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What's the point of putting five face-up train cards next to the deck in Ticket to Ride?
What is the point of placing 5 train cards face up next to the deck? I can't see any purpose for this apart from taking 5 cards out of the deck!

ATTN voters-to-close: the question is neither unclear nor opinion-based. It's clear what the issue is and the rules are clear on what the answer is.

Agree with @L.ScottJohnson. While the question could be asked better, the meaning is very clear if you have any familiarity with ticket to ride, and it is not opinion based, since the function of those 5 cards is explicitly spelled out in the rules.

@L.ScottJohnson to me the question sounds like the author is implying that those 5 cards are useless. This is certainly opinion based and depend on your strategy. If you do consider them useless, you are completely free to ignore them and draw from the deck only. Also note, that the author did not accept the well-written and clear answer by  GendoIkari, This indicates to me that the question might be an attempt to bash the rules, rather than a genuine question. The curt tone of the question corraborates that. Any clarifications or rewording from the author are welcome!

@andrewsavinykh One problem with that interpretation is that if you know the rule; I don’t see an even potentially reasonable argument that the cards add nothing to the game. Even the most novice beginner would immediately recognize the value in getting to choose the color of your cards instead of just taking random ones. And answers go in-accepted often; especially when asked by new users who likely aren’t familiar with the working of the site.

@GendoIkari, fair enough. I'm not certain I'm right but it appears to me that the question could be improved in any case.

Rules, even in games as simple as this.  I once met people at a games club who didn't realised they could not take cards AND claim routes on same turn, when this was pointed out they looked shocked.   I only found out a year ago I'd been playing Snowdonia incorrectly for about 5 years.  People do over look rules, no matter how seemingly obvious.  Unlike some questions on here I don't feel like the asker is trying to twist rules to there favour someway and looking for back up.

@AndrewSavinykh If you think the question can be improved, the "edit" button is right there beneath it :-)

@AndrewSavinykh Is it in fact clear that the author "did not accept" the answer? Given they're a new contributor, they may simply be unaware that this is an action they can and should take, nor are there any comments on it indicating that they have actively rejected it.

@StartPlayer A big example is Monopoly.  That seems to have different house rules everywhere, and those seem to rarely match with all of the rules of the game.

@AndrewSavinykh answers not being accepted is far from proof that the answers do not address the question. There are answers with dozens of votes that clearly answer questions, but some users never actually accept answers to their questions for whatever reason. And could be improved isn't a valid close reason for SE questions, if the question being asked is clear enough, improve it or comment on it, closing is for unclear or unsalvageable questions.

@JMac Or the common "$500 on free parking" and "No auction" house rules that most people have grown up with and assume are actually in the rules - Most people learn games based on how the people they play with play them, rather than by reading the rules, that leads them to assume house rules are actual rules.

When you are taking cards on your turn, you can choose to take any of the face-up cards rather than taking one from the deck. Taking a face-up card this way lets you choose what you are getting (as long as the color you want is available).
From the rules:

Draw Train Car Cards – The player may draw 2 Train Car cards. He may take any one of the face-up cards or he may draw the top card from the deck (this is a blind draw). If he draws a face up card, he immediately turns a replacement card face-up from the deck. He then draws his second card, either from the face up cards or from the top of the deck. (See Train Car Cards for special rules for Locomotive cards).

This is a key part of the game, so I recommend reading through the rulebook again; if you missed that you likely missed other important things.
One rule that's related (though not all versions have this, Nordic Countries doesn't):

If a Locomotive card is one of the five face-up cards, the player who draws it may only draw one card, instead of two. If, after having drawn one card the replacement card is a Locomotive, the player cannot take it. If at any time, three of the five face-up cards are Locomotives, all five cards are immediately discarded and five new ones are turned face-up to replace them.

To summarize; if you are choosing to draw 2 train cards on your turn; you choose one at a time to draw the top card of the deck, or one of the 5 face-up cards. If you chose a face-up card, you replace it with one from the top of the deck, and then you make the choice a second time. If you choose to take a locomotive card from the face-up cards, then you can only take 1 card that turn instead of 2.

Furthermore, it opens up strategy of tracking what colors people pick (or choosing to obfuscate your own colors by randomly drawing, even if it puts you at a disadvantage).

In some Ticket To Ride variants (I've seen in the Nordic version) the "Only one if a locomotive" rule is not applicable.  This is explicitly mentioned in the rules as being different from the original game.

To make it entirely clear - when drawing one of the face up cards, you know exactly what you will get, unlike a blind draw. This avoids filling up your hand with cards you can't use.

If players could not choose cards that they needed and it was left completely to the random chance of pulling the cards they needed from the deck, the game would take forever and would get frustrating. Having a few cards of known color to pick from helps speed the game along, and ensures that it is still fun to play.


text:
Identify board game with a 1 ┤SK token
Can someone please help identify the board game which has this token?

