text:
What do you do with the Basement Entrance Exploration card if you choose not to discard it?
The Basement Entrance card says after entering:

Add 5 cards from the Exploration deck to the Unknown deck.
Also, you may discard this card during this Scavenging to draw and resolve 2 Findings cards instead of one.

I do not understand this wording.  Are you supposed to be able to, by default, discard these cards to get 1 Finding?  The quote above seems to imply there is some base behavior allowing me to obtain 1 Finding.  Further, What do you do if you choose not to discard it?  Doesn't it get discarded at some point anyway?

The Basement Entrance gives you more time to search stuff (+5 cards) and you get to find more stuff if you find any (resolve 2 Findings).
When you find the Basement Entrance just set the card aside, add 5 cards to the Unknown deck and later, when you get to resolve a Finding, you can decide to discard the Basement Entrance to resolve two Findings instead of one. You don't need to discard the Basement Entrance on the first Finding, you can decide to hold the car for another Finding.


text:
Should reckoning effects be executed as if done simultaneously?
Should reckoning effects on an investigator's conditions execute as if done simultaneously (starting with the lead investigator)?  Two situations come to mind here.
1) If a reckoning effect condition results in my investigator being blessed or cursed, should I wait until after resolving all other reckoning effects to put the new condition into play (i.e. not benefit from the blessed condition for the other reckoning effects)?  This would effectively be like executing the effects simultanously.
2) If a reckoning effect kills my investigator, should I resolve all other effects as well? The case I am thinking of is where the Dark Pact roll of 1 results in another investigator being devoured.  If my investigator dies from another effect, does he get to skip resolving this dark pact or should he have to resolve it as well, as if simultaneous?
Mostly, this question came to mind because of a recent answer on this question.

You get to determine the order of the effects, but they do not happen simultaneously. Thus you should resolve curse effect first and blessings last.
Do note that if a reckoning effect causes you to gain a card (e.g. a condition) that has its own reckoning effect you do not have to resolve the reckoning effect on the new card until the next reckoning. Basically, once the reckoning phase begins, the list of things that need to be resolved is fixed for that round.
And, yes, if your investigator is defeated by a reckoning effect you do not continue to resolve reckoning effects on his/her remaining items. Defeats are immediate and you discard all conditions and tokens (except clues) and set the investigator aside along with his possessions. This can also cause you to loose beneficial effects that the investigator was providing to others on his/her space. So, again, the order you resolve the effects does matter, even between investigators where you are supposed to follow the usual turn order (lead first, then clockwise).
The case in the question you cite is a little less clear. Lost in time and space usually allows an investigator to skip resolving reckoning effects. But you could argue that all the items' reckoning effects are triggered already (just unresolved) and since you take the items with you you have to continue to resolve them all.

I could not find any definitive documentation that  implied how to treat that case where you are sent into Lost in Time and Space midway through the reckoning. I was tempted to say the condition and state of the player nullifies all further reckoning just by virtue of being in limbo, but can't really justify it fully because I don't really understand what it is that causes a reckoning. Is it the alignment of the stars over Earth? I feel I may have to default to simply picking the option that makes it harder for the investigators to win, something I read somewhere to do as a fallback.

Yes, that is a good, all purpose, fallback.

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text:
How does Finn Edwards' passive ability affect the directive to discard Unique Assets?
There is a specific faq item regarding Finn Edwards' passive ability as pertaining to Unique Assets, but I cannot make heads nor tails of it.

Q. What happens if Finn Edwards uses his passive ability to not discard a flipped Unique Asset when instructed?
A. After resolving the effects on the back of a Unique Asset, if the investigator was not instructed to “flip this card”, the Unique Asset is shuffled into the Unique Asset deck.
This occurs most frequently when Finn Edwards uses his passive ability to choose not to discard a Unique Asset, but it could occur in other ways. Regardless of how it occurs, the Unique Asset is shuffled into the Unique Asset deck.

First of all, my understanding of double-sided cards is that any time they are discarded, they are shuffled back into the deck (as mentioned under the Discarding section on page 5 of the Reference Guide).  If this is the case, how is Finn "using" his passive ability doing anything other than what would normally happen anyway? If the answer is implying that he cannot use his passive ability to any positive result, then I would expect the answer to start by saying "Finn cannot use his passive ability to change the outcome etc."
I have read other comments suggesting that any item that is discarded is not done so technically as part of the item 'effect' so the discard doesn't fall under the power of Finn to affect it.  The answer in the faq here still seems unclear to me.  What is it saying?

According to the Reference Manual for Eldritch horror, when Discarding:

When a double-sided card is discarded, it is immediately shuffled
  back into its respective deck.

Nothing supersedes this rule in updated manuals or contradicts this anywhere else within the guide.  For instance, under the Discarding heading in the Reference guide:

Whenever a card is discarded, it is placed in a faceup discard pile
  by its deck.
Double-sided cards, such as Spells or Conditions, are immediately
  shuffled back into their respective decks when discarded.

This still indicates that while the general rule is a separate discard pile, double-sided cards as a whole are immediately reshuffled.
For how this works with Finn's passive ability, we need to look at the specific wording of his character sheet

Effects cannot cause [Finn] to discard ...

Emph added.  So this means that a game effect (e.g. an encounter causing Finn to become delayed and lose an ally) can be ignored.  The natural consequences of using an item, however, cannot be ignored.
Thus, Finn's ability would prevent someone from stealing his Old Journal, but if said journal were aged and worn, he would be unable to prevent its falling apart upon reading it.

I agree with everything you said, and this leads me to feel the answer, and even the question, quoted in my question doesn't really make sense. It just seems wrongly stated and answered, not typical of what I've seen in their faq. Shouldn't it say "Can Edwards use his passive ability to etc" and then be answered by "No, etc", or am I misreading some detail where they are explaining how it actually is used in some way?

I agree with you that it's a bit ambiguously worded, but I think that it can be parsed in a way that's in keeping with both making sure that the character's ability is meaningful, without being game-breakingly amazing.  I think that an attempt to prevent being overly-descriptive in wording is causing a bit of lingering ambiguity.
From the sound of the FAQ answer listed above, the game intends for resolving a card to be an atomic action.  That is, as you resolve the effects of a card, everything happens at once -- the spell fires AND the words fade from the investigator's mind.

I think the intent behind the phrasing above is to remove the word "discard" from the reading of the backs of cards.  This makes the use of a Unique Asset card the following steps:

Test a skill.
Check results versus the Back
Return card to the Unique Assets supply (note: separate from a discard) UNLESS the card specifies that the investigator may keep it.

I agree that it seems poorly-described, but FFG tends to try to look forward to future expansion possibilities with card wordings, and occasionally causes issues with understanding.

The ruling you have quoted, is essentially a patch for incompatible effects.
At the back of every Unique Asset, after resolving the other effects, you'll find instructions to either flip it or to discard it.
When it's a discard then there are no instructions to flip it (like "flip this card then discard it") as this would be redundant and pointless in most situations.
However when Finn's ability prevents this kind of discard, you would be left with a Unique Asset that is useless. All the effects on the backside were triggered when the card was flipped - without the instruction to flip it back you will not be able to use the card again.
To avoid such weird game state, the faq(/patch) tells you to shuffle the card back into the Unique Asset deck, even though the discard was prevented.

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text:
Once the last mythos card is flipped, is there no point to continuing the game?
Since most (or all) mysteries require waiting until the end of the mythos phase to officially complete them, is there no point to continuing the game once you flip the final mythos card and still haven't solved the first 3 mysteries? The exception to giving up the moment the final mythos is flipped seems to be when the ancient one awakens and you stand a chance to defeat him in that final round after the last mythos is flipped (since defeating the ancient one appears to end the game immediately in the investigators' favor).
Do I have this all correct?
Edit: from the Eldritch faq under Ancient Ones & Mysteries

Q. Do effects that “advance the active Mystery” or “solve the active Mystery” affect the Final Mystery?
A. No. The Final Mystery is not a Mystery card nor is it considered the active Mystery. As such, effects that advance or solve the active Mystery do not affect the Final Mystery.

This seems to support the treatment of Final Mysteries as a different entity as normal Mysteries.

Since most (or all) mysteries require waiting until the end of the mythos phase to officially complete them, is there no point to continuing the game once you flip the final mythos card and still haven't solved the first 3 mysteries?

In theory this is covered by the Reference Guide under Winning/Losing.
In practice, the official errata explains this:

If a Mythos card cannot be drawn during the Mythos Phase, the Mythos Phase ends. Then, if investigators have not won the game, investigators lose the game.

So what happens is:

The mythos phase skips right to the end.
At this moment a Mystery card may say "At the end of the Mythos Phase, if [something], solve this Mystery". If this was the last mystery, the Ancient One sheet will usually say that you win.
Anyway, the rules say that you lose, because you had no Mythos Card to draw.
BUT the rules also say that if you win and lose at the same time, you win. (again Winning/Losing in the Reference Guide)

So if that was your last mystery, and you solved it, then you win.
As for beating the Ancient One: Yes, if a game component says you win, then you win. That's the actual rule (Winning/Losing).
Waiting until the end of Mythos Phase to win, is not a rule - it's written separately on each Ancient One's sheet.

I don't see any mention in the Winning/Losing section that says if you win and lose, you win. The errata faq says outright "No. If investigators lose the game while resolving a Mythos card, the Mythos Phase has not yet ended, and the third Mystery is not yet solved."

@Joey

The game immediately ends when an effect specifies that “Investigators
win the game,” or “Investigators lose the game.”

In the rare circumstance that both of these effects happen at the
same time, investigators win the game.

@Joey If you lose while resolving the myhos card, then the mythos card haven't ended yet. It ends after resolving the card.

