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How can I effectively use the Coppersmith?
I've been playing Dominion with the base + Intrigue sets for a while now, and one of the cards that I have trouble wrapping my mind around is the Coppersmith.
In theory, it seems like a great card - it allows you to make all of your Coppers into Silvers, and in some cases - even Gold. However, I have yet to build a deck during a game that manages to use it effectively. Usually, in order to pull one or two of those with a nice amount of coppers, I need to create a deck that relies heavily on +cards/+actions chains, and the Coppersmith only begins to affect the game too late for it to be of any use.
So... how can this card be effectively used?

Coppersmith is a terminal action, therefore you have to design your deck to accomodate it.  (By terminal action, I mean it is an action that does not give +Actions).  As others have noted, Coppersmith is a card used to give you an early start.  By managing your deck, you can extend that advantage well into the mid game.
Here is a chart showing how 1 coppersmith can accelerate your early game.  I've compared it to Smithy, and "Big Money"  (A base strategy of buying only silver/gold/provinces which is actually fairly strong)

                    Chance of Having (once during the two turns)
Turn 1-2 purchases   $6      $7     $8       Notes
---------------------------------------------------------------------------------
Coppersmith+Treasure 50.91%   0.00%  21.21% # 1 in 792 that 2 golds can be bought
Smithy+Silver        42.60%  19.39%   2.12% #Not possible with a 5/2 copper split
2 Silver             33.59%   8.84%   0.00% #Not possible with a 5/2 copper split

This shows a few things

you have a vastly higher chance of affording a province.
The coppersmith gives you a 70% higher chance of being able to afford a gold than "Big Money"
The coppersmith works with a 5/2 copper split on the first 2 turns, while the other two do not.
Coppersmith will not give you possibly annoying $5 and $7 hands.

Is buying a province on turn 3 a good move?  Maybe not, but it could be.  My point is that if you buy coppersmith, you stand a higher chance of getting an early gold.  This advantage will carry over to future turns.  Coppersmith will help you hit the financial high points of $6 and $8 more often, while avoiding $5 and especially $7.
Every set of 10 cards in Dominion is different.  Coppersmith won't always be a great card, but it can be used to great effect.  Try playing it with wishing well!

In addition to the odds laid out by Pat Ludwig, there are some cards that will make Coppersmith more or less useful.
Cards that could make Coppersmith more useful.  Actions are terminal unless otherwise noted:

Adventurer (B) - Flip cards from the top of your deck until you've revealed two Treasure cards.  Put the Treasure cards in your hand, discard all other cards.  Discard all other flipped cards.  Only useful if you have two or more actions.
Apothecary (A) - Draw a card, then flip the next 4 cards.  Put any Copper or Potion cards you flipped in your hand, discard all other cards you flipped.  Also has +1 action.
Counting House (P) - Take any number of Coppers from your discard pile and put them in your hand.  Only useful if you have two or more actions.
Golem (A) - Flip cards until you get two action cards, then play them in either order.  Useful if you manage to get Coppersmith with one of the other actions mentioned here.
King's Court (P) - Increase the value of each Copper by 3 when used with Coppersmith.
Throne Room (I) - Increase the value of each Copper by 2 when used with Coppersmith.
Any cards with +Cards and +Actions, such as Village (B), Laboratory (B), Market (I), Bazaar (S), Caravan (S), Haven (S), Alchemist (A), City (P), Grand Market (P), etc... provided you have the Coppersmith in your hand already.

Cards that could make Coppersmith less useful:

Loan (P) [Treasure] - Flip cards from the top of your deck until you flip over a Treasure card, then trash or discard that card.  Discard all other flipped cards.
Pirate Ship (S) - Other players' attacks can cause treasure cards to be trashed.
Swindler (I) - Other players' attacks can cause next card to be trashed.  If it was Copper, you'll likely be given a Curse back.
Thief (B) - Other players' attacks can cause treasure cards to be trashed or stolen (attacker's choice).

Cards that could help or hurt:

Venture (P) [Treasure] - Flip cards from the top of your deck until you flip over a Treasure card, then immediately play it.  Discard all other flipped cards.

The power of this card is in the early game. Buy it on your first or second turn to get some extra buying power early on before you have a more Silver- and Gold-heavy deck.  This can get you a Gold as early as your third or fourth turn.
You'll probably want to try to trash it later, since it provides little benefit after mid-game or so.

Not mentioned thus far, but an important aspect of this card:  it stacks.  You need extra actions, of course - City, Laboratory, Worker's Village, etc. - but playing two Coppersmiths in a turn can be very strong.

I think that you should basically ignore the coppersmith unless there is a really good way to draw a lot of coppers and still have actions left.  It only pays off if you have more than 2 coppers in hand.  In practice, this means I will only consider it if there is a tactician or apothecary.  In my opinion, coppersmith is the worst 4 cost card in the game.
Here's why:  It's only sometimes better than silver early, and late game, it's much worse.
Assume a standard 5 card hand, with a deck of mostly coppers, some estates, and a coppersmith:  If you buy a coppersmith instead of a silver early game, it will give you likely anywhere from +1 to +4 coins.  If it gives you +4 coins (meaning you have $8), it's probably not better than +3 or, since an early gold is almost always better than an early province.  If it had been a silver, you'd still hit $6, and you can still buy that valuable $6.  If it gives you +3, that's pretty nice, you can get an early gold or other useful 6 cost card, and no other card could fill it's particular role.  If it gives you +2, you were better off with buying a silver, since the silver does just as well there, and is much better late game.

Seems like what they call in Magic design a "skill-testing" card to me.  For the price of this card, you could have bought a Silver instead.  A Silver will be better than a Coppersmith in your hand unless you have, what, 3 Coppers?  Crucially, a Silver won't cost you an action.
Sure, there are probably some scenarios people can come up with where you draw a ton of cards including Coppers, and extra actions, and Coppersmiths, and you generate a fortune.  However!  If you can assemble that kind of drawing power, you could probably have assembled a different, better kind of deck instead.  Simply put, the Coppersmith is going to pay off once in a blue moon, when you draw Coppersmith plus four Copper in the same hand and buy yourself a Province.  The rest of the time it seems more likely to sit around and clog up your deck.
Do yourself a favour and buy a Silver instead!

Coppersmith + Countinghouse + a card that gives extra actions (Village,Shanty, etc) is pretty ridiculous, especially if you also have something that gives extra buys.

I think the only circumstance where a Coppersmith will be likely to be better than a silver is if you're stuffing your deck with lots of copper, playing a lot of +Buy cards and using all the spare buys to get copper. A Gardens / Council Room kind of strategy might appreciate a Coppersmith or two. I've never seen anyone put lots of Laboratory together with +Buys for Gardens and copper, but if someone did that, then that could work with Coppersmith.

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Is drawing resources required?
Players draw resources, commodities (Cities and Knights), and fish tokens (Fishermen of Catan) during the course of various Catan games. It is almost always advantageous to draw your earned "goods" whenever possible, but is it required?
Forgetfulness
Players sometimes forget to draw the goods they are entitled to. If play passes to the next player before they remember, we often forbid the forgetful player from drawing what they forget. They should learn to pay attention, although I am very lenient with new players.
Strategy
I was recently in a situation where purposely not drawing all my goods would have been to my advantage had a certain number been rolled. A player may do this to avoid having more than 7 resources in hand when a win is otherwise clenched, or to avoid drawing the Old Boot in Fisherman of Catan (which requires the owner to have one additional victory point to win). Had someone reminded me to draw all my goods, I might have suspiciously declined.
What do the rules say (or imply) about this? Are players required to draw all their goods? Are forgetful players allowed to gather previously forgotten goods?

You must draw all of your goods.
From the FAQ:

Resource Cards - Can I voluntarily take less resources or none at all after the production roll?
Answer
No.

I don't believe there is any official guidance on what to do if draws are forgotten about.  If it is noticed fairly quickly, the draws should be made up.  But if you've gotten to a new round of trading, or a 7 has been thrown, it is probably too late.
You can consult this past question that gives general advice on Take Back Privileges that might be helpful in resolving the situation.
The answer stays the same for Cities and Knights

It's really important that everyone draw all their goods.  One of the core elements of Catan in my mind is the need to keep your hand size to 7 or below as much as possible, to ward off the attentions of the robber.  This "problem" is in fact hugely beneficial to the flow of the game, because it encourages people to trade with their opponents, when they might otherwise just play solitaire.
Also, if I have a plan to steal your resources with a Monopoly or the like, you'd better keep on drawing them, or I'll be angry!
I would recommend employing the services of a "banker" who is responsible for giving out resource cards to every player after every die roll.  Letting people draw (or forget to draw) their own resources is a recipe for disaster!

Great suggestion!

There are some games (chess comes to mind) where every turn is a "must move" situation (if legal). "The right to make the next move" is usually an advantage. But not always! Sometimes it can be a distinct liability, and you have to accept that liability as part of the game. This situation chess is called a "Zugzwang," and part of the game is maneuvering your opponent into one.
Other games such as Go, are more "constructive." Having the move is always helpful (or neutral) meaning that it's "your privilege" NOT to move.
Because having too many resources is actually bad if the robber is moved, Catan seems to fall into the first category--you must move because it's possible to be in a situation where not moving may be beneficial. (But the "newbies" who fail to draw hurt themselves more often than not.)

Why the downvote?

@greg: If you like the answer, you can "upvote" it and basically cancel the downvote.

-1 because there's no rationale for why Catan falls into the first category. As far as I know forced draw is not in the actual rules (like taking a jump if you have one in checkers). Please add why you think Catan is a game of the first type, or why it should be.

@adam: Why Catan is s/b game of the first type? Because "bad" things happen to players occasionally, (like "blowing up,)" which they might be able to avoid if they were allowed to "pass." I call it a "chutes" and ladders phenomenon. All ladders, everyone should roll every turn. But if there are a few chutes, it might be advantageous to refrain from rolling, until the other guy hits a chute.

Well I buy that, but the site won't let me change my vote unless you edit your answer.

@adam: I added a sentence about "Zugzwang" as an edit. Thanks for engaging me the first time, and changing your vote when I gave you a satisfactory answer.