It looks like what I have seen in several hidden role games where you have two different roles take actions based on them or bluff on what you claim to have but I can't remember the name.

This is the currency from Coup.

That is the game that I could visualize but not name!

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sort: votes


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Why should a player reveal his victory points if someone else wins?
In the base Catan rule book it says 'keep victory points cards hidden until you have 10 points in your turn and you can declare victory. (You should also reveal them if someone else wins.)
Why should a player reveal his victory points if someone else wins?
Can you please clarify what this means?

You should reveal everything when the game ends so that

You know who came second
You know all the bits end up back in the box
You know someone hasn't made a mistake and failed to claim victory.

The rules state.

If you have 10 or more victory points during your turn the game ends and you are the winner.

Which can be interpreted to mean you don't have to 'say' you have won and people can keep rolling and building, but you are already the winner.
Though this is controversial. And Here, and here.

Pretty much exactly what I was about to write up, except with a different link that I'll leave here for reference: https://boardgames.stackexchange.com/a/21107/16063

@TheThirdMan a much better link. Thanks. Everyone Feel free to edit in more if you find them.

It also stating that the points in your hand counts if you are tracking scores between games.  I I have heard of tournaments where amount of point scored affected your placing.

@BrianMifsud Please note that "Thank you"-styled comments are actually discouraged for the purpose of keeping the information down to a usable minimum. Instead, if you feel like your question has been sufficiently answered, please consider accepting an answer, which doesn't only mark this question as solved, but also awards you and the answerer a small number of points!

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What does the "Appoiintment with the Street of Peace" Monopoly chance card mean?
Can anyone please help me with this card, it is so confusing.

What version/edition of Monopoly do you have; is it in a non-English language?

Appoiintment with two "i"s is non-standard English, for sure.

@L.ScottJohnson Can't believe I missed that!

This card appears to be poorly translated into English from a French version of the game. In the original English, it would be "Advance token to Boardwalk", so you simply move you token forward until you get to Boardwalk, and then deal with it normally (either pay rent, or purchase it, etc).
In the official French version, the card says "Rendez-vous à la Rue de la Paix". The Boardwalk space is called "Rue de la Paix", which translates to "Street of Peace". And "Rendez-vous" is literally "go to", but generally translated as "Appointment" because that's also how it is used.

Not sure what the current wording is, but in the 1970s this card in American sets read "Take a walk on Boardwalk." along with an admonition to follow usual rules once there ("If it is unowned, you may buy it." or something of the sort).

Interestingly it has a copyright from Hasbro, so it's not some random knock-off.

I think the translation of appointment might stem from a mistranslation of "rendez-vous (à)" the French imperative for "present yourself (at)". As a noun "rendezvous" also means appointment.

@Barmar My thought is it could be a counterfeit copy.

Next thing you'll tell me is that Monopoly money isn't legal tender.

Note that it says appoiintment, there's an extra i there, which lends further credence to your bad translation theory. What a shoddy piece of work!

My hovercraft is full of eels.

@QueensKnight It's indeed the formulation used in French Monopoly, at least on my (pre-euro) edition. This card reads "Rendez-vous à la Rue de la Paix".

@AmiralPatate Thanks, I've added that info to the answer!

@Barmar: Typing © Copyright, 2003 Hasbro is pretty easy to do.  So is saying Monopoly™ is a trademark of Hasbro, Inc..

I will not buy this community chest, it is scratched.


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Is there an instant win condition by abusing a loophole in Bang?
I apologize for not using the exact names of the cards as I did not play the English version of the game.
In a gaming session with some friends the following situation arose:

A player had the character which allows to draw a new card when he remains with no cards.
We were playing with the expansion pack which contained random effects whenever a stagecoach was activated. The special effect in this round was "every card can be used as a Bang card, and Bang cards can be used as missed"
That player acquired a volcanic pistol, allowing the use of unlimited bang cards in a round.

Isn't that an instant win, by being able to eliminate all players, one by one? If not, what rule did we miss?

The cards in question are: Showdown

Showdown - All cards may be played as they were BANG!. BANG! cards may be played as they were Missed!.

The designer notes what the card should read.

Actually, the correct text should be that BANG! card must be played as Missed!, otherwise Suzy Lafayette armed with a Volcanic could shoot the whole deck!

and the cards Volcanic and Suzy Lafayette

Volcanic - (Weapon) Max distance: 1. You may play as many BANG! cards as you wish.
Suzy Lafayette - As soon as she remains with no cards in her hand, she draws a card from the deck.