My point is that the actual win after completing the last mystery doesn't occur until the 'end of the mythos phase', so you clearly can't win because you won't reach the end of the mythos phase before you lose from having no draw pile. A large number of steps fall between drawing that mythos card and reaching the end of the mythos phase, so I find it a stretch to consider these two happening 'at the same time.'

Solving any mystery happens at the end of the Mythos phase. Let's say you have only one card left only two things can happen:

If you have solved the final mystery, then you have to survive that Mythos card event.  If you do, you win the game.  If not, you lose.
If you have not solved the final mystery, then you have lost the game. There is no point in doing the last round as you cannot draw a new mythos card.

The "final" mystery is stated as being solved after certain conditions and doesn't explicitly say "at the end of the mythos phase."  This omission strongly implies to me that it doesn't require investigators to survive a mythos after completing it and seems to make thematic sense as you've directly beaten an active awake ancient one, so the threat is directly neutralized. Do you have documentation referring to the final mystery requiring a completed mythos after the fact?

I view the final mystery as a super set of the normal mysteries thus the same rules that apply to mysteries apply to the final one, including that it is resolved at the end of the mythos phase. However, I have no backing for my reading of the rules beside the first rule of cooperative games which states that if a rule is ambiguous, its reading is the least favourable to the players.

I found a good reference in the faq that clarifies the difference between mystery and final mystery which I'll edit in the main question. It states final mysteries are not mysteries, so perhaps implies they should be treated differently here.


text:
In Arkham Horror, if I lose a turn and then am drawn into a gate, am I then only delayed?
I am not sure what 'losing a turn' is. Is it identical to being lost in space and time in that nothing, even monsters, affects you or fights you? Does being sucked into a gate that appears then convert me to simply being delayed?
Edit: Here is some documentation I found in the Dunwich Horror rules which implies that being drawn into a gate nullifies and previous delays and lost turns and 'resets' it in a way while retaining any possible benefits earned for the previous cost/penalty of delay/turn-loss. It likely means the accepted answer's latter section on what to do in these more complicated situations should be modified.
A link to the Dunwich Horror rules (part of interest on page 10)

Q: If an encounter, when failed, tells the investigator he must “stay here for 2 turns and gain 2 Clue tokens” but before the 2 turns have ended a gate appears at the location and draws the investigator into an Other World, does the investigator lose the 2 Clue tokens?
A: No. The investigator immediately receives the Clue tokens when he or she agrees to meet the requirements of the encounter. If a gate opens in the meantime, the investigator is sucked through the gate and becomes delayed in the Other World. He no longer has to miss two turns if sucked through on the first – the new condi- tion (being delayed in the Other World) is in effect.

Can you give an example of how you would "lose a turn" before a gate appears?

There was a location encounter that allowed me to choose to lose a turn to gain some benefit. A gate then opened as a function of the mythos card and sucked me in.

Losing your turn is different to being delayed. Being delayed means you cannot do anything during your movement phase
If you have lost your turn, you skip all phases (including upkeep) except for the Mythos phase.
From the FAQ:

Q: If a card tells you to “lose your next turn,” does this mean skip
all phases, including Upkeep?
A: Yes, all phases except the Mythos
Phase. Note that this is different from being told to “stay here next
turn,” which means you are delayed.

For your original question, based on the above, I would say you can only get sucked in to the gate that appears as part of the Mythos phase and then you have both lost a turn AND are delayed.
On your next turn, you cannot stand as that normally happens in the movement phase. You also do not have any encounters. The turn after that, you get to stand and have encounters. The following turn will be when you can actually move. This way, you still have at least 2 encounters in the other world (unless you get lucky - but to be in this situation, you need to be really unlucky)

On page 10 of the rulebook in the Mythos phase section: "If a gate opens at a location that contains an investigator, he is immediately drawn through the gate to the first area of the corresponding Other World. As a result of the sudden disorientation caused by the gate swallowing him, the investigator is delayed. Place his investigator marker on its side."

Will fix my answer,  but the premise is the same... U Do nothing other than get sucked in to a gate and get delayed as per usual

So what you're saying is the delayed side effect compounds with the lost turn and essentially makes me miss two movement phases in all?

Yes... Who said arkham was supposed to be any fun :p

Could you please provide a link to the FAQ in your answer?

The encounters aren't all carefully worded; while some say "lose your next turn" or "stay here next turn"/"delayed", I think at least a couple say "spend your next turn [doing something]". You might end up making a flavor ruling instead, depending on your play group. For example, if an encounter said "Spend your next turn reading the book. Gain 1 Skill." and you get drawn into R'lyeh, I'd say you probably put down the book, so you're not delayed, but you also don't get the skill.

@Jefromi well, if you're drawn in and normally get delayed, you probably don't get the luxury of choosing to put down the book to avoid it, unless what you meant was the additional delay of the lost turn.

Yes, I said you'd lose the delay and the reward, not avoid being drawn in. If you're sitting there for a turn (a day?) to read a book and gain something from it, and all of a sudden a portal to another world opens up and sucks you through, you're probably not just going to sit there and keep reading.


text:
Microbot Alpha: how far does it consider others Microbots?
Microbot Alpha says "All of your minions are considered Microbots."
Does this mean while this card is in play, every single minion in play and in discard and in my hand are considered Microbots?

For every Microbot Fixer in play, does every single minion I have in play now get +1 because they are all microbots?
If Microbot Archive is in play, does every single minion I have now cause me to draw a card when destroyed?
If I play Microbot Reclaimer, can I shuffle any minions from my discard pile into my deck (by considering them all microbots)?

It looks like this link addresses the first item above: Smash Up: Microbot Fixer + Microbot Alpha
And it looks like this link addresses the 3rd item, implying that minions in the discard are not minions because they are out of play:
Do actions that affect all cards also affect cards in your hand.?

Yes, if the card says all your minions are microbots, then all your minions are microbots

See also: Questions on Microbot Alpha.

Yes. All minions are considered Microbots. So everything that applies to Microbots applies to all your minions in play.
However, as included in the Pretty Pretty Smash Up rules, page 10:

"A minion" or "minions" means any minions in play unless stated
  otherwise.

So minions in your discard pile are NOT affected.
See also this discussion on Boardgamegeek.

Interesting link that sounds like it throws a wrench in possibly multiple cards, for instance the reference to the war raptor and questioning whether it no longer is a war raptor but now a microbot.

The fun part is that there is a whole section in the manual about conflicting rules.

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text:
Can I gain a random Ally card from the reserve if the deck and discard do not have one?
When instructed to gain a random Ally card, if the asset deck and discard do not have any ally cards (and the reserve has 1), can I take the Ally from the reserve?  The rulings of interest are from the reference guide
Under Gaining Possessions and Conditions, the guide says:

Gaining a Card with a Specific Trait: Some effects instruct an investigator to gain a card with a specific trait (for example, “Gain a Madness Condition”). The investigator searches that card type’s deck then discard pile for the first card matching the specified trait and gains that card. Then he shuffles the deck.

Note that this does not say you can take a card from the reserve, but it also is speaking of not just the asset cards but others, such as conditions, spells, etc, so the omitted mention of the reserve could be a candidate for clarification.
The guide says, for specific cards:

Gaining a Specific Card: Some effects instruct an investigator to gain a specific card by name (for example, “Gain an Axe Asset”). The investigator searches that card type’s deck then discard pile for the first card matching the specified name and gains that card. Then he shuffles the deck.
•  If the named card is in the reserve, the investigator gains that card instead.

So for a specific card the reserve is game, but to the letter of the rules, the directive was to gain a card with a specific trait, not a specific card.
My house rule on this was to allow the reserve to provide the ally card, and if there were more than one, for it to be drawn randomly.

Nikki from FFG kindly provided an official answer:

I believe you’re asking:
If an effect says “gain 1 random Ally Asset from the deck,” but there are no Ally Assets in the deck or discard pile, can I gain one from the reserve?
To which the answer would be no. When an effect instructs an investigator to search a deck, he also searches the discard pile if the card cannot be found in the deck. If a card that matches the search criteria is not found in the deck or discard pile, the search is finished without finding a card.
If looking for a named card, that card is taken from the reserve to prevent unneeded searching and shuffling.

So, no Ally card from the reserve.

I can't find anything specifically about this in the rules or in any official rule clarifications. Given that I'd go with a strict reading of the rules and only allow a random ally to be drawn from the deck or (if none in the deck) the discard pile as per the section you quote.
When in doubt, applying the rules literally is usually the correct option. Especially, if it makes things harder for your Investigators.

The strict reading would mean to search the discard pile, not the reserve. Is that what you meant?

Yes. Fixed. Sorry about that.

I agree with the literal reading combined with the 'making things harder for the investigators' guidance. This particular case seems to be dripping with 'oversight' and not intentional omission of mention of the reserve for the case of the asset cards. Is there a general best place to get the attention of the designers, such as the ffg forum or boardgamegeek?

The golden rule: If in doubt, the interpretation on any rules must be whatever most hinders the players.


text:
Is a monster taking damage and a monster having its toughness reduced the same thing?
This post I authored months ago questions the use of the term "toughness".  I recently found a reference in the Strange Remnants expansion that states:

If the active Mystery requires an Epic Monster to be defeated, place two Health on the card. The Epic Monster's toughness is reduced by one for each Health on the active Mystery.

The act of placing health markers strongly implies to me that it is to be treated as if one did 2 damage on the monster, yet the wording here clearly states that the markers reduce the toughness of the monster and do not necessarily do damage.  This makes me question whether my original post regarding whether toughness was a measure of max or current HP was concluded incorrectly.  The only way it could be valid is if this treatment of Epic Monsters when advancing mysteries is explicitly different from the normal course of doing damage to monsters, which seems less likely, given that I haven't found any actual supporting documentation that clarifies that toughness is max HP.  Thoughts?