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How many cards should be collected using the Mining Progress card?
The Cities and Knights expansion to Settlers of Catan has two cards that have caused us a bit of disagreement.

Take the Mining Progress card for example.  It says "...collect two ore for each mountain hex adjacent to one of your cities or settlements."  If I only have one settlement and one city and they are both adjacent to the same ore hex, should I collect two cards or four?  If you are to count mountain hex then it should be two, but if you are counting cities/settlements it should be four.
The same questions exists for the Irrigation Progress card.

You only collect two ore.
The 4th edition cards(pg. 15) contain updated text which makes this clear.

You may take two ore cards from the bank for each mountain hex which is adjacent to at least one of your cities or settlements.

You can purchase an updated card deck.  I think this is all you need to update your set to the latest edition.

It would count twice (so four ore) in your example if the rules would have stated for each mountain hex adjacent to each of your cities or settlements, unfortunately it doesn't and you have to pick one of your cities or settlements to process the cards for, which one is up to you.

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Can you build the same building more than once in Puerto Rico?
Is one player allowed to build the same building more than once in Puerto Rico? If so, do the bonuses stack, so that a player building both Large Markets would get +4 for trading? I wasn't able to find anything in the rulebook that specified this either way. My play group ruled that you can't build the same building multiple times, on the grounds that some of the bonuses seem overpowered if you can double up on them (e.g. Harbor), and other bonuses make no sense if applied more than once (e.g. Construction Hut). Is there any official rule on this?

According to the rules for the online version on BrettspielWelt, "no player may build a building more than once."
And you can confirm this by looking at the official rules from Rio Grande, page 8:

For all buildings:
  - Each player may build each building only once.

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Can you Twisp off-turn?
My opponent attacks, and I have a Baby Ninja and a Twisp in hand.
I play the Baby Ninja, and she attacks again. Can I play Twisp to duplicate the Baby Ninja? It seems like you should be able to, but there is an explicit list in the rule book, that doesn't  cover PVP Defense.
Edit for clarity: This question relates to Penny Arcade: Gamers Vs Evil and NOT to the Penny Arcade Card Game.

No. You don't play PVP Defense cards, you use abilities on those cards. Twisp doesn't mention copying PvP Defense abilities. This is supported by the fact that play is only used in reference to a player's own turn. From the correct Rulebook.

PVP ATTACKS AND DEFENSES - Some cards allow you to make a PvP Attack against the other players in the game. When you play a card with a PvP Attack ability, each other player has an opportunity to block the attack with a card that has a PvP Defense ability. A player using a PvP Defense card’s ability negates the attack only for that defending player.
Twisp: Catsby’s special card Twisp simply copies all abilities of any other card played in a turn, including Tokens, Power, card drawing, and PvP Attacks. If the card would be deleted, then Twisp is deleted.
Baby Ninja - {2}, +1 Power, PVP Defense: Discard this card to block a PVP Attack. If you do, draw a card.
Jim Darkmagic: Jim’s special hero power must be used before you take any actions or play any cards for your turn.


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What Are The Official Rules for Carcassonne: The River?
I played a game of Carcassonne last night with some friends, using amongst other expansions, the river.
Their "house rule" was that both the river Source and Lake tiles do NOT wrap around: on each tile there was a corner that they designated a boundary between one pasture and the next.  I'm pretty sure that in my edition of the rules it specifies that these tiles DO wrap.  But I'm not a rules lawyer, I'm perfectly happy to play a variant, as long as it's clearly stated from the outset!
They had another rule for playing with the river, though, and in this case I realize I have no idea whether it's "official" or not: neither the Source or the Lake tiles counted as a "turn", so in both cases no meeple could be placed, and the player whose turn it was drew another tile.
Does anyone know for sure whether the Source and Lake tiles "count" for the purposes of taking a turn?  (I can't even remember whether the starting tile from the basic game "counts", right now...)

Lance is right on the starting tile -- according to the rules (emphasis mine):

Remove the special starting tile from the game.  Instead, begin the game by laying the spring that starts the river.  Set aside the lake tile and shuffle the remaining 10 river tiles face down and draw from these before drawing from the tiles from the normal game.  These tiles are played and followers may be played as in the normal game with one exception: a river tile may not be placed so that the river makes a "U" turn.  Followers may not be placed on rivers.  Once these 10 river tiles have been played, the next player plays the lake and then play continues with the normal tiles.

In regards to the Source and Lake, we've always played that they connect in all directions.  The following links seem to back this up:
According to the PDF rules from ModernJive:

The statement from Hans im Glück is "Officially, the farm goes around the spring. So it is a connected farm. I personally prefer to use it as a division, since otherwise, despite the river, there is still only one large farm."

And then from Carcassonne Central is this (although the confirmation link in the thread is no longer valid):

Yes, officially the farm does go around the spring.

It's worth noting that in the River II Expansion, which replaced the original River expansion, the One Giant Farm issue is addressed by having several river tiles that break up the farm. Also see this question.


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Can You Shoot Yourself in Bang?
Our group does like a bit of Bang!, but one of the problems we have with it is this: there's no reliable way to guess how long a game of it is going to take, which, combined with the fact that players can get knocked out very early on, means that sometimes it goes on and on long past the point where some people in the room or even still playing are fed up with it.
As such, can your gunfighter elect to do the decent thing and shoot themselves in the head to spare themselves the misery?  More importantly, if you shoot yourself from behind a Barrel, do you get to draw to see if your barrel protects you from damage?
(It's been a while since I last asked a question here, so I'm glad to be getting back into the fray with an issue as complex and important as this one.)

Why don't you just quit playing the game at that point and either start over or move on to a new game. Games are meant to be fun. If it ceases to be fun, then its not worth continuing. I have no problem with the question and in some cases might have strategic value, but for your specific reasons for the question, it just makes sense to me to just end the game.

@Charles: Presumably it's fun for some of the players at the table.  If it's fun for nobody, then yes, they should all quit.  If just you are having no fun, then maybe finding a funny, in-game way to leave is the best option, rather than just walking away?

I'll give it to you in that scenario, definitely.

Since we're talking about custom rules, I'd safely say that if you allow to shoot yourself, of course the barrel will not protect you! The barrel protects you because you are hiding behind it to protect you from other people firing at you, which clearly does not apply in this case.

@Lo'oris: Though you could be hiding inside it, in which case it would be hard to manipulate your gun effectively, no elbow room etc...

You can get yourself shot by playing a Duel and then not responding if the person you dueling with discards a BANG! Besides that and Dynamite, I don't know of other ways to hurt yourself with your own cards.
However, there are other ways to solve your problems:
(1) Problem of Player Elimination: Add the unofficial BANG! expansion, Death Mesa, to the game. In this expansion, dead players continue on as ghosts, draw from a specialized deck, and influence who wins (all ghosts, regardless of role, can still achieve victory). Visit http://bangcardgame.blogspot.com/p/death-mesa-expansion.html
(2) Problem of BANG! Dragging On: Give BANG! an internal clock through adding one of the following house rules:
-  Play only with 3 life point characters. If you have Dodge City, this gives you 7 characters; The Bullet, 8 characters; with Wild West Show, 9. Or make all 4 life point characters have 3 life points and only play with them.
- Players can shoot as many players as they want per turn (given distance constraints), but they can only fire 1 BANG! at each player per turn.
- If things are really getting slow, you can impose a new rule after a certain amount of rounds: Beer and/or Missed! are not reshuffled into the deck. They are placed in a separate discard pile and cannot be reused.
If you are interested in different house rules, visit http://bangcardgame.blogspot.com/p/fistful-of-bang-variants.html

These are wonderful and could really revolutionize our Bang! games.  Thank you very much for creating them :D

No,
From a message board post by the Designer of Bang!

As I said, the "no-killing-yourself" rule is the official one to prevent unfairy behaviours in tournaments.
However, if you sure to play with fair people, you can allow self-shot, which adds strategic elements to the game as well (think of Bart Cassidy, for example).

What a spoilsport!  But okay, if Bart Cassidy is hiding in a barrel in a game where self-shooting is allowed, is it harder for him to hit himself?  Inquiring minds need to know...

Also, I find the idea of a serious, high-level Bang! tournament quite an amusing one!

@the - I'd play in a Bang! tournament :)

@thesun I guess the question is really whether or not blue cards, like Barrel, are optional. Obviously Jail and Dynamite can't be optional, so something would have to differentiate the other cards. If I had to make a decision right now, I'd say they are not optional.

Could you not just use the High Noon Expansion which limits the game to around 13 to 15 ish rounds and has the ghost town card so players come back for a single round?


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Timing Rules in Seven Wonders
There may or may not be a clear definitive ruling on this in the rulebook, but it's something that keeps confusing me in the actual play, so I'll ask anyway.  Here are several questions that, when put together, should illustrate the points I'm having difficulty.

I play a card that gives me one coin for each brown card in front of me or either of my two neighbours.  I personally keep forgetting whether this counts cards that came into play this turn, but a quick check of the rules summary clarifies that yes, these cards are counted.
I mistakenly play a card that I (and my neighbours) have insufficient resources to build, when I look closely.  Obviously I cannot play the card - but I take it I can still discard it for 3 coins, or to build a section of my Wonder?
I mistakenly play a card that I (and my neighbours) are short of resources to build: let's say short by one stone.  However, it fortuitously happens that one of my neighbours has built a one- or two-stone quarry this turn.  Can I pay him the appropriate fee and build the card?  Or can I still only discard it for coins or to build a section of Wonder?  My first suspicion was that all building happens simultaneously, so I can't use materials that aren't available at this point - but if the card mentioned in my first example derives benefits from cards played this turn, why can't this?

We could get all MtG on this and break it down into discrete phases: the buy-resources-from-neighbours-phase, the build-with-resources phase, the reap-benefits-from-just-built-cards phase.  But that all seems a bit anal for Seven Wonders.  Is there an easy solution to the problem of looking left and right to see if your neighbours have the resources you need, and having to ask "errrrrr... so did you play that quarry this turn, or not"?  (Apart from not being so scatterbrained, obviously!)

For the 3. question, see: Can I buy resources that were built by my neighbors this turn?

The 2. question is similar to the situation when someone mistakenly builds a building twice.

