/59) * 100 = 88.1356%
this value is then multiplied by the probability of drawing the desired card now, which, having returned the 7 first cards in the deck, but now having to draw 6, is (6/59) * 100 = 10.1695%
Consequently, the probability of one mulligan, and then of finding the card sought in the six cards of the second draw is (52/59) * (6/59) * 100 = 8.9629%

If the card you are looking for hasn't arrived at this point, then, as specified in the question, you will mulligan for the second time.
So, this probability is obtained by multiplying the value of the probability of not drawing the card among the first seven (52/59), multiplied by the probability of not drawing the card even the second time (53/59), multiplied by the probability of draw the desired card now, but drawing only 5, or 5/59.
Consequently, the probability of mulligan twice, and then of finding the card sought among the five cards of the third draw is: (52/59) * (53/59) * (5/59) = 6.7%
At this point, the total probability of finding a desired "key-card", being able to make two mulligans, is given by the sum of the three possibilities of finding the card:
(possibility of finding the card after the first draw) + (possibility of finding the card after the first mulligan) + (possibility of finding the card after the second mulligan) =
11.8644+ 8.9629+ 6.7 = 27.5273%
Well ... that's all ... at least I think.
I hope there are no typing and/or math errors !!!

Odds for Brawl using the Core 2020 mulligan rules
The result shown above works only if, in making Mulligan the first time, the player draws 6 cards (not 7), and, in making Mulligan the second time, if the player draws only 5.
In addition, the result is valid only without looking at the cards that are placed at the bottom of the deck.
But, as I suspected, this does not accord with the current Mulligan rule...
I think I will leave this answer here just the same, also because, even if it refers to a mulligan rule that is not now applied, its procedure is correct.
To prove it, I will verify that his calculations are correct, rewriting them taking into account my own procedure, but considering the correct Mulligan rule.
The probability of having a desired card in the hand, at the first draw of seven cards, is, by removing the Commander, (7/59) * 100 = 11.8644%.
The probability of having the card at the first mulligan is:
(52/59)*(7/59)*100 = 10.4567652973%.
The probability of having the card at the second mulligan is:
(52/59)(52/59)(7/59)*100 =9.21613212646%.
(This means that the player is able to, according to the current Mulligan rules, look at 7 cards, and choose for himself the one that will be placed at the bottom of the deck, according to his needs. And it is always for this reason that the calculations agree based on the fact that, making Mulligan, for the purpose of searching for a "key card", it is as if the player was still drawing seven cards).
The probability of having the card at the third Mulligan -
(or when, granting the answer to the fact that the asker claims to want to stay,at most,with 5 cards in hand, after the Mulligans, which, however, are 3, and therefore not 2) -
is:
(52/59)(52/59)(52/59)*(7/59)*100=8.12269272162%
(And this also means that the player is able to, according to the current Mulligan rules, look at 7 cards, and choose for himself the two that will be placed at the bottom of the deck, according to his needs).
At this point, the total probability of finding a desired "key-card", being able to make three mulligans, is given by the sum of the four possibilities of finding the card. And is:
11.8644% + 10.4567652973% + 9.21613212646% + 8.12269272162%= 39.6590927216%.
This result is absolutely close to the correct one, already calculated in the right answer, higher.

If so, you are right Ikegami. Even if this really leaves me perplexed ... I can't believe it: can a player see all seven cards even in the two mulligans after the first draw, and choose the 5 or 6 ones to hold in his hand? That's why this game has lost charm ... in this way it is practically always worth taking Mulligan ...

Anyway, I think Ikegami's 40% is really too high ... It is equal to take almost 24 cards,but in only one draw...while here we are talking about taking 21 cards,in three draws,and in the best case! But...Ok...Ikegami has right: it is equal to take 28 cards,in 4 draws, and because the first mulligan means to draw 7 cards, just as it was the first time. It is this way...but it is also unbelievable...These new and honeyed in syrup rules are killing me ...Mulligan should be an exception, but this way it becomes the rule instead, because it is too convenient to do it!

But no!Your calculation is absolutely right, Ikegami.But mine is too,and you can still leave it here with the new editing, because now the steps are well specified and can be followed well by anyone in this way.And then, if the current Mulligan rule were to change, there would be all the procedure specified step by step. Anyway, you are right, Ikegami:you calculated everything before and correctly, because you know the rules of the game much better than me! You have to admit, however, that kind of Mulligan is the rule,not the exception.You first say:-"Ok, players:make your hand as you please!"

Re "But mine is too", Well, now they are.

Re "This result is absolutely close to the correct one, already calculated in the right answer, higher.", It's the same answer; I simply rounded to the nearest %. No point in being more precise.

Your answer came first,and has been rightly accepted.My incorrect answer makes yours stand out.I carried out the calculations independently of your answer, and in the end,having ascertained thanks to you that the Mulligan rule to which I referred was incorrect,I showed how the procedures were still correct both.And then,in this way I point out that this Mulligan rule is in my opinion excessive. Ok, that's the rule: but I think that having the chance to always see all the 7 cards,at each Mulligan,and choose the ones that serve the most,is excessive. But I'm only an useless dreamer...

However, there is another good reason why I think this answer may still have reason to be left where it is.

The OP may know the current Mulligan rule; after Ikegami's answer,certainly yes; but,before asking the question,he might not even know it at all, OR would not want to refer to it. Based on the question,the asker demonstrates - it is true - that he knows that the discarded Mulligans'cards go to the bottom of the deck. But, assumed that he talks about 2 Mulligans instead of 3, as he does in the question,he could actually talk about a personalized Mulligan procedure to be carried out in the manner that the asker himself describes in the question.

Then,the data that I used during the first part of the answer,agree exactly with the Mulligan as described by the OP,which may not be the one in force at all.

Let us continue this discussion in chat.