This response differs from the publisher's answer, or lack there of:

Q: Shootout. There are questions about how this interacts with player abilities: Calamity Janet ... Willy the Kid..... Suzy Lafayette. If she has a Volcanic card during a Shootout, she can keep drawing new cards and using them as Bang!s until she kills all the other players, if she wants. Should there be some way to limit these situations??
A: About your questions: please consider that unpredictability is a BANG! feature and it is part of his style. We want BANG! to be characterized by two sides: strategy (bluff) and luck. Characters in Wild West Show are powerful, and sometimes the interactions between two of them can be "lethal" for the opponents, but we think this "chance component" is what makes this expansion so crazy and exhilarating! Have fun with BANG!

The publisher's response sounds like they just didn't carefully check the rules, and assumed that the question was correct about the shoot everyone scenario.

I think that while this issue is resolved, the publisher's answer says clearly that situations may arise that "break" the game, with some many possible combinations.

@jsbueno I disagree with your assessment of their answer. I think they meant it as "Look, some combinations of cards will allow one player to completely take out another player." Not "Look, some combinations of cards will allow one player to win the game in a single turn." I just don't think they realized 'broken' aspect of it. The person that answered was probably a customer service rep, not someone who actually understands non-poker-deck card games.

@jsbueno, I would take the designers word over the publisher.

I'd say this combination (player is Suzy, has a Volcanic and Showdown is active) is rare enough that in our group, we'd let it go just for laughs. If someone was specifically aiming to get this combination, then we would weaken it, either by adding the Bang -> Missed constraint, or tweaking the character selection process (e.g. by eliminating any choice the player has)

I'm struggling to follow if three listed scenario changes due to your clarification...?

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In 2-3 player Citadels, can your second character buy a destroyed building using Graveyard?
In Citadels you have a card that permits you to buy buildings (Graveyard) destroyed by the Warlord. But the Warlord cannot buy the destroyed building. In the 2-3 players variant you select 2 characters.
Can your alter ego buy the destroyed card then?

you wouldn't by chance be referring to Bruno Faidutti's city building card game Citadels, would you?

No. Machiavelli, a.k.a. Citadels, has a purple building called Cemetary/Graveyard. The Warlord or (Condotaire) is not allowed to buy the destroyed building. You don't really have an alter ego, you have two turns during each round. You can only destroy a building during your Warlord turn, and during that turn you are the Warlord and will not be able to use the Graveyard/Cemetary ability.

Graveyard - When the Warlord destroys a district, you may pay one gold to take the destroyed district into your hand. You may not do this if you are the Warlord.
In a two- or three-player game, all players play with two characters each. The game is played normally, except that each player has two turns each round (one turn for each character). Players do not have to separate their gold or their districts between their characters, as the players still only have one city.
Each character has a special ability, also called its power. You may use your character’s
power once during your turn.

Thx for clearing this out.

Good question, and good answer!

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Is the Merchant ability mandatory if you have two identical tiles?
In Libertalia, the character card Merchant says:

Discard two identical booty tiles to gain 3 doubloons, or 3 identical booty tiles to gain 5 doubloons.

Is the ability mandatory? (I.e. If you had 2 chests and 3 treasure maps, would you have to discard at least two of one type?)
The rules note that booty tiles in your den remain hidden until needed. If you don't discard any identical booty tiles, are you required to reveal all your tiles to prove you have no duplicates?
An official FAQ, or designer response is what I am looking for.

Yes. If you are able to perform the Merchant's action, you must. This does involve a certain amount of trust, since you do not have to reveal your tiles to prove that you cannot perform the action. (On that point I would paraphrase the Love Letter rulebook on a similar point, which says "Don't play with cheaters".)
From the rulebook, page 5:

Using a special action is mandatory unless otherwise written on the card. If the action is impossible, it is ignored.

authors: คนทำนา ตัวดำ


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In 4-player Unspeakable Words, what is the likelihood of more than one player not going insane?
Unspeakable Words is a word game where each player tries to score 100 points without going insane. Each player begins with 5 tokens that represent their sanity. Whenever a player scores a word, they roll a d20, and if the roll is greater than or equal to the points you scored, you don't lose a sanity token. If you lose all your sanity tokens, you lose the game.
What are the odds that 2 or more players would survive to 100 points in a 4-player game?
(Note: Assume players can always score words of the point value they wish between 1-20, and will not attempt actions that leave them with a less than 50% chance of winning)

What is the range of wow many points can be scored per word? (Otherwise, every player scores a single 100-point word, takes a single hit to sanity, and wins. Similarly, if you can score as little as 1 point, you just score 1-point words until the end of the game.)

The link you provided includes the chances given a point goal, so what point goal would you assume players are aiming for? Do you assume an equal probability of 1 to 10 points each turn? Or would you want to go back to the actual deck of cards to determine what cards would likely be available?

This would be more appropriate for math.stackexchange.com, and it's an interesting question. Come to Math.SE, we have cookies! And LaTeX.

The link you provided establishes 1 as the lowest possible score, but what's the highest?