Reducing toughness is very different from dealing damage.
Toughness tells you how much damage can a monster withstand. From the Reference Guide, "Monster" (page 8)

When a Monster has lost Health equal to or greater than its
toughness, it is defeated.

Toughness almost never changes. Several monsters (mainly epic ones) have variable toughness. Barring that, advancing the mystery is the only effect that can reduce a monster toughness.
An official response from FFG (Nikki Valens) on this matter:

A Monster’s toughness is the printed value in the top-right corner on the token’s back (surrounded by green tentacles). A Monster’s toughness is unaffected by the amount of Health the Monster has lost.
In the case of advancing a Mystery that includes an Epic Monster, the Epic Monster’s toughness is reduced by a number equal to the number of Health tokens placed on the Mystery card. This effect specifically reduces the Epic Monster’s toughness; it does not cause the Epic Monster to lose Health. In this way, advancing the Mystery can bypass any restrictions the Mystery might have for harming the Epic Monster.

Reducing toughness would also prevent a monsters capability to heal as damage can (conceptually) be healed up to maximum toughness.  Same for investigators really=)

I had a ruling come down from FFG and felt it better to put the details in the accepted answer than extend the original question.


text:
What do the colors mean on Better Settlers?
Answers to this question about tile layouts for Catan suggests Better Settlers. When I open Better Settlers, the page shows a board setup for Catan. What do the colors in the setup stand for?
My guess:

maroon/dark red: brick
dark green: wood
light green: wheat
yellow: sheep
grey: ore

Is that correct?

The colors surely match the colors of the resource cards and the actual hexes:

maroon - brick
dark green - wood
light green - sheep
yellow - wheat
gray - ore

And the blank orange hex would be the desert.

(clockwise from bottom left, that's sheep, wheat, ore, brick, and wood;  the center is desert, and there's an extra wood cut off on the bottom)

I am the creator of Better Settlers and I approve this correct answer
:)


text:
Do I "control" creatures in my graveyard?
Cloudshift has

Exile target creature you control, then return that card to the battlefield under your control.

Are creatures already in my graveyard still under my control? Can I return one of them to the battlefield?
(No, I don't think that one answers my question. It's not "permanent", but "creature you control")

In the future, you should probably wait a little while before accepting an answer, so that you can evaluate it against other answers.

For the reason people can easily see this as a duplicate, see murgatroid's and diego's answers, which clarify that "creature" means "creature permanent." Unfortunately there are probably a half dozen possible duplicates for that half of your question as well...

Indeed, one distinction that trips people up is that there are never creatures in your graveyard. Only creature cards. The only place you find creatures is on the battlefield.

Rule 109.2 says

If a spell or ability uses a description of an object that includes a card type or subtype, but doesn't include the word "card," "spell," "source," or "scheme," it means a permanent of that card type or subtype on the battlefield.

In its targetting clause, Cloudshift says "creature" without any of the specified words, so it's referring to a creature permanent on the battlefield.
In general, you don't control anything in the graveyard. Rule 109.4 says

Only objects on the stack or on the battlefield have a controller. Objects that are neither on the stack nor on the battlefield aren't controlled by any player.

Effects that deal with cards in the graveyard would refer to their owner.

Giving this one my +1, due to being the first correct answer, given that two are essentially the same now, and the other is just plain misleading.

No. You only control creatures on the battlefield and creature spells on the stack. Anywhere else they are creature cards that you own, but do not control.

109.4. Only objects on the stack or on the battlefield have a controller. Objects that are neither on the stack nor on the battlefield aren't controlled by any player.

Though if you control the cards in your graveyard isn't really material, Otherworldly Journey still couldn't target a creature in a graveyard, this is because if a spell or ability doesn't have any sort of qualifier on it (like Creature spell, Land card) it means a permanent on the battlefield.

109.2 If a spell or ability uses a description of an object that includes a card type or subtype, but doesn't include the word "card," "spell," "source," or "scheme," it means a permanent of that card type or subtype on the battlefield.

This isn't really the right reasoning. It's true that you can only control stuff on the battlefield, but the "you control" clause on Cloudshift isn't the reason that you can only target stuff on the battlefield.

@murgatroid99 You're right, while this does answer the question in the title, it isn't exactly answering the one in the body of the question.

Butt... you say you still own the creatures in your graveyard.. so Trostani Discordant returns cards from the graveyard?

@JohanSellberg No, because rule 109.2 applies to Trostani Discordant's ability as well.


text:
Pilgrimage World with Consume 2x action--how many points?
When consuming 4 goods on the Pilgrimage World using the 2x VP consume action, what is the total number of points that should be received: 6 or 7?
The Card: "Discard all remaining goods to gain that number -1 VPs"
The Instructions: "Discard all goods (possibly after using other consume powers) to gain VP chips equal to one less than the number of goods discarded"
Consume Card: "Receive twice as many VP chips as you would have normally received in this phase"
My interpretation

Six points. Determine how many victory points that you will receive normally, and then double them. You would normally receive 3 VP, so it should now be 6.

My wife's interpretation

Seven points. You would normally take 4 VP, then return one. So now you should take 8 and return one.

I'm smart enough to know that my wife's interpretation is right. But who's actually right?

Yeah, the second interpretation is wrong because it's added in a step of taking the VPs and then returning them. The consume ability doesn't say take 1VP for each, then return one; it says take one less than the number. You won't get 7 because you wouldn't have received 8 in the first place.

The rules say (page 6):

Players who chose Consume: 2x VPs double the number of victory point
  chips — but not any other effects — that their consume powers provide.

The consume power on Pilgrimage World defines the number of victory points you get for consuming a number of goods.  Consuming one good is zero points, two goods is one point, three goods is two points, and so on.  Then, the Consume 2x doubles the amount provided by Pilgrimage world.  In the situation you described, the player consuming 4 goods would get 6 points.


text:
When exactly does the black pawn gets placed on the board?
As I mentioned in another question, I recently got the On-the-Brink expansion and have been a bit puzzled with the BT challenge gameplay. I was looking for gameplay videos but haven't found any..
My question is about when the BT/black -pawn gets placed on the board. The rulebook says it's placed on the board whenever another player is at a city BT is in. So far each time I've played Pandemic, we play the actions as a sequence and not really one-by-one. So while the "good guys" are playing their actions, does the BT have to jump in and put his pawn on the board? What happens if one "good guy" happens to pass by the city BT is in, en route to some other city?
Example:

Player1 starts her turn in city A and wants to treat disease in city D.
BT resides in city C.
Player1 moves from A to D; via B and C.

Does BT jump in the gameplay to place his pawn on C, only to remove it again immediately?
Edit: As I stated in a comment below, how the goodies move becomes non-trivial for as the path from one city to another, may involve passing through different cities, even though the number of actions required might be the same.
Example: If a player announces that she's going from Cairo to Karachi, does the BT have to ask "via Baghdad or Riyadh?". Doing so, sort of invalidates the point of secrecy for the BT.

"If a player announces that she's going from Cairo to Karachi" - even before the bioterrorist existed, everyone I've played with would have at least counted along the board to show the number of moves it is, if not moved their pawn along the path. That part of the question seems like kind of a non-issue.

Yes, as soon as the player moves through the city you are in, you should interrupt them and inform them that you are in that city.  They then have the option of continuing with their original plan, or deciding to stop in that city and capture you instead.

does that mean the "good guys" need to break up their moves to individual actions? It becomes not trivial as there are several spots on the map where there are multiple possibilities to get from A to B, e.g. From Cairo to Karachi, a player can go via Baghdad or via Riyadh.

They must specify which cities they are moving through.  They should be doing this anyway when they physically move their pawn, but it is especially important when playing the Bioterrorist variant.  You should enforce it at all times, whether you are in one of those cities or not.  If they still forget to do so regularly, I suppose that you can house rule that the BT does not have to reveal at intermediate cities if the player never specifies they are going through it.  That should give them incentive to develop better habits.

@posdef It's not so much "break up their moves" as "say what they're actually doing" - they can still do it all at once, they just have to quickly tap their pawn through the cities. Besides, if they're playing against the bioterrorist, surely they want to consider which path to follow because they have some guess where the bioterrorist might be.


text:
Isn't secrecy quite pointless for Bio-terrorist?
My buddies and I have been hooked on Pandemic lately and I just picked up the On-the-Brink expansion for Pandemic.
I was reading the rulebook yesterday and I'm not quite sure if I get the Bio-terrorist (BT) challenge right. So the BT is off the board, and plays mostly on pen-&-paper, and the pawn comes to the board only when BT happens to be in the same city as one of the players (timing of putting the pawn on the board is also interesting, but I'll ask that separately).
According to the rulebook the BT has 2 actions and a move, so the BT can essentially:

spread disease locally (L)
spread remotely (R) by discarding
travel remotely (F) by discarding
sabotage (S) by discarding
draw card (D)

I don't quite get the point of secrecy (writing on paper), especially considering that the actions have pretty obvious consequences:

purple cube gets placed on the board
BT discards a card and a purple cube is placed on the board
BT discards a card (destination is the city on the card), BT announces that he was seen at an airport.
BT discards a card, a station is removed from the board
BT draws

So unless I misunderstood something, all BT actions are obvious to any alert player. What's the point of secrecy and note-taking?

For what it's worth, I think you might want to just try a game. It'll probably make it more obvious than all our explanations.