On the other hand, if the question is structured in this way,in the sense that the OP specifies clear that he wants to do 2 mulligans - and besides,even the editing of Ikegami himself does not correct what he himself calls as an "OP's error",and still reports the fact of running a maximum of two(not three) mulligans remaining with 5 cards in hand,even in the edited question - I do not see what is so inappropriate in responding even taking into account the possibility of a "customized mulligan",or regarding the past rules.Then,Ikegami is right:OP certainly means the actual Brawl's Mulligan Rule

Are you seriously claiming the fact that I think the OP wants to do 3 mulligans is a problem? If someone wants to calculate the odds for a different number of mulligan, they just need to replace the 4 in (52/59)^4. Remember, a good answer shows how to do something rather than just giving the answer for this very reason. (That's also why I calculated the odds in terms of not finding the card. Your approach only works because every card is different.)

No, seriously Ikegami, your answer is the best. But I did many things: 1)
I reviewed the Post; 2) I compared the Post with those to which it had been wrongly associated; 3) I found that it is not a duplicate, as other users had stated - and on this point, Ikegami, it seems to me that we have always agreed, and we only; 4) Then, I answered the question, doing all the calculations,but with a procedure very different  from yours, to show an alternative way of answering.

I believe that for all these reasons, and others too, it is right that my answer should be considered as a good attempt. Considering also that my questions have always been subject to dubious duplications, and that they are always well-edited to eliminate any type of error, it seems to me that I have done well what I have been asked to do. Then, it's true ikegami: your answers are always excellent, and it is very difficult to overcome them, for anyone ...

I'm confused. Who said it wasn't a good attempt? And why would anyone care?


text:
Scrabble premium word score question
In the following scenario, does a double word premium apply to both words made on the same turn? Player 1 plays the word "book". Player 2 plays the word "shot", placing the "s" on a double word score space, and also forming the word "books".

Yes.
From the rules,

When two or more words are formed in the same play, each is scored. The common letter is counted (with full premium value, if any) for each word.

(Emphasis mine.)

Thanks. Maybe I'm taking the rule too literally, as it says that the "common letter" is counted with full premium value for both words. Can I interpret that as saying the premium applies to both words (not just the "common letter")?

The premium value of the letter in your example is to multiply the points for the word built. Or as the rule I quoted says, each of the words built. There is nothing to support a theory that a Double Word square sometimes acts as a Double Letter square


text:
Checking a +1/+1 counter requirement
Abzan Battle Priest has an ability that reads

Each creature you control with a +1/+1 counter on it has lifelink.

Does it grant lifelink to creatures with two +1/+1 counters?

What do you mean by "becomes a +2/+2"? Are you talking about cards like Dauntless Onslaught that say a creature "gets +2/+2", or cards that put two +1/+1 counters on the creature?

@dopplegreener, First of all, he's clearly talking about Abzan Battle Priest since he copied the text verbatim, and there's nothing else remotely similar. So both the card and the question mentioned counters. Contrary to what you say, this is definitely a question about counters, whatever he meant by "becomes say a +2/+2". I have no problem editing an unanswereable question into a question whose answers answer his question whatever his question might be. He can always clarify or ask a followup question later. The sites of this network work better that way.

@doppelgreener, You seem to have stopped reading my comment halfway through.

@Ikegami: While you are correct that the question is clearly about counters, your edit loses an important distinction.  It is possible - likely even - that the asker and is confused by what happens if the creature has a single +2/+2 counter (for some reason), as opposed to two +1/+1 counters.  Answers to the question in its new form may not discuss that case.

@Tynam, I know. I explained that. You also seem to have stopped reading my comment halfway through.

@ikegami I read it.  I disagree: I don't think answers to the edit will answer his question.  We should close and request clarification - this situation is exactly what close-as-unclear is for.

@Tynam, won't? It was. Closing as unclear is a completely stupid alternative over actually answering the question. Nothing stopping the OP was clarifying if need be. Really guys, think for a second about what's best for everyone.

A creature with two +1/+1 counters has a +1/+1 counter, so Abzan Battle Priest's static ability will grant that creature lifelink.
Note that being granted +1/+1 and having a +1/+1 counter are very different. If you control Dauntless River Marshal, Abzan Battle Priest and an Island, the Marshal won't be granted Lifelink.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How does Blood Sun interact with Cavern of Souls?
How does Blood Sun interact with Cavern of Souls? Does the interaction depend on the order in which the cards are played? If I use the cavern's second mana ability and I never chose a creature type, what happens? Does the Cavern produce mana that can't be spent on anything?

If Blood Sun was on the battlefield already when Cavern of Souls entered, then the third ability will produce mana that can't be spent on anything. Otherwise, the third ability will function as normal.
As of the 2017 Ixalan rules update*, Blood Sun's ability will stop Cavern of Souls' first ability from working as it enters the battlefield:

614.12. Some replacement effects modify how a permanent enters the battlefield. (See rules 614.1c–d.) Such effects may come from the permanent itself if they affect only that permanent (as opposed to a general subset of permanents that includes it). They may also come from other sources. To determine which replacement effects apply and how they apply, check the characteristics of the permanent as it would exist on the battlefield, taking into account replacement effects that have already modified how it enters the battlefield (see rule 616.1), continuous effects from the permanent’s own static abilities that would apply to it once it’s on the battlefield, and continuous effects that already exist and would apply to the permanent.

If you never choose a creature type, then this affects Cavern of Souls third ability because it is linked to the first ability:

Linked Abilities

607.1. An object may have two abilities printed on it such that one of them causes actions to be taken or objects or players to be affected and the other one directly refers to those actions, objects, or players. If so, these two abilities are linked: the second refers only to actions that were taken or objects or players that were affected by the first, and not by any other ability.