Now that I think more clearly about the question, it isn't as interesting as I thought. It might be more inteesting to figure the average turns for a game. Assuming a MAX =20 per turn, a 4p gam will have two players 21% of th0e time that didn't lose sanity for the first two tuns scoring '11' pts. (the more correct calculation uses 10 pts and is closer to 30%)

This question's been unanswered for a while, so I thought I'd take a crack at it.  First up, I should say that I haven't looked at the four player case, just the two player game. The main reason for this is that it makes presenting the data a lot easier. I've put some graphs later, and you can see that they're already hard enough to understand for the two player case!  The methods I've used here could be applied to the four player game, the program would just take longer to finish as there are many more game states.
Assumptions
I've assumed a two player game, where players can choose to score between 1 and 20 points on their go. I've ignored the letter distribution, and whether it might be hard to score a large number of points in one turn.  I've assumed both players start from 0 points and 5 sanity, and have to get to 100 points without losing all their sanity.  I've accidentally alternated between the use of "sanity" and "lives", because this game is rather like a computer game where you start with five lives.
The script
I wrote a program to search through all possible game states for two players, and for each determined the best move.  This isn't as hard as it sounds, the positions near the end of the game are easy to find the best move, and then the positions earlier in the game quickly reduce to a position we've already considered.  The source code is here.
The output of the program is a lot of lines like:
2 3 3 13 [3, 0.16920657742104506]

This line means that if it's player 1's turn, and they have 2 sanity left and 3 points, and their opponent has 3 sanity and 13 points, then the best move is to make a word of length 3, which will give them a 17% chance of winning.
The results
The output of the program is 250000 lines long, which is not very accessible. I've converted this into a couple of graphs.  The first looks like this:

The graph is divided into 25 squares, representing the number of sanity tokens the two players have left. Each square is 100x100 pixels, representing possible pairs of scores. Each pixel represents a possible game state, and the colour shows the length of word player 1 should make to maximise their probability of winning.  Red represents states player 1 should score one point, green represents states player 1 should make 10 points, and blue represents states player 1 should make 20 points.  Colours in between represent values in between, e.g. the states where player 1 should make 3 points is shown by a reddish brown. The black areas are regions where player 2 can force a win, and so it makes no difference what player 1 does.
Another thing I looked at was the probability of player 1 winning in each state.  This second graph shows that:

In this graph white represents a forced win for player 1, and black represents a forced win for player 2.  The graph is not symmetric, as it's player 1's turn.
Conclusions
Some interesting features of the results above.
Steps
Within many of the inner squares there is a step-like pattern. This is clearest in the bottom right square. This is caused by a player being able to force a win. A player can force a win when they have more sanity left than the number of turns it will take them to get to 100.  The bottom right square is where the players have the most lives, and so a forced win is more common.
Rainbow Stripes
In the first graph, at the bottom of many of the inner squares, there are 'rainbow stripes'. These represent points at which player 1 can force a win in one turn. If they are on 99 points then they can force a win with a score of 1 (red), if they are on 80 points then they need a score of 20 (blue).
Dotty Red/Blue Regions
Towards the left of the first graph are some dotty red/blue regions, e.g.:

I'm not completely sure about why these exist. It could be a bug in the program, although I did check it through.  It might be related to the fact that when the probability of winning with different moves is the same, my program picks the largest score. The other thing I noticed was that the best moves, as shown by the link in the question, are the extreme scores.
First Player Advantage
My program gives the first player probability of winning for the starting position as 53.9%. It advises that the best move is to risk a 20 point score (if you don't lose a sanity then you have a forced win by continuing to make 20 point scores).


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Why isn't the minimum bid for a property its mortgage value?
It seems rather odd that the minimum bid for an unmortgaged property in Monopoly is $1, instead of it being the mortgage value (or mortgage value -10% approximately).

Bidding may start at any price.

You can mortgage a property at any time, getting its mortgage value out of it. The only point at which I can see this making a difference is when all remaining players except 1 have less than the mortgage value in assets, which is very unlikely since the maximum mortgage value is $200.
Imagine this scenario. I only have $20 in cash and the only assets I own are Park Place (Mortgage Value $175). Boardwalk is up for auction. The maximum bid I can make is $195, which is cash on hand and Mortgage Value of Park Place. If I can buy Boardwalk for $180 or less, I should do so. The worst case scenario would be that I would have to mortgage Park Place for $175, and pay an additional $5 when I win the auction. I could then mortgage Boardwalk for $200, use $18 to unmortgage Park Place, leaving me with $18 which is pretty much back where I started. While this calculation isn't perfect as it depends upon how many properties you have to sell to get enough cash to win an auction, it would seem reasonable to begin the auction at/near Mortgage Value - 10%.

Keeping the Rules Short
The rules as written follow the KISS principle. This avoids the necessity of worrying about edge cases where no one is in a position to buy a property without selling houses or other property that cannot be easily recouped by mortgaging the newly Auctioned Property.
Simple strategic play should keep most Auction Property from selling for less than its break even point.