@Jefromi you're probably right... I'm just too psyched to play this role :D

To answer the question in the title: No.  Without secrecy, it would be trivial for the BT to be caught, making them mostly ineffective.  Even a little bit of doubt as to where the BT is will often be enough to keep the players from attempting to capture them, since the moves would be wasted if they're wrong, and capturing often isn't strong enough to risk that.
In terms of why it isn't actually trivial to know where they are, you left out an action, which is probably the most common one.  The BT may drive between locations just like a player does.  This does not reveal where the BT is, and is your primary method as the BT of keeping the players from knowing where you are.  Also, when you discard a card, the players do not know if you are using it for a Direct Flight (going to the city) or a Charter Flight (going from the city).

Hmm, isn't it so that for DF the discarded card is the destination, and for CF it's the origin? So for example: if BT have spread disease in Johannesburg last turn, and haven't discarded a card, and in this turn he announces that he was at an airport then discards a card; if the card is Johannesburg, Khartoum or Lagos; he's likely on a Charter flight (thus could be anywhere), and otherwise he's flown to the city that is on the discarded infection card?

You can certainly make inferences like that if you know a recent location.  But if it has been a turn or two, you may not know for sure.

@posdef It's not like you have unlimited cards; you'll definitely have plenty of normal movement, not just flights. Besides, part of playing well as the bioterrorist is keeping conscious of exactly what the other players are able to learn from your actions, and what they can do about it. Even a single uncertain step is huge; it's not like they can just waste turns trying all four cities you could potentially be in. On the other hand, in many cases it's fine if they know where you are: they may be too far away or too busy with other problems to do anything about it.

@posdef Also keep in mind that they don't know which of your actions the flight was.  You might have ended your turn with the flight, or you might have done it and then driven elsewhere.

@Jefromi so the answer to the OP is; "the secrecy doesn't really matter much anyways?"

@posdef No! The answer is that sometimes it's not that important, but sometimes it matters a lot, and knowing the difference is part of playing it. The amount of information you're forced to reveal is, I think, fairly carefully balanced to make this all work; it's possible to track down the bioterrorist (so the others have a chance) but it's also possible for the bioterrorist to be pretty evasive (so they have a chance).

I could see trying to mislead people by the information you give, by spreading remotely in the city he is in.


text:
MTG: What set is Seelenruhe (Tranquility in German) from?
I recently picked Magic: The Gathering back up after a 15 year lapse. Currently making my way through what seems like a thousand cards from my childhood and entering them into a card database at DeckBox. I happened upon a German card by the name of Seelenruhe and I'm trying to figure out what set/edition this card is from.

I Google searched and found that seelenruhe means tranquillity in German. An image search, via Google, gave me an image of the same Seelenruhe that I have (plus a Jolly Roger added in?).
However, the image is from an Ebay account that is no longer hosting the image or product, and the Ebay website is written in what I think is German, which I can't read. Therefore, I can't figure anything out about the set/edition this card comes from.
Do cards written in a language other than English, belong to a set different than their English counterpart's set?

- Other Research -

According to CardKingdom, Seelenruhe most resembles Tranquility from the following sets/editions: Alpha - Beta - Collectors Edition - Collectors Edition International
Following How can I tell the difference between early Magic: The Gathering card editions?

    Black Border -> No Right-Side Center Symbol -> Corner Curve Sharpness?
        Sloped -> Alpha
        Sharp -> Not Titled "Reconstruction" -> Beta

Obviously it is from Alpha or Beta, the corners look sloped to me, so I'd say Alpha.

@Vilmar I thought that at first, but Seelenruhe has a 1994 copyright date. "Limited Edition cards have no expansion symbol, no copyright date, and no trademark symbols; the text on the bottom left consists only of an artist credit." - Wikipedia: Limited Edition

This card is a Revised German Tranquility. For English sets without a set symbol only Alpha and Beta have black borders. But when Magic started out whenever a Core set was printed for the first time in a language it had black borders. The first printing of German Magic cards was Revised Edition, Strangely it seems that German Revised had at least 2 printings, one in black border and one in white.
Reference: Revised German Tranquility on Scryfall

The set is commonly called Foreign Black Bordered.


text:
If I have two things that activate at the beginning of the end step, what happens?
What happens if I have two cards that have an effect that happens at the beginning of my end step? Which of them activates first, or do they activate at the same time.
I am asking because I am using:

Both of them have an effect that activates at the begining of the end step, but could I exile Obzedat before the Revolt effect activates?

Unfortunately, you don't get the servo.
Hidden Stockpile has an "intervening if" trigger, which means it checks the state to see if it even goes on the stack.  If a creature had not left before the end step, then it won't ever go on the stack.

603.4. A triggered ability may read “When/Whenever/At [trigger event], if [condition], [effect].” When the trigger event occurs, the ability checks whether the stated condition is true. The ability triggers only if it is; otherwise it does nothing. If the ability triggers, it checks the stated condition again as it resolves. If the condition isn’t true at that time, the ability is removed from the stack and does nothing. Note that this mirrors the check for legal targets. This rule is referred to as the “intervening ‘if’ clause” rule. (The word “if” has only its normal English meaning anywhere else in the text of a card; this rule only applies to an “if” that immediately follows a trigger condition.)

None of these abilities "activate", they trigger. In particular, Hidden Stockpile only triggers if a permanent you controlled left the battlefield this turn. Obzedat, Ghost Council exiling itself in the end step won't happen in time to meet this condition. This is due to what's called an intervening if clause. If no permanent you controlled left the battlefield this turn, only Obzedat will trigger.

603.4. A triggered ability may read “When/Whenever/At [trigger event], if [condition], [effect].” When the trigger event occurs, the ability checks whether the stated condition is true. The ability triggers only if it is; otherwise it does nothing. (...)

Assuming another permanent did leave the battlefield, both abilities trigger at the same time - at the beginning of the end step. Since you control both abilities, you choose the order in which they are put on the stack.

603.3b. If multiple abilities have triggered since the last time a player received priority, each player (...) puts triggered abilities he or she controls on the stack in any order he or she chooses. (...)

Okay, I see. They would both trigger at the exact same time, so Revolt would not get activated.


text:
Mathematically / logically determine unit alignment for close combat
** This question involves knowledge of computer programming and/or math and algorithms. **
I am working on writing a Warhammer combat simulator in Java and have hit an interesting snag.
Pg 20 BRB:
"Once the charging unit contacts the enemy unit, it must perform a second, bonus wheel if required to bring its front facing into flush contact with the facing of the enemy unit that has been charged, maximizing the number of models in base contact on both sides"
Aligning the units in this way is absurdly simple when looking at them, but trying to write the code to do the same thing is proving difficult. I tried posting the question to Stack Overflow, but I think the concept is too vague.
Basically, I'm trying to simulate combat between a unit of spearmen (20x20mm bases) and a unit of horsemen (25x50mm bases), and programatically determine how many models will get attacks during the combat.
Any help would be appreciated.

I'm pretty sure this would get a much better answer on GameDev.SE, since it's related to the math of a design instead of an actual board/card/table-top game.

The linked question on Stack Overflow actually describes what you're trying to do - that seems like a much better way to get answers, and honestly, it seems like it's a better fit for Stack Overflow. There might be a few more people who know how Warhammer works, but I think most of us don't so you're going to have to explain anyway to get an answer, and as long as you're explaining, you're just asking a programming question, so SO is a good place.

Note: The question has been asked, and answered on SO.  It's a neat analytic geometry problem.  http://stackoverflow.com/questions/29035291/programmatic-analysis-of-geometric-shapes/29036305#29036305

@John Has it? I see a lot of attempts at answers, and the OP seems to have a lot of trouble making the question clear.

Yes, PaulDouglas's answer is spot on.  It requires some more math to get to code, but if the OP can't do that math, then I think this project is beyond his ability.

I'm voting to close this question as off-topic because it is only tangentially related to gaming

get the number of unit on the face being charged, see how many of the charging units front of base length fit in it then * that by the number of units being changed are open on that side.
if you have uneven formations you would need to keep the current shape of a unit in an object, then find all the lengths along the charged side and use some sort of packing algo to put the charging units along them


text:
Can I play blessed alliance after damage resolves to make him sac a creature I didn't kill?
I was reading over various questions and it occurred to me that I might not be using Blessed Alliance as effectively as I could be. Am I able to cast Blessed Alliance after damage has been resolved, but still during combat, to make my opponent sac a creature that has been declared as an attacker? I believe I can, as 510.3 says active player gets priority and 510.2 is damage is dealt, and state based actions will be checked before active player gets priority.
Therefore, the specific example I'm asking about is, if my opponent attacks with two creatures, one I can kill with my blocker and one I cannot, can I block the one I can kill, and after it dies cast Blessed Alliance to make him sacrifice the other? I believe 510.3 means that it is still the combat phase, and his other creature is still "attacking" as defined by the rules of the game, but I am unsure.
Can instants be played after I decide not to block?
I believe this question is also relevant, as the linked answer states that you can play instants during each of those phases, and end of combat is one. Are creatures that attacked still "attacking creatures" when I get priority at the end of combat?

Yes this works how you want it to. Note however that the creature will die after it has already dealt its combat damage, so if you are trying to stop it from hitting you this would not work.
The combat phase has 5 steps:

Beginning of Combat
Declare Attackers
Declare Blockers
Combat Damage
End of Combat

A creature becomes an 'attacking creature' during the Declare Attackers step and remains an 'attacking creature' until the Combat phase is over (or it gets removed from combat some other way). Thus you are able to use Blessed Alliance during either the Combat Damage or End of Combat steps to have your opponent sacrifice a creature that survived the damage dealing part of the Combat Damage step.

508.1k Each chosen creature still controlled by the active player becomes an attacking creature. It remains an attacking creature until it’s removed from combat or the combat phase ends, whichever comes first. See rule 506.4.

This is correct, however we should note that it is impossible to prevent the second attacking creature dealing damage.

