text:
Ticket to Ride house rule to help players with short destination tickets
The game balance in Ticket to Ride (in my social not highly competitive environment) seems to favour the person with the longest destination tickets and getting short tickets is almost automatically a loss.
I have been wondering about game balance in Ticket to Ride and thought a good idea might be to allow a player to draw a new ticket after completing a mission and being allowed to throw it back if they don't want it.
Perhaps we play in too friendly a fashion and the person with small tickets should be more disruptive of long routes but what are people's thoughts on the game balance and this suggestion?

not an answer, but you are playing it too social. if nobody blocks another one routes, then the long routes will almost win.

Sometimes we just do what TTR: Europe does, and take all the routes with value > X (usually 16 or 18), shuffle them separately and deal 1 to each player as the first of their starting hand.  The rest are discarded or shuffled back in after everyone has all their starting tickets, depending on how we feel

The Europe map might be better suited for you in this case: it guarantees that one of the four "ticket" cards drawn at the beginning will be a long route.

The goal in TTR is not "complete the cards you're given", it's "get the most points". There are plenty of ways to do this with short routes (e.g. by focusing on building only length 4+, not going for longest route, and picking the spots where everyone wants to go).
If no one in your group wants to adjust their play style to match the cards they're dealt, then you obviously don't want to play the game as intended. A rule change might be an outcome, but this perceived unfairness might be part of the process of getting to know the game mechanics.

I (...) thought a good idea might be to allow a player to draw a new ticket after completing a mission

This rule comes with a drawback. You would have to show your completed routes, which is something you would normally do at the end of the game. I would not recommend this as it would upset yet another part of the game.

All true - I just wanted to share a thought.

We've had the same kinds of discussions when playing TTR:Europe, complaining about certain impossible long routes. Then one person decided to discard his long route and just wreak havoc across the map. It won him the game and briefly pissed some people off, but it had no long term effect on social interaction within our group.

I disagree that the player who draws the longest routes has an advantage. A winning play in Ticket to Ride (TTR) doesn't boil down simply to that. Here are a few reasons why I don't think changing the rules is necessary (according to my many many TTR games played):

Many games can be won by completing only small routes, and by using connections that are essential to connecting certain parts of the map that many players often want to use in a game (for example Omaha to Kansas City on the America map). Choosing strategic routes (not simply the shortest one to get from on city to another) makes a big difference in TTR.
There is also the possibility to take smalls gambles when you think you have an idea of where someone wants to go. Here's an example: if you think a player is trying to connect Houston to New Orleans in the America map, a move I have often used is to play that connection even if it has nothing to do with my objectives. As I said, it is a gamble, but more often than not you can somewhat judge the direction a player is heading in.
Another good strategy is to draw three more route cards on your first turn, and sometimes three more on your second turn. You will be behind other players in map building cards, but you can potentially have 5-6 route cards that you can manage to connect with 25-30 trains.

My opinion is that in TTR there are many ways of playing to avoid the winner to often be the one with the longest route cards drawn at the beginning.

How much have you played?
I agree that someone who gets lucky and gets several long tickets that are all along the same route will have a big advantage.
However, other players can disrupt the player by blocking his routes... in which case he loses a lot points.
Also - the 1910 expansion has a 'most tickets' complete bonus -which gives a bonus 15 points.

Yes, without the risk of beïng disrupted, the long routes are a winner. I'm not sure if your solution helps. But I have a few of my own:

Disregard the points for the routes.
Divide the routes in three stacks (long, middle and short). First deplete the long routes, then the middle and then the short. That way each player has a similar chance for long routes.

In my experience, I don't think starting with short tickets is an automatic loss. In fact, I've been very successful with shorter tickets. When I start with short tickets, I try to complete them fairly quickly, but keeping an eye on the long routes on the board. Completing a 6-long route gives you 15 points. Your routes do not need to connect to each other, so just going around gobbling up the longer routes is a viable strategy, especially since it will be fairly disruptive to the other players. In addition, the 1910 expansion gives 15 points to the player with the most tickets.

I think you could mitigate this by having a hand drafting of initial routes, ala 7 wonders/agricola.  Let's say you seed each player with a minimum of two long routes and two short.  On top of that, people will have partial knowledge of routes in play which may lead to more screwage.

Yes, I've also observed that to be the case in friendly games of the original TTR.  Probably it means that you have to be more aggressive in disrupting each others routes.  Other boards, like Netherlands and Africa, don't have the super long routes that TTR, TTR Marklin and Switzerland have, perhaps this change was in part to fix that imbalance.

Don't lose sight of the bonuses for long route segments. For instance, in the Europe version there is only one eight route segment worth as much as a long route all by itself.  Folks can get so caught up in the routes that they lose sight of the points available by simply grabbing long segments.  There is an added benefit that grabbing random long segments might mess up an opponents route segment plans for their formal routes.

I also feel as if the longer routes become an automatic win, what ends up happening with the group I play with, is one player hoards the cards and therefore I am left with the cards I am dealt with. The issue arises when there are three other players to deal with. A solution would be to divide tickets into three categories (short, medium and large) and when exchanging a card, the player must return one or more cards of the same category

One player taking all the route cards should be a straightforward victory to heir opponent, every turn taking a card is a turn not claiming routes or picking trains, and every incomplete route reduces the final score.

What do you mean returning /exchanging cards? You can’t exchange routes once you’ve taken them.


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Has Hive already been completely 'solved' for AI?
According to some calculations, Hive has not too many positions to check them all out through programming, and basically make the best possible AI for it. Has it been done yet?

I don't know Hive, but most people in that thread are only talking about the placement of the first three pieces, which probably isn't enough to solve the game.

@RemcoGerlich oh, now I see. And next pieces' placements raise total number exponentially, I guess.

No, just the total number of placements is far too huge to "solve", and that doesn't even start to take into account movements occurring between placements.

source: Stack Exchange

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

sort: votes


text:
Can the Powder Tower be used while its owner is bewitched?
I played as Witch and bewitched Thief. The Thief has previously built the Powder Tower and would like to use it at the turn he's bewitched.
Can the Thief or the Witch use the Powder Tower this turn?

No one can use the Powder Tower this turn.
The Thief can't use the Powder Tower, because he's bewitched he loses his turn after taking his action (choose 2 gold or 1 card).
As the Witch, when you bewitch another character you still play with your own city. Since the Powder Tower is not in your city, you can't use it.
A quote from the rules that best describes this:

When the bewitched character is called upon, its player must show his character card, take an action, and then immediately end his turn. [...] You now resume this player’s turn as if you were playing the bewitched character,
  using all the character’s powers in your city. You still build districts from your hand of cards and pay with your gold (including taxes).

What makes you saying that ? An official rule ?

Yes, this is all based on rules. I will add these rule citations.

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


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Are there quality custom boardgame card printers?
I'm designing a custom boardgame and am looking to get a few stacks of custom cards printed.
I'm looking for a high quality card printing service but a little worried about simply using Google to find one.
Which specifics should I look for in a card printing service to ensure a high quality result? And are there perhaps comprehensive lists of printing services available that I can use for this comparison?
Thank you.  :-)

I have found the most reliable way is to print prototypes yourself.

I'm voting to close this question as off-topic because it is asking us to promote our favorite printer at best and at worst asking us to advertise for specific companies whose effectiveness to satisfy the OP's desired experience, may not be completely objectively measurable. Also there are plenty of sites that rate businesses, I don't think we need to do it here.

Edited the question to better suit the Q&A format. Tried to stick to the principles behind this blog: https://blog.stackexchange.com/2010/11/qa-is-hard-lets-go-shopping/

possible duplicate of What is the best way to print cards for my game?

@GendoIkari I don't think it is a dupe, that one is asking 'how do I print cards', and this one is 'how do I find someone else to print cards for me'

Possible duplicate of Where can I find a printer for a budget game board?

Possible duplicate of On-Demand Cards Deck printing service?


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is your turn over after playing an action card and doing what the card says
The Action card "Draw two, play two" says to draw two cards and play them. Then you're supposed to continue your turn.
But isn't your turn already over? Because you've already drawn a card and played the action card and completed the action.

In Fluxx what you do on your turn all depends on what the rules currently are so there is no one clear answer that can be given. If there is a rule in play that requires you to play multiple cards then the card you are talking about counts as just 1 of your actions and you need to continue, if there is not a multiple card play rule in effect then you are probably done.

To contextualize this answer a bit to the question: ASSUMING the state of the game is the initial "Draw 1 per turn, Play 1 per turn" state, you would be playing the Draw 2 Play 2 action card and then playing and resolving both of the drawn cards AS THE ACTION of the "Draw 2, Play 2" card. If by the effects of these cards the "Play 1 per turn" rule is overridden (by, say, a Play 4 rule card), that comes into effect after the Draw 2 Play 2 card finishes BUT BEFORE YOUR TURN ENDS, so you would still need to play 3 more cards [or, if you don't have that many, the rest of your hand one at a time].

Playing the "Draw Two, Play Two" card counts, in itself, one play. Cards that are played as a result of this action do not go towards your Play count for your turn. Similarly, the two cards that are drawn do not go against your Draw count for the turn.
After that, I echo Joe W's answer. Your turn is over based on the current rules after the Action is played; if the rule is still Play 1, then yes, your turn is over. If not, and that Action was not your last play, then your turn is not over.


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Where can I find the rules?
Many years ago, I bought a game called 'Elixir', in which the players represent wizards trying to obtain the ingredients for the Elixir of Life.
Unfortunately, I've lost the rules, and I can't find any reference to the game or its manufacturers on the Internet.
The main thing I need to know is the table that tells you the result of an attempt to brew a potion, based on the value of gems in the mix, and the die roll you throw.

This might help determining the manufacturer: https://boardgamegeek.com/geeksearch.php?action=search&objecttype=boardgame&q=elixir

http://www.bombjack.org/commodore/amiga/games/Elixir_of_Life_Manual.pdf
The table is at the bottom of page 2. When brewing, add the cost of your gem(s) to the dice roll. On a 3 or less, the brewing fails and you lose the ingredients. On a 4 or 5, nothing happens but you can attempt to brew again on your next turn. Finally, on a 6+, the potion is successfully brewed.