From the official rules (emphasis mine):

BANKER… Select as Banker a player who will also make a good
  Auctioneer. A Banker who plays in the game must keep his/her
  personal funds separate from those of the Bank. When more than five
  persons play, the Banker may elect to act only as Banker and
  Auctioneer.

As many people who play auctions in Monopoly know, starting a typical English-style auction at a $1 minimum bid and allowing increments of $1 each makes for painfully slow auctions, especially for higher value properties.  However, the Auctioneer's primary purpose isn't to make auctions fast, profitable, or even interesting; he's there to sell property.  Selling property moves the game along, and the faster the property gets into the hands of the players, the faster everyone can get to the interesting trading and development part of the game.  There should never be any cases where any property remains on the board after someone lands on it.
However, the rules explicitly state that the Banker needs to be a good Auctioneer.  Any Auctioneer who lets all of his auctions run forever for no good purpose doesn't really qualify as "good".  Adequate, sure.  Competent, maybe.  But not good.
Regardless of the starting bid, any decent auction will result in the property selling at a reasonable market price; the only variable is how long the auction runs before this price is reached.  If the opening bid is too high, nobody will bid until the Auctioneer lowers it.  If the opening bid is so low that everybody's bidding, he can make larger increments to weed out the cheapskates.  He has many tools at his disposal to figure out what this reasonable market price is, and exactly which tool is needed depends on the situation.
An Auctioneer who knows how to control an auction as efficiently as possible to sell property is a good Auctioneer.  An Auctioneer who does nothing but lets his players control the auction is just a Banker.
As the opening question states, properties will rarely if ever sell at below the mortgage value, if for no other reason than because other players don't want their opponents to get free money.  As such, starting an auction at or near the mortgage price is perfectly reasonable in most situations.
However, most situations are not all situations.  There will be cases where nobody actually wants the property at mortgage value; it could be because all cash is tied up in developments, it could be because people are saving up for that last property in a colour group, it could just be because nobody likes utilities.  By having no fixed minimum bid, the Auctioneer thus has the option to sell the property at a loss, just to sell the property.  That is his job, after all.  Sell property.
Breaking one of the tools in his toolbox makes for a less efficient Auctioneer, and we don't want that.
The chances of all players being unwilling to spend $1 on a property, which they can immediately mortgage for what is effectively a full refund, is negligible; if the price is right, any property should sell.  Fixing a minimum bid in the rules would open up too many chances of a property going unsold, which would inevitably slow down the game each time someone lands on that square and the whole process starts up again.
Even a property that's immediately mortgaged can eventually be paid off and activated, or traded to other players.  An unsold property is just useless.

We've had worthless properties before in games; that is their expected loss on income tax exceeded their expected gains on rent. If they fell to a bankers auction they might not get bid on for even $1. Buying a property to mortgage is actually another loser. There's a reason why late in the game I start unmortgaging all these singles I happen to still own.

@Joshua That makes no sense if you run the numbers, property is never worthless. A free property immediately earns you the mortgage value of the property, but income tax only costs 10% of that value. You could land on income tax 10 times and you'd still break even! This doesn't even account for any rent you could collect if the property is left unmortgaged. Rent and income tax are ~10% of the purchase price, so it would be very unusual to pay more in income tax than you earn in rent. You would have to land on income tax more often than everyone else combined landing on your property.

One reason comes to mind:
Keeping the game short.

Monopoly is noted as a long game; when played by the rules as written, it's not overly so — 1-4 hours.
there are points, especially on the first circle through, where the players will not have enough cash to buy everything at mortgage cost, especially Boardwalk and Park Place.
the auctions are to get as many properties into player hands as quickly as possible; increasing the minimum bid lengthens the game by slowing this down.
the meat of the game is the management once all properties are owned. Getting there quickly prevents the game from stagnating into a 6-8 hour grind.

Why you shouldn't mortgage properties to buy others if you can avoid it:

Mortgaged properties, by not generating cash exchange, lengthen the game.
Mortgaged properties can't be improved, and prevent having houses/hotels on the rest of the color group as well, due to the requirement for balanced development (a difference of 1 house is allowed)
Mortgaged properties are essentially additional "free parking" spaces unless and until you either are forced to sell them or you hit a tax space or card.

You should go ahead an mortgage a property if you win a bid you can't afford to complete a color group; mortgage something from a different color group.

No, you've missed the point: setting a minimum bit higher than 1 dollar lengthens the game.

@user1873: It's only "giving money away to other players" if no other player raises the bid up to the breakeven point.  If absolutely nobody is willing to counterbid for Boardwalk while it's such a bargain, why should the first bidder not get it?

The point is that if NOONE hash enough cash to pay the morgage cost, it can still be bought (and thus come into the game).  So basically, no matter what, when someone first lands on a property, it will be bought by someone.  Unless the player are being silly.