607.2. There are different kinds of linked abilities.

607.2d If an object has an ability printed on it that causes a player to “choose a [value]” and an ability printed on it that refers to “the chosen [value],” “the last chosen [value],” or similar, those abilities are linked. The second ability refers only to a choice made as a result of the first ability.

When you produce mana with the third ability, the list of things it can be spent on is {creature spells of the chosen type}. That list happens to be empty, as you never chose a type, so there's nothing you can spend it on.
However, if Blood Sun only entered the battlefield after Cavern of Souls had; then it would not affect the third ability. Even though the Enters the Battlefield ability would be gone now, the creature type was already chosen; that choice is still "a choice made as a result of the first ability".
*Previous to the rules update, Blood Sun's ability wouldn't have effected Cavern of Souls until after it entered the battlefield, when the choice would have already been made.

Might be worth clarifying that Cavern will work as normal if it came into play with no Blood Sun on the field.

Indeed, since that's a major component of the question. I would go so far as to say this answer is misleading in its current form; I'd want to see it edited to make it clear that the statement in the first paragraph (up through the first paragraph after the rules quote) only applies if Blood Sun enters first.

@DavidZ Indeed, I actually misread the question and didn't realize it specifically asked if the order mattered. For some reason I read it as asking about the case where Blood Sun entered first.

Ah, that makes sense. Anyway it's definitely better after the edit.

Does rule 706.7a apply here? If it does, is it possible that the second mana ability never generates any mana in the first place?

@user3490 The last sentence, "If an ability refers to an undefined choice, that part of the ability won’t do anything" may apply, although the previous paragraph seems clear that it is only talking about copying objects. Also, I am not sure on what a "part" of an ability is... one could interpret this to say that the "spend this many only..." is the part of the ability that doesn't do anything, meaning that the mana can be spent normally. I'm almost sure that's not correct, but I don't know how to define how much of the ability is the "part" that refers to the choice.

@user3490, When 706.7a says "If an ability refers to an undefined choice, that part of the ability won’t do anything", it's just an application of 609.3 that's equally applicable in your scenario.

@ikegami is there a clear way to know that 609.3 (combined with 706.7a) isn't saying that because the choice is undefined, that the "spend this mana only..." will simply do nothing? (Meaning you could spend it however you want).

"Spend this mana only..." is not an action. You never do or not do it. It's a restriction on another action (actually spending the mana). 609.3 doesn't apply there. In fact, 609.3 is rarely going to be applicable since effects rarely have you spend mana. If an effect has you spend mana, you wouldn't be able to spend the produced mana because of 609.3 (whether the choice was defined or not, because it can only be spent on casting costs). (Yes, that means my earlier comment is wrong, at least misleading.)

Is Blood Moon a typo in the answer? I guess it works for both cards (Sun and Moon) tho.

@JamesRisner Thanks, fixed.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Translation Software
Which software is best for translating Yu-gi-oh cards from Japanese to English without changing the original art.

This question appears to be off-topic because it is about translation software recommendations.

Can someone just answer me ?

I don't understand the question. Are you looking to identify what a Japanese card is, given the clear inability to understand the name? Or are you looking to do something to cards that don't exist in English?

Im trying to edit the cards that dont exist in english

Software recommendations are off-topic since 1) they have nothing to do with board or card games, 2) they are primarily opinion based, 3) they are specifically disallowed on the sister sites that do deal with software for numerous reasons.

Hello @Leopold, welcome to this site. I'm afraid this is not the best place to ask this question as we probably don't have the required expertise to give you a really helpful answer.

Two steps are needed

Erase the existing text. This is done by finding a card with little text and construct a square you can copy and paste over the existing text of the card you want to translate.
Add text.

Every single image manipulation program can perform those operations (copy, paste, move and add text), including MS Paint, a little plaything that comes with Windows.

Could you recommend a software ?

@Leopold, Any will do.


text:
Will Exquisite Blood trigger even if the life loss eliminates the player
We are trying to workout a ruling for a multi player game of mtg involving 8 players and a 1 player with an Exquisite Blood.
So lets say there are eight players. Player #3 has Exquisite Blood. Player #5 hits player #8 for 100 damage. Player #8 dies and Exquisite Blood triggers (For 100).
The confusion comes in for when Exquisite Blood resolves. As the opponent that lost 100 life not longer exists, will Exquisite Blood still resolve?

I edited your question a bit to collapse the list. I'm confused though: player #3 is the one with exquisite blood, and nothing happened to them. Did you mean to suggest player #8 has exquisite blood?

Player #3 has the Exquisite Blood. Nothing happens to them. The confusion is; does player #3 gain all 100 life or the fact that the opponent that lost the life is dead at the time the spell resolve make any difference.

Oh, right, yeah, this makes sense now.

Fixed the title a bit. It's still misleading because the EB triggers when the life loss happens, which is before the player is eliminated by the SBAs.

Related: http://boardgames.stackexchange.com/questions/4749/in-mtg-what-happens-when-a-player-loses-in-multiplayer (which is kind of a more general version of this question)

Player #3 will gain 100 life.
The fact that #8 is no longer in the game has no effect on Exquisite Blood's triggered ability. All of its effect was put on the stack when #8 lost 100 life. It does not require #8 to be in play when it resolves, because it does not check for any current information about that player. Even if it did, rule 800.4g would take effect:
If an effect requires information about a specific player, the effect uses the current information about that player if he or she is still in the game; otherwise, the effect uses the last known information about that player before he or she left the game.
Only effects that require a target are affected if that target is no longer legal or does not exist anymore; they would be countered in that case.

interesting side question, although not  really appropriate for this format - is player #8 a terrible person if he scoops before damage is dealt to prevent the life gain?

I can only confirm that conceding would prevent the damage and therefore the trigger. It doubt it would be unsporting conduct either even in a multiplayer match where his decision would influence the rest of the game, because every player may concede at any time, immediately removing him from the game. Whether or not that makes him a "terrible" person is indeed beyond the scope of this site.

thank you for your answer. Maybe not the best place for this, yet still related, would conceding stop a Exquisite Blood + Sanguine Bond from killing all players?