1) Nothing to say for me here ;-)
2) I'm not sure about the actual rules, but I think the penalty of disallowing everything else would be too harsh, as the player loses a complete turn. One could argue if only the three coins should be allowed or building a part of the wonder as well. Personally, I would allow both.
3) Again, not 100% sure: as you cannot use coins that you got this turn, you shouldn't be able to use any resources that have become available in the same turn. My thinking is, everybody builds at the same time and hence completes their building at the same time, so you can't get anything out of another building to complete your own.

Answer #1 - Yes, any brown cards played by neighboring players count toward earning coins this way.
Answer #2 - This question is unanswered and not clarified by the official rules. You can only change the action you intended for the card you selected if you can't perform the action you intended for it. In your case, since you can't build the structure as you intended, you can choose between building the next stage of your Wonder or Discarding it for three coins. You cannot change your mind & apply a different action to that card by choice, reason being that a player could change his intended action in response to a card played by his neighboring players. This problem can be avoided by having each player verbally declare what action they intend for their chosen card before the cards are simultaneously turned face up, then turn them face up and apply those assigned actions accordingly.
Answer #3 - No, you cannot count your neighboring players newly acquired resource this turn toward the construction of your chosen card. This question is answered in the FAQ section on the official website of the game developers here, http://rprod.com/index.php?page=download-2

Bear with me here, I'm new so I'm not sure it's the best place but I just find your situation 3 a very interesting premise for a new way of playing 7 Wonders.
Personally, I've never encountered a situation where someone couldn't play a card that happens he could play it because of a new reveal card from a neighbor.
But, allowing this kind of dynamic you could had a great deal of bluff to the game especially with smaller groups (3 or 4 players). Since you know after 3-4 rounds what cards are in your neighbors hands you could anticipate the cards they might play and play something you couldn't. On the other hand, your neighbors would be in the same situation and could adjust their strategy to prevent you from playing your card.
Could be interesting to give it a try.


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Monopoly: Strategy in trading --how do I compute the expected value of a trade?
I love playing Monopoly with my friends and I'm known for giving extra money or trading a cheaper property with a more expensive one and even paying some amount on top of it.
Obvioulsy I love the smaller cards, as I can buy more houses in a short amount of time and earn more money in the long run. This strategy did work really well most of the time, but lately I'm paying too much for those properties.
I wonder if there is a quick method to figure out when a trade gives me positive expected value (+ev).

How do you know you're paying too much?  Could it be that your friends are on to your strategy? ;-)

in the last game my friend used that money to build houses on his now complete property, and I lost ;) I guess I did a bad deal that day

I don't think there is a quick method, but there are a few simple rules to consider when trading your cards.
1) How much is the card worth to you. Therefore, what can you do with the money you gain as opposed to what could you do with the card you are selling
2) How much is the card worth to your opponent. If you are selling a card that allows an opponent to complete a row, you have to understand how likely they are to build houses/hotels and the likelihood of it affecting your game plan further into the game.
3) If selling to an opponent who doesn't really need the card, consider the likelihood of them selling the card on, and the impact this will have, as per point 2.
For buying a card, the rules are pretty much the same as above, but inverse. You always need to weigh up the benefits you gain as opposed to what your opponent may gain, and think several moves ahead.
Also, take a look at this thread about the best ROI for properties. Which property group colour gives the best ROI , undeveloped or fully developed, in Monopoly?. This will help you to calculate the real values of the properties in the game.

I actually do have a couple of quick methods that I use when offered a trade, and the one I use depends on the stage of the game we're in.
Early game
The first few trips around the board. No one has a monopoly, most color groups have 1-3 properties available.

Do the properties contribute toward possible monopolies equally? (If she has an orange, I'm not trading her an orange unless I get a second property in another color in return. This applies to railroads as well: I'm not giving someone a second railroad for a property in a "new" color, for the most part.) Note: even if they do, make sure you keep an eye on the third property in each group. This can be especially important if additional opponents are prone to completing monopolies through trade, even if it hurts them to do so.
If that's not applicable, are the properties roughly of equal total value? (I'm not giving him Boardwalk for Indiana Avenue.)
If they're not roughly equal, can money be added to make them so? (A rough estimate is all you need; early in the game, it's hard to tell which properties are key.)

If it's "no" to all three, then I decline the trade. Otherwise, it might be worth it ... getting a chance at an early monopoly or a set of railroads can make a big difference early on. $200 isn't nearly as big of a deal when you have $4000 in assets as it is when you have $1900.
Mid to late game
A good number of properties have been purchased; one or more opponents may have monopolies, both utilities, or all four railroads.

Does everyone but me have a monopoly? Then yes, I make the trade. You simply can't build up enough cash without a monopoly in the late stages.  Conversely, if everyone but my opponent has a monopoly, then no, you don't get one.
If there are several possible monopolies left, and we're setting each other up, are the monopolies roughly equivalent? If not, then no trade. (Remember, light blue, orange, and dark blue are your targets. Don't give one unless you get one. Greens aren't really an asset unless you can afford to build them up immediately and still have an emergency fund. While they're a good initial value, Baltic/Mediterranean start to lose value once your opponents can pick up 2x to 8x that rent in one roll.)
If it's not going to be for mutual monopolies, am I helping out my opponent too much? Keep in mind that you have no control over your opponents' trades: picking up Boardwalk for New York Avenue seems like a decent late-game trade, but you may not get Park Place out of the person who has it, and if someone else passes the last orange to your trading partner, suddenly they have a new monopoly and you've got nothing.

How successful is that? It's hard to say, given the luck involved and the difference in opponents you'll face. Even with that handy ROI table next to you, you still can't be sure you'll win more games in the long run, and it's always possible that developing a reputation as a shrewd trader will hurt you within your group: people may actually trade with you less often, even if your trades help your opponents as well, just because you're good at it! But that philosophy seems to work for me.

Basically, I like to trade when I have a lot of money. I don't like to trade when I don't have money (and my opponents do). Unless you're trading for Baltic and Mediterranean, it's not a question of who has the better properties, but who can develop them fastest. (And I once lost a game landing on a hotel on Baltic, even though I had the maroon and green groups.)
For this reason, I don't like to give money in trade. Much more inclined to accept money as the "difference" in a trade of say, the dark blues for the maroons. But once I traded the greens for the maroons "straight up" and won, because I had over $1,000, and my main opponent (in a three way game) was almost broke.

That trade wasn't a bad one anyway. Although the maroons don't have a good ROI (with the exception of St. Charles), the greens are even worse: they're the most expensive to develop (because there are three as opposed to just Park Place/Boardwalk) and have no cards that force an opponent to land there.

@dave: One opponent had the dark purples (and four houses on each). The other had two out of three oranges (and would get later the third). I had the most money and was desperate to get ANY monopoly, because I had none. They (brother and sister) laughed at me. Until I won.

That actually reminds me of something that I consider when a trade offer is made ... thanks for jogging my memory!

It may not be a quick method. But for a full answer to property expected value, I recommend the following book:
Winning Monopoly
This book has includes the results of doing the math for the expected value of the different properites in key stages of the game.  The book is not just tables, but includes solid explainations and good rules of thumb for evaluating most decisions in the game.
Some key factors to consider in evaluating each side of a trade are:
1. Having a winning Monopoly that can bankrupt an opponent(s).  Including ownership and the ability to buy improvements.
2. Number of turns left to play in the game, if a time limit is explicity or implictly in place.
3. Likelihood of getting rent in addition to amount of rent.
Understanding the concepts in the book and the data presented really helps estimate the value of each property to each side of a trade, in the current game situation.

Can you summarize the important bits on trading from that book? I can't tell if this is a good answer or not, because I'd have to buy the book to find out.


text:
If I rez Rasmin Bridger after an ICE, does the Runner have to pay its bit immediately?
If I rez Rasmin Bridger after the Runner has passed a piece of ICE, does he get to pay its bit immediatly, or does it take effect only after the following ICE?

you switched your accepted answer. Any reason why? Did you find an official source that corroborated the new accepted answer?

@user1873: Thomas' answer used your same source, reading it differently. I re-read them both and decided his interpretation was more likely right.

my source included the PDF chart that shows when you can Rez an upgrade, and when Rasmin Bridger triggers. How does that chart square with what you believe is the correct interpretation?

I don't know of an official ruling on this, but I believe that they will pay a bit.  You can rez Rasmin Bridger "during a run while the Runner approaches a piece of ice and while the Runner passes a piece of ice" (one of the times the rules lists as the Corp being allowed to rez cards) and since the runner is still passing the piece of ice, they will have to pay a bit /after/ passing.

I think it only takes effect on the following ICE. While I also have not seen any official ruling on this card, it hinges on several different rules. When does Rasmin Bridger's ability trigger?

After Runner passes each piece of ice on this fort, Runner must pay 1 or end the run.

From this chart, step 32, the ICE is passed and this is where the effect triggers if Rasmin Bridger had already been rezzed. If he had not been rezzed, the next opportunity to rez him would be after the runner decides to jack out or not (step 35).
The rulebook only mentions that a Rezzing a Corporation Card can occur (among other times):

during a run while the Runner approaches a piece of ice and while the Runner passes a piece of ice;

The rules don't really clarify the timing of while. The closest thing regarding timing is the Run Protocol - Overview section. It seems to indicate that Upgrades must be rezzed during the approach or just after the runner doesn't jack out.

If the Runner passes all of the pieces of ice (or if there is no ice to be passed in the first place), he or she gets one last chance to jack out. If the Runner doesn't jack out, we have one last opportunity to rez any of the nodes and upgrades inside the fort, and unless these end the run, the run is considered successful, and the Runner accesses the appropriate cards.


text:
Do I get to declare HQ Mole before watching the first cards I'd access anyway?
If I make a successful run on the HQ or R&D, do I have to declare I'm going to use a Mole before watching the cards I would access, or can I access them, and then decide if I want to access more (HQ or R&D Moles) or trash them (Mercenary Subcontract)?

There is no official rulings, because an official (Wizards of the Coast) hasn't yet ruled.
The last official rules errata was v1.7 from March, 2000.

March 11, 2000 - Netrunner Rulings, v1.70 - Author: Top Runners' Conference   List of official rulings used by the Top Runners' Conference. It is compiled primarily from rulings made on the Netrunner mailing list by representatives of Wizards of the Coast. Covers Limited, Proteus and Classic.