@user1873 The players can start at any bid; setting the minimum above 1 creates a very real possibility of no one being able to bid at all.  Even at 1, it's very possible for one player or another to be unable to bid without mortgaging, but it's far rarer. There is no requirement for sequential bids, so your claim of a $1 min lengthening things is bogus - players raising sequentially are being slow. Nothing in the rules prevents you from making your opening bid the mortgage value. Or even the cash value.

Slightly off the question's topic, but I do have a slight philosophical disagreement with the idea of not mortgaging properties to acquire more. I'm not a Monopoly strategy expert, but the best players I know all liberally mortgage properties and use the proceeds to drive up their rents elsewhere; one landing on a medium valued hotel will often bring in enough cash to unmortgage most it. I agree with your caveat that this should definitely done to complete a group, but I think there are enough other situations (trade bait etc.) that I wouldn't characterize it as a bad idea overall.

One purpose of the auction rule is to speed up the game. Basically, the designers wanted property to get in the hands of players as quickly as possible. So if the player who lands on a property won't buy it for list price, then the designers want it auctioned.
It could be that all (or all but one) players are very short of cash, to the point where they couldn't pay even the mortgage value for the property. (This is particularly true if there is a house rule that says that players can mortgage their own property at the beginning of their turn.) Then someone would get a "steal" by buying the property for less than mortgage value, then mortgaging it on a later turn.

Newest standard edition (post 2008) rules say "Bids start at M$10".


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In Dungeon Run, is there ever a reason to assist another player if you have full health?
In Dungeon Run, when another player is on your tile, you can assist them. If you do, you are allowed to rally which means to remove a single damage from your character.
Since any other player will eventually be an opponent, does it make any tactical or strategic sense to assist them if you don't get a benefit?

I can think of a couple of situations where assisting another player without being able to rally might be beneficial to you.

If not doing so will most likely lose you the game - If an opponent has the summoning stone, and is rolling many more dice than another player, it may be useful to help that player damage the summoner. You will most likely be wounding your opponents, with no possibility of hurting yourself. If you cannot do enough damage by yourself or others to kill the summoner, then you will have no chance of winning the game without getting the bonus dice from assisting.
If another player is out of Actions and a Monster has a treasure - The monster will drop the treasure, and you will be able to pick it up during your turn (thanks to the other player).

Sure.  There are several reasons you might want to do this.  Obviously if there's a monster that will threaten you, and you're worried about it, and your opponent can't kill him on his own, you may want to assist.
On a more strategic level, you may want to cooperate, especially if there are more than 2 players.  See http://en.wikipedia.org/wiki/Tit_for_tat for a game theory perspective on building trust.  If you and another player are cooperating, you both get a benefit relative to the other players (as the next time he'll assist you, even if he doesn't have damage).

+1 for important general theory.  Even in purely-competitive games, if there are more than two players then any two who cooperate have an advantage over all people who don't.

@Tynam, General Theory is great an all, except when it doesn't apply. Tit for Tat works in a non-zero-sum game which is played over a period of time, and it works because both parties win more by cooperating. Without receiving a free rally, I don't know if this intangible "trust" will help you level up faster, or remove the Summoning Stone from the dungeon. If I help someone else defeat a monster, I actually only benefit that player relative to other players including myself. That player is half way to leveling up, if I help them escape I cannot receive help to escape myself.

@user1873: Your comment is correct, but only if you're only playing once.  If you and another player are cooperating well over multiple games - even when it doesn't help you in one game - then the trust advantage lets you both win more games than others who don't cooperate.  In Dungeon Run this is a small consideration, but it's not quite zero.  (As you correctly say in your answer, in a single game the only real reason is to help remove a monster, or winning player, that you want gone.)


text:
Can you only buy one extra die with the The Calm Before The Storm destiny square?
On the Board of Destiny for the Demon player, there is a destiny square that allows you to use any die to buy an extra die next turn. Can you only buy only one extra die with this ability, or can you use multiple dice to buy more than a single die?

You can only buy a single die, per Threat Phase.
The Destiny Board says:

Any Die Score: Roll an extra Die of Destiny during the next Threat Phase.

Rules page 8

(red) The Calm Before The Storm (red) allows the Demon player to roll an extra Die of Destiny during their next Threat phase.
Abilities printed in (red) can only be selected once per game Threat Phase (Errata: claustrophobia-theboardgame.com/fr/?p=441). There are no limitations for the Abilities printed in (black).

From the errata (en Francais) at http://www.claustrophobia-theboardgame.com/fr/?p=441 red abilities are limited to once per threat phase, not once per game.