As I wrote, conceding before the life loss happens prevents Exquisite Blood from triggering, and therefore Sanguine Blood as well.


text:
What happens if I have two different cards that give me an extra turn after this one?
For example, I play Emrakul, the Aeons Torn. That gives me an extra turn after this one. Then I activate Ral Zarek's ultimate and only get one extra turn (one) heads. Will the Emrakul's extra turn meld with Ral Zarek's so I only get 1 extra turn, or will I get two?

You get two extra turns.

500.7. Some effects can give a player extra turns. They do this by adding the turns directly after the current turn. If a player gets multiple extra turns or if multiple players get extra turns during a single turn, the extra turns are added one at a time. The most recently created turn will be taken first.

How did you turn the card names into hyperlinks?

@BolasStone, [mtg:Card Name]. (Fails for cards with apostrophes because of a bug in Gatherer.)


text:
What happens when two players both play Control Magic on the same creature?
For example, I control a Pathraizer of Ulamog. One of my opponents cast Control Magic on it to take control of it. Then my other opponent casts the same spell. How can they both control it?

I think that an older question, Perplexing Chimera + Hypnotic Siren. Do I control all of my opponent's spells?, should be closed as a duplicate of this newer question. This one is worded so much better.

Even if this question were asked first, I would still leave that question open to handle the interaction between different types of control change effects (aura vs unattached continuous effect).

The last effect created takes precedence.
Section 613 resolves interaction of Continuous Effects. The rules relevant to this specific question are the following:

613.6. Within a layer or sublayer, determining which order effects are applied in is usually done using a timestamp system. An effect with an earlier timestamp is applied before an effect with a later timestamp.
613.6a A continuous effect generated by a static ability has the same timestamp as the object the static ability is on, or the timestamp of the effect that created the ability, whichever is later.

Continuous effects are continually reevaluated. This means that destroying the second Control Magic gives control of Pathraizer to the controller of the first Control Magic.


text:
Is Bioshift affected by Doubling Season?
I just want to know if Bioshift is affected by Doubling Season. The text states that Bioshift moves counters from a target to another, but is that the same as "putting them on" a target?

Actually, I made a deck last year that was based on this.

I am working on a Hydra Deck at the moment, where bioshift is going to play a vital part, after I had this issue clarified!

Yes it does. See the rulings on Bioshift:

To move a counter from one creature to another, the counter is removed from the first creature and placed on the second. Any abilities that care about a counter being removed or placed on a creature will apply.

The ruling stems from the following rule:

121.5. If an effect says to “move” a counter, it means to take that counter from the object it’s currently on and put it onto a second object. If the first and second objects are the same object, nothing happens. If the first object has no counters, nothing happens; the second object doesn’t get a counter put on it. If the second object (or any possible second objects) is no longer in the correct zone when the effect would move the counter, nothing happens; a counter isn’t removed from the first object.


text:
Vastwood hydra + Cathars' Crusade + Mercy Killing
I have a question about a combat trick involving the cards mentioned in the Title. Situation is:
I control a 4/4 Kalonian Hydra and a 4/4 Vastwood Hydra. I also have Cathars' Crusade on the battlefield as well.
On my turn I declare both of the Hydras as attacking creatures. As such the Kalonian Hydra doubles the number of +1/+1 counters on both Hydras, making both of them 8/8 creatures.
My opponent decides to block my Vastwood Hydra with multible blockers, wich would do  leathal damage to it.
So when blockers are declared, before combat damage is resolved, I cast Mercy Killing on my Vastwood Hydra. As a result 8 Elf Tokens comes into the battlefield, and triggers Cathars' Crusade (and makes all the Elf tokens 8/8 Elves, but that is not important in this situation)
It also puts 8 +1/+1 tokens on my Kalonian Hydra, making it a 16/16 Hydra. And when the Vastwood Hydra goes to the graveyard (because I sacrificed it) I distribute all of its 8 +1/+1 counters to my Kalonian Hydra and makes it a 24/24 Hydra.
Combat damage resovles and the kalonian Hydra kills off my opponent.
Is this how it would work, or have I missed something?

What part are you unsure about? This looks correct to me

The elfes are 9/9 (1/1 with 8 counters) instead of 8/8 :)

That's right :)... even better!

Vastwood Hydra is already in the graveyard at the point where you say "And when the Vastwood Hydra goes to the graveyard". But everything else is correct.

You declare your attackers. [Kalonian Hydra's third ability triggers.]
Kalonian Hydra's third ability is placed on the stack, then resolves:

Add 4 +1/+1 counters to Kalonian Hydra.
Add 4 +1/+1 counters to Vastwood Hydra.

Your opponent declares his blockers.
You cast Mercy Killing targeting the Vastwood Hydra, then it resolves:

Sacrifice Vastwood Hydra. [It goes to the graveyard here. This triggers its second ability.]
Put 8[1] 1/1 Elf Warriors on the battlefield. [Cathars' Crusade's ability triggers 8 times.]

At this point, a player (you) is about to gain priority for the first time since some abilities have triggered, so triggered abilities are placed on the stack. You can place your abilities on the stack in any order you want. It doesn't really matter here.

Place Vastwood Hyra's second ability on the stack.
Place Cathars' Crusade's ability on the stack. (1)
Place Cathars' Crusade's ability on the stack. (2)
...
Place Cathars' Crusade's ability on the stack. (8)
Cathars' Crusade's ability (8) resolves.

Place a 9th counter on Kalonian Hydra, and one on each of the 8 Elf Warriors.

Cathars' Crusade's ability (7) resolves.

Place a 10th counter on Kalonian Hydra, and a 2nd on each of the 8 Elf Warriors.