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Epaminondas puzzle
In this description of Epaminondas with puzzles at the bottom, what is the solution to the first puzzle?

The solution is for white to move their row of 3 one space forward. There are a bunch of possible moves for black but white can guarantee a win from all of them.

If black moves away from their back line:
If black moves their leftmost piece away from the line, then red pushes forward. Black is forced to use their remaining 2 to capture red, who captures again. Black can't capture the remaining piece.

This is similar to the situation with the rightmost piece:

If black moves the middle piece away, then they will have to reform the broken 2 in order to defend the back line. Red aims a phalanx there:

If black moves the middle piece to the side to avoid this diagonal phalanx, red goes too far away on the other side:

If black stays on the back line
If black is staying on the back line, then their first move cannot be to the left. Any distance to the left allows red to take the first capture and keep pushing to win. Thus black moves right. If black moves more than one space to the right, then they're too far away to defend from the single piece moving in. If black moves one space to the right, red forces the black row to capture their piece, lining up the black phalanx to be split in half:

That said, this question is probably more suited to the Puzzling Stack Exchange.

And probably this answer is oversimplified - I've been playing with having black make the best defensive move each time and it's tough!

You cannot choose how black moves; you should show that it cannot play in any way so as to prevent the win by white. Also to win white only needs one of its pieces on black's back rank (since black is nowhere near white's back rank).

@hkBst Right! I think I've got it, just working through the possibilities.

@hkBst Okay, I worked it out over lunch with some board game parts :P I'll write it up now!

From the help: "Questions about Go problems or Chess problems are fine, even if they are static puzzles, since they are related to a dynamic game." Can be extended to Epaminondas so this question is certainly on-topic here (as well as on Puzzling).

NOPE I GOT IT WRONG AGAIN AAAAAAAAAAAAAAAAAAAAAAAAAAARGH

Okay, I've actually got it! Finishing this up.

I don't think these 'forced captures' are mandatory, sorry.

If the middle black piece moves out of line, red can immediately move forwards for a win, because black cannot recapture.

@hkBst Right you are, but I this way gets a win in exactly three moves :D

@freekvd Every move I label as "forced" is one which, if black doesn't take that move, black loses immediately.


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In Secret Hitler, what happens when exactly 50% of players vote "Ja!"
In a game of Secret Hitler with an even number of players, what is the result of the election if exactly 50% of the players vote "Ja!"?
Does the government get to pass a policy, or does it count as a failed government?

The Kickstarter page says the following (emphasis mine):

Each round, the President selects a Chancellor. The whole table votes "Ja!" or "Nein!" on their ticket; if a majority of players vote “Ja!,” the elected President and Chancellor take the following actions to pass a new policy:

Since 50% is not a majority, the election failed.

That is precisely the wording that confused me - not being a native English speaker, I was not sure what it meant I case of a tie. Also, I just read the rules (for backers) and they spell out explicitly that, in case of a tie, the government does not get elected. Thank you.

@JeanHominal Sorry for the short answer then. 'Majority' basically means 'more than half'.

The answer is absolutely fine! Had I wanted more in that language vein, I would have asked on ell.stackexchange.com :)

Wikipedia tells us that a majority is "A majority is a subset of a set consisting of more than half of the set's elements". It is better to go by the rules rather the Kickstarter page. Unfortunately, the rules are contradictory. As noted by Kthaeh:

"The back of the President placard says at least 50% of the players
  have to vote for a proposed government in order to get them elected.
  The printed rules say more than 50% of the players have to vote for
  the government to get them elected."

However, there are four reasons to resolve this contradiction in favour of the rule that a government needs more than half.

Firstly, the rules also say "If the vote is a tie, or if most
players vote nein!,  the Presidency placard moves clockwise". Since
this is the only place in the rules that explicitly considers a tie,
this carries the most weight.
Secondly, the very end of the strategy
guide in the rules states that "The Fascists can win ... by
rejecting Candidates until the board fills with fascist policies.".
This suggests that the intended play style makes it easy-ish to
reject candidates. Although ambiguous this is perhaps the most important reason, as the rules should serve the game rather the the other way round.
Thirdly, Kthaeh notes that similar games such as Avalon and Resistance require
more than half, so by principle of least surprise, we should be
more inclined not to have a different voting rule in this game.
Finally, as noted by freevkd,
the Kickstarter page says a majority, and we could at least
give the Kickstarter page a tie-breaking vote between the rules and
the President's placard.


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is party bridge the same as rubber bridge?
I am in a group that says we play party bridge and the rules for honors are different than for rubber bridge. What exactly are the differences between party bridge and rubber bridge, and why are they different?
For example, are the honors scored the same in both party bridge and rubber bridge?

I know of no rules for Party Bridge named as such, but there are other  scoring table besides standard Contract Rubber Bridge; it is possible that your friends are referring to what is best known as Chicago Scoring. Also known as Four-Deal Bridge, Chicago scoring frequently, though not always, omits scoring points for Honours; I played it both ways while learning the game many years ago.  To the best of my knowledge the rules for scoring Honours are always only On or Off - I know of no variations beside making it optional.
Chicago Scoring puts more points in play per hand than Rubber Bridge scoring, and is consequently also frequently preferred for high-level money games - the greater number of points in play further reduces the luck element over a session, and more easily accommodates rotating a fifth or sixth person in and out of a game.
This same dynamic is valuable in a party setting, because each table should finish a four-deal round in under 30 minutes. Players can then socialize for a few minutes before receiving their new seating assignment, typically determined either round-robin or progressive. A 6-round session should easily play in under 4 hours while still allowing frequent opportunities for socializing, or even in close to 3 hours for a more serious-minded group.
However if you attempt to break in to a new money game and find Rubber Bridge scoring being proposed, remember the old Poker adage - if you don't know who the sucker is, it's probably you. Amateur players will sacrifice far too often and frequently to save the rubber, and good players will rapidly fleece any person or partnership that exhibits this behaviour.


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Counting in Blackjack using Uston APC
Note: Please see my edit to see how I am doing it now.
I am making a simple blackjack game that only uses 1 deck and I read that Uston APC is the best one for just 1 deck. There is no doubling or splitting, it's a very simple game for a class.
I read here: http://www.countingedge.com/uston-apc.html and a few other sources and it seems the way I implement it is keeping a running count based on this chart:
2   1
3   2
4   2
5   3
6   2
7   2
8   1
9   -1
10  -3
J   -3
Q   -3
K   -3
A   0

I will call this running count "running_count". I also keep track of how many ace's ive seen, which I will call "ace_count".
When it comes time to make a bet, my decision for how much to bet is based on running_count - (3 * ace_count), that is for every ace i've seen subtract 3 from running count. This new count I will call "temp_count". If "temp_count" is 2 or higher, then bet the minimum bet multiplied by the temp_count. So if temp_count was 5 you would bet 5x minimum bet. If temp_count is anything lower than 2 then just do minimum.
After betting has taken place, I disregard "temp_count" and go back to my "running_count" and "ace_count".
Is this how Uston APC works? is there a table on how I should play based on dealer card and my total compared to the running count? Right now I am using:
Hard:

•  totals 11 or less, he always hits.
•  totals 12, he stands if the dealer shows 4, 5, or 6; otherwise he hits.
•  totals between 13 and 16 inclusive, he stands if the dealer shows a 2 through a 6 inclusive; otherwise he hits.
•  totals 17 or greater, he always stands.

Soft:

•  totals 17 or less, he always hits.
•  totals 18, he stands if the dealer shows a 2, 7, or 8, otherwise he hits.
•  totals 19 or greater, he always stands.

But I wasn't sure how these might change based on the running count of Uston APC.

Edit:
A bit more about the strategy, dealer stands on 17 (soft or hard)
To make my programming a bit more simple, because there is just 1 deck it kind of seemed silly to count aces separate from the normal counter so instead I just had it do count = count - 3 when it saw an ace. Effectively using -3 instead of 0 in the table described above.
I found his matrix for APC but it seems to talk about a "running count" and a true count. My understanding was a running count is as it seems, the count based on cards seen, and the true count is the running count divided by the number of decks used thus far. Some people even use half decks for it, so because I have just 1 deck, I decided to make my true count always running count / 2. Based on this I am now using this strategy as an exception to the one given above:
if hand value is 16 dealer upcard is TEN and running count is  >= 0
    stand

if hand value is 16 dealer upcard is NINE and true count is  >= 6
    stand

if hand value is 15 dealer upcard is TEN and true count is  >= 4
    stand

if hand value is 13 dealer upcard is TWO and running count is  < 0
    hit

if hand value is 13 dealer upcard is THREE and true count is  < -1
    hit

if hand value is 12 dealer upcard is TWO and true count is  >= 4
    stand

if hand value is 12 dealer upcard is THREE and true count is  >= 2
    stand

if hand value is 12 dealer upcard is FOUR and running count is  < 0
    hit

if hand value is 12 dealer upcard is FIVE and true count is  < -1
    hit

if hand value is 12 dealer upcard is SIX and running count is  < 0
    hit

This seems to do lose money at around 5000 hands but at 50,000 hands it can turn 10,000 into 15,000

I accepted the only reply mostly because solution isn't needed anymore. However, I added to what I was doing incase someone else trying to make a simple game strategy happens to come across this in the future

The above rules are good, but there are a couple of things you can do to improve your odds.
With a total of hard 16, against a dealer's T, it is very close whether you should stand or hit. You should hit if the count is at all negative, because you will make a winning hand often enough to compensate for the times that both you and dealer would have busted.
With a hard 13 against a dealer upcard of 2, you should hit if the running count is -1 or lower. Dealer is a favorite to improve, and with a low running count, you need to "keep up."


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Are the boats you build in Stone Age visible to other players?
In Stone Age, when you build a boat you receive a Civilization card. This card gives you an immediate bonus, as well as some form of additional points to be calculated at the end of the game.  I've always played that when these cards are purchased, they are placed face down on the player's individual board.
I recently saw a game played where the cards are placed face-up in front of each player.  This changed the dynamic of the game as players could immediately see which cards where more beneficial to certain people.
What is the correct placement for these cards and are other players allowed to look at previous cards you've purchased?