@Vlad274 I'm pretty sure given the rules they quoted that they understand that damage will have been dealt, but I added a note explicitly stating that


text:
What's the probability of having a combo on the first turn in MTG?
I'm creating a combo deck, and I'm interested in the probability of having a certain combo on the first turn.
For this combo, I need 4 different cards. There are 4 copies of each of these cards in the deck. The deck has 60 cards.
What is the probability of seeing these 4 cards in the starting hand of 7 cards?
I read How do you calculate the likelihood of drawing certain cards in your opening hand?, but my question is a little different and I don't know how to calculate this probability.

My guess is that besides these 4 cards you also need a land or other mana source right? if that is the case you also need to put in the equation the amount of lands you have.

@IvoBeckers yeah you're right, i need a land .. so become 5 cards in start hand. i'm playing 18 lands

I guess you should better ask it at mathSE, because this question is about probability and chances

@Novarg also question in link that i've posted is on probability, but no one said that the question is in the wrong section ...

I think that this question is on topic. It's based on mathematics but also has practical usage in Magic. Someone with a solid understanding of magic may be more thorough.

You probably also want to take into account mulligans.  But I still think the odds are going to be ridiculously against you.  Even getting a specific two card combo isn't all that likely.

wow!!! two down vote in a question about a game! and without comment too..! please smile

This question is on topic.  Just because you don't know how to do the math, doesn't mean I don't....  :)

If I would have to guess, the reason why people downvote (I didn't) may be that the probability of this happening seems so low that it sounds like a brainteaser more than a concrete, directly applicable in-game question of strategic relevance.

@xLeitix thanks for your comments man. But before doing this question, i didn't know the probability :) so I thought someone here could help me :)

Probability drops to about 0.97% when you need one each of 4, 4, 4, 4, 18.

You can do this calculation using the multivariate hypergeometric distribution. The setup is as follows:
The deck of 60 cards consists of: 4 cards of type A, 4 cards of type B, 4 cards of type C, 4 cards of type D, and 44 cards of type E (other).
Your criteria are that a hand of 7 cards contains at least 1 card of type A, at least 1 card of type B, at least 1 card of type C, and at least 1 card of type D.
For a given hand arrangement, you can calculate the probability using the formulas in the link. As an example, the probability of the hand (1 card of type A, 1 card of type B, 1 card of type C, 2 cards of type D, and 2 cards of type E) is:
(4 choose 1) * (4 choose 1) * (4 choose 1) * (4 choose 2) * (44 choose 2) / (60 choose 7) = ~0.000941.
Note that this probability is for this specific hand, and there are many that meet your requirements. You will want to make a table of all realizable hands and sum up the probabilities. (or alternatively a table of all hands that don't meet the criterion, and subtract the sum from 1).
=== table of realizable hands ===

1,1,1,1 of (A, B, C, D), 3 of other

(4 choose 1) * (4 choose 1) * (4 choose 1) * (4 choose 1) * (44 choose 3) / (60 choose 7) = ~0.00879

1,1,1,2 of (A, B, C, D), 2 of other [4 variants]

4 * [(4 choose 1) * (4 choose 1) * (4 choose 1) * (4 choose 2) * (44 choose 2) / (60 choose 7)] = ~0.00376

1,1,2,2 of (A, B, C, D), 1 of other [6 variants]

6 * [(4 choose 1) * (4 choose 1) * (4 choose 2) * (4 choose 2) * (44 choose 1) / (60 choose 7)] = ~0.000394

1,2,2,2 of (A, B, C, D), 0 of other [4 variants]

4 * [(4 choose 1) * (4 choose 2) * (4 choose 2) * (4 choose 2) / (60 choose 7)] = ~0.00000895

1,1,1,3 of (A, B, C, D), 1 of other [4 variants]

4 * [(4 choose 1) * (4 choose 1) * (4 choose 1) * (4 choose 3) * (44 choose 1) / (60 choose 7)] = ~0.000117

1,1,2,3 of (A, B, C, D), 0 of other [12 variants]
12 * [(4 choose 1) * (4 choose 1) * (4 choose 2) * (4 choose 3) / (60 choose 7)] = ~0.0000119

Total Sum of above = ~0.01307

60 - 16 = 44, not 46. I changed the numbers, but you may want to give a new result.

@HaoYe so i've < 1%  to draw this start hand??

@Ilario There is a reason combo decks tend to run a lot of card filtering and tutoring

@diego you're right i've some cards for tutor my combo, but this question its only when i can do the combo on the firs turn... I thought they were a little bit higher odds ..

To put diego's statement into a question: Does this answer does take into account card filtering, drawing, and tutoring?  If not, then I'd say it's not very comprehensive.

@Rainbolt The question is about the chances to see 4 specific cards in the first 7 you draw, it doesn't ask about anything else including mulligans, card draw, or tutoring. Therefore there is no reason for the answer to include information about that, if they wanted to include it they could, but I don't feel it necessary (especially for card draw/tutoring since the question doesn't say the deck is running any)

@diego You're right - the answer technically addresses the question as it was written by the OP. But the answer has almost no practical usage, so I clicked the "This answer is not useful" button.

@Rainbolt, I fully disagree. Not only goes it give the answer, it explains how to get it. You act as if the question seeked deck-building help, but that's not the case.

@Ilario, Slightly less than one in a thousand (0.1%), actually.

@Rainbolt Wow, that's probably the most liberal use of the "not useful" flag I have seen lately on SE.

I think everyone here misunderstood me. The downvote button says "This answer is not useful" when you hover over it. I did not flag anything. @xLeitix Look at the available answer flag reasons and tell me your last comment was not a total lie. Seen lately? How could you have "seen lately" something that doesn't even exist?

I wrote a simple simulation which confirms your result.


text:
Deriving statistics from original card games
I have drafted a game that uses a French Deck with 52 cards, 4 suites (diamonds, clubs, hearts, spades), and no jokers.
I'm interested in profiling game play along a few axes:

How many turns are there in a typical game with different play strategies?
If I change the winning condition from, say, getting all four queens to getting all four jacks, how much longer or shorter is play. This would make an actual difference in my game, but I'd like to quantify that difference.
How frequently will players have matching ranks for the cards they play if I force them to play a random card from their deck versus selecting which card to play?

Besides those, I could think of other ways to tweak the game, and I want to know what effect that has without manually playing the game several times in real life.
It would be possible to get this information by using some ad-hoc scripting on my computer to calculate values for the specific cases I mention above. However, adding more measurements, could require changing or improving my model of the game which takes more work. What would be better is a suite of computer programs that would allow me to change rules and gather new statistics without, in general, needing to add additional instrumentation to the game engine. Does such a suite of programs already exist?
I have seen this question which is similar to my own, but the poster requests statistics for a much more specific condition. I am looking for a general tool. I am hoping that the situation has changed since the original negative answer.

According to wiki, there is no standard size for a French Deck- which size are you using?

52 cards, 4 suits (H,D,S,C), no jokers.

Sorry I don't see how your question is any different from the one you're citing. This is strictly a DIY programming exercise. You could look into object oriented programming (eg Java) and model the game yourself so you can run simulations.

Thanks for taking the time to read over my question. I readily acknowledge that the questions are similar, and that's why I cite it. I've modified my question to show how it differs from the other. Will I end up making my own solution? Probably, if I get a chance. Is there any harm in asking if someone has already? Nope.

I think "I am hoping that the situation has changed since the original negative answer" is pretty key here. If the situation has changed, it would be appropriate for there to be a new answer to the original question, and for the community to curate the information there.

A university computer science professor wrote a program for UNO in Java. The only "easy" way to do this for your game is to do something similar. Create objects for players, games, hands, decks, points, etc. and then create strategies for individual players. Make sure to keep track of important information you want to track. The UNO simulation on a reasonably powerful processor can play tens of thousands of games in the time you could play 1 game.

If you have a small amount of experience programming, you could achieve your goals here pretty easily, and as customized as you need.  If you want graphs and other 'pretty'  output, I suggest R.

This answer is going to repeat the answer to the question you cited, but add emphasis on one point:
Running simulations is the key. You do what Adam's answer said: trying out random combinations and just reducing the results to some numbers you can compare.
The relevance to your question is this: these simulations are, by definition, general. The method involves trying out everything, and basing the result on some defined requirements you have specified (step 3 in Adam's answer). The results at the end are determined by those requirements.
If you want to be general for any requirement, just change the requirement. You still run a simulation, and you different numbers at the end because they're based on different conditions.
Of course, you still need to write out the function (and it is a bitch, trust me) and you need to write every requirement you're going to use, but you'll get what you're after.
Tan's answer still holds too: some languages make statistics far easier to display than others. If you were more comfortable in a language other than R, you could export your results to a text file from one language you are confident in, and then import them using borrowed R code, if you want to be lazy. Which is never a bad thing.


text:
Probabilities: Backgammon - final move
So I was playing a lot of backgammon lately, and I noticed something. It seemed like I was "going out" on doubles more often than not. And my initial reaction was "Shouldn't that be, at best, a 1 in 6 chance of getting doubles on that final roll?"
But then my math/statistics side kicked in and I realized it's probably slightly more complicated than that.  So being out of practice, I figured I'd drop this hear to get a "proper" answer, and correct my math :) I also wasn't sure whether I should ask here, or over in : https://math.stackexchange.com/questions  But figured I'd start here. If it get's too "mathy" perhaps I'll re-ask over there :)
Here's my thoughts so far:
our end states are:
5+ checkers
4 checkers
3 checkers
2 checkers
1 checker