...
Cathars' Crusade's ability (1) resolves.

Place a 16th counter on Kalonian Hydra, and an 8th on each of the 8 Elf Warriors.

Vastwood Hyra's second ability resolves.

Place 8[1] counters on Kalonian Hydra for a total of 24.

Notes:

Last known information (LKI) is used. When it last existed, it had 8 power and 8 +1/+1 counters.

I can provide rules numbers/quotes or extra details if you tell me where to focus.

Thank you for the detailed explanation of the situation. My opponent was not quite sure, that all of the aforementioned triggers would resolve before the Combat damage step. I guess I have something to show for it now! Thanks

Moving to the next step happens when everyone passes when the stack is empty. There can't be triggered abilities on the stack left to resolve.


text:
What is this game called?
I was recently taught a game with the following rules.
Deal 5 cards face down to each player in a line and then 5 cards on top of the face down cards. Then deal 3 cards to each player for their hand.
Play begins with the player to the left of the dealer. The player selects a card from their hand and plays it face down. The next player plays a card with the following rules. The card must be of equal rank or higher, aces high. If a set of 4 of the se rank is ever on the stack, the stack is cleared and all cards in the stack are removed from play and the last player to play plays again. 2 and 10s can be played on any card. 10s also clear the stack and give the playing player a turn again. If a player cannot play a card they take the stack into the hand. Only once all cards in your hand are played can you play the face up cards in front of you. And only when all face up cards have been played can you play the face down cards.

The game you describe seems to be a fairly typical example of the Shithead / Palace / Shed / Karma / China Hand / Ten-Two Slide (Wikipedia) family of card games.  These games are known by a variety of names around the world, and have many local variations.  The game you describe does, however, contain most of the basic elements:

Players are dealt three initial sets of cards: one set in their hand, one face up on the table, and one face down.
The goal of the game is to get rid of all your cards, both those in your hand and those on the table, by placing them on top of a common discard pile.
You must first play the card in your hand, then the face-up cards from the table, and finally the face-down cards, which must be played blind without looking at them.
On each turn, you may place one or more cards of the same rank onto the pile, but only (with some exceptions) if they equal or exceed the rank of the current topmost card on the pile.
The most common exceptions are that a 2 or a 10 may be played at any time; a 2 allows the next player to play any card, while a 10 (or, commonly, any four cards of the same rank together) will clear the whole pile and allow you to restart it from any rank you want.
If you cannot play any card (or fail to play a blind card), you must pick up the entire discard pile into your hand and let the next player restart a new pile.


text:
Can I control a planeswalker with Confiscate?
Can I control a planeswalker with Confiscate? Is a planeswalker a permanent that can be the target of Confiscate?

If you ever actually do this, remember to use your priority wisely. If your opponent has a way to deal with enchantments at instant speed, you will have one and only one shot to activate the planeswalker's abilities at sorcery speed.

Yes. All cards and tokens on the battlefield are permanents, including Planeswalkers.

110.1. A permanent is a card or token on the battlefield. A permanent remains on the battlefield indefinitely. A card or token becomes a permanent as it enters the battlefield and it stops being a permanent as it’s moved to another zone by an effect or rule.
110.4. There are five permanent types: artifact, creature, enchantment, land, and planeswalker. Instant and sorcery cards can’t enter the battlefield and thus can’t be permanents. Some tribal cards can enter the battlefield and some can’t, depending on their other card types. See section 3, “Card Types.”


text:
How does Imprinting and the affinity work together?
I'm currently working on a artifact deck (casual). I am thinking of imprinting Somber Hoverguard on Soul Foundry. But how do I dertermine the imprint cost, and will the activation cost of Soul Foundry change as the numbers of artifacts change on my battlefield?

You don't get any benefits from Affinity.

The activation cost of Soul Foundry's activated ability is based on the Converted Mana Cost (CMC) of the imprinted card, and Somber Hoverguard's CMC is 6.

202.3. The converted mana cost of an object is a number equal to the total amount of mana in its mana cost, regardless of color.
202.1. A card’s mana cost is indicated by mana symbols near the top of the card. [...]

Affinity doesn't affect the mana cost. It's a cost reduction that reduces the cost to cast the spell, not its mana cost. To paraphrase 601.2e,

cost to cast a spell
  = mana cost or alternative cost
  + additional costs and cost increases
  - cost reductions


text:
Goblin Brigand and Norn's Annex
My opponent has Goblin Brigand on the battlefield. It says: "Goblin Brigand must attack each turn if able"
I control a Norn's Annex. My opponent is unable to produce white mana, but can (of course) pay for the attack with his life.
The question is:
Is the Goblin still "able to attack" since my opponent has no other option (in this particular game) than to pay for the attack with his life? Is my opponent "forced" to pay 2 life for the attack?

From rulings of Norn's Annex you can clearly see your answer. The answer is:

If you control Norn's Annex, your opponents can choose not to attack with a creature with an ability that says it must attack.

You are mistaken about the lack of another option: He has the option of not paying the cost, since it's an optional cost ("unless you")[1].
If he doesn't pay the cost, the goblin isn't able to attack, so it doesn't attack[CR 101.2].
You are never forced to meet any requirements. If you don't have enough mana to cast a spell, you don't have to play that land that would allow you to cast it. If Norn's Annex prevents your Goblin Brigand from attacking, so be it. This is supported by a ruling

If you control Norn's Annex, your opponents can choose not to attack with a creature with an ability that says it must attack.

He might also have the option to sacrifice the Brigands, to destroy the Annex, etc.

The effect of Norn's Annex can be interpreted as follows:

Creatures can't attack you or a planeswalker you control.
A creature's controller may pay (W or 2 life) at a given attack step for that creature to be exempt from this restriction for that attack step.