The next official rules errata v1.8 has been stalled, you can read the discussion thread here. There is errata concerning the Moles cards, but I do not believe that WotC has provided an official ruling regarding the matter within their mailing list.

D. all topics concerning errata and clarifications discussed on netrunneronline, but maybe not completely validated.
D10a. HQ Mole - ... Clarification: HQ Mole triggers during the “access cards” phase, and can be used anytime during this phase access. When Runner has HQ Mole installed, he or she may first access the card(s) normally accessed without using HQ Mole, processing any immediate effect that would occur due to this access (like the 2 Net damage of Setup!). After that, he or she may still trigger HQ Mole to access the additional cards.
Similarly, when Runner has multiple HQ Moles installed, he or she may use them one by one (and processing for each Mole any immediate effect). In other words, Runner is not required to declare at once how many HQ Moles will be used, but can decide after each HQ Mole used, if he or she will trigger another one.

It is possible that WotC hasn't ruled on the matter, but netrunner online (not official), may have come to a decision. This can be observed on their rules page.

67 HQ Mole - When Runner has HQ Mole installed, he may first look at the card(s) that he would access without using HQ Mole, and after that, he may still trigger HQ Mole to access the additional cards. In the same sense, when Runner has multiple HQ Moles installed, he may use them one by one; i.e. he is not required to declare how many HQ Moles he will use already when his run is successful. When Runner has finally decided how many HQ Moles he uses and after looking at all the cards stored in HQ that he will access, he decides in which order he accesses the cards. As always, he may intermingle with upgrades accessed from HQ.(–Netrunner Online)

I don't know of a specific official ruling on this, but I believe you can access the cards you would normally access, and then decide if you want to use HQ or R&D Mole to view more.
This is based on the fact the runner does have the ability to use cards (such as Mercenary Subcontract) that affect the cards currently being accessed, and that the Runner can intermingle card draws with accessing upgrades
("If you are accessing multiple cards from HQ, you can choose the order in which you take picks, and intermingle them with accessing upgrades as you wish.  (Sparky, Netrunner-L, 11/18/96)", from errata), which seems to imply an ability to use abilities while accessing , even though it's not listed as one of the times the Corp can rez cards ("Either player can use special effects during the same times the Corp can rez cards.  However,if the effect requires a cost of one or more actions, it can only be used during that player's turn.  (Netrunner FAQ v1.0, 5/23/96)", also from the errata; "We can only rez other cards at the following times:
at the start of either player's turn;
after each action;
during a run while the Runner approaches a piece of ice and while the Runner passes a piece of ice;
just before the Runner accesses a fort's contents, whether or not the Runner encountered ice during the run." from the rules.)
Edit: This appears to be a subject of debate on boardgamegeek.  I agree that it would simplify several things to errata the card to "Use this ability only when you are about to access cards from R&D.", but as written it seems to allow use after accessing cards.  The main point against it is this bit from the errata: "The Runner chooses the order in which he or she will access cards, including any draws or face-down cards as well. That's the order in which they affect the Runner. If the Runner survives accessing all the cards there, the Runner can then score agenda and pay to trash nodes and upgrades. (Sparky, Netrunner-L, 6/12/96) The effect of each accessed card must be dealt with separately and immediately, before accessing the next card. (Sparky, Netrunner-L, 11/18/96)".  This seems to suggest that while the Runner can choose the order, they may not be able to initiate further accesses after accessing some cards.  I would probably suggest making a house errata as suggested above, in order to simplify things.


text:
Dominion - Using Stash
How do you use this card strategically?  Since you don't know where any other cards are in your deck, how do you choose where to put this one?  Are there other cards this combines well with?

Normally you would want to put it on top of your deck.  Getting a silver (essentially) is generally a pretty good move.
If you create a deck that draws a lot of cards you will be shuffling in the middle of your turn fairly often.  In that case, you definitely know whether you want to draw more treasures now or not.

Sometimes an attack card such as Thief or Spy can trigger a reshuffle.  In these cases, you would usually not want to put stash on top of your deck because it would get hit immediately.

As far as combinations go, this card is good with anything that makes you reshuffle your deck  more frequently - Chancellor, obviously, but also Chapel or Trade Route or anything else that's commonly used to keep your deck size down.

The Dominion Strategy Blog has a nice post about leveraging Chancellor and Stash together.

It's fun w/ Wishing Well. Once you have 4 of them you can stack 'em together ensuring Province every deck-cycle. If there are other deck revealing/attack style stuff in play (Thief, Pirate Ship, Swindler, etc) you can put them near (but not directly on) the top of your deck so they aren't affected and you'll still draw them next turn.

Putting them below the top would not help much with Thief, Pirate Ship, etc. because generally you shuffle with intent to immediately draw 5 more cards for your next hand.

If you have a lot of Expands, it can be good to distribute them evenly in your deck so that you guarantee each hand has at least one 5-cost card you can turn into a Province. Granted, this is a pretty rare case.


text:
What factors should I consider when choosing card sleeves?
There seem to be a lot of options when it comes to card sleeves.  What factors do I need to consider about different sizes, materials, and quality of sleeves when deciding which to purchase?

How much larger should the sleeves be compared to the cards?
What different materials are available, and how do they effect how well the cards are protected and shuffling?
What sleeves seem to be the best quality?

related - http://boardgames.stackexchange.com/questions/793/worn-out-deck-of-cards

The most important aspect of sleeves is of course size. If the sleeves are to small then obviously your cards won't fit. But the sleeves being to big can also have a number of negative consequences:

sleeves can get in the way during suffling
harder to nicely stack the cards

Probably most common are the sizes for the dimensions used in trading card games but there's also an increasing selection for board and cardgames.
Next on the list is thickness and rigidity. While thinner sleeves tend to be cheaper too they will also be a bit harder to shuffle and offer somewhat less protection. If you play on clean tables that's not an issue but rougher surfaces (i.e. outdoor) can wear through the thin sleeves faster.
Finally there are many designs to choose from - especially among the more common sizes used for ccg's. Anything from a clear plastic on both sides, over single colored backs to elaborate patters and themes can be found. Sleeves with simple opaque backs are favored in tournaments because they make it harder to mark cards.
As a small addition I'd like to point out that shuffling with sleeved cards can be very different from what you are used to. Stripping can be difficult because some sleeves tend to cling to each other. Standard riffle shuffle is also harder because the sleeves do not grant the same fine control over the cards. Probably the easiest way to shuffle is by separating the deck in two stacks and pushing the corner of one stack into the side of the other. If the sleeves are sufficiently rigid it will be very easy to slide them together producing the equivalent of an almost perfect riffle shuffle.

A side note specific to Dominion: Do not use sleeves with opaque backs if you ever plan on getting the Stash card, as it intentionally has a different design on the back.

What, me, Dominion? :checks his question history: ... :blush:

Well. You could simply mark those sleeves...

Get polyurethane sleeves, not polypropylene.  The former are firm, strong, and have sharp edges that make them easy to shuffle simply by pressing two piles of cards together at their edges.  The latter are flimsy, floppy bags with cards inside them, and they're just as easy to manage as it sounds like they would be.
The brand I've used the longest is Ultra Pro, which manufactures card sleeves with opaque solid-color backs that are appropriate for just about any game that doesn't require you to see the backs of your cards (and is handy for those unfortunate CCGs whose card backs have changed over the years).  The only downsides to these are that they're not cheap, they tend to split with heavy usage over time (I can't tell if it's a function of age or not, though I bet it is), and every model I've seen has a little holographic dot on the front lower-left corner, which is less than ideal placement for games that use that space.

I sleeve the cards for my Card Driven Wargames, but be aware they never fit in the box right after that, and shuffling is way different.


text:
"Best of the Best" and Heavy Raiders?
There is a card in the Battlestar Galactica: Exodus expansion called "Best of the Best".  It says:

Play before a piloted viper makes an
  attack.  Instead of rolling the
  attack, roll the die and destroy that
  many raiders in the viper's space
  area.

Question is:
Can this card be used to destroy heavy raiders in the viper's space area?
Can it be used to destroy heavy raiders in the viper's space area if a 7 or 8 is rolled on the die?
Can it be used to destroy heavy raiders in the viper's space area if Samuel T. Anders' "Longshot" ability is used to declare that the die result is an 8?

The card cannot be used to destroy Heavy Raiders. The word "Raiders" refers to normal Raiders only and not Heavy Raiders.
This is similar to the question about whether or not nukes can destroy Heavy Raiders (the card also only refers to Raiders). The FAQ's answer is that nukes only destroy normal Raiders, not Heavy Raiders.

Just to clarify, the restriction on nukes mentioned doesn't apply to the result that destroys all ships.


text:
Unique Item Cards in Bang?
Can you play an item card (like a Barrel) if you don't already have one, but someone else does?

@LittleBobbyTables -- Quite correct, and so marked.

Certainly you can.  Per the rules:

no player can ever have two identical
  cards face up in front of him.

There is no restriction on an identical card being face up in front of another person!

Doh!  We were reading "in front of him" as "on the table"...

Actually there is a character from an expansion that causes other players to discard an item if he plays the same item.


text:
Character Specials Triggering on Non Bang Cards?
Several of the characters in the game have special rules that allow them to do things when they are damaged.  For example, one character, when "damaged by another player" may then draw a card from their hand.  Do these abilities trigger only when directly damaged by another player (i.e., Bang!) or do they also trigger off of effects like Duel! (which deal damage, but lack the Bang! icon) or even effects like Indians (wherein every player who does not discard a card loses a bullet)?

In the copy of the rules I've consulted, El Gringo's rules text begins:

Each time he loses a life point due to
  a card played by another player...

I think it's pretty clear from that that, as long as a card was played that can be held responsible for damaging him, El Gringo's special power will trigger.  Whether or not the damage was "direct".


text:
Clarification on the Royal Garden from Puerto Rico expansion 2
The Royal Garden states:

At game end, the owner of the occupied royal garden takes 1 VP for each noble on his island.  As each noble is normally worth 1 VP (whether on a building, plantation or in San Juan), with the occupied royal garden each is worth 2 VPs to that player".