Now with 5+ checkers left, we cannot go out, so whether we roll doubles or not is irrelevant, it will leave us with 1, 2, 3 or 4 checkers, at which point we could go out if:
1) 4 checkers:
   a) roll doubles of a value equal to or larger than the Runner (the checker farthest in)
         (ie if we have a checker on each of 2, 3, 4 and 5 point in your home, we need to roll double 5's or double 6's to go out)
2) 3 checkers:
   a) roll doubles of a value equal to or larger than the Runner (the checker farthest in)
         (ie if we have a checker on each of 2, 3 and 4 point in your home, we need to roll double 4's, 5's or 6's to go out)
   b) roll doubles of a value equal to half or more of the Runner, and equal or larger than the other 2.
         (ie if we have a checker on each of 2, 3 and 4 point in your home, we can go out with double 3's: half of 4 is 2. next largest is 3. so double 3's puts us out)
3) 2 checkers:
   a) roll doubles of a value equal to half or more of the runner.
         (ie if we have a checker on 2 and 4 point. We can go out with double 2's)
4) 1 checker:
   a) roll doubles of a value equal to a quarter or more of the runner.
         (ie if we have a checker on the 4 point, we can go out with double 1's)

I'm not sure these are all the cases .. but it's showing that there's a lot of variations .. and I'm not sure how to go about calculating all the variations accurately :)
So back to my question :
How to calculate the odds of going out by rolling doubles ?
(let's ignore the chance your opponent goes out first at any time ... or interferes in any way ... )

As in all math problems, it is often best to start with concrete/simple examples, then start to generalize.  I would approach this problem, by looking at the case where all the checkers are on the 1.  (My back of the envelope gives a probability of going out with doubles of 11/42 in that case.)  Then, I would start generalizing, and making sure my results were consistent with the above.  My gut is that the probability in general is closer to 1/3 than it is to 1/6.  It's definitely above 1/6.

If you like the math, here's an actual mathematical paper on the subject.

@ghoppe: Neat ... thanks.  I browsed through it, I'll have to ponder it more deeply with alcohol this weekend  ;)

John's figure sounds close to what I'd intuitively expect - it's near 1/3 because in most games you'll have two attempts at finishing with a double, once when you have 3 or 4 men remaining, and once when you have 1 or 2 remaining.

Let's set a lower bound on the likelihood of winning with doubles, by simply ignoring all cases where it is impossible to win without doubles.
Assumption: All board positions considered are equally likely. This is probably not true, but will approach truth in longer games.
Consider the case of two men only left on the board, both in the home court, and not both on the 6-point (so that it is possible to win without doubles). There are 20 possible board positions for the two men, of weight 2 when on separate points and of weight 1 when on the same point. Likewise there are 30 possible non-double dice rolls and 6 possible double rolls. Averaged over all (the assumed equally likely) board positions under consideration, a simple counting exercise gives a probability of winning without doubles of 550 out of 1050, and a likelihood of winning with doubles of 168 / 210. (Note that the total possibilities = 1050 + 210 = 1260 = 35 * 36 as required.)
These are probabilities of 53.9% and 80% respectively, with a ratio of double-win to total wins of 23.4%.
This ignores the relatively small case-set of single blot board positions and the significantly larger case-set of board positions with 3 or 4 pieces or one or more pieces out of the home board. The net affect of these will clearly increase the likelihood of winning with a double roll.
This analysis can be extended over the additional case-sets, but it is clear that the likelihood of winning with a double roll far exceeds the 16.7% chance of rolling doubles.
                   **Wins**       **Wght * Wins**
Wght  Position  Single  Double   Single    Double
 1      1 - 1      30      6       30         6
 2      1 - 2      30      6       60        12
 2      1 - 3      28      6       56        12
 2      1 - 4      26      5       52        10
 2      1 - 5      24      5       48        10

 2      1 - 6      22      5       44        10
 1      2 - 2      20      5       20         5
 2      2 - 3      20      5       40        10
 2      2 - 4      18      5       36        10
 2      2 - 5      16      5       32        10

 2      2 - 6      14      5       28        10
 1      3 - 3      12      5       12         5
 2      3 - 4      12      5       24        10
 2      3 - 5      10      5       20         8
 2      3 - 6       8      4       16         8

 1      4 - 4       6      4        6         4
 2      4 - 5       6      4       12         8
 2      4 - 6       4      4        8         8
 1      5 - 5       2      4        2         4
 2      5 - 6       2      4        4         8
----------------------------------------------------------------
35                                550/1050   168/210

guess it's been long enough .. I should accept the answer :) hehe  sorry, lost track of this and had forgotten about it ... was a nice re-visit and re-read though.  I think it actually made more sense to me today on re-reading it than it did when I first read your answer ... shrug  Thanks!


text:
Does "new city" mean a city no other players are infecting?
There is an Event Card which has the text:

Massive Bird Migration
You may move 1 of your disease cubes from a city you are currently infecting to a new city.

Does new city here mean a city which isn't already infected by another player?

A new city means one that you are currently not infecting.
From the infect a city infection rules,

You can infect a new city OR a city you are currently infecting.
new city: Discard 2 cards whose color must match the city you
  wish to infect. Place your disease cube(s) on the uppermost
  available row (without any cubes).
City you are currently infecting: Discard 1 card whose color must
  match the city you are currently infecting. Place your disease
  cube(s) next to those already on the City card.

I concur. New city means new to you.


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Talking about Go: Using "he" and "she" for Black and White
It comes quite handy to assign different genders to the two abstract players "Black" and "White" and than use gendered personal pronouns in order to refer to them.
But who should be male, and who should be female?
Berlekamp and Wolfe in their book Mathematical Go use Wright White (male, "he") and Bella Black (female, "she") for the two rôles, but in this answer  Using the edges of the board in Go by Benjamin Cosman it seems that the opposite assignment is used. Where does it originate from?
Which assignment is preferable?
EDIT: I just noticed that Sensei's Library seems to use the White/she convention.

I don't know of any convention one way or another - my assignment in that answer was quite arbitrary.

Employing some search engine tesujis, I found the page Player Gender on senseis.xmp.net. It clearly states my anecdotal finding, that Sensei's Library uses White=she and Black=he as a convention.
The same convention is used in several influential books, e.g.,   Whole Board Thinking In Joseki by Yi-Lun Yang, Essential Joseki by Rui Naiwei and in the works of Janice Kim (co-author of the Essential Go series).
Note that traditional Yin/Yang symbolism would support the other assignment that is also used by Berlekamp and Wolfe in their book Mathematical Go. From all what I can see, this is the minority position.
My conclusion is that—if one want to use gendered abstract players at all—the White=she and Black=he convention is used in the majority of influential sources and therefore preferable over the opposite convention.

Chinese languages are generally  gender-neutral ( The use of letter 'She' only started in early twentieth century and in written form only). I find it rather odd to assign gender to stone colours.

There have historically been a few different ideas how to approach this issue, as you have discovered. Nowadays, if genders are assigned to colors at all, usually white is female, and black is male. As a source I can only name my personal experience on Go servers and discussion boards, so I'm curious to hear other experiences.
I'm not sure where this convention originated - myself, I read about this in a book translated from Japanese, though I can't remember which one. Most people however prefer to just use the colors themselves as nouns ("Black made a mistake here").
As a side note, if either player is a computer program, and you want to gender it, I encourage you to use it to address it. There have been moves to assign a female gender to computer programs, though that was, I believe, not well thought through, and leads to a wide range of related problems, some not too obvious.

I severely dislike the idea of assigning gender to a color under any circumstances, unless the only reason you want to do this is for personal use as some kind of a mnemonic that you wouldn't share with anyone. A couple of reasons why:

There will always be people who associate themselves with the color that you assigned to the opposite gender, and your (semi-)arbitrary choice may frustrate them.
There are just too many negative connotations with the colour black that could come out as offensive to some people, such as "evil", "bad", "sinful", etc.
There are people who dislike the idea of thinking in the binary sexual terms.

With this out of the way, it is important to remember that from the physical point of view, white is the colour formed by a multitude of electromagnetic waves of different lengths, while black is nothing but the lack of any visible light and colour.
Colour white can "give birth" to any colour of the visible spectrum by means of refraction, colour black cannot give birth to anything.
Colour white is also the colour that our Sun gives to Earth to bring and maintain life, while colour black is the color of lifeless space.
Colour white = fecundity, colour black = lifelessness.
From the above, it seems more logical to consider colour white more feminine than colour black.

Well, we are talking about the game of Go: Like checkers or 9-men-morris it is traditionally played with Black and White stones, so the colours are given.

@jknappen Colors are given, genders are not. I see absolutely no need to sexualize the game. If for some obscure reason you still want to do that, read the second part of my answer.

Your conclusion agrees with the majority of sources I have seen so far.

If by my "conclusion" you mean choosing one of the two colours for each gender, there can only be two "conclusions" in this problem. ;-)


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Activate ignition effect from hand?
I have 2 questions, one generally-scoped and one specific:

Can I activate the effect of Disenchanter while it is still in my hand?

In general, can one activate monsters' ignition effects from the hand?

Disenchanter:

Once per turn: You can target 1 face-up Spell Card on the field; remove 1 Spell Counter from anywhere on the field, and if you do, return that target to the hand.

Quite simply, no. It is common knowledge that a monster has to be on the field or revealed from the hand/graveyard/banished zone (if the card permits it) to be able to resolve ANY effect, given the conditions (if any) are met, unless the card says otherwise.
You can always play Disenchanter on the field,  THEN activate its effect.

Could you please add the exact wording from the rulebook? I would upvote this, but I want to make sure that your answer is accurate and the best way to do that would be to quote the relevant text straight out of the rulebook instead of "the Rulebook specifically says..."

@Thunderforge , I would, though I have lost the rulebook. I can only remember it. =)

Well, the rule book is online. Perhaps you can find the rule there?