So as far as the game is concerned, Goblin Brigand is not able to attack. The fact that a player could choose to pay a cost which would make Goblin Brigand able to attack is not relevant.
The comprehensive rules that support this interpretation are

508.1d ... If a creature can't attack unless a player pays a cost, that player is not required to pay that cost, even if attacking with that creature would increase the number of requirements being obeyed.

and more generally,

117.12a Some spells, activated abilities, and triggered abilities read, "[Do something] unless [a player does something else]." This means the same thing as "[A player may do something else]. If [that player doesn't], [do something]."

I think it's a fairly general rule of thumb that players are not required to pay costs to make things possible.


text:
Reverse the Sands and Sanguine Bond
If I have Sanguine Bond on the battlefield, and I play Reverse the Sands, will the redistribution of life trigger Sanguine Bond if I choose to distribute less life to my opponent than he originally had?
ex. I have 10 life left, my opponent has 14 life left. I play RtS and distribute life totals so I end up with 23 life and my opponent with 1 life, would that be considered a "loss of life/life gain" situation?

Note that Sanguine Bond triggers when you gain life, NOT when your opponent loses life.

Yes, Sanguine will trigger, but you can't set his life total to 1.
Changing life totals is treated as gaining life or losing life to reach the specified life total[CR 118.5]. So that will trigger the bond in this case. (It wouldn't if you started at 14 and he started at 10.)
However, you've misunderstood what Reverse the Sands does. The specific gatherer ruling on this for the card is "You can't split up a life total when you redistribute it". You can only make your opponent's total 10 and yours 14 , or make no changes.
So you end up with the same life totals on different players, and then resolve any following effects. In this case he loses 4 and you gain 4, and then Sanguine triggers off your lifegain and he loses another 4. If you choose not to redistribute, Sanguine won't trigger.

The good news is that if he has more than twice as much life as you, you immediately win!

@bwarner Not if he can respond to the trigger and gain even more life.

I guess it was the word: redestribute that threw me off. Why not just use the wording: exchange life totals?

@user7268, An exchange is an A for B swap, but Reverse the Sands can manipulate the life of more than two players. Note that it's life totals being redistributed, not life points.

Also, I think Exchange is reserved for permanents/spells, not values, because you need specific rules. To the comp rules!

@deworde, Nope, see Tree of Redemption: "Exchange your life total with Tree of Redemption's toughness." Not only are neither of the things involved in the exchange permanents or spells, they're not the same kind of thing!

@BrianS Do you know, I'd forgotten Tree of Redemption's official rules text. God, I love this silly, silly game.


text:
Can I cast a Slaughter Pact on a Spellskite after I played Lightning Storm and I already discarded some lands?
My opponent has a Spellskite on board. He is at 17 life, has one blue mana and one land in hand. I just resolved Ad Nauseam + Angel's Grace and I have my entire deck in my hand.
I have 2 Slaughter Pacts, 15 lands, and a Lightning Storm in hand.
Now because I'm a dumb I forget to cast Slaughter Pact before I play Lightning Storm.
After playing Lightning Storm, when I'm deciding how many lands I'm going to discard to add charge counters, is there any way to win the game? Let's say I'm in the situation where I can't play any more Slaughter Pacts in answer to Lighting Storm, but Lighting Storm is still on stack.

So, I'm confused. If you have 2 slaughter pacts in hand, what is the situation where you can't cast any more slaughter pacts in response to the lightning storm?

I allready placed charge counters discarding a land and then we stopped and started thinking to the rules. But from the other answers i can understand that i when he changes the target of the spell with spellskite i can answer with a slaughter pact!

Given the situation you described (2 Slaughter Pacts and 15 lands in hand, and a Lightning Storm on the stack), you can definitely win.
I assume that you started by targeting your opponent with Lightning Storm. Then you still have priority, so you cast Slaughter Pact targeting the Spellskite. You also discard 7 lands to Lightning Storm's ability (failing to choose new targets) so that it has 14 charge counters and still targets your opponent.
Your opponent gets priority after each of those abilities goes on the stack, so they can activate either Spellskite's or Lightning Storm's ability to redirect Lightning Storm, but each of your activations redirects it back to targeting them, so that doesn't matter. After you have resolved the ability 7 times, pass priority with only Lightning Storm and Slaughter Pact on the stack. Your opponent should then activate Spellskite's ability once to change Lightning Storm's target to Spellskite.
After that ability resolves and your opponent passes priority, you should pass priority and allow Slaughter Pact to resolve, then activate Lightning Storm's ability once to redirect it to target your opponent. Your opponent should then activate the ability using the one land card in their hand to redirect it back to you. You can then use one of your remaining lands to activate Lightning Storm's ability one final time.
At this point in the game, your opponent has no non-land permanents, no mana, and no cards in hand. You have 1 Slaughter Pact and 6 lands in hand. The stack has a single Lightning Storm with 20 charge counters on it targeting your opponent. Once you both pass priority, Lightning Storm resolves and deals 23 damage to your opponent, who then loses the game.


text:
Question about interaction: levitation, then gravity sphere, then oblivion ring on each, then tranquility. Who flies away?
Here are the cards involved:

Levitation

Creatures you control gain flying.

Gravity Sphere

All creatures lose flying.

Oblivion Ring

When Oblivion Ring enters the battlefield, exile a nonland permanent (card).
When Oblivion Ring leaves the battlefield, return that permanent (card) to play under its owner's control.

Tranquility:

Destroy all enchantments.

So, I play a Levitation, and all of my creatures gain flying. Next turn, my opponent responds with a Gravity Sphere, and so, since the Gravity Sphere came into play more recently, it overrides Levitation, so all of my creatures would then lose flying. So, I play Oblivion Ring on his Gravity Sphere, my creatures gain flying again. He then plays Oblivion Ring on my Leviation, they lose flying. I then play Tranquility, which simultaneously kills both Oblivion Rings and brings back the Levitation and Gravity Sphere. So at this point, does everyone fly, or do they all stay on the ground?