Thanks for the catch. We were playing this wrong for a while. It became especially clear on the mission where you delve into the sewers towards the Demon with 10 HP, and the Demon player has nothing better to do than buy events and extra dice (for 12+ die totals, instant damage). The demon player was rolling 4 extra dice one turn (incorrectly of course).


text:
 With Print&Play board games, should the focus be beauty or simplicity?
Many Print & Play (P&P/PNP) board games come from a designer's desire to get people to play their game. If the intent of designing a P&P board game is getting the most people to play it, should the components be designed to look the most beautiful (full color images, art, etc.), or should the components be designed to be functional and less ink intensive (grayscale/b&w images)?
Does the need for a large amount of printed components scare too many people away? If functionality isn't affected too greatly, can substitution components be used?
Example: a randomizer without replacement (a deck of cards), with just a chart/look-up table showing what each card represents vs. creating card art and rules text for a brand new deck of cards?
Basically, do people who PNP prefer a good looking game, or a cheap and easy to create game?

Is "both" an option? One set of high-quality images, one set of ink-saving line art?

I'd like to hear more about this print-and-play concept...

Don't conflate beauty and functionality too much. They can go hand in hand, but whether to use rules text or a look-up table is more of a usability question than a beauty question.

@AlexP, both is always an option. I was just wondering what is most important to people who actually P&P. I realize that the question is subjective (and probably needs some cleaning up), but the only people who can answer this question are people that P&P, and their opinion about what gets them to print one game over another is what I am looking for. Do they care if the game requires a lot of printing of components if the components look good, or do they want the minimal amount of work to determine if a game is worth investing their time/ink for.

I haven't printed a game in a good long time, but I just downloaded one, and my first thought was "wow, this looks like crap" and I was far less inclined to print and play it.  Take that for what it's worth.

Generally, the best solution is to have a very functional design first, with a "pretty version" using the exact same underlying design, but with full color pretty images, including both in the package. The simple functional version will often be the trial - it needs to be low ink, and preferably grayscale.
If the underlying design isn't functional, the game won't get played, no matter how pretty...
As an aside: for a P&P, if people wouldn't play the cheapass games edition of it, they're unlikely to play the fancy versions, either. Note that several of Cheapass Games' could-have-been-laser-printed games later had much pricier full color versions sold. Some examples include Cartagena, Give Me The Brain, and Lord of the Fries.
Hell, people will play some DAMNED F'UGLY games... so long as the play is good.


text:
Can an opponent make me discard a Miracle card before I cast it, after it is revealed?
For example, if my hand is empty and I draw and reveal Temporal Mastery, can an opponent cast Funeral Charm and make me discard it before I have a chance to cast it for its Miracle Cost?

Yes. Miracle triggers upon revealing the card when you draw it. You cannot cast the card until the triggered ability resolves and you pay the cost. Until that time, the card is in your hand and subject to the effects of discard.

702.93. Miracle
702.93a Miracle is a static ability linked to a triggered ability (see rule 603.10). "Miracle [cost]" means "You may reveal this card from your hand as you draw it if it's the first card you've drawn this turn. When you reveal this card this way, you may cast it by paying [cost] rather than its mana cost."
702.93b If a player chooses to reveal a card using its miracle ability, he or she plays with that card revealed until that card leaves his or her hand, that ability resolves, or that ability otherwise leaves the stack.

Won a game once by using Surgical Extraction on Bonfire of the Damned in response to my opponent revealing a second Bonfire. Good times.

It depends on whether the Miracle card is a Sorcery or Instant
It would be correct to say: Your opponent can make you discard a Miracle card before you can cast it for its Miracle cost, as user1873 already pointed out.
If the card is a sorcery, you can't usually play it outside the main phase or with a non-empty stack, so for your specific example, he could make you discard before you can cast it in any way.
If it's an Instant, however, you can still play it for its regular cost in response to the Funeral Charm.

Indeed, I was wondering about paying the Miracle Cost. I have clarified the question to reflect that.

If you are casting it as a Miracle the Avacyn restored FAQ says specifically that you ignore timing restrictions.  I am not sure the Sorcery vs Instant applies to this question.  Restored FAQ

@Akedren We already established that the Charm is faster than the Miracle alternative cost. The original question asked whether the card can be played at all, and if the Miracle card is an Instant, it can be played for its regular cost rather than its Miracle cost, in response to the Charm. You can choose between Miracle and regular cost, you are not forced to play it for its Miracle cost if you reveal the card.


text:
Can I Wish for a card outside my Commander's colors?
I have been told that Wishes work differently in Commander than in normal casual Free-For-All (FFA) mtg. Do the comprehensive rules cover how wishes work in Commander? Do the rules for wishes differ from normal FFA multiplayer?

Wishes are non-functional in Commander, unless your group specifically allows them and defines where they work.
The comp rules define the basics of the format but mostly point you to mtgcommander.net for the up-to-date official definition. Per mtgcommander.net,

Abilities which refer to other cards owned outside the game (Wishes, Spawnsire, Research, Ring of Ma'ruf) do not function in Commander without prior agreement on their scope from the playgroup.