Puerto Rico defines "island" as the twelve spaces at the bottom of the board to place plantation tiles; the Royal Garden states that you get 1 VP for each noble on his island (which I would assume means "nobles on plantations") but then goes on to state that nobles in San Juan and buildings are worth 1 extra VP as well.
So... which is it?  Do nobles on island tiles get the extra VP, or nobles on your entire board get the extra VP?

In my opinion, a player's island is their game board, unless explicitly referring to plantations.  Its semantically tricky, as the island on the board contains San Juan which is implicitly the player's city.
I admit, that's kind of shaky reasoning.
With regards to the Royal Garden explicitly, an unofficial translation of the original rules was:

At game’s end the owner of the Garden
  gains 2 VP for each Noble he owns
  instead of the usual 1VP/Noble. This
  is true wherever his Nobles are:
  building, plantation, San Juan, etc...

(source)
And it makes more sense in terms of design as the rest of the expansion clearly wants you to place Nobles on buildings.

Interesting note about the translation!  I agree, it's shaky, and I'm torn between "rules as written" versus "intent", but I'm inclined to agree that it's a mistranslation.  Sure makes Villa + Jeweler + Royal Gardens nasty if you can pull it off.

You win 2VP per noble in the island and 1VP per noble in the city or san juan IMO. This is more balanced.


text:
Can I play a Beer to no effect in Bang?
Is it legal to play a Beer card in Bang! even if I am at full life?  Some scenarios where I might want to do this:

I don't want someone else to "Panic" it from my hand
I'm Suzy Lafayette and I want to empty my hand so I can keep drawing more cards

You can Panic from other people's hands!?!?

Absolutely!  From their hand or the cards played in front of them.  When pulling from their hand, it's done blindly, though - you don't get to peek. :-)

To the best of my knowledge...
You can play a beer card on your turn, even if it has no effect.
You can only play a beer card off-turn if you've taken a lethal hit.

We had this question at a game a few months ago, and settled it exactly like that. There's nothing really hinting one way or another in the rules book (at least in the French translation!), but it seems reasonable to me.

Yes you may, but only on your turn.
Playing this way is not considered a discard. Emiliano Sciarra (the game's creator and designer) posted some answers on his site as follows:

Q. Can I play a Beer if I am at full life points?
A. Yes: the Beer will simply have no effect at all.
[...]
Q. Can I play a Beer card if there are only two players alive?
A. Yes, but it has no effect at all.
[...]
Q. Can I voluntarily discard my cards?
A. No, neither cards from your hand nor cards in play in front of you. ... [T]he character Sid Ketchum can discard exactly 2 cards when he chooses so to regain one life point.

Also, the creator explained that other cards with the  icon are not the same as a Beer card. They can be used to restore life points when only two players are in play, and can also be played without effect when your life points are full. However, they can't be used to restore your last life point (or played outside of phase 2 of your turn for any other reason).

I think yes, because it's like using a Panic on a player with no cards. Here is why:
Beer has an effective range of [Casting Player]. It automatically hits the person who plays it without fail. Since it is not out of range, it can be played.
Panic has an effective range of [1]. If you are going to compare the two, you must assume that the player being targeted is also in range, just as the Beer casting player is in range of the card "Beer".
This is a more fair comparison than Tycho's because both cards can technically be played at this point, but neither will be able to have its intended effect; Beer wont be able to restore the health point since the player is at max and Panic wont be able to draw a card from a player because that player doesn't have a card to draw.
The new question is "Can you play cards that have no effect?" The answer to this is most likely yes. One in game clue to this is that the character "Apache Kid." He is unaffected by diamonds, yet this does not mean you cannot play a diamond on him. The diamond card will simply have no effect. If we carry this rule on to normal play, one should be able to Beer them self when at full health.

I actually don't believe you can play a beer if you cannot take the effect just as you could not play a panic if both players next to you had an appaloosa.  By my reading of the rules the only way to willingly discard a card is to have more cards than your hand limit at the end of your turn.
Also, Kevin is correct that you may only play a beer out of turn on a lethal hit.


text:
How many spins does it take to complete a game of Hi Ho! Cherry-O?
I recently picked up this game for my 2-year-old daughter, who is learning how to count. I find this game fairly infuriating to play by the actual rules. It contains a spinner with the 7 following outcomes:

Gain 1 Cherry
Gain 2 Cherries
Gain 3 Cherries
Gain 4 Cherries
Lose 2 Cherries (x2)
Lose all Cherries

Play is to 10 Cherries.
Question: On average, how many spins will it take to finish the game?
(Edit: I had another question about "expected # of Cherries gained from a spin," but it turns out to be rather difficult to define due to that pesky "lose all your Cherries" option.)

Might actually be better suited to math.stackexchange.com. But, the lose all cherries option might make the questions needing more clarification. For instance, is the expected number even well defined?

FYI, this is a question about probability, not statistics. I've re-tagged.

@Aryabhata By "for a given spin" I meant: "I have 3 Cherries - how many would I expect to gain?" This should be well defined for all Cherry counts 0-9.

Can you clarify your request for expected # of cherries? I don't think it's quite defined enough to give a mathematically precise answer. As you note, the expected value is heavily dependent on the number of cherries pre-spin. Averaging together these expected values wouldn't be useful, IMHO. A table of expected cherries as a function of cherries could be tabulated, but that would be heavily influenced (in a non-helpful way) by the possibility of losing all. I think much more useful would be the average change if you don't lose them all, as a function of # of cherries pre-spin.

I don't know that this is easy to verify, but each of the seven outcomes is equally likely (same width on the spinner), right?

Looking at the work that's been done so far, it seemed to me that this is basically just a Markov chain, and in looking for ways to calculate Markov chains, I found that someone has been there and done that.
The data from his work matches the simulations here, which should be no surprise given that the rules are very easy to code:

Minimum game length: 3
Expected game length: 15.8
Maximum game length: Unbounded
50th percentile (median): 12
95th percentile: 40
25th percentile: 7
75th percentile: 21

He also posts the Matlab code that he used to calculate the chain ... it's actually pretty simple, comparatively speaking.
The Markov chain transition matrix shown below lists the probabilities of transitioning from one state (i.e. number of cherries pre-spin) to another (number of cherries post-spin).
          # of cherries after spin
       0   1   2   3   4   5   6   7   8   9  10
     --------------------------------------------
  0 | 3/7 1/7 1/7 1/7 1/7  0   0   0   0   0   0
  1 | 3/7  0  1/7 1/7 1/7 1/7  0   0   0   0   0
# 2 | 3/7  0   0  1/7 1/7 1/7 1/7  0   0   0   0
  3 | 1/7 2/7  0   0  1/7 1/7 1/7 1/7  0   0   0
b 4 | 1/7  0  2/7  0   0  1/7 1/7 1/7 1/7  0   0
e 5 | 1/7  0   0  2/7  0   0  1/7 1/7 1/7 1/7  0
f 6 | 1/7  0   0   0  2/7  0   0  1/7 1/7 1/7 1/7
o 7 | 1/7  0   0   0   0  2/7  0   0  1/7 1/7 2/7
r 8 | 1/7  0   0   0   0   0  2/7  0   0  1/7 3/7
e 9 | 1/7  0   0   0   0   0   0  2/7  0   0  3/7
 10 |  0   0   0   0   0   0   0   0   0   0   0

+1 for Markov chains, which is really the only way to analytically model this type of problem, and for finding prior art. I notice everyone else with a correct answer used a Monte Carlo simulation that they had to code up themselves. I wonder what was the fastest, C, Python, or Google. :)

Good call on finding the paper. I was just about to try and invert the 10x10 matrix.

The mathematics to take into account the fact it cuts off at 0 and 10 (i.e. spinning -2 when you are on 1 will leave you on 0, not -1) and the "loose all" result are quite complex.
Instead I've opted for a simulation. This was run with 10,000,000 simulations, which should be more than enough to get a good result.
Here is a histogram of the result. The x-axis shows the number of spins, while the y-axis shows the number of results in the simulation that took that number of spins.

The average (mean), shown by the red line, is 15.8.
Here is the C# code I used to generate the results.
using System;
using System.Linq;
using System.Collections.Generic;

public static class CherryO {
    private static Random rng = new Random();

    const int NumRuns = 10000000;
    const int NumSquares = 10;

    public static void Main() {
        int total = 0;
        Dictionary<int, int> totalTurns = new Dictionary<int, int>();

        for (int i = 0; i < NumRuns; i += 1) {
            int turns = GetNumTurns(NumSquares);
            total += turns;
            if (!totalTurns.ContainsKey(turns)) {
                totalTurns.Add(turns, 1);
            } else {
                totalTurns[turns] += 1;
            }
        }

        int max = totalTurns.Keys.Max();
        for (int i = 0; i <= max; i += 1) {
            if (!totalTurns.ContainsKey(i)) {
                totalTurns.Add(i, 0);
            }
            Console.Write("{0} ", i);
        }

        Console.WriteLine();

        var list = from kvPair in totalTurns
                   orderby kvPair.Value
                   select kvPair.Key;
        foreach (int i in list) {
            Console.Write("{0} ", i);
        }

        Console.WriteLine((double)total / NumRuns);
    }

    public static int GetNumTurns(int target) {
        int position = 0;
        int turns = 0;

        while (true) {
            int spin = GetSpin();
            turns += 1;

            if (spin == 0) {
                position = 0;
            } else {
                position += spin;
            }

            if (position >= target) {
                return turns;
            }

            if (position < 0) {
                position = 0;
            }
        }
    }

    public static int GetSpin() {
        int result = rng.Next(0, 7);

        if (result == 5 || result == 6) {
            return -2;
        }

        return result;
    }
}

This is a good way to show the answer to part 2, now we just need a simulation for part 1.

It looks like your mean is actually the expected value multiplied by 100.

I wrote something to enumerate the relevant states and then print out the probabilities it found that way. Somehow a simulation felt like a cop-out to me :)
I believe this game is for more than one player? And it ends when ANY player gets 10 cherries, right? The code and results below assume that.