You shouldn't use priority to refer to interactions between continuous effects. "Priority" in Magic specifically refers to when each player has the opportunity to play spells and activate abilities.

so what would you call it then? precedence?

It's technically the order of application of continuous effects. But you might have noticed I already removed all mentions of priority from your question without using any jargon that heavy.

Short answer: your creatures lose flying.
Since both Oblivion Rings are destroyed at the same time, their triggered abilities "When Oblivion Ring leaves the battlefield, return the exiled card to the battlefield under its owner's control." are put on the stack. They are put on the stack in APNAP order:

603.3b If multiple abilities have triggered since the last time a player received priority, each player, in APNAP order, puts triggered
  abilities he or she controls on the stack in any order he or she
  chooses. (See rule 101.4.) Then the game once again checks for and
  resolves state-based actions until none are performed, then abilities
  that triggered during this process go on the stack. This process
  repeats until no new state-based actions are performed and no
  abilities trigger. Then the appropriate player gets priority.

Since we are in your turn (because Tranquility is a sorcery), the ability of the Oblivion Ring you control is put on the stack, then on top of that is put the ability of the other Oblivion Ring. First we resolve the one your opponent controls which returns Levitation to the battlefield. Then we resolve the one you control, and return Gravity Sphere. Since Gravity Sphere entered last on the battlefield, all your creatures lose flying.
Of course, if you somehow managed to destroy both rings on your opponents turn, then the result would be reversed and your creatures would have flying.

When you asked the question, you thought Levitation and Gravity Sphere were entering the battlefield simultaneously. You now know that's not the case here, but you probably still wonder what would happen if they did enter the battlefield simultaneously.
The order in which the effects are applied are defined in the section of the rules entitled "Interaction of Continuous Effects". As you've indicated, it comes down to timestamps in this case (whichever has the latter timestamp wins out). So which one has the later timestamp if they both entered the battlefield simultaneously?

613.6g If two or more objects would receive a timestamp simultaneously, such as by entering a zone simultaneously or becoming attached simultaneously, the active player determines their timestamp order at that time.

Being the Active Player, you would decide the timestamp order.

If you decide that Levitation has the lower timestamp, it gives the ability then Gravity Sphere takes it away, so none of your creatures can fly.
If you decide that Gravity Sphere has the lower timestamp, it task away the ability then Levitation grants it, so all of your creatures can fly.


text:
does gaining control of an aura gaining control of a permanent give me the permanent?
If my opponent steals a creature from me with Control Magic, I could steal it back by casting another Control Magic on it and (I think due to timing?) I would control the creature.
If instead I cast Steal Enchantment on the Control Magic, does the "You" become "me," and would I now control the creature?

Yes, if you take control of your opponent's Control Magic, you also take control of the creature it enchants.
The "you" on objects always refers to the controller of that object. If you take control of that object, you become the new "you". In the case of Control Magic, that also means that you take control of the enchanted creature.

109.5. The words “you” and “your” on an object refer to the object’s controller, its would-be controller (if a player is attempting to play, cast, or activate it), or its owner (if it has no controller). For a static ability, this is the current controller of the object it’s on. For an activated ability, this is the player who activated the ability. For a triggered ability, this is the controller of the object when the ability triggered, unless it’s a delayed triggered ability. To determine the controller of a delayed triggered ability, see rules 603.7d–f.

If you cast your own Control Magic after the opponent played his, you would also take control of the creature. However, it's not due to timing, which governs who can take action and when, but because of the more recent timestamp of your Control Magic vs. your opponent's Control Magic timestamp, meaning the latest timestamp "wins":

Interaction of Continuous Effects

613.1b Layer 2: Control-changing effects are applied.
613.2. Within layers 1–6, apply effects from characteristic-defining abilities first (see rule 604.3), then all other effects in timestamp order (see rule 613.6). Note that dependency may alter the order in which effects are applied within a layer. (See rule 613.7.)
613.6. Within a layer or sublayer, determining which order effects are applied in is usually done using a timestamp system. An effect with an earlier timestamp is applied before an effect with a later timestamp.

There is a MtG-specific meaning of "timing" that doesn't apply here, but this is due to "timing" in the general sense of "when things happen". Of course, if Control Magic were an interrupt, and a second one were played before the first resolves, then the player playing the first one would end up with control of the creature.

@Acccumulation OP specifically linked to the CR meaning of timing, so OP didnt mean it in the colloquial sense.

If my opponent steals a creature from me with Control Magic, I could steal it back by casting another Control Magic on it and (I think due to timing?) I would control the creature.

Yes, but not due to timing. CR 116, Timing and Priority, is a section of the rules that covers when players can cast spells and activate abilities.
In this scenario, you have two continuous effects (one created by the first Control Magic, and one created by the second Control Magic) trying to change control of an object. This is handled by CR 613, Interaction of Continuous Effects.
Both of the effects are applied in layer 2 (Control-changing effects)[CR 613.1b], and neither is a characteristic-defining ability, so they are applied in timestamp order[CR 613.2], so the last effect created wins out in this case[CR 613.6].

613.1. The values of an object’s characteristics are determined by starting with the actual object. For a card, that means the values of the characteristics printed on that card. For a token or a copy of a spell or card, that means the values of the characteristics defined by the effect that created it. Then all applicable continuous effects are applied in a series of layers in the following order:
613.1b Layer 2: Control-changing effects are applied.
CR 613.2. Within layers 1–6, apply effects from characteristic-defining abilities first (see rule 604.3), then all other effects in timestamp order (see rule 613.6). Note that dependency may alter the order in which effects are applied within a layer. (See rule 613.7.)
613.6. Within a layer or sublayer, determining which order effects are applied in is usually done using a timestamp system. An effect with an earlier timestamp is applied before an effect with a later timestamp.
613.6a A continuous effect generated by a static ability has the same timestamp as the object the static ability is on, or the timestamp of the effect that created the ability, whichever is later.
613.6c An object receives a timestamp at the time it enters a zone.