You're playing it right. From the rules (page 6):

Each player places their [Civilization] cards face-down on the corresponding space of their player board.

In other words, no one can see which cards you have. The only exceptions to this rule are the one-use tool cards. They are kept open until they are used.

One-use tool cards

It is worth noting that with the exception of the card that gives you an immediate bonus of the top card in the pile, all cards gotten are seen at some point, so it is trackable who has which cards, and being able to memorize this information is useful.

@GendoIkari You could even way that playing with open cards is a way to disadvantage those with good memory, or those whose strategy involves collecting a lot of cards. I'd edit all this in, but I really prefer short and simple.


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Blocking with First Strike + Deathtouch
Rule 509.3 says:

Third, for each blocking creature, the defending player announces that creature's damage assignment order, which consists of the creatures it's blocking in an order of that player's choice. (During the combat damage step, a blocking creature can't assign combat damage to a creature it's blocking unless each creature ahead of that blocked creature in its order is assigned lethal damage.) This turn-based action doesn't use the stack.

Rule 702.2b says:

Any nonzero amount of combat damage assigned to a creature by a source with deathtouch is considered to be lethal damage, regardless of that creature's toughness.

And in Unhinged there are creatures with 1/2 power(for example Little Girl)(so basically it looks like there's no minimum damage a creature can deal).
Now let's imagine I have monstrous Hundred Handed One on the battlefield(so it is a 6/8 creature that can block up to 100 creatures).
Now assume I gave my monstrous Hundred Handed One First Strike and Deathtouch. During my opponents combat phase he declares 10 3/3 creatures without any abilities as attackers.
So the question for this scenario:
Would I be able to block them all dead? Or will I have to assign 3 damage to first creature and 3 damage to second creature? Or can I only assign minimal of 1 damage per creature that would kill 6 of the attacking creatures my opponent controls?

CR doesn't cover silver-bordered cards. You should rule as desired in those cases, or ask on Maro's tumblr.

107.1a You can’t choose a fractional number, deal fractional damage, gain fractional life, and so on. If a spell or ability could generate a fractional number, the spell or ability will tell you whether to round up or down.

Normally, you must pick an integer number of damage. So in a normal game, a 6/8 Hundred Handed One with First Strike and Deathtouch only kills 6 of the blockers. The remaining four 3/3 survive to deal 12 damage to Hundred Handed One, killing it.
In an Unhinged/Unglued game, 107.1a is modified.

For right now (and I reserve the right to change this with future Un sets), whenever you choose a number, if that number is written as p/q in lowest terms, q must be 1 or 2. If you stopped taking math in high school, for any fraction the greatest legal denominator is 2. And if you've mentally repressed all math-related knowledge, the only fraction you can choose is {1/2} (and yes, 1{1/2}, 2{1/2}, etc. is fair game).

Does this mean that while playing silver-bordered cards I can break up Pyrotechnics to deal {1/2} damage to eight creatures (trust me, I can come up with a reason)?

Yes. Yes, you can.

Hundred Handed One can now deal 1/2 damage to 12 creatures. In a Unhinged/Unglued game, and only in such a game, HHO will survive the scenario unscathed. It would take 15 3/3 attackers to kill HHO.

...or 15 3/3 attackers that he's blocking. :)

@Neo, Fixed. You know, I seem to remember writing "attackers" and thinking, "wait, that makes no sense, since you can only block one creature". But of course, it makes sense here.

How much damage must be assigned to each attacker?
With deathtouch, 1 damage is lethal. Therefore, you must assign at least 1 damage to each attacking creature. You can assign more if you choose. For example, you could:

Assign 6 damage to 1 attacking creature
Assign 1 damage to 6 attacking creatures
Anything in between

That much is covered by rules you already quoted, CR 509.3 and CR 702.2b.
Can I assign damage in increments smaller than 1? For example, 1/2?
No, because of this comprehensive rule:

107.1. The only numbers the Magic game uses are integers

There is one exception to this rule that is not present in the comprehensive rules. Mark Rosewater, the Head Designer of Magic, stated in an article that if you are playing with silver bordered cards, all cards can suddenly deal fractional damage. This only applies to casual play - silver bordered cards are not tournament legal.

Question: Does this mean that while playing silver-bordered cards I can break up Pyrotechnics to deal {1/2} damage to eight creatures (trust me, I can come up with a reason)?
Mark Rosewater: Yes. Yes, you can.

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text:
How can I handicap in Suburbia?
I have Suburbia, and I quite like it. However, some of the people I play with, don't quite have the cutthroat edge to win. It makes not so fun to play if I win every game.
What's an effective way to handicap it?
I was thinking I could simply start with less money, in which case, what would be idea starting amount be?

Maybe play a different game.
Also, a general strategy in Suburbia is to favor tiles that raise Income early on, while keeping your reputation low.

Generally, score handicaps are the way to go. You don't have to worry about playing games and experimenting in order to find a balance, you can just use the knowledge you already have of how much you tend to win by.
Money handicap could possibly work, but I think it'd be pretty tough - you only have so much room to adjust before it basically becomes a "you have waste one early turn to build a lake right away" handicap.
That said, you can also just not make the handicap formal: try to do the best you can every game, and accept the fact that you're striving for a larger margin of victory instead of worrying about winning.

An alternative is to handicap the premium costs for early development
This could be implemented for stronger players as:

+$1 or +$2 for every spot
+$1 or +$2 for every spot except the last free spot
+$1 or +$2 for every spot except both free spots

Or for weaker players as

$1 or $2 less for every spot
$1 or $2 less for every spot except the last free spot
$1 or $2 less for every spot except both free spots

Pick one of these.  I think it would be simplest


text:
Older Pokemon cards
I can tell by checking sites that older Pokemon cards are most likely not allowed in a tournament setting.  Would that include energy and training cards?  They should seem to be generic and okay.

The list of sets legal for tournament play in the 2015-16 season are spelled out here
http://www.pokemon.com/us/pokemon-news/pokemon-tcg-format-rotation-for-2016/
Basic energy is always allowed.   Special energy depends on the set in which it was released.
Pokemon allows for older cards that have been reprinted exactly and are currently legal for play.
Pokegym maintains a handy list that explains which older cards are still legal for play and why
http://pokegym.net/forums/view.php?pg=xpdformat


text:
What gets Destroyed?
Given

Hang in There - Play on one of your minions. Ongoing: If this minion would be destroyed, move it to another base and destroy this action instead.
Nukebot - power 5 - Ongoing: After this minion is destroyed, destroy each other player's minions on this base.
Sacrifice - Choose one of your minions. Draw cards equal to its power. Destroy that minion.

I am have a bit trouble figuring out when destroy triggers.
I play Sacrifice on a Nukebot that has Hang in There, what happens?  Which cards get destroyed and why?

Stepping through it in order.
Start - Nukebot with Hang in There are on a base
Play Sacrifice

Choose one of your Minions - Choose Nukebot
Draw Cards equal to its power - draw 5 cards
Destroy that minion - prepare to destroy Nukebot, check Ongoing triggers
Hang in There triggers on If this minion would be destroyed
Move the Nukebot to another base
Destroy Hang in There

The trigger on Nukebot never fires because it is never destroyed.  The Hang in There prevents it.


text:
Help Determining a Card Game w/ 4 Stacks and Hidden Card Bidding
I played a game about 5 to 8 years ago with the following rules:

The deck was similar to a Skip-Bo deck (Numbered cards with no suits)
Each player was dealt a hand and 4 "piles" were created
On each round the players would each secretly bid a card and reveal it all at once.
Starting with the (don't remember either highest or lowest card), you get to play your card in increasing numbers on one of the 4 stacks.  If you cannot play, you have to keep your card.  Object to to clear through all cards in hand.

Thanks for your help!

Please don't put the answer in the question.

But the title of the question doesn't give enough information on the details of the question.  The purpose of various SE sites is to have a searchable repository for future users.  If someone searches for "Game I played which was called Take something" it would grab that in the title.

The title of the question could be more descriptive, but the title of the question is about the question, not the answer. You could modify the title to say that the game is a card game, or add other details, but the title of the question should only describe the question. Plus, search (both Google and site search) looks at the content of both the question and answer unless you specify otherwise.

I belive the game is 6 Nimmt! Each card has a big picture of a bullhead on it and it seems to fit the description.
Here are the rules from http://www.gamecabinet.com/sumoRulesBank/Take6.html