@Thunderforge, I didn't mean that sort if Rulebook; I meant one of those quick start guides

to my understanding unless the card states that it can be played from the hand then it cant, pretty sure if you asked a konami judge they wouldnt allow this to be done from the hand

No
Not unless it explicitly says so.

See this link for more details.

Can you add more info in your answer about what the link contains, it is encouraged that a questions and answers has more content.


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How could you change what creature a creature with soulbond is paired with?
I'm trying to make a custom card. It's a creature with soulbond. I would like it to be able to tap to change what it's paired with. I was thinking:

{T}: Unpair this creature, then you may pair it with another unpaired creature you control.

Is this the correct formatting? Is there a better way to do it? Does this idea even work within the rules?

Side note, would the card be more interesting if it just flickered itself? You're paying a hefty price (tapping) for a pretty situational ability.

@corsiKa I think that I'll end up doing that, because it makes the card flow much better

Currently the Soulbond rules don't have a specific precedent for ending a pairing of two creatures like this, but we can work one in.
We need to format your ability slightly differently to take care of some things:

Make sure this ability can only be used while it's paired, if that's what you want.
Make it clear to players that the other creature gets unpaired as well.
Use the portion of the Soulbond ability templating that's missing.

First, point 3. I'm going to paraphrase part of rule 702.94a here slightly for the sake of readability:

702.94a Soulbond is a keyword that represents two triggered abilities. “Soulbond” means:

When this creature enters the battlefield, if you control both this creature and another creature and both are unpaired, you may pair this creature with another unpaired creature you control for as long as both remain creatures on the battlefield under your control
Whenever another creature enters the battlefield under your control, if you control both that creature and this one and both are unpaired, you may pair that creature with this creature for as long as both remain creatures on the battlefield under your control.

We'll need to take some of that boilerplate; the "as long as" stuff is important.
It seems like you might want this unpair-repair ability to only work if the creature's already paired, but I could see it either way. We could ensure it only works if it's paired by borrowing Deadeye Navigator's ability-gain wording, but all those cards with similar wording grant the same ability to the paired creature. Because that doesn't happen here, using that wording would be confusing and should be avoided.
I think templating from Hanweir Battlements is more appropriate.
A version which works whether it's paired or not:

{T}: If {this card} is paired to another creature, unpair them. Pair {this card} with another unpaired creature you control for as long as both remain creatures on the battlefield under your control.

If you want it to work only when it's already paired, add this clause:

Activate this ability only if {this card} is paired.

An option for leveraging existing unpairing mechanics
Currently 702.94e says:

702.94e A paired creature becomes unpaired if any of the following occur: another player gains control of it or the creature it’s paired with; it or the creature it’s paired with stops being a creature; or it or the creature it’s paired with leaves the battlefield.

You have some options here for fiddling around with this card to leverage existing un-bonding rules:

Give this creature an ability that has it leave the battlefield temporarily and come back sooner or later. This also lets the Soulbond ETB do the rest of the hard work. For example, flicker it (similar to Deadeye Navigator, but using Acrobatic Maneuver's wording): "{1}{W}: Exile {this card}, then return it to the battlefield under its owner's control."
Enable it to turn into a noncreature (a land, enchantment, or artifact).
Give away control of the bonded creature to an opponent for some proportionate benefit. This is probably a Red or Black effect based on past cards that give away control. Cards like Humble Defector and Harmless Offering have your back here for templating.

The ability that turns this card into a noncreature, or which gives away control of its paired creature, should reminder text similar to the following: ({this card} and the creature it is paired with both become unpaired.)
In the second and third cases, you'll need to give it a means to re-pair itself again. Options include:

{T}: If {this card} is unpaired, pair {this card} with another unpaired creature you control for as long as both remain creatures on the battlefield under your control.

For either of the second cases, add this to the end of the ability that turns this card into a creature again or gives away control of the bonded creature:

Then you may pair {this card} with another unpaired creature you control for as long as both remain creatures on the battlefield under your control.

For the becomes-a-noncreature option, something like the following (suggested by diego):

When {this card} becomes a creature, you may pair {this card} with another unpaired creature you control for as long as both remain creatures on the battlefield under your control.

@diego I've expanded the last section a bit and included your suggestion.

You don't need the 'for as long as both remain creatures on the battlefield under your control.' part since that is already true of all pairs.

@diego Since Soulbond explicitly says that, I figured there was not existing rules infrastructure to make that text automatic. But I suppose any homebrew that introduces this card could also introduce rules to cover that, plus the rules are sometimes redundant e.g. in the case of Ashcloud Phoenix's update.

It does... I thought that was only the reminder text not part of the actual ability. I still don't think it is necessary, but redundant rules that help improve clarity aren't necessarily a bad thing.

That definitely works within the rules. Generally speaking, cards use the actual card name instead of "this creature" the first time it's referred to, so your ability should probably read as:

{T}: Unpair {CARDNAME}. Then, if this is unpaired, you may pair it with another unpaired creature you control.

That also adds in some protection in case, for whatever reason, attempting to unpair this card fails.
Another way of handling this would be:

{T}, unpair {CARDNAME}: Pair this with another unpaired creature you control.

This forces the card to be paired initially in order to change who you're paired with. Not sure if you want that or not, but if you do there you go.

Your first one also requires the creature to be paid to work. If it isn't it can't unpair to get the 'If you do...' part of the effect. If you wanted to protect against the unpair failing (which I don't think is something that can fail, so I don't think it is worth the extra characters) you would want something like '{T}: Unpair ~. Then if ~ is unpaired you may pair it with another unpaired creature you control'

You are entirely correct. Edited to work as intended.

"Unpair {cardname}" as a cost is a great idea. Good thinking.

It may be worth noting that the "unpair as a cost" wording also opens up a window for the opponent to interact while {cardname} is unpaired that none of the other wordings in either this or @doppelgreener's answer have.  This may or may not be desirable depending on your design goals.


text:
What is the name of this solitaire variant?

I recently learned a new solitaire from a friend, but we have no idea what the name of it is. Can anyone help?
You use two decks of cards, and start by placing 14 rows of cards. The middle 4 rows are laid horizontal and are locked. The rest of the cards are placed vertical and are open. All of the cards are placed face-up. The rest of the cards are left in a pile with the face down and drawn one-after-one as you need them. (The bottom card is also open, but not sure if that's a rule my friend started with or not.)
The game is very similar to Klondike as you're supposed to build from kings down to ace with different color for each card (red on black or black on red). Then you gather all the aces and build up on them. The difficult part of this solitaire is that the locked cards in the middle have to be placed in one of the ace piles before you can use it up in the main game again.
I hope my explanation isn't too difficult to understand. If anyone knows the name, please tell me.
Thanks in advance.
Apologize for the bad picture, but hope you get it.

Take a look through this list

can you provide a picture?

Will provide picture as soon as I have card decks available.

In official terms, there is a 10 by 5 main tableau, a 4 by 5 secondary tableau, eight foundation spots, a stock, and a discard pile.  The goal is to build up in sequence by alternating color.  Cards from the secondary tableau must first be played directly to a foundation.  (Or something like that.)

It's possible that this is simply a two-deck version of some other variant, with two "locked" piles and 5 unlocked ones.

It looks like stalactites. ( Wikipedia, Double Deck Version )

Welcome to Board & Card Games! Your answer would be much better if it contained a (short) explanation of the rules; right now it has been flagged as low quality. Links to external sites may break, that's why.


text:
In Star Fluxx, how does the game end with Play all?
I know that there has been complaints about other cards being overpowered, but after playing a few games with my sister, who had never played before, she won almost every game by strategically using the play all card.
Her strategy was to remove the hand limit, increase the draw limit and decrease the play limit so that she could keep her cards in her hand. She would then wait until she got the play all card and a goal that matched two of the keepers in her hand. Then she would win the game by playing the play all rule and then playing only the cards she needed to win.
I thought that perhaps the rules required for all cards to be played first and then determine if she had won, but she reminded me that I had told her "The game ends when the current goal conditions are met".
So what's the actual verdict on this situation?

This strategy doesn't rely on the "play all" card at all... she could do it with "play 3" also, or by playing the keepers and the goal one at a time with "play 1". Plus, this strategy requires her to have the ability to increase the draw, decrease the play, and have no one else change the rules she plays... Any player who happens to end up with "play all" in their hand, along with 2 keepers that match a goal they have would win as well... that's not a particular strategy, that's just playing the cards you draw correctly.

Your sister is correct. The Fluxx rules do say that "as soon as someone meets these conditions, they win!"
I don't think this should be an issue, it requires her to find the "play all" (or play 3, etc.) card and there are plenty of ways you can disrupt her strategy:

Change the rules so that there is a hand limit
Change the rules to increase the play limit
Use an action or surprise card to disrupt this strategy: Trash a new rule, Trade hands, rotate hands, rules reset, That's Mine! etc.

I figured as much, however I did try these strategies but she still thrashed me. I'd play a hand limit, she'd trash the rule, I'd discard the play all card, she'd use the time portal to get it back, I traded hands, she reassigned creepers. Here I was thinking it was a game of luck.

Flux IS a game of luck... there's some basic tactics that you can do to increase your chances of winning, and you need to be smart enough to play the right cards in the right order to win when you can, but ultimately it's 95% luck.

If she was able to trash your hand limit rule, and get the Time Portal to get Play All, and reassign creepers, etc... those are all things that she was very lucky to be able to do. And once again, "play all" has very little to do with what you are describing.

She is right, the game ends instantly when someone meets the current goal requirement, even if that person isn't the current player.

The game continues until someone meets the conditions of the current Goal. That player wins instantly, no matter whose turn it is!