For the two-player case, I got a mean end turn of around 9.56. Since I was counting turns as every player spinning once, that's two spins per turn, so let's say that in the mean case, it takes between 19 and 20 spins for the game to end when two people are playing. It looks like the median is lower: about 15 spins, so half of the two-player games should end on or before the 15th spin.
The original question makes it sound like the one-player case is what was being considered so here is the info for that one.

For one player: the mean game length is 15.8 spins and the median game length is between 11 and 12 turns (half end on or before 11, half on or after 12 spins).
Here is the code to generate the data used above. Warning: It gets really slow as you add players (exponential runtime in number of players, but only quadratic in terms of the number of turns). You can change the values of turn_cap and num_players at the top of the file to match what you want to find out. It only enumerates up to turn_cap, because you probably don't care beyond that. Games that end later ARE still taken into account, it just doesn't spit out data for them.
(Python 2.5):
from copy import deepcopy
from sys import stdout
num_players = 2
turn_cap = 50 # I assume everybody is sick of it by this point
print_merges = False
print_num_states = False
print_turn_progress = True
ignore_score_of_finished_games = True

def flatten(list_of_lists):
    r"""helper for flattening a list of lists into a single list"""
    return [item for sub_list in list_of_lists for item in sub_list]
def count(iterable,p=bool):
    r"""return for how many items in iterable the predicate p evaluates to true"""
    n = 0
    for item in iterable:
        if p(item):
            n += 1
    return n
def lists_the_same(l0,l1):
    return len(l0)==len(l1) and \
           all(l0[i]==l1[i] for i in range(len(l0)))
def split_list(items,p):
    r"""split the list "items" into two lists based on a predicate "p"

    returns a tuple. (items where the predicate is true, items where the predicate is false) """
    good, bad = [], []
    for i in items:
        if p(i):
            good.append(i)
        else:
            bad.append(i)
    return good, bad
class World:
    def __init__(self, num_players=2, probability=1.0):
        self.player_scores = [0 for x in range(num_players)]
        self.probability = probability
        self.turn_number = 0
        self.ended_on_turn = None
        self.num_players = num_players
    def check_win(self):
        if any(( (score>=10) for score in self.player_scores)):
            self.ended_on_turn = self.turn_number
    def gain_cherries(self, player_index, num_cherries):
        self.player_scores[player_index] += num_cherries
        self.check_win()
    def lose_cherries(self, player_index, num_cherries):
        self.player_scores[player_index] = max(0, self.player_scores[player_index] - num_cherries)
    def lose_all_cherries(self, player_index):
        self.player_scores[player_index] = 0
    def run_turn(self):
        r"""if not ended, it splits the world into 7 new worlds based on each player's spin"""
        if None==self.ended_on_turn:
            self.turn_number += 1
            worlds = [self]
            for i in range(self.num_players):
                worlds = flatten([w.run_player_turn(i) for w in worlds])
            return worlds
        else:
            return [self]
    def run_player_turn(self, player_index):
        new_worlds = [deepcopy(self) for i in range(7)]
        for w in new_worlds:
            w.probability = self.probability / 7.0
        new_worlds[0].gain_cherries(player_index,1)
        new_worlds[1].gain_cherries(player_index,2)
        new_worlds[2].gain_cherries(player_index,3)
        new_worlds[3].gain_cherries(player_index,4)
        new_worlds[4].lose_cherries(player_index,2)
        new_worlds[5].lose_cherries(player_index,2)
        new_worlds[6].lose_all_cherries(player_index)
        return new_worlds
    def mergable(self, other):
        if world_ended(self) and ignore_score_of_finished_games:
            return ( (self.ended_on_turn == other.ended_on_turn) and
                     (self.num_players == other.num_players) )
        else:
            return ( lists_the_same(self.player_scores, other.player_scores) and
                     (self.turn_number == other.turn_number) and
                     (self.ended_on_turn == other.ended_on_turn) and
                     (self.num_players == other.num_players) )
    def merge(self, other):
        w = deepcopy(self)
        w.probability += other.probability
        return w
    def __str__(self):
        return ( str(self.player_scores) + ", " +
                 "p:" + str(self.probability) + ", " +
                 "ended on:" + str(self.ended_on_turn) + ", " )
def collapse_duplicate_worlds(world_list):
    r"""look through the list of worlds, find ones that can be merged
    and merge them, modifying the list in place"""
    i0 = 0
    while i0 < len(world_list):
        i1 = len(world_list) - 1
        while i1 > i0+1:
            w0 = world_list[i0]
            w1 = world_list[i1]
            if w0.mergable(w1):
                world_list[i0] = w0.merge(w1)
                if print_merges:
                    print "merging",w0
                    print "with",w1
                del world_list[i1]
            i1 -=1
        i0 += 1
def dump_stats(worlds):
    print 'turn\tON\tBY'
    p_on = [0.0 for i in range(turn_cap+1)]
    p_by = [0.0 for i in range(turn_cap+1)]
    for w in worlds:
        if world_ended(w):
            turn = w.ended_on_turn
            p_on[turn] += w.probability
            for i in range(turn, turn_cap+1):
                p_by[i] += w.probability
    for i in range(1,turn_cap+1):
        print str(i)+'\t'+str(p_on[i])+'\t'+str(p_by[i])
def world_ended(w):
    return None!=w.ended_on_turn
def main():
    worlds = [World(num_players)]
    finished_worlds = []
    for i in range(turn_cap):
        if print_turn_progress:
            print "turn#:",i+1
        worlds = flatten([w.run_turn() for w in worlds])
        collapse_duplicate_worlds(worlds)
        done, worlds = split_list(worlds,world_ended)
        finished_worlds.extend(done)
        if print_num_states:
            print len(worlds),"unfinished world states"
            print len(finished_worlds),"finished world states"
    finished_worlds.extend(worlds)
    dump_stats(finished_worlds)

if "__main__"==__name__:
    main()

I'm not sure what's wrong with your code, but it doesn't appear to be giving the right results. It sounds like you were attempting the same type of Markov chain approach mentioned in another answer. Perhaps the referenced paper can help.

How did you determine that I got the wrong results? Please note that the Markov Chain linked does not take into account that there are multiple players. Hi Ho! Cherry-O is a 2-4 player game.

You know what? Good point. The other simulations and the Markov chain example are all answering the question of how many spins it takes to amass 10 cherries, and now that I think about it, that is a totally different question than how many turns (i.e. each player spinning once) it takes to end the game. In fact, I think you're the only person to answer what is really the actual question being asked. I think we could validate your sim by running it with num_players = 1 and seeing if the results match the others.

Bug fix for the code above. Setting num_players up at the top wasn't being respected. The first line of main should be worlds = [World(num_players)] (I can't edit w/o losing the image because my reputation isn't high enough)

If you make the edit to the code and also put the link to the image in the body we'll edit the post for you. I'd recommend two runs: one for 1-player to prove your code matches the Markov result, and one for 2-player.

This is actually a pretty interesting question. I also decided to write a simulation. The results are quite interesting, and tend to show a bit about how careful you have to be to get a good simulation. For those who care, here's a bit of code:
[Edit: replacing code due to stupid mistake]
[Edit2: added code to compute average gain per spin]
#include <iostream>
#include <time.h>
#include <vector>
#include <numeric>
#include <algorithm>
#include "bounded.h"

static const int win = 10;
static const int games = 20000000;
static const int init[] = { 1, 2, 3, 4, -2, -2, -10};

int rand_lim(int limit) {
    int divisor = RAND_MAX/(limit+1);
    int retval;

    do {
        retval = rand() / divisor;
    } while (retval > limit);

    return retval;
}

int main(){
    srand(unsigned(time(NULL)));
    std::vector<char> spin_vals(init, init+7);
    std::vector<char> gains;

    std::vector<char> spins;
    unsigned total_spins = 0;

    for (unsigned game=0; game<games; game++) {
        unsigned current_spins = 0;
        bounded<int, 0, win> current_game_val;
        while (current_game_val != win) {
            ++total_spins;
            ++current_spins;
            int current_spin = spin_vals[rand_lim(spin_vals.size()-1)];
            int old_val = current_game_val;
            current_game_val = current_game_val + current_spin;
            gains.push_back(current_game_val - old_val);
        }
        spins.push_back(current_spins);
    }

    double mean_spins = (double)total_spins / games;

    std::cout << "Mean spins/game: " << mean_spins << "\n";

    std::vector<char>::iterator median = spins.begin()+games/2;

    std::nth_element(spins.begin(), median, spins.end());

    std::cout << "Median spins/game: " << (int)*median << "\n";

    double total_gains = std::accumulate(gains.begin(), gains.end(), 0.0);

    double mean_gain = total_gains / gains.size();

    std::cout << "Mean gain per spin: " << mean_gain << "\n";

    return 0;
}

For output, I get:
Mean spins/game: 15.7967
Median spins/game: 12
Mean gain per spin: 0.633044
The gain per spin seems to fit with the number of spins per game -- if it takes ~16 spins to get to 10, then the average gain per spin should be ~10/16.

Your "lose all" option is always -10 cherries.  I think the line adding to current_game_val needs a max in there to ensure it doesn't go negative.

The distribution of the random number generator doesn't seem to be what causes the difference between our results. I've swapped in a cryptographic strength random number generator, as well as randomly ordered the sequence of the items chosen - it makes no difference.

@Mag That's achieved by the bounded type.

Shouldn't rand_lim(spin_vals.size()) be rand_lim(spin_vals.size() - 1)? If I make that the results of your code matches with mine :) The reason it worked when you replaced it with return rand() % limit is that it should actually be return rand() % (limit - 1), so both the errors cancelled each other out.

@ICR: Yes, you're quite right, it should be. Posting corrected code now...

The following calculations are based on having an equal probability of being at any particular number (0-9), when the spinner is spun.
The expected number of Cherries per spin can be calculated as follows:
Contributions of Spin (1/7 probability ea):

   1:  (1*10)/10                  = 1
   2:  (2*9+1)/10                 = 1.9
   3:  (3*8+2+1)/10               = 2.7
   4:  (4*7+3+2+1)/10             = 3.4
  -2:  -(2*8+1+0)/10              = -1.7
  -2:  -(2*8+1+0)/10              = -1.7
-all:  -(0+1+2+3+4+5+6+7+8+9)/10  = -4.5

                            Total = 1.1

Total/7 = 0.157143 <-- expected value

So theoretically, the average game should take 10/0.157143 = 63.5 spins.
Note with much more precision, it comes out to 63.636363.... spins.