6 Nimmt! (Take 6)
  Translated by Michael Schloth.
Goal
  Take no cards. If you must take cards, take as few Cattle Heads as you can. Each Cattle Head is worth (- 1) point. The winner is the player who at the end of the game has taken the least number of Cattle Heads.
Preparation
  You'll need paper and a pencil.
Shuffle the deck and deal each player ten cards. The players keep their cards secret and they should arrange them numerically. Four cards are taken from the deck and placed face up on the table. Each of the four cards is the first card in a row. Each row may never contain more than 5 cards. The illustration (Abb.1) on page 2 shows the 4 rows at the start of a game.
The Course of the Game
  Part I: Playing Cards
  All players choose one of the cards from their hand and place it face down before themselves. They are revealed at the same time. Whoever played the lowest card goes first. That player places his card into one of the four rows, then the player who played the next higher card places his card into one of the four rows etc. until all players have placed their chosen cards. The cards are always placed in a row side by side. This process is repeated until all players have played their 10 cards.
How are cards played?
There are two rules:
Rule 1. : Ascending Numerical Order
  The card to be placed in a row must be greater than the last card already placed in the row.
Rule 2. : Lowest Difference
  The value of a row is the numerical value of the last card in the row. If a card may be placed in more than one row (by Rule 1.), then compare its numerical value to the value of each eligible row. The row with the closest value gets the card.
Example: (page 3)
  The last cards in the four rows are 12, 37, 43, and 58. Four players play the following cards: 14, 15, 44, and 61. The "14" card is lowest and so is placed first. By Rule 1, it can only be placed in the first row next to the "12". For the same reason, the "15" must be placed next to the "14". The "44", by Rule 1, may be placed at the end of rows 1, 2, or 3. However, because of Rule 2 it must be placed in the third row next to the "43". The "61" is then placed in the fourth row for the same reason.
Abb. 2: Shows the 4 rows after the first round.
Part II: Taking Cards
  Each of the four rows may hold only 5 cards maximum. What happens when a row must receive a sixth card? What happens when a card needs to be placed but it is lower than the last card of all four rows? In both cases the player will have to take cards because of the last two rules.
Rule 3.: Full Row
  A row has 5 cards and a player, whether by Rule 1 or by Rule 2, must place the sixth card in the row. That player must take all five cards from the row and his sixth card now becomes the new first card for that row.
Example: (page 4.)
  The same four players now choose the following cards to play: 21, 26, 30, and 36. The "21" and the "26" are placed in the first row after the previously placed "15". The first row now has 5 cards. The next card to be placed is the "30". By rule 1, it too must be placed in the first row after the "26". Because this will give the first row six cards, the player takes cards "12", "14", "15", "21", and "26". He places his "30" as the first card in the first row.
The "36" is then placed in the first row after the "30".
Rule 4.: Lowest Card
  Whoever plays a card that is too low to be placed at the end of any of the rows must take cards. That player selects any one of the rows and takes all of the cards from that row. His card now becomes the first card in the row.
Example (page 5.):
  The same players now choose the following cards: 3, 9, 68, and 83. The "3" is the lowest so it must be played first. It is too low to be placed at the end of any of the rows. To make it fit, the player of the "3" must take one of the rows. He decides to take the second row as it contains only one card. The "3" is now the new start card for the second row. Next the "9" must be placed. This player is lucky as now it can be safely placed next to the "3". The "68" and the "83" are both placed in the fourth row.
Remember: When a player must take a row because his card was too low to fit any of the rows, that player decides which row to take. As you do not want to take Cattle Heads if you can help it, you will usually take the row that has the least number of Cattle Heads.
HORNS = NEGATIVE POINTS.
Each card has a certain number of Cattle Heads (from 1 to 7).

>All Cards have at least 1 Cattle Head.
Cards with values that end in "5" have 2 Cattle Heads.
Cards with values that end in "0" have 3 Cattle Heads.
Cards with values that are multiples of 11 have 5 Cattle heads.
Card value "55" is both a multiple of 11 and ends in "5". It has 7 Cattle Heads.

Each player keeps the cards they had to take because of Rules 3 and 4 separate from the cards they are trying to place into the four rows. These "Take" cards make up your "Horn Pile". (I guess because you are being gored by them?).
When all players have played their 10 cards, the round is over. Each player counts the number of Cattle Heads in their "Horn Pile" and the scorekeeper records the amount. All of the cards are gathered and reshuffled. Each player is dealt 10 cards and another round is played.
The game is over when one or more players reach 66 Cattle Heads. The player with the least number of Cattle heads wins.
Game Tips
  Example 1
  This example emphasizes the point of Rule 2. You MUST place your card using the Least Difference rule. Here the player needs to place "45" and would love to place it at the end of the third row but Rule 2 forces him to place it in the fourth row as the sixth card. He must take up the fourth row into his "Horn Pile".
  Example 2
  This example emphasizes the point that you should not take anything for granted. Player 1 chose "62" to place thinking that it would be an easy placement in the first row. Unfortunately for him, Player 2 chose "29" to place. As "29" is lower than any of the last cards in any of the rows, player 2 must take one of the rows. He selects the first row. Now Player 1 has lost his safe play. "29" is the last card in the first row so by Rule 2 player 2 must place his "62" in the already full fourth row and take that row into his "Horn Pile".
Variant
  All previous rules hold. There are two new ones:

All players will know exactly what cards will be in the round. The rule is: Cards in the Round = ((# of players) x 10) + 4.

example:
3 Players
      Use Cards 1 through 34.
  4 players
      use cards 1 through 44.

Each player takes his own 10 cards himself. The cards are turned face up and spread on the table. In turn, each player takes one card. When everyone has 10 cards, there will be 4 cards left. These 4 cards are the start cards for the 4 rows.

If you're still not sure if this is it or the rules seem confusing you can play it online here:
http://www.onlinebrettspiele.de/6nimmt/Default.asp?l=en


text:
Can I build a settlement if I don't have the building piece for it?
Each player has 5 settlement tokens in hand at the beginning of the game. After placing the 5th settlement, considering I have not built a single city, can I build a 6th settlement?

This question makes me irrationally angry.

The way to place it on the board is a city on its side, but you can't build like that. It can only happen if a city gets reduced.

This is super clear in the rules.
Under Build in the Game Rules section:

You cannot build more pieces than what is available in your
  pool—a maximum of 5 settlements, 4 cities, and 15 roads.

Then, later, under Build (Building) in the Almanac:

You can build as many items and buy as many cards as
  you desire—as long as you have enough resources to “pay”
  for them and they are still available in the supply.
  [...]
  Each player has a supply of 15 roads, 5 settlements, and 4
  cities.

(Emphasis added.)
and again later, under Settlements:

Note: If you have built all 5 of your settlements, you must
  upgrade 1 of your settlements to a city before you can build
  another settlement. You will then have the settlement in your
  supply, so you can build another settlement.

No.
Supplies are finite. If you need more settlements, you have to upgrade some to cities. As soon as you upgrade one settlement to a city, the settlement returns to your supply.
The same goes for roads and cities, they too run out. As such, the longest possible road has length 15 (or 30 if you include ships from the Seafarers expansion).

And, for completeness, for cities too — you can't upgrade 5 settlements.

@mattdm: Huh? Our understanding was there was no need for the fifth city counter because that nets 10 points which wins the game immediately.

@Joshua Maybe that's deserving of its own question and answer, then, but again I think the rules are pretty clear. See the quotes in my answer: you cannot build more than 4 cities. (So, you can't win that way — which, by the way, makes the end game a lot more interesting!)

@Joshua In my experience, 10-point games end too fast. We usually play to 12.

For three players, 12 is good. For four, it can make the endgame very boring and arbitrary, essentially betting against a particular roll of the dice that will give the lead player the last of a resource he needs for that one final point.

No, unless you have lost a settlement. Check the contents list of the game to see if you have every settlement pawn for that color.


text:
What types of printers or printing techniques are used to produce commercial cardboard pieces?
I am interested in the topic of printing game pieces, specifically, cardboard tokens and pieces. Usually these pieces come on cardboard sprues that you push out. I am curious about exactly what equipment commercial manufacturers of board game pieces use. I want to learn specifics about the names of the device families (not specific printer models) and techniques used by commercial printer. For example there  might be a technique called 'fusion printing' (fictional name) that I could use to search for various manufacturers of 'fusion printers' on my own.
When I searched for this topic, I found two answers on how to make these game pieces. One is arts & crafts which uses printing, sticking and adhering, usually paper to cardboard or wood. The other is to use a commercial printing service online.
I am not interested at all in the arts & crafts solutions - any kind of amateur DIY production as this has been asked and answered all over the internet. I'm asking this question here because I could not find a scrap of information on what commercial manufacturers use, or what you might start looking for if you wanted to print pieces as professionally (and commercially) yourself but on a smaller scale.
For example, I can see through research something called Dye Sublimation Printing and there are cheap printers that can be used at home that can do this. But I only found it relating to the methods of producing plastic ID cards. I don't know if it's the correct technique for cardboard game pieces, but it does seem a step closer to what I'm looking for than using a standard laserjet printer.
I am sure there will be a specific technique or techniques used with pieces and I am curious to learn about them.
To state my question specifically:
To produce cardboard printed game pieces to the same or close to the same standard as commercial board games, what types of printer and/or techniques are used to create them?
I ask close to the same because I am also interested in both a large industrial and smaller scale perspective and there may be slight differences.
I am also interested how the pieces are cut to shape within each technique. Commercial games almost always use cardboard sprues and can cut into any shape. Cutting circular or irregular pieces cleanly looks complicated. It would also be a bonus for any other information about each printing technique, such as how pieces are finished.

As someone who really likes boardgames and has a career in the graphical industry I can tell you from experience that flatbed UV-printers (Industrial Epsons, Rolands an Arizona's) and CNC table cutters (like ElCeDe, Zund and Kongsberg) can get you real close to the real thing.
These (usually large format) printers can print directly on thick board, often with a white primer if the substrate (The material you are printing on) isn't white or opaque and finishes like varnish/gloss effects.
CNC-cutting machines can cut the paper or board in the shape you wish, even making creases and half-cuts (for folds) and perforations.
They are great for prototyping games.
Some printing shops which also make stand-ups or signage will have both of these machines, so they may be a good place to start.
It's industrial material. Not many people can put one or both in the garage.

Great reply, flatbed UV printer looks like a great avenue to explore, as does a CNC machine. Versions are available that put themself into the home enthusiast price range and it looks like they'd produce the desired result. Many thanks!

Have a look at blogs or mediasites of Game Companies: for example this one
https://www.youtube.com/watch?v=6BIgDpvooA8 "How board games are made" by  Czech Games Edition, makers of Codenames and Mage Knight.
Videosites like youtube have more similar videos, search for "board game manufacturing" or "boardgame factory visit" an so on.
Also on https://boardgamegeek.com/ have a look at the developer diaries and the forums and you'll likely find more good information.
If you want to have your own Game components, ask your local printer, they may have a collaboration with a card manufactory offering, also for manufacturing small series of your own card/boardgame there are some shops online. I don't want to link to anyone since I don't have personal experience with any of them, but have a search for "custom boardgame" or "own cardgame" you'll find some. Use your native language for search words and/or add your city/region/country to find something close by.

Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the video link for reference. Can you [edit] your answer to summarize important parts of the video to answer the question (by naming the commercial printers and describing the process)?

@ryanyuyu: Basically i agree with you, but its takes to much of my time, and on this network, i experienced way to often that somebody deleted/closed something I put effort in - for no real reason other than disagreeing. Until this policy changes its not worth my lifetime.