Listed under "winning" in the rules for normal Fluxx, and I'm pretty sure Star Fluxx is no different.


text:
What are the odds of being dealt particular melds in Double Deck Pinochle?
This is a followup to another question: What are the odds of getting a double run in four-handed pinochle?
I'd love to know the odds of being dealt particular melds in Double Deck Pinochle. I found a table of odds for Single Deck Pinochle, but nothing similar for the Double Deck variety.
Does anyone know how to calculate this?
Double Deck Pinochle is played with 80 cards: 4 of each rank (A 10 K Q J) for each suit. That's four Ace of Hearts, four Ace of Spades, etc. (Unlike the single deck variety, double deck Pinochle does not include 9s.) All 80 cards are dealt so that each player has 20 cards.

This question appears to be off-topic because it is about programming, and "write my code for me".

Seriously, if you had just asked for "how can I calculate the odds" instead of asking "can you change this code for me" it would have been on topic and a stats guru would have crunched the numbers for you, and probably shown you how. So close. SO close.

Thanks.. I think I'll repost in a programming stack. I hadn't thought of that. But I'll reword the question here to exclude the programming aspect.

The odds of a given player being dealt hands with those melds in them are roughly as follows:
|  MELD TYPE  |   NONE   |  SINGLE  |  DOUBLE  |  TRIPLE  |   QUAD   |
| Runs        | 0.522113 | 0.476676 | 0.001211 | 0.000000 | 0.000000 |
| Marriages   | 0.055983 | 0.716652 | 0.221239 | 0.006113 | 0.000013 |
| Pinochles   | 0.528127 | 0.411803 | 0.058548 | 0.001521 | 0.000001 |
| Ace Arounds | 0.791536 | 0.206261 | 0.002203 | 0.000000 | 0.000000 |
| Roundhouses | 0.967973 | 0.032027 | 0.000000 |    [Not Possible]   |

These results are probably only accurate to one or two significant figures.
To understand the table - in each row the probabilities add up to 1. Taking the first row as an example - there is roughly no chance a hand will contain four or three runs in a single suit, 0.1% chance it will have a double run in a suit (i.e. ATKQJATKQJ in a single suit), 52% chance it will contain no runs, and 48% chance that it will contain a single run, but no double runs.  Note that if someone is dealt a hand containing exactly one of each of the 20 cards this would be counted as a single run in the above table, because it contains no double runs or better.
Method
Rather than calculate the exact probabilities (as the C program linked to in the question does), I simulated 1,000,000 deals and counted the number of each type of meld in the hand.  The script I used is here in case anyone wants to modify it.  1,000,000 hands took a few minutes to simulate.
To decide what to look for in the hand I used this double pinochle scoring chart.  I didn't calculate the probabilities of royal marriages because they are identical to the "Pinochles" row.  The other "Around" hand probabilities are identical to Aces Around.  The single deck probabilities contained an entry for "exactly 7 aces", which I didn't include as it's not a meld (and because it wasn't obvious what the analogous hand would be).
A roundhouse is a marriage in each suit, i.e. eight cards total.  Consequently it's not possible to get more than a double roundhouse in a twenty card hand.
Verification
To check the validity of the method I made some small changes in order to simulate the single deck 4 player game.  I ran it with 100,000 iterations and the results from this are very similar to the probabilities given by the C program:
|  MELD TYPE  |  NONE   | SINGLE  | DOUBLE  | TRIPLE  |  QUAD   |
| Runs        | 0.96159 | 0.03841 | 0.00000 | 0.00000 | 0.00000 |
| Marriages   | 0.40970 | 0.57963 | 0.01067 | 0.00000 | 0.00000 |
| Pinochles   | 0.81385 | 0.18380 | 0.00235 | 0.00000 | 0.00000 |
| Ace Arounds | 0.97302 | 0.02698 | 0.00000 | 0.00000 | 0.00000 |
| Roundhouses | 0.99973 | 0.00027 | 0.00000 | 0.00000 | 0.00000 |

Note that the triple and quad results are impossible with a single deck, and a double roundhouse is impossible with only twelve cards.
Suggested improvements
It would be interesting to calculate the exact probabilities, particularly for very rare hands.  The exhaustive search method of the original program has to consider about 5*109 hands for four players and 6*1010 hands for three players.  For twenty cards chosen from a deck of eighty it would need to consider 3*1018 hands. Since the original source said that the three player game was analysed in 12 minutes, we can extrapolate that double pinochle would be done in about 1140 years!
This doesn't mean that the probabilities can't be calculated exactly, just that a more clever method would need to be used. The symmetry of the deck should help with this, but the results above were enough to convince me that I wasn't going to get dealt a quad-run any time soon.*

* The chance of a quad-run is about 1 in 1018, as there are exactly four possible quad-run hands (one for each suit), and there are about 3*1018 hands in total.

Wow, thanks for your thought on this. I have a question about the code. It looks to me like the card deck was only 40 cards instead of 80 cards = [(r,s) for r in ranks for s in suits]*2 or do I misunderstand how that is being initialized?

@Clumsy Thanks - you're exactly right.  I re-ran the updated program and we now get triple and quad marriages, which we didn't before (hard to get with only a single deck!)  I verified the bulk of the script by calculating probabilities for single deck pinochle, and I've added a section about that.

Fantastic! I really appreciate your help. I was way out of my league trying to do this myself. According to your results the odds of a double run are approximately 1:825. That's one of the melds I was most curious to know about.


text:
Do I have to "choose" when not targetting?
This is related to this question, but without the targeting. Let's say I have Penance and Narset, Enlightened Master on the battlefield, my opponent having nothing but (colorless) lands. It's my turn, combat phase, I attack with Narset, and with her triggered ability still on stack I want to activate Penance in order to put a non-creature, non-land from my hand on top of my library to free cast with Narset.
Do I have to choose Narset for Penance's damage prevention?

If Narset is the only Red or Black source you will have to choose her when the ability resolves if you activate Penance. The only way you could not make a choice is if there are no Red or Black sources to choose, but you would still be able to activate the ability. Note that the source you choose doesn't have to be something that will ever deal damage, it can be something like Doom Blade or Aggravated Assault.

608.2d If an effect of a spell or ability offers any choices other than choices already made as part of casting the spell, activating the ability, or otherwise putting the spell or ability on the stack, the player announces these while applying the effect. The player can’t choose an option that’s illegal or impossible, with the exception that having a library with no cards in it doesn’t make drawing a card an impossible action (see rule 120.3).

119.7. The source of damage is the object that dealt it. If an effect requires a player to choose a source of damage, he or she may choose a permanent; a spell on the stack (including a permanent spell); any object referred to by an object on the stack, by a prevention or replacement effect that’s waiting to apply, or by a delayed triggered ability that’s waiting to trigger (even if that object is no longer in the zone it used to be in); or, in certain casual variant games, a face-up card in the command zone. A source doesn’t need to be capable of dealing damage to be a legal choice. See rule 609.7, “Sources of Damage.”

You have to make the choice when the ability resolves, right? You say you have to choose her "when you activate".

Also worth emphasizing that the thing you choose doesn't have to be something that'd ever deal damage, so it's easier to avoid being stuck choosing Narset than the OP might think. For example play another creature pre-combat that won't be able to attack and choose it, or play a red or black spell that doesn't deal damage (even an instant in response to Narset's trigger!) and choose it. The Oracle text for Penance makes this clear, since it says "the next time...would deal damage, prevent that damage" rather than just "prevent all damage from...".

Rule 119.7 mentions what you're allowed to choose as a source, and offers a couple more possibly interesting options besides permanents and spells: "any object referred to by an object on the stack, by a prevention or replacement effect that’s waiting to apply, or by a delayed triggered ability that’s waiting to trigger (even if that object is no longer in the zone it used to be in); or, in certain casual variant games, a face-up card in the command zone."

@Jefromi wait, in EDH games I can choose commanders in the command zone as sources of damage?

@Mephy Maybe? Not sure if EDH is meant to be included; it'd just be a placeholder so that if in the future they make a commander that can ping from the command zone (like how Oloro lets you can life) it'll be preventable. Other casual variants might make more sense.

@Jefromi Planechasem Archenemy, Vanguard and Conspiracy all use the command zone for the Plane and Phenomena, Scheme, Vanguard and Conspiracy cards respectively


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Does reduced casting cost affect casting of cards with variable casting cost?
I know the title's wording is a bit confusing, so here's an example scenario:
I have Etherium Sculptor in play and I cast Chimeric Mass for 0 mana. My understanding (this post and my understanding of the rules) is that Chimeric Mass will have one charge counter on it as opposed to zero. Is this correct?

@diego It's not quite a duplicate, though I know they are similar. In my question I'm also concerned about how many charge counters are on Chimeric Mass if I technically paid 0 mana for it.

What part of your question is not answered by the answer to that question?

@player3 The mass doesn't care how much mana you paid, it cares what X is

I suppose it may be answered by that question, but if so it wasn't clear to me. Seeing as I didn't technically pay any mana for Chimeric Mass, does it still get a charge counter for the one mana covered by Etherium Sculptor?

Did you even read the answer to the question you linked? It tells you exactly how X spells work with cost reduction and what the resulting value of X is.

I did, but like I said it wasn't especially clear to me. But diego just confirmed my understanding, and I realize that what he said is essentially the same as the answer in the linked post.

Unless you somehow thing it matters that you're talking about an Artifact instead of a Sorcery -- it doesn't -- this is the exact same question as the one to which you linked. You don't calculate X based on how much you payed. You pay based on what value you chose for X.

@player3 for future reference, if you have a question which is marked as a duplicate but you really don't think it is, the best way to handle that is to edit your question to make it explicitly clear what it is that you're asking which is not covered by the other question. Once you edit it, it goes into a queue to be reviewed for reopening. But it sounds like you got your answer in this case, which is good. Welcome to B&CG Stack Exchange!