I've now run a simulation to figure out the probabilities for what your score is when you spin and get the Lose All result.  I ran 3 10,000,000 simulations and averaged the results for the ALL, and 1 10,000,000 simulation for the others, so here are the new calculations:
Contributions of Spin (1/7 probability ea):

   1:  (1*10)/10                  = 1
   2:  (2*9*0.967145 +
        1  *0.032855)/10          = 1.74415
   3:  (3*8*0.923705 +
        2  *0.043412 +
        1  *0.032884)/10          = 2.22886
   4:  (4*7*0.865895 +
        3  *0.058240 +
        2  *0.042263 +
        1  *0.033602)/10          = 2.45379
  -2: -(2*8*0.563137 +
        1  *0.079143 +
        0  *0.357720)/10          = -0.90893
  -2: -(2*8*0.563137 +
        1  *0.079143 +
        0  *0.357720)/10          = -0.90893
-all:  -(0 * .358065 +
         1 * .078969 +
         2 * .092976 +
         3 * .094948 +
         4 * .107013 +
         5 * .069808 +
         6 * .064553 +
         7 * .057246 +
         8 * .042941 +
         9 * .033480)             = -2.75975

                            Total = 2.84919

Total/7 = 0.40703 <-- expected value

So theoretically, the average game should take 10/0.40703 = 24.56839 spins.

It's also going to be slightly off because of the limits. +2 on 9 is going to make it to 10, and -2 on 1 is going to cap at 0. The effect of these are going to be pretty small though.

Sorry, have to downvote just because the things not modeled (i.e. impact of lose all cherries and not being able to have negative cherries) make this result mathematically incorrect.

@ICR, answer updated, though it doesn't agree with your simulation.

The probability of being on a particular number are far from being evenly distributed. They're going to be quite heavily skewed towards lower numbers.

@ICR, that would increase the probability of 1 thru 4, and increase the magnitude of 5 and 6 (making more negative), and decrease the magnitude of 7 (making less negative).  So overall it would increase the expected value a little, and decrease the number of spins needed.  Any ideas on how to calculate the probability of what number you are on when you spin?

The probability of being on a particular square is dependant on how many spins have happened. It's trivial to calculate the probability of being on a given square after a given number of spins (enumerate all the combinations it takes to get there), but translating that into the probability of being on a given square regardless of the number of spins is difficult and I've not yet worked that out.

@ICR, ok, updated to the next level.  Note, I'm also getting the 15.8 - 15.9 when running simulations for length of game, so there may still be some probability thing we need to take into account.


text:
When drawing Destination Tickets, can you discard any Destination Tickets or only those you just drew?
When picking up destination cards, after the initial deal, does the card(s) which the player chooses to discard, have to be from the three he picked up at that time, or can he discard a card which he picked up from a previous turn?

The cards you discard/return must be from the three you drew this turn.
Here's how the rules phrase it:

The player draws 3 Destination Tickets from the top of the deck. He must keep at least one of them, but he may
  keep two or all three if he chooses. Any returned cards are placed on the bottom of the deck.

It's not so much that you "discard" destination tickets, but rather that you draw 3 and keep from 1 to 3 new destinations.  This means that any destinations you were already holding are totally unaffected - you're stuck with 'em.


text:
Can a route be included in more than one destination ticket?
The rules that come with the game do not specifically state the a route (between two cities) can be included in more than one destination string, but this is assumed when we play.
Is that a correct assumption?

Hi!  Welcome to BCGSE.  You might want to try and make the title of your question have more information about the question itself, instead of just the name of the game.  You had two questions that looked identical on the main page because they had the same title.

Yes. The section about scoring says:

Players should then reveal all their Destination Tickets and add (or subtract) the value of their Destination Tickets still in hand, based on whether
  they successfully (or not) connected those cities together.

So the only thing that matters is whether those cities are connected to each other. It's very possible that all your routes will share a single link across the middle of the country, and that's fine.

I'd go even further: potential reuse of the routes for multiple tickets is main strategical objective when you draw tickets

To put it another way: all that matters is that at the time you score the route card, your network of track connects the two cities on each card. In some versions, it's possible to have the same route multiple times; they all can use the same chunks of track. In versions with stations, the station has to be used for one specific chunk of track, but all routes can use that one track "borrowed" using the station. We place the station over it to indicate its use during the "proving" of one's tickets' completion.


text:
In Puerto Rico when using the "residence" how is "occupied island spaces" interpreted?
When playing Puerto Rico and using the residence building it states that it is based on the number of "occupied island spaces".
Does this mean plantation spaces with a farm and a colonist? Or does this mean plantation space with or without a colonist?

It means just a plantation, no colonists are required.
From the rules:

Residence The owner of the occupied residence earns, at game end,
additional victory points for the plantations and quarries he has
placed on his island. For up to nine filled island spaces, he earns 4
VP, for ten filled island spaces, he earns 5 VP, for eleven filled
island spaces, he earns 6 VP, and for all twelve spaces filled, he
earns 7 VP.
Example: at game end, the owner of the occupied residence
has filled 10 of his 12 island spaces with plantations and quarries:
he earns an additional 5 VP.

No requirement for colonists.

I just find it odd that it specified that the "occupied residence" means a colonist on the residence but the "occupied island spaces" does not mean a colonist on the island spaces.

@alex, they definitely could have been clearer, this question comes up a lot for new players.

I'm not sure Lance Roberts' answer is correct. For the Mayor, the rules say this: "If there is at least one colonist on a tile, it is considered occupied. Only occupied tiles may use their functions; unoccupied tiles never function!" (emphasis theirs).
Given that emphasis on the word 'occupied' on the residence card, and the word 'never' in the mayor description, it could be interpreted as island spaces without a colonist do not contribute towards the residence card's ability, as they are not occupied and do not function.
This ambiguity could be fixed by changing the word 'occupied' in relation to the island tiles on the residence card.

The other answer is definitely correct, but your reasoning is reasonable given that the rules don't use the clearest language. The difference is that the rules you are quoting are discussing the definition of an "occupied tile", while Residence deals with "occupied island spaces". And while unoccupied buildings "never function"; they DO count for points at the end of the game.


text:
In Sorry, can I be forced to move out of my "safe zone"?
For example, if I draw a 4 (move 4 spaces backward) and the only piece that could make the move is in my "safe zone", do I have to make the move?

Yes. From a PDF version of the rules, you are required to move if a legal move is available:

... if at any time you can move, you must move, even if it's to your disadvantage.

and from the description of the Safety Zone, moving backward out of the zone is legal:

... however, a pawn may move backward out of its SAFETY ZONE and on subsequent turns move back into the ZONE as cards permit.

Note that trading places on an 11 is an exception to this - you may decline to switch places with another pawn if you like - but yes, in your situation, you must move a pawn out of your safety zone.

Also later in the same pdf "...however a pawn may move backward out of its safety zone..."

@tttppp, I think I'll add that to clarify that this specific movement is "permitted".

It's the 'may' part of the second sentence that caused some confusion in the household. Kind of contradicts the 'must' part of the first.

@basilard, that's understandable. In this case, I think "may" refers to permission rather than choice; the rule probably could have been written to emphasize that.

On the "trading places", you're saying you can swap two of your own pieces, not throwing an opponent's piece into your safe zone, correct?

@Allen, correct; I was mentioning it as an exception to "if a legal move is available" (meaning that if you have the opportunity to swap, you can decline it, whereas you cannot decline the opportunity to move a pawn if it's available) rather than as a way to move out of the Safety Zone.

@Allen, actually, I didn't answer the other part of your question correctly. No, you cannot swap two of your own pieces, but trading pieces in any situation is optional rather than required, and you cannot swap a piece (yours or someone else's) in a Safety Zone.

Confusion arose when there was a disagreement about whether you move within the safety zone.  One says you must move, while another says once in the zone, you cannot move unless you draw the exact number To arrive in your home.  Since the rules specify having to move out on a four, it stands to reason that you must move if possible, but you cannot enter home on a number greater than required to arrive there.

Hey Sam!  Welcome to the site.  Do you have any rules citations to back up your statements?

Or is this meant to be a comment on the other answer?  I'm a bit confused


text:
Cribbage score going out
i.e. Player needs 4 points to go out.  The hand value is 3.  Does the Player get to go out with the added 1 point "Go"?

It's fine to go over in standard cribbage; the game is won by the first player or partnership to reach or exceed 121 points.
But remember the way things are scored: as they happen.
If that three points you mention was in play, such as opponent leads a 9, you play a 5, opponent plays a 6, and you play a 7, scoring three for the run in play, then the go from the opponent puts another 1 point score on the pile.
If it's a run in hand, however, it's scored only after the last card, starting with the player to the dealer's left and going left around the table (same as play order). If you are at 117 points, have a 3 point hand (say AJQK and a 2 up), and get the last card for 118, and the other guy is at 110 and gets 16 (AA23), it is determined by who scores hand first. If you dealt, you lose; if he dealt, you win. In three or four player non-partnership, if he's to your right, you've lost.
And remember: muggins can't be claimed until after the peg has been moved. Only on an overpeg would it prevent victory.
Also remember: if your scoring sheet shows last card separate from Go, there is no "go" on last card.
There is a common variant, called exact out, where you have to hit 121 exactly, and any score that would take one over is lost (but must be called out to prevent it being muggins-claimed).

In your example, if the opponent lead Q and you played a 5, then you'd get 15-2 right there, plus 3 for the run in a minute, thus ending the game. I know that's not what you were trying to show in your example, so perhaps Q followed by a run of 4-5-6 could make the same case without the player getting 2 points for the 15.

Good catch, Adam. Fixing... changing the Q to a 9... slightly less likely t get a go, but I've floated a 24 for a go more often than I care to think.

A player pegs out at soon as they are able.  You would score the points for playing the hand (Go's and anything else) first during the play of the hand, and then score your hands, non-dealer first.  So in your case, since the Go was part of the play, you'd score the point, and then score your hand and peg out.