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


text:
Which, if any, colored cards can be made colorless after being cast?
Which, if any, colored cards can be made colorless after being cast?
I'm not interested in cards that use some factor to determine color at moment of casting (such as Devoid).

Care to share why you are interested in doing this?

@John - Got inspired by the question on Prismatic Lace: https://boardgames.stackexchange.com/questions/53414/can-i-select-zero-colors-for-prismatic-lace

I have cleared out the conversation about whether Magic: the Gathering is on topic on this site. Community consensus on that decision is already well established. If you would like more information about what is on topic here, please see the [help/on-topic]. If you would like clarification, or to request/suggest a change to the rules, or you believe the rules have been applied incorrectly, you should ask a question on [meta].

No spell has an inherent way to become colorless after it has been cast while it's still on stack. There are ways to do this however, but it requires using other cards. This means that almost all spells can be made colorless, the only ones that can't are spells that can't be targeted or prevent using the stack until they resolve (split second).

Change target spell to colorless:

Ersatz Gnomes is an artifact creature that a tap ability to turn a targeted spell on the stack colorless.
Moonlace is a one cost blue spell that will turn a target spell on the stack colorless.

Permanents that change themselves to colorless. (all red creatures)

Ancient Kavu 2 cost activated ability to turn itself colorless.
Blazing Blade Askari 2 cost activated ability to turn itself colorless.
Raging Spirit 2 cost activated ability to turn itself colorless.

Turns another permanent colorless.

Ersatz Gnomes has a second tap ability to turn a targeted permanent colorless until end of turn.
Moonlace can also target a permanent to turn it colorless. This effect is permanent.
Xathrid Gorgon is a black creature that has a tap ability costing 3 mana that will permanently turn a target creature into a colorless artifact and gives it defender (flavored turn to stone gorgon ability).

Change everything to colorless.

Thran Lens this 2 cost artifact makes all permanents colorless, this applies to all permanents controlled by all players.
Mycosynth Lattice this changes everything to colorless, in play, on the stack, in the hand, in the deck, etc.
Ghostflame Sliver this 2 cost black and red creature doesn't apply to everything, but it applies to all slivers, regardless of who controls them.

So, if I'm reading this right, there's no way to change creatures to colorless permanently, and still have it be useful. Unless there's a way to un-disable a Gorgon target and remove its Defender?

@Malady there are ways to let a creature with defender attack, like Aracdes, the Strategist, I also doublechecked moonlace and it is permanent. I'm editing the answer now that I understand your question better.

Thanks! ... Doesn't Mycosynth Lattice not apply to things on the battlefield?

@Malady "All [,..] permanents are colorless" the only place something is called a permanent is when it is on the battlefield. Lattice applys everywhere. It also makes those permanents into artifacts.

I have cleared out the conversation about whether Magic: the Gathering is on topic on this site. Community consensus on that decision is already well established. If you would like more information about what is on topic here, please see the [help/on-topic]. If you would like clarification, or to request/suggest a change to the rules, or you believe the rules have been applied incorrectly, you should ask a question on [meta].


text:
Cheapest mandatory loop in Magic the Gathering
What is the cheapest combination of cards (in terms of total mana value of cards used for the combo), that, when played, result in an infinite loop with only mandatory actions - resulting in a draw?

You cannot go infinite in mtg. Many magic youtubers get it wrong. If there is a mandatory loop then the game ends in a draw. If you have demonstrated a loop that is not mandatory then you chose an arbitrary amount of iterations of the loop and the game continues.

@NeilMeyer but the OP is asking for a mandatory action loop that ends in a draw?

Perhaps you didn't see my comment on the answer. It is not acceptable to edit a question after there are already answers in a way that invalidates existing answers. I have rolled your edit back again. Please do not roll it forward again.

@murgatroid99 I disagree with you here. I think it is pretty obvious the intended question is about total mana cost of producing the loop, and that specifying mana value was an oversight.

It doesn't entirely matter what was intended. The answer is in accordance with what the question actually said. Practically speaking, they got what they wanted either way: my answer likely provides the actual minimum setup cost.

You can create a mandatory loop spending only 3 mana:

Cast Careful Study for 1 mana and discard Worldgorger Dragon and another card that is not a creature.
Cast Animate Dead for 2 mana targeting Worldgorger Dragon.

This creates an infinite loop: Worldgorger Dragon, when it enters, exiles all other permanents, including the Animate Dead. Then Animate Dead's triggered ability makes you sacrifice Worldgorger Dragon. That triggers Worldgorger Dragon's last ability, returning all permanents to the battlefield including the Animate Dead, which can then reanimate Worldgorger Dragon again to continue the loop.
This loop is only mandatory if there are no other creatures in graveyards, which can be accomplished for no additional mana cost with one Bojuka Bog or Tormod's Crypt per player.
If activating abilities is allowed, the cost of this combo can be reduced to 2 mana by replacing Careful Study with Bazaar of Baghdad

It can also be reduced to 2 mana if you draw up to max hand size and discard Worldgorger Dragon that way.

Two mana
Play Mirrorpool (0 mana).
Play Mirrorpool (0 mana).
Cast Pact of the Titan (0 mana)
Activate Mirrorpool’s second ability, targeting the Pact
Cast Pact of Negation (0 mana), targeting Pact of the Titan
Let Pact of Negation resolve, countering Pact of the Titan.
At this point, the only thing on the stack is the Mirrorpool's activated ability.  It will fizzle when it resolves, but it doesn't know that quite yet.
Cast Vantress Visions (2 mana). Its only valid target is Mirrorpool’s second activated ability.
Activate the second Mirrorpool’s second ability.  Its only target is the Visions.
When the copy of Vantress Visions resolves, it’s only valid target will be the original Mirrorpool ability. That, in turn, will need to copy an instant or sorcery, and the only valid target for it is Vantress Visions. Because neither can copy anything but the other, neither the original Mirrorpool ability nor the original Vantress Visions will never actually resolve, since each time something above them resolves, it will add a new spell or ability copy to the stack.

I originally tried to form this loop with two Forks, but that doesn't work because you can choose the resolving Fork as the target of a newly made copy Fork in order to force a fizzle.

Could you elaborate on what makes this loop? It isn't clicking for me.

I agree that this describes the setup, but could be expanded on for the actual loop.

Isn't this ignoring the 3 mana it takes to activate Mirrorpool?

@MartinEpsz Yes, but the question as asked at the time the answer was written does not technically care about the cost of any activated abilities involved (it asked for an answer in terms of CMC, and ability costs do not, strictly speaking, contribute to CMC).

@MartinEpsz the OP defines "cheapest loop" in terms of total converted mana cost.  If you think that measure is flawed take it up with the OP, but this is a 2 mana loop by their definition.

@doppelgreener I definitely should, but it was late and I was typing on my phone. One I get to my computer I'll fix it up

@ArcanistLupus Right, I didn't consider this - I changed the question to "total mana cost to initiate", I hope you don't mind

I reverted that. It is not OK on these sites to make edits to the question after answers have been posted that invalidate any of those answers.

I think this hits a slight variation of the issue with using two Forks. When the copy of Vantress Visions is created, the Mirrorpool ability that created it is still on the stack because it's still resolving, so you have the option of terminating the loop by choosing the second Mirrorpool as the Vantress Visions copy's target to make it fizzle.

@murgatroid99 I don't think the question is very meaningful without the edit, but if you insist I will ask it again with the intended meaning (total mana cost to initiate)

I think the answer I posted on this question is also the answer to that question. So I would say that you already have the answer you are looking for, without making that edit or posting a new question.

I'm actually wondering, because the wording on Vantress Vision says "you may choose new targets", wouldn't that mean that you can choose not to change the targets even if the original target isn't there yet, so it makes a copy that also fizzles?


text:
If you have 2 "s" in your rack, do you score statistically higher playing the extra "s" immediately or saving it?
Although it doesn't happen often, when I have 2 "s" in my rack, I'm not sure if I should try to use one immediately to score an additional 2-7 points, or save both s's for bingos/parallel plays.
Like any duplicate letter, I seem to have trouble making high point moves when have duplicate letters in my rack including 2 "s". I'm leaning towards using 1 s immediately if appropriate like if it would help balance the consonant/value ratio on your rack more than if you didn't use it.

The answer is probably "it depends". A useful tool to evaluate specific racks is the Cross-Tables leave evaluator: https://www.cross-tables.com/leaves.php

As noted in the comments, it very much depends on the board. In general, S is useful for tacking on various plays your opponent makes - especially punishing them for thinking that stopping a word short of the TW row is safe. Two S tiles lets you possibly tack two words that end in the same row or column, and plenty of words (in English anyway) have double-s.
Given the general utility of pluralizing a word already on the board to create space for yourself - I would personally look to use each S in its own move, but only after confirming that I don't have any options to pluralize two words in one play - scoring three decently long words is pretty strong.

I disagree with the advise @William Walker III. In my opinion, it is best to save both S's for later in the game when you can get a bingo. Here's why:

An S can be tacked onto a majority of words, giving you a point where you can build from.
If you use the 's' as the end of the word, you only need to have a six-letter word that can be pluralized. (Example: Streak, which can turn to Streaks)
Most importantly, if it is a close game, the extra 50 point bonus would be a lot more than almost any other move with a single 's', which will usually give south of 50 points.
If something does come up, with an s or multiple s's, you will still have them instead of wasting them getting "2-7" additional points. (Which, in advanced level play, is an extremely small margin.)


text:
If I resurrect a previously enchanted creature, does it get its old auras back?
If my enchanted creature dies, and I then bring the creature back to the battlefield does said creature come back with the same enchantment on it?

ConMan's answer is very good, but it goes explaining auras, their effects and characteristics but I think this rule is more important to explain the answer to your question.

217.1c An object that moves from one zone to another is treated as a new object. Effects connected with its previous location will no
  longer affect it.