If instead I cast Steal Enchantment on the Control Magic, does the "You" become "me," and would I now control the creature?

Yes.
"You" refers to its controller[CR 109.5]. Since you control the enchantment, and since it gives its controller ("you") control of the enchanted creature, you control the enchanted creature.

CR 190.5. The words “you” and “your” on an object refer to the object’s controller, its would-be controller (if a player is attempting to play, cast, or activate it), or its owner (if it has no controller). For a static ability, this is the current controller of the object it’s on. For an activated ability, this is the player who activated the ability. For a triggered ability, this is the controller of the object when the ability triggered, unless it’s a delayed triggered ability. To determine the controller of a delayed triggered ability, see rules 603.7d–f.

To be specific, there is an interaction between the control changing continuous effects of Control Magic and Steal Enchantment, and I didn't realize at first that the question was not asking about that part of the interaction between those two cards.

There is no interaction. There are two effects, but they affect different objects, so there is no interaction, so 613 doesn't enter into play.

There is no requirement that two effects apply to the same object for them to interact. You can see in rule 613.7a the set of criteria for continuous effects to have a dependency relationship. And there has to be some relationship between the abilities for you to know that you have to apply Steal Enchantment's effect, which changes what "you" refers to on Control Magic, before applying Control Magic's effect, even though that's the opposite of timestamp order for those effects.

"613. The values of an object’s characteristics are determined by starting with the actual object. [...] Then all applicable continuous effects are applied in a series of layers in the following order". 613 documents how to apply effects to an object. You do each layer for each object. You don't do each object for each layer. Effects that effect Control Magic aren't relevant when applying effects to the creature. As such, there's no dependency since you can't have a dependency when there's only one effect that affects the object.

That is simply an inaccurate interpretation of that rule. Another example of a continuous effect on one object affecting continuous effects it applies to others is Blood Moon and Urborg, Tomb of Yawgmoth: "this is an example of dependency: both Blood Moon and Urborg want to apply in the same layer, but applying Blood Moon first will change what Urborg affects (since the Blood Moon will remove Urborg’s ability of making lands into swamps)"

No, that doesn't disagree with what I said. BM and UTY could potentially affect an object, so 613 comes into play. Furthermore, it's a dependency because the order in which the effects are applied prevents the other from being applied.

By the way, even if 613 did come into play in the Steal Enchantment + Control Magic scenario, there would be no dependency because of the (b) condition of 613.7a ("applying the other would change the text or the existence of the first effect, what it applies to, or what it does to any of the things it applies to") wouldn't be met.

Of course the (b) condition would be applied. It changes which player "you" refers to, which changes which player Control Magic grants control to.

No. Steel Magic's effect always applies to Control Magic, no matter the order in which it's applied. Control Magic's effect applies to the creature, no matter the order in which it's applied.

If you wish to ask a question, please Ask a Question.

@ikegami the explanation of my "sub question" in parenthesis question was super helpful, thanks!

@ikegami It looks like you were right: http://magicjudge.tumblr.com/post/157884125436/i-have-a-question-about-layered-control-change

I'm afraid I believe both of the previous answers to be wrong - though I haven't evaluated a copy of the rules closer to their posting, so that might be a matter of timing. Both answers rely on the effect of the first control spell being reevaluated upon the second spell coming into effect.
Somewhat unintuitively, the "you" in the text on the card is not "you" in the text of the continuous effect. According to the comprehensive rules...

611.2c If a continuous effect generated by the resolution of a spell or ability modifies the characteristics or changes the controller of any objects, the set of objects it affects is determined when that continuous effect begins. After that point, the set won’t change. (Note that this works differently than a continuous effect from a static ability.) A continuous effect generated by the resolution of a spell or ability that doesn’t modify the characteristics or change the controller of any objects modifies the rules of the game, so it can affect objects that weren’t affected when that continuous effect began. If a single continuous effect has parts that modify the characteristics or changes the controller of any objects and other parts that don’t, the set of objects each part applies to is determined independently.

Effectively, this rule causes "You" to be permentally evaluated as "Bob" in the effect - where Bob is the caster of the first control spell. Even when the control spell becomes controlled by "Sue" (where Sue is another unique player), the first control effect gives Bob control of Target [creature].
If it wasn't for this, then of course Hackworth's answer would be correct, because the You in the effect would apply.
With ikegami's answer, again, continuous effect Layers might be applicable in determining the order in which two control effects are evaluated (I don't think they do in this case, but it's irrelevant), however the effect created by the first control effect still applies to Bob.
Thanks!

This is not true. The continuous effect that rule describes doesn't apply to auras, only "one-shot" effects that change control (eg. Confiscation Coup).

I don't really know how to argue with that. It simply doesn't say what you claim. The text reads, " If a continuous effect generated by ... ability modifies the characteristics or changes the controller of any objects, the set of objects it affects is determined when that continuous effect begins. After that point, the set won’t change."
The static ability on an aura does create a continuous effect.
Nothing I quoted states "one-shot" effect.
This is literally from section 611 of the comprehensive rules - continuous effects.
One shot effects are handled by section 610.

I wonder if this has confused you.
"610.1. A one
-shot effect does something just once and doesn’t have a duration. " Note, this implies instantaneous effects, unrelated to this issue. As 611.1 says "A continuous effect ... for a fixed or indefinite period." One-shot effects have nothing todo with the OP or 611.2c. Sorry.

The thing that shows your mistake is in the rules you quoted, immediately after the part you bolded: "Note that this works differently than a continuous effect from a static ability." The effect from the aura is a static ability. It is not  an "effect generated by the resolution of a spell or ability".