No you have to get the exact number to be able to win. If not you let the other person go until one of you gets the right number.


text:
You are a sympathetic Cylon
I'm still pondering over our Cylon's greatest loss in the last 6 player game we played.
In turn 3 the first Cylon player got executed by an Admiral Chooses crisis card (Kain-player resolved that one true to character). Soon thereafter the 2nd Cylon revealed himself, turn 6 if I remember correctly, which was quickly followed by the Sleeper Agent phase. Now, what happens at this moment is that one of the Cylon players is dealt the You are sympathetic Cylon loyalty card. Which he, according to the rules must reveal immediately, and he becomes revealed Cylon. This seemed and felt a bit redundant, and quite a disadvantage for the Cylons.
From the base game rules:

If a player receives a You
  are a Sympathizer card, then he must immediately reveal
  and resolve it. If he is a revealed Cylon player, then he may
  first give the card to any other player (who then immediately
  resolves it).

Pegasus rules instead say:

When a human player receives this card (You are a Sympathetic Cylon) during the Sleeper phase, he must immediately reveal it. However, rather than following normal Sympathizer rules, this player becomes a revealed Cylon and draws an Agenda from the Sympathetic Agenda deck.

What I would like to know is if we interpreted the Pegasus rules correctly in the way that the part of the rule rather than following normal Sympathizer rules also affects passing on this card to another player as was done in the base game.
For those interested, Humans won with 4 fuel, 8 Food, 5 Morale and 9 Population, two civilian ships where destroyed while in space (Blind Jump by Kain), only one civilian ship was left behind on New Caprica. All the Raptors were accounted for and only one Viper was damaged. Only three Jumps were made to reach New Caprica.

A revealed Cylon player must always pass the You are a Sympathizer and You are a Sympathetic Cylon cards to another player. The Pegasus rule you quoted is for "when a human player receives" the card. See page 7 of the Pegasus rulebook:

Sleeper Phase: When a Cylon player receives Loyalty Cards during
  the Sleeper Agent phase, he looks at all of his facedown Loyalty
  Cards and gives them to one human  player of his choice. If the Cylon
  player receives a “You Are Sympathizer” or a “You Are a Sympathetic
  Cylon” card, he does not reveal it.  Instead, the human player he
  passes it to must immediately  reveal this Loyalty Card as if it had
  been dealt to him.

So the original rule in the base game hasn't changed. It's just been expanded to include the You are a Sympathetic Cylon card.

Thanks! I completely missed that as I did not look in the correct place in the pegasus rulebook. gee.. it even has a big header named Handing off Excess Loyalty Cards


text:
Does Wrath of God really destroy creatures with shroud?
I heard - (from an official MtG judge) - that Wrath of God ("Destroy all creatures") destroys creatures with the Shroud ability, because it doesn't target a single creature, it targets all creatures.
Is this really true? In my opinion it is very illogical.

That phrasing — "targets all creatures" — isn't correct, and if the judge used it, I can understand your confusion. Wrath of God doesn't target creatures at all; it wipes them out in a grand swath.

In terms of flavour, shroud can represent a creature being "magically" elusive. You can't shoot at a guy if you don't know where he is in a building, but he'll still die if you blow the whole building up.

Yes, Wrath of God will kill all creatures with shroud.  It will also kill all creatures that have protection from white.  It won't kill creatures that are indestructible.
To understand why this is, you have to look at the definitions of the terms.
Shroud means: "this creature can't be the target of spells or abilities".
Protection means (among other things): "this creature can't be the target of spells or abilities of the colour in question".
Wrath of God just says "Destroy all creatures".  No mention of targets at all.  And I think this is where your confusion is arising.  Wrath of God doesn't "target all creatures and destroy them".  It doesn't target anything.  It just destroys them, whether they're slippery customers or not.  Boom!
Indestructible creatures survive a Wrath because their ability ("cannot be destroyed") specifically interacts with the word "destroy" on the card Wrath of God.  Creatures with Shroud only interact with, and care about, the word "target" on a card.  If you can't find "target" written anywhere on the card, then Shroud won't have any effect on what that card can do.
As an interesting side point, take a look at a card like Copy Enchantment.  This puts into play a copy of an enchantment already on the battlefield.  But note that nowhere on the card does it say "target".  This means that if you copy an Aura on the battlefield, you can put that copy onto a creature with Shroud... even if the original Aura would have been unable to enchant that creature when it was cast!  Sometimes Magic does some pretty counter-intuitive things, but it's very internally consistent.
Hope that's clarified things more than it's confused them, for you!

Huh, that's an interesting note about Copy Enchantment. Auras do target the object or player you're going to enchant when you cast them as a spell, but not when they enter the battlefield by any other means, according to rule 303.4f in the Comp Rules. Neat!

Note that the same enchantment shenanigans can be performed with a card like Lost Auramancers to put Aura Enchantments on creatures with Shroud or Hexproof.

Good unasked for tidbit about protection I didn't know.  Thanks.

Not only Wrath of God is possible of doing this: Damnation, Hallowed Burial, Infest and Pyroclasm for instance can do it as well. Because they don't target.

Another thing to note is that all the cards mentioned except for Pyroclasm also kills creatures with protection from < color >. Pyroclasm doesn't kill them because protection, besides from making it untargetable, also prevents damage.
I've seen a lot of people missunderstanding this.

Useful clarification - I didn't want to go too heavily into protection in my answer, in case I strayed too far from the point, but you're absolutely right.  A creature with protection from all colours dies to a Wrath of God, and also from an (untargeted) effect giving it -X/-X where X is equal to or greater than its toughness; but no amount of (coloured) damage will kill it, not a million or a million million!

First of all, there is a little rule everyone is forgetting and it is the legal target rule which simply states before a permanent can be destroyed or removed from play it has to be a legal target of a spell:

608.2b [During resolution, if] the spell or ability specifies targets, it checks whether the targets are still legal.

All spells target, including those with "all" in their text. If a spell states that it does not target, then it cannot destroy anything according to the legal target rule. For a spell to affect a permanent, the permanent must be a legal target for said spell.
For example, Noxious Ghoul reads:

Whenever Noxious Ghoul or another Zombie enters the battlefield, all non-Zombie creatures get -1/-1 until end of turn.

Zombies do not get the -1/-1 counter because they are not legal targets of Noxious Ghoul's ability. Basically, because there are legal targets for every spell, spells with "all" target "all" as well.
Another basic common sense example is Crux of Fate. Crux of Fate offers a choice of either destroying all dragons or destroying all non-dragons, which means means it will either target and destroy all legal target dragons or target and destroy all legal target non-dragons.
This is basic common sense - if a spell does not target anything then it does not do anything.

so creatures with hexproof or shroud cannot be legal targets of spells therefore are not destroyed by wrath of god

it is a fact all instants sorceries target something otherwise you get this misconception that wrath of god doesn't target anything since if it doesn't target anything then it destroys nothing according to legal target rule now on the otherhand if wrath of god targets all creatures then if they are legal targets they will get destroyed creatures with hexproof or shroud can not be targeted so will not get destroyed basic common sense

This answer is wrong. Wrath of God does not target anything. Neither does Noxious Ghoul or Crux of Fate. Just because a spell affects something does not mean it targets that thing.

"All spells target" is where you've gone wrong. Not sure where you got that idea, but it's definitely NOT correct. The only spells that target are ones that have the word "target" in their text, and Aura spells because the rules for Aura spells specifically say they target.

The common sense answer is wrong in this case because "target" is a keyword with a specific definition in the context of M:tG. Also, I would be hesitant to argue with a judge unless I myself had been certified as one - those nerds know what they're talking about. :)

You will notice that  608.2b states that "IF the spell or ability specifies targets...". Obviously, not all spells or abilities have targets, and these are not covered by 608.2b. I can legally cast Wrath of God when there are no creatures on the battlefield (I don't need to specify targets), but I can't cast Terror when there are no non-artifact, non-black creatures on the battlefield (I can't put it on the stack with a legal target).

These are horrible answers, so I'll just kill the Q&A and answer everything quickly
The confusion here is the word "target".  The key words MTG really wants you to look at is "a" and "the".  Shroud and hexproof prevent single targeting spells. IE: [can't be THE target]   Creatures with shroud and hexproof can still technically be "a" target, they just can't be singled out(although the MTG rules don't seperate, have, or recognize the difference because "Target" is only used on spells that single out permanants).
This game really puts your language reading skills to work, but if you take your time and read up on the rules, it'll make sense.
Keep rocking the cards, guys.

No, creatures with shroud and hexproof can't be targets, whether there are other targets ("up to two target creatures") or not ("target creature"). They can be affected by things that don't target at all ("destroy all creatures"), but that doesn't make them "a" target.

Absolutely wrong.


text:
Principato: availability of cubes
In Principato two basic resources are gold and food. When 4 people play the game the bank runs out of these cubes pretty quickly. The rule book says nothing about this.
First question: I assume that it is so on purpose to give players more options to play against each other. Is that correct? Or should we have as much food and gold as we want?
One observation I've made is that it becomes very different when 3 players play. Availability of cubes is much better and I expect that with 2 players it's not a problem anymore. It changes the mechanics of the game.
Second question: Shouldn't cubes' availability be limited in some way when only 2 or 3 players play? I haven't found anything on this in the rule book but my feeling is it would make sense.

There is no FAQ from the designer, and he has never answered definitively on BGG, but without any rule in the rulebook to the contrary.
The cubes are lmited.
For example, take:

Monopoly the houses are limited, but cash has a specific rule that it isn't.
Puerto Rico the goods and buildings have a limit, but if victory points run out keep track on paper.
Race for the Galaxy the VP points if you run out, you are told specifically to keep track some other way.

Unless a rule tells you otherwise, I would assume that components are limited by their physical natuire unless otherwise noted.


text:
Are starting cards in Principato unbalanced?
In Principato each player randomly gets one card that allows them to build a facility: a house, a bank, a field or a farm. However, only houses and fields yield resources needed to expand your principality, forcing players that drew banks or farms to exchange them quickly. Additionally, it makes these players late starters in a run for farms and houses, which often happens in 4-player setup.
Is there something I'm missing here, or are the starting cards unbalanced?
Note: I know that you can draw a more or less fitting set of starting action and goal cards, but I'm asking only about the 4 starting action cards that allow you to build something.