Answer to your question: No, it will not have any auras or equipment attached to it.
This is because every time an object moves from one zone to another, this is a new object. Whether this new zone is graveyard, battlefield, exile, library or hand.
A good and relevant example of this is a card by the name of Animate Dead with the following text:

Enchant creature card in a graveyard When Animate Dead enters the
  battlefield, if it's on the battlefield, it loses "enchant creature
  card in a graveyard" and gains "enchant creature put onto the
  battlefield with Animate Dead." Return enchanted creature card to the
  battlefield under your control and attach Animate Dead to it. When
  Animate Dead leaves the battlefield, that creature's controller
  sacrifices it.

Very convoluted text for a simple, return creature to the battlefield isn't it? The gatherer explains why:

Animate Dead is an Aura, albeit with an unusual enchant ability. You
  target a creature card in a graveyard when you cast it. It enters the
  battlefield attached to that card. Then it returns that card to the
  battlefield, and attaches itself to the card again (since the card is
  a new object on the battlefield). Animate Dead itself never moves into
  a graveyard during this process.

As Animate Dead attaches to the creature and enters the battlefield, the creature is still theoretically in the graveyard, the effect from Animate Dead then brings the creature to the battlefield which means that as it is now a new object, Animate Dead has to attach itself to it again.

From the Comprehensive Rules (note that an Aura is an enchantment that attaches to another permanent):

303.4c If an Aura is enchanting an illegal object or player as defined by its enchant ability and other applicable effects, the object it was attached to no longer exists, or the player it was attached to has left the game, the Aura is put into its owner’s graveyard. (This is a state-based action. See rule 704.)

So when the creature leaves play and is no longer a valid target for the Aura, the Aura is put into its owners graveyard. If the creature is later returned to the battlefield, the Aura has already forgotten that it was attached to the creature and does not come back.
The one exception to this would be if the creature regenerates:

701.14a If the effect of a resolving spell or ability regenerates a permanent, it creates a replacement effect that protects the permanent the next time it would be destroyed this turn. In this case, “Regenerate [permanent]” means “The next time [permanent] would be destroyed this turn, instead remove all damage marked on it and tap it. If it’s an attacking or blocking creature, remove it from combat.”

Note that regeneration causes an effect that replaces the destruction of the creature, so it never actually goes to the graveyard, and hence the Aura stays attached to it.
EDIT: As pointed out, Totem Armor is another related situation:

702.88a Totem armor is a static ability that appears on some Auras. “Totem armor” means “If enchanted permanent would be destroyed, instead remove all damage marked on it and destroy this Aura.”

In this case, if the creature is enchanted with an Aura that has this ability, then the Aura gets destroyed instead of the creature. So here the creature does not get destroyed, but the Aura does.

No, the creature will not be enchanted.
There are a variety of reasons for this, but the most basic reason is that the a resurrected card doesn't know it was ever previously enchanted.  And cards only do as much as they say, so the resurrection spell won't enchant the creature unless it says otherwise.
According to the rules (my notes)

400.7. An object that moves from one zone to another becomes a new object with no memory of, or relation to, its previous existence. There are seven exceptions to this rule (none of the exceptions apply)


text:
Can effects (like Majesty's Fiend) prevent Exodia from winning?
Majesty's Fiend is on the field.  It says

Cannot be Special Summoned. Monster effects cannot be activated.

If all Exodia cards are collected, does Exodia's player win even though Majesty's Fiend prevents Monster effects from happening?

Possible duplicate of How does the new Sangan errata affect Exodia?

I think the duplicate target is too narrow since the question is very focused on Sangan.  I think reversing the duplicate target is better since this question (now edited) is more general.

If all Exodia cards collected, does Exodia's player win?

Yes, the player will win.
The reason is because Exodia has a win condition not an effect. From its rulings, emphasis mine:

When you add the fifth piece to your hand and declare that you win the Duel, you are using a victory condition. You are not activating a monster's effect, so "Divine Wrath" and "Gladiator Beast War Chariot" cannot be activated

This means that Majesty's Fiend, who negates effects, won't be able to prevent the player from wining. From its text, emphasis mine:

Cannot be Special Summoned. Monster effects cannot be activated.

I've been playing YuGiOh for years and this is the first time I know about "victory condition"!

There are a few cards with such conditions. See this answer of mine for a list of those (I even happen to mention Exodia at the end). Welcome to Board & Card games BTW :D


text:
Infect vs first strike
Let's say a creature with infect attacks and the opponent blocks with a creature that has first strike. How does the combat resolve? Does the blocking creature with first strike still get the -1/-1 counter?

The two relevant rules for this are 702.7b and 702.89c; 702.7b does not stop 702.89c from occurring unless the creature with infect dies before it can deal damage itself.
Essentially if the infect creature either has first strike itself, or survives the first strike damage step, an appropriate number of -1/-1 counters will be added to the creature with first strike.

702.7b If at least one attacking or blocking creature has first strike or double strike (see rule 702.4) as the combat damage step begins, the only creatures that assign combat damage in that step are those with first strike or double strike. After that step, instead of proceeding to the end of combat step, the phase gets a second combat damage step. The only creatures that assign combat damage in that step are the remaining attackers and blockers that had neither first strike nor double strike as the first combat damage step began, as well as the remaining attackers and blockers that currently have double strike. After that step, the phase proceeds to the end of combat step

702.89c Damage dealt to a creature by a source with infect isn’t marked on that creature. Rather, it causes that source’s controller to put that many -1/-1 counters on that creature.

If a first strike creature kills the creature it is blocking/blocked by, and the other creature does not have first strike and/or double strike, the other creature does not deal combat damage.  Abilities that modify how a creature deals combat damage (such as infect) don't change this fact.


text:
Is there a penalty for constantly calling a revoke to disrupt the play?
A player in duplicate bridge constantly calls a revoke to disrupt the play. Is there a penalty to this player's side for this discourtesy?

Do you mean this player doesn't follow suit when possible, or that they accuse others of doing so?

There is no reason for this question to be closed as Needing Clarification. My answer properly addresses, I believe, all issues; and if another contributor was able to add a better or supplementary answer so much the better for the site.

@ForgetIwaseverhere It's not clear what the OP is referring to - e.g., the question by steenbergh.  Is the player repeatedly revoking, or is the player repeatedly claiming someone else revoked, or are they doing something else?

@Joe: What part of "... calls a revoke to disrupt the play." is unclear on being the latter? If they were repeatedly revoking themselves that is either readily corrected before establishment, without disrupting the play, or is readily penalized under the normal rules for penalizing an Established Revoke.

This looks like a pretty objective question to me (and answer well by FIWEH). Voting to reopen.

Absolutely! This type of disruptive behaviour is in no way acceptable by any player.
The Laws of Duplicate Bridge (2017) provide mechanisms under which the Director can act in such a circumstance of disruptive play.

LAW 90 - PROCEDURAL PENALTIES
A. Director’s Authority
The Director, in addition to implementing the rectifications in these Laws, may also assess procedural penalties for any offense that unduly delays or obstructs the game, inconveniences other contestants, violates correct procedure, or requires the award of an adjusted score.
...
Law 91 - PENALIZE OR SUSPEND
A. Director’s Powers
In performing his duty to maintain order and discipline, the Director is empowered to assess disciplinary penalties in points or to suspend a player or contestant for the current session or any part thereof. The Director’s decision under this clause is final (see Law 93B3).
B. Right to Disqualify
The Director is empowered to disqualify a player or contestant for cause, subject to approval by the Tournament Organizer.

Where I have seen this procedure used (Matchpoints only), the escalation sequence was as follows:

First offense
A specific warning on the conduct, specifying the penalty (see below) to be applied on a subsequent offense.
Second Offense (of the same kind)
A penalty of 1/4 of a board assessed to the player (not the partnership). A warning as above given that a third offense will result in suspension from the event.
Third Offense (of the same kind)
Player suspended from the event for cause. The Director also files an Ethics Report with the ACBL in regards this player.

However, I suspect that the player involved is also not adhering to the proper procedure for Inquiring About (Law 61), Correcting (Law 62), Establishing (Law 63), and Making Adjustment For (Law 64) a failure to follow suit. More details on the precise circumstances, and timing and sequence of events, would allow a more precise determination of correct and appropriate action.
Note that under Law 61, any inquiries concerning a revoke are limited as follows:

B. Right to Inquire about a Possible Revoke

Declarer may ask a defender who has failed to follow suit whether he has a card of the suit led.
  69
(a) Dummy may ask declarer [but see Law 43B2(b)].
(b) Dummy may not ask a defender and Law 16B may apply.
Defenders may ask declarer and one another (at the risk of creating unauthorized information).

Any violation of Law 61 A will result in an immediate penalty as outlined in that law and all referenced laws concerning Unauthorized Information. Any such behaviour, when repeated, also remains subject to Procedural Penalties under Laws 90 and 91.
Note that if a Defender explicitly points out an Established Revoke prior to the conclusion of play, whether by his Partner, Declarer, or Dummy, and whether correct or not, this almost certainly constitutes Unauthorized Information to his Partner. The appropriate time and means to point out an Established Revoke is at the conclusion of Play, by calling the Director before the score is agreed or the hands shuffled and returned to the Board.
Also, competed tricks can only be inspected as described by Law 66C:

LAW 66 - INSPECTION OF TRICKS
A. Current Trick
So long as his side has not led or played to the next trick, declarer or either defender may, until he has turned his own card face down on the table, require that all cards just played to the trick be faced.
B. Own Last Card
Until his side has led or played to the next trick, declarer or either defender may inspect, but not expose, his own last card played.
C. Quitted Tricks
Thereafter, until play ceases, the cards of quitted tricks may not be inspected (except at the Director’s specific instruction; for example, if necessary to verify a claim of a revoke).
D. After the Conclusion of Play
After play ceases, the played and unplayed cards may be inspected to settle a claim of a revoke, or of the number of tricks won or lost; but no player should handle cards other than his own. If the Director can no longer ascertain the facts after such a claim has been made, and only one side has mixed its cards, the Director shall rule in favor of the other side.

Thus there is no value to claiming a revoke prior to conclusion of play, as quitted tricks cannot be inspected until after play completes.


text:
Name of solitaire game with layout like US flag?

The layout is as shown in the picture.  I call it the “flag version”of solitaire, but that is my own made up name.  My grandmother taught it to me.
Stacks are moved to make descending order stacks but in the same suit, not red on black / black on red.  For example, the four of diamonds may be moved to be on top of the five of diamonds.  The cards on top of four move with it.
The three leftover cards may be played at any time.  Kings can be moved to empty columns.  The cards in the upper left may be turned face up one at a time when revealed.
Winning condition is every suit is stacked alone in order, King to Ace.

You've described a version of Scorpion Solitaire, but with more cards face up than the usual version.

Set up
Scorpion Solitaire uses one deck (52 cards). You have 7 tableau piles. Three rows of 7 cards are dealt - the first 4 cards face down and the next 3 face up in each row. Then four more rows of 7 cards are dealt, all face up. The remaining 3 cards are put face down as a reserve.
The object of the game
To build four sequences down in suit within the tableau, each running from King to Ace.
The rules
You may build tableau piles down in suit. Groups of cards can be moved regardless of any sequence. Any face up card in a tableau can be moved to make a build. All the cards covering it are moved together as a unit. If, during play, any closed cards become the top card of a pile it turns face up.
Empty tableaus may be filled with a King or group of cards headed by a King.
When you have made all the moves initially available, click on the reserve pile to deal the 3 remaining cards face up to the three leftmost piles.


text:
Can hexproof remove an opponent's aura already enchanting my creature?
If a creature I control is being targeted by an enchant creature my opponent controls, can I equip it with an artifact that grants hexproof to counter the existing enchant creature?
Example: My opponent casts Kasmina's Transmutation on my Serra's Guardian. Several turns later I equip Swiftfoot Boots. Does the "hexproof" ability make Serra's Guardian an invalid target for Kasmina's Transmutation?

If a creature I control is being targeted by an enchant creature my opponent controls, can I equip it with an artifact that grants hexproof to counter the existing enchant creature?

Yes, but note that equipping an artifact goes 'at sorcery speed'; normally you can only do it in your own turn when the stack is empty. See the Comprehensive Rules:

702.6a Equip is an activated ability of Equipment cards. “Equip [cost]” means “[Cost]: Attach this permanent to target creature you control. Activate this ability only any time you could cast a sorcery.”

You'll need an equipment that specifically has an ability to attach it to a creature you control (e.g. Cranial Plating, but then an equipment that grants hexproof) or another instant that is able to attach equipments.

Does the "hexproof" ability make Serra's Guardian an invalid target for Kasmina's Transmutation?

Not for the Kasmina's Transmutation that's already enchanting it; an aura only targets when it is played:

115.1b Aura spells are always targeted. An Aura’s target is specified by its enchant keyword ability (see rule 702.5, “Enchant”). The target is chosen as the spell is cast; see rule 601.2c. An Aura permanent doesn’t target anything; only the spell is targeted. (An activated or triggered ability of an Aura permanent can also be targeted.)

(emphasis mine)

I find the first "Yes" and the later "Not for the Kasmina's Transmutation that's already enchanting it" contradictory (or confusing).  Maybe the stuff you quoted, I think I see the intent.


text:
How do Yarok, the Desecrated and Doubling Season interact?
If you control Doubling Season and Yarok, the Desecrated, how do the triggers work for +1/+1 counters from ETB effects and loyalty counters from planeswalkers? Counters in general?

Does this answer your question? Does Doubling Season affect planeswalkers?

Not a duplicate because the question is about the Doubling Season/Yarok interaction

Doubling Season and Yarok, the Desecrated have no direct interaction. Doubling Season's ability creates a replacement effect. It is not a triggered ability on which Yarok's ability could trigger.
As per this Q&A, Doubling Season affects permanents as they enter the battlefield with counters already placed. These static abilites create replacement effects, which are different from triggered abilities. Yarok only affects affects triggered abilities.

603.1. Triggered abilities have a trigger condition and an effect. They are written as “[When/Whenever/At] [trigger condition or event], [effect]. [Instructions (if any).]”

Replacement effects have a variety of wordings. The most common wordings that apply to permanents are:

614.1c Effects that read “[This permanent] enters the battlefield with . . . ,” “As [this permanent] enters the battlefield . . . ,” or “[This permanent] enters the battlefield as . . . “ are replacement effects.
614.1d Continuous effects that read “[This permanent] enters the battlefield . . .” or “[Objects] enter the battlefield . . .” are replacement effects.
614.1e Effects that read “As [this permanent] is turned face up . . . ,” are replacement effects.

These are affected by Doubling Season if they involve placing counters on a permanent, but do not trigger Yarok.
If counters are placed on an object due to a triggered ETB ability, Yarok will double that ability, and Doubling Season will double both effects, for example if you choose the +1/+1 counters on Knight of Autumn

Doubling Season does not directly interact with Yarok, becasue it creates a replacement effect, but does not trigger at anything. They do each apply independently of each other.
Example 1: Barkhide Troll is successfully cast. Doubling Season sees the creature trying to enter with one or more counters, so it doubles the counters as it enters. Yarok sees no triggered abilities and does not interact. Barkhide Troll enters with 2 counters.
Example 2: A Planeswalker is successfully cast. Doubling Season sees the Planeswalker trying to enter with one or more counters, so it doubles the counters as it enters. Yarok sees no triggered abilities and does not interact. The Planeswalker enters with twice its starting loyalty.
Example 3: Hada Freeblade is successfully cast. As it enters, it triggers it's own ability. Yarrok sees an ETB ability, and makes it trigger an additional time. As each of the triggers resolve, Doubling Seasons sees that one or more counters are being placed on a creature, and doubles the number of counters being placed on the creature. Hada Freeblade has 4 counters after all triggers resolve (2 triggers * 2 counters per trigger = 4 total counters).


text:
Freecell puzzle unsolvable?
I think this freecell puzzle (739671 on Odesys) has no solution but not 100% sure.
Using this solver could not solve it. Is there a better solver, or does this really have no solution?
This is the starting layout:
8C JD TS 6S 9S 2C 7H
8H 4D TC 8D 9H AS JS
QC AH 9D 5S 5D 3D 2H
QS 8S 6H JH 3C AD KS
4C 6C 7S 6D KC JC
4H 4S 2S KD TD KH
TH 7C 3S 2D AC 7D
QD 3H QH 9C 5H 5C

I just tried it on that website with the default solving method and it solved the puzzle.

It solved for me also using all the available solution methods.

Ouch! Right, no idea how THAT puzzle got in there. Wrong one. See edit for the right puzzle.

Why don't you just remove the wrong puzzle entirely? The revision history records all earlier versions of the post already.

@Nij I took an edit :) you could have done the same though...

That is the top result I get when searching for freecell solvers on google and appears several times in the results so guessing you are not going to find anything better.

And how would that help a new user to practice doing things properly? @DarkCygnus

@Nij there is some method to the madness: I didn't want the comments by Mosquite and Joe W to appear erroneous (unless one looked at the edit history). Anyway, that one seems unsolvable. Anybody knows of proofs of unsolvability except for evidently unsolvable freecell puzzles.

@Nij learning by example :)

That's what the revision history and "no longer needed" flags are for. Once improvements are made, the comments should be removed and a flag by the question asker helps make that happen. @GilbertWeinstein

Related, possible duplicate (I'd say it's a duplicate except that the answer doesn't address specific games past the original 32000: https://boardgames.stackexchange.com/questions/15287/is-there-any-configuration-of-free-cell-that-cannot-be-solved


text:
Do I have to pick destination cards?
I just claimed my final route and now have 2 cars left, so everyone gets one last turn, including me. Must I pick new destination cards on my final turn and then lose those points in the final tally?
Are you required to pick new destination cards when you have fulfilled all your routes, no matter what turn you are on?

You are never required to obtain additional destination tickets cards, even if you've completed the ones you have.
You have three options available to you:

You could claim a route. Remember, routes are worth points, and nothing prevents someone from claiming routes once their destination tickets have been completed.
You could pick new destination ticket cards. You might get lucky and get one or more that's already completed! If you don't think you're winning, this can be very effective.
If you're unable or unwilling to take the above actions, you can effectively "pass" by drawing more transportation cards.

While I used the terminology of New York in this answer, this answer applies to all of the following:

Ticket to Ride
Ticket to Ride: Asia
Ticket to Ride: Europe (You also have the option to build a train station.)
Ticket to Ride: Germany
Ticket to Ride: India
Ticket to Ride: New York
Ticket to Ride: Nordic Countries
Ticket to Ride: Pennsylvania
Ticket to Ride: Switzerland
Ticket to Ride: United Kingdom (You still have the option of buying a technology card before your normal action.)

I have no experience with the other Ticket to Ride titles.

This is great info. We wanted to start out easy with Ticket To Ride New York. The instructions are brief. Maybe it is ore clear with Ticket To Ride full game. Thank you.

One thing I didn't mention is that one usually only needs to keep one of the new tickets when getting more tickets. So as long as one of tickets is already completed, you're golden!

No, you do not have to claim new destination cards. There are other options available, such as claiming routes to block others or pick up the points.
Deliberately aiming to complete the minimum destinations early and claim routes arbitrarily to block the opponents can even be a viable strategy depending on what you first pick up at the start of the game.

Unless New York changes this from all other TTR games, you can also just pick up more train cards in your last turn, even though it will not help at all. It could still be your best option.

Thank you GendoIkari. That's what I wanted to do, but didn't know if you were allowed to do that since there was no point as you had no more destinations to claim.

Starting with the youngest player and proceeding in clockwise order,
  players take turns until the game ends. On your turn, you must do one
  (and only one) of the following three actions: draw Transportation
  cards, claim a Route, or draw Destination Ticket cards.

https://ncdn0.daysofwonder.com/tickettoride/en/img/tt_ny_en_rules_2018.pdf
There's nothing there about having to claim new Destination Tickets. Destination Tickets are purely a matter of end-of-game scoring; they have no direct effect on the game state during play. You don't reveal what Destination Tickets you have until the end of the game, so other players have no idea whether you've completed them or not (although skilled players will be able to make deductions from your play). So if there were a rule saying you have to draw new Destination Tickets once you've fulfilled your current ones, it would rely on players being honest (and observant; if you're focused on other things, you could not notice that you've completed all your Destination Tickets, especially if it's a circuitous route). Moreover, it would punish players for finishing their Destination Tickets, and finishing Destination Tickets is a major part of the game. Game makers tend to try to avoid rules that punish players for achieving the goal of the game, reward dishonest players, or result in players that are not paying attention looking like they deliberately broke the rules.


text:
Identify this peg game?

I'm looking for help with identifying this game. The pegs are lost but can be replaced. I can't find the directions or anything on the internet in shopping or otherwise.
The board sort of looks like tic-tac-toe with lines between connected holes, but it also has the seven peg holes on each side. There is also a reservoir cup for loose pegs or maybe dice.

If the numbered holes on the side were not there, it would be a board for tapatan, aka 3 men's morris.

It can be a tic-tac-toe variant in which you can (re)move pieces. With 7 pieces on the board you always have a winner. You can place 6 without 3 in a row.

how many pegs does it come with for each side?  perhaps the 1-7 track is just for keeping score rather than part of the game?

I'd guess it to be a tic-tac-toe variant in which you can (re)move pieces. Players play normally with seven pegs, but three in a row or some other pattern allow you to capture an opponent's piece which goes into the hole.  Player with most pegs on board wins.
Couldn't find anything exactly like this at BGG.
https://www.boardgamegeek.com/geeklist/16886/games-played-3x3-grid

Or a varient of https://shop.crackerbarrel.com/toys-games/games/travel-games/peg-game/606154 ?

@joedragons - No, that is a single player puzzle. You start with one hole, then use one peg to jump another peg. The jumped peg is removed. The idea is to end up with only one peg left on board. (Easy to solve if you start with reverse problem. Start with one peg, then jump an empty hole to add a peg back to the board.)


text:
When there are multiple "enters the battlefield" effects but "lasts until end of turn" is said once, do they all last until end of turn?
For example, Cavalry Drillmaster has two enters the battlefield effects separated with an and:

When Cavalry Drillmaster enters the battlefield, target creature gets +2/+0 and gains first strike until end of turn.

Do both effects last until the end of the turn or just first strike? Where is this defined in the rules?

They all last until the end of the turn, not just first strike. Since it's hard to keep track of these abilities for a longer period, permanent boosts (or curses) almost always use 'counters', like a +1/+1 counter, which stay on the card as a visible representation. (There are a handful of exceptions like Figure of Destiny.)
So if the +2/+0 boost would be permanent, it would have read something like this:

When Cavalry Drillmaster enters the battlefield, choose target creature. Put a +2/+0 on that creature. That creature gains first strike until end of turn.

Note that the effect has now been split in two sentences, to indicate the difference in duration.


text:
Perplexing Chimera and control changing enchantments
Perplexing Chimera can steal spells but gives control of itself to that player.  But if I enchanted with an enchantment that says "You control enchanted creature" (such as Hypnotic Siren) would that enchantment let me retain control of the Chimera after stealing spells and basically have infinite uses of the chimera's spell stealing?

A better plan would be to use Homeward Path, as long as you're not stealing other player's creatures. As a bonus, it's a counter against other players stealing your creatures.

Related: https://boardgames.stackexchange.com/questions/44686/how-does-homeward-path-work-with-cards-that-enchant-to-steal/44689#44689

No, this is not how it works.  This is perhaps unintuitive since Perplexing Chimera's ability appears to be a "one-shot" effect, but by rule all control-changing effects are continuous effects, and these effects use a timestamp system to determine the final outcome (with the most recent effect winning out).  The control-changing effect from the aura can be overridden by Perplexing Chimera's own ability that also changes control.
See also Djinn of Infinite Deceits interacting with other control-changing effects which describes some other other unintuitive corner cases with control-changing effects.
Relevant rules (emphasis mine):

611.1. A continuous effect modifies characteristics of objects, modifies control of objects, or affects players or the rules of the game, for a fixed or indefinite period.
613.1. The values of an object’s characteristics are determined by starting with the actual object. For a
  card, that means the values of the characteristics printed on that card. For a token or a copy of a spell or card, that means the values of the characteristics defined by the effect that created it. Then all applicable continuous effects are applied in a series of layers in the following order:

613.1b Layer 2: Control-changing effects are applied.

613.6. Within a layer or sublayer, determining which order effects are applied in is usually done using a timestamp system. An effect with an earlier timestamp is applied before an effect with a later timestamp.
613.8. One continuous effect can override another. Sometimes the results of one effect determine whether another effect applies or what another effect does.

Example: Two effects are affecting the same creature: one from an Aura that says
  “Enchanted creature has flying” and one from an Aura that says “Enchanted creature loses
  flying.” Neither of these depends on the other, since nothing changes what they affect or
  what they’re doing to it. Applying them in timestamp order means the one that was
  generated last “wins.” The same process would be followed, and the same result reached, if either of the effects had a duration (such as “Target creature loses flying until end of turn”) or came from a non-Aura source (such as “All creatures lose flying”).

As an aside, using an aura (like Hypnotic Siren's Bestow) to regain control of perplexing chimera you gave away is probably a bad idea.  You'd likely regain control of the chimera, but only because your opponent would steal your aura spell along with another of your permanents.  And pre-enchanting a chimera you already control is a waste since of resources since the timestamp system will override the aura.

As the other answers, you can't do it due to the fact that it acts on layer 2, as a continuous effect and timestamps are taking into account (last effect prevails)

613.1b Layer 2: Control-changing effects are applied.
613.6. Within a layer or sublayer, determining which order effects are applied in is usually done using a timestamp system. An effect with an
  earlier timestamp is applied before an effect with a later timestamp.
613.8. One continuous effect can override another. Sometimes the results of one effect determine whether another effect applies or what
  another effect does.
Example: Two effects are affecting the same creature: one from an Aura that says “Enchanted creature has flying” and one from an Aura
  that says “Enchanted creature loses flying.” Neither of these depends
  on the other, since nothing changes what they affect or what they’re
  doing to it. Applying them in timestamp order means the one that was
  generated last “wins.” The same process would be followed, and the
  same result reached, if either of the effects had a duration (such as
  “Target creature loses flying until end of turn”) or came from a
  non-Aura source (such as “All creatures lose flying”).

Layers and timestamps are something to learn to understand this.
What you can do is use things allow you to regain control of it, for example you can use eldrazi displacer to flicker it back to your control and then you can use it again. There are also creatures  like Trostani, that allow you to regain control of your creature.
Regarding your example, the enchantment if played by your opponent can be neutralised with Perplexing Chimera as you change control of it for the control spell, when it resolves you target chimera and regain control immediately.

That's not how this works. Control changing effects like that work in order of timestamp, the order they were put onto the creature. Because Perplexing Chimera's ability to trade control is put on after the bestow of Hypnotic Siren, it will be the last one active and your opponent will be given control of the Chimera, still enchanted with your Siren.

Thanks for clearing that up for me, I'm glad I don't have to worry about a combo lile this but kinda sad I can't use it


text:
How do Jokers change the bidding in Spades?
In the game of Spades, there are several popular jokers variations which add the two jokers, named big joker and small joker (BJ, SJ) to the ♠ suit while removing the 2♡,2♦.
In the simple jokers variation:

♠ suit contains 15 cards, sorted from high to low: BJ, SJ, A, K, ..., 2.
♥,♦ contain only 12 cards: A, K, ..., 3
♣ have 13 cards as normal A, K, ..., 2

What are the major differences in the bidding of the simple Jokers variation, compared to the normal Spades game?

This is a difficult question, but by no means "too broad"? An earlier question on Spades I was able to answer based on my Contract Bridge experience, but I have never played Spades itself. However the specific strategic consequences of there being two more trumps and two 12-card suits is by no means unanalyzable.

The main complexity is that instead of 4 suits of 13 cards each, the deck now comprises one 15-card suit, one 13-card suit, and two 12-card suits. Thus the relative value of cards, in the three side suits particularly, is subtly different. The biggest direct play difference probably hinges on the change in the increased  a priori probability of a red-suit singleton or void compared to the baseline probability for a 13-card suit (now just Clubs).

I haven't rolled the mathematics yet, but in particular I suspect that these two probabilities jump significantly when one holds 4 or 5 cards in a red suit compared to the same holding in Clubs

right, in order to evaluate the value of high cards I run Monte Carlo simulations.  needed just small tweaks to get good evaluation for the 12 and 15 suits.

what about changes in the play? now it is more likely to get a hand with e.g. 8 spades.

Such extremes are irrelevant - When one picks up one's cards it is already apparent whether one is holding an 8-card Spades suit or not, and bidding is subsequent to that revelation. Of greater consequence would be the probability of partner or an opponent having a 5+ card Spade suit given one's own (known) length holding.

P.S. Note my Meta question on the ethics of questions being closed by contributors with zero experience in the game in question, or even in any related game.

I'm curious about your Monte Carlo simulations. (1) Are you familiar with the 4-3-2-1 Work Point Count used in Bridge? (2) What do your simulations show is the relative weight for honours in red suits compared to Clubs?

I am familiar with the 4-3-2-1 Work Point Count used in Bridge, however I don't use it.  Instead I estimate the strength of the Ace, King, Queen by assuming they only lose if someone trumps that suit.  So Ace gets the trick if no opponent is void, King gets the trick if no opponent is void or singleton and queens gets it if no opponent is void / singleton or doubleton.

We call that Losing Trick Count in Bridge. It works well for highly distributional deals with good fits, quite poorly for balanced deals and mis-fits. Experience in Bridge suggests that it is more accurate as a fine-tuning of Work Point Count than as a stand-alone evaluation method. Of course Bridge offers multiple rounds of bidding, which is a big deal.

Have you seen this Wikipedia article on Hand Evaluation in Contract Bridge? You might consider a Quick Trick evaluation in preference to Losing Trick Count.