Basically the Rules Committee's reasoning was that using wishes in their original, casual-play way (searching your entire collection) wastes too much time in play1. So they used to restrict you to searching your sideboard, which is how wishes work in competitive MTG. However, Commander's sideboards are "optional" and Sheldon has generally come out against them, so the default is that wish cards don't work at all.
Sideboards still have to follow all the normal deck-construction limitations, like in any other format; for Commander, this means you can't have break the one-of restriction or violate color identity just because a card is in the sideboard. Thus, if you group agrees to play with wishboards, you'll still be limited to your commander's color identity.
(Note also that the official sideboard size is 10 cards, not 15 like 60-card constructed.)

1 – Which is also why Shahrazad was finally banned after some time as the only Vintage-banned card legal in Commander.

I got 99 cards but a Wish ain't one... sorry, I had to.

I have done a search for all Commander references in the CompRules and have found nothing specific. There are several points to consider when determining this question in particular:

Commander is a semi-official format. It's covered in the official rules, but it has been invented and is maintained by regular players. So questions specific to the format that are not answered by the CompRules should be probably be asked and answered there.
Going by the rules, I would argue that Wishes work the same as in regular games: Since a Commander game is casual only, you are allowed to get any non-ante card from your collection. In sanctioned events, Wishes can only get cards from your sideboard.
Going by the philosophy of the format, Wished cards should probably be restricted to cards in the Commander's color identity. But that is more a house rule, as the CompRules don't regulate that particular point.


text:
What first turn kill in multiplayer FFA provides the best ratios?
People love to figure out first turn kills in Magic: the Gathering, and have been around since the beginning, when someone threw together a deck with Channel, Black Lotus, Fireball, Mountain. Pulling off a first turn kill in multiplayer is much harder, because you need to do much more damage to your opponents.
Assume the following:

You have 4 opponents.
Legacy Banned List.
You get to determine every card you draw (including starting hand).
Your opponents take no actions. (goldfish).

What is the maximum amount of total life loss that is possible? Not sure of the best way to rank submissions, perhaps fewest cards used, least mana used, most life loss, or least expensive cards. Perhaps an average of those values.

I don't like "you have perfect luck" thing in puzzles, because it allows you to ignore one of the biggest constraints in deck design. So my vote goes to 60 copies of Relentless Rats with the "reveal" ability of Chancellor of the Dross painstakingly pasted onto each one of them. >.>

@AlexP, CotD gets you more than half way there. You only need 8 more life lost in 4 cards. That shouldn't be too difficult. When you are talking about first turn kills, you pretty much have to have perfect luck.

False Cure deck - I used to have a Vintage version of this deck, using proxies for all the expensive cards in the deck.

Play Overgrown Tomb ($14)
Cast 1xDark Ritual ($1)
Cast 1xFalse Cure ($1.50)
Cast 4xSkyshroud Cutter ($0.10) for alternate casting cost.

First Skyshoud Cutter causes opponents gain 5 life (25 life), False Cures and opponents lose 10 life. Remaining 3 Skyshroud Cutter's do the same.

7 Cards (AVG 22.9 life/card). Cheating a little since we are giving opponents life
6 Mana (AVG 55.3 life/mana), or using CMC instead 8.42
40 * 4 life = 160 total
$14 + $1 + $1.50 + 4x$0.10 (approximate $16.90 total, AVG 9.47 life/$) (Prices by TCGPlayer.com, 5th lowest Near Mint price)

Not that big of an improvement, but you can make this cheaper (in monetary price) if you trade 1 Dark Ritual and 1 False Cure for 2 Skyshroud Cutters. Obviously, you can also trade de Bayou for an Overgrown Tomb :)

@rahzark, Thanks for the improvements. I should have remembered the pain lands (huge $ improvement). I will update the post with your suggstions. I wonder if any T2 solutions exist?

Geth's Grimoire deck - Win through discard. Half as cheap, with a better AVG life loss per dollar.

Swamp ($0.05) {B}
Dark Ritual ($1) 3x{B}
Priest of Gix ($0.25) 3x{B}
Priest of Gix ($0.25) 3x{B}
Culling the Weak on Priest of Gix ($0.50) 6x{B}
Culling the Weak on Priest of Gix ($0.50) 9x{B}
Geth's Grimoire ($0.25) 5x{B}
Rotting Rats ($0.05) 3x{B}

8th card played, opponents discard down to 6. Draw 4.

Culling the Weak on Rotting Rats ($0.50) 6x{B} #Threshold#
Cabal Ritual ($0.75) 9x{B}0
Liliana's Caress ($1.20) 7x{B}
Lilianas Caress ($1.20) 5x{B}
(Unearth) Rotting Rats 3x{B}

Opponents discard down to 5 and -4 life (16). Draw 4.

Cabal Ritual ($0.75) 6x{B}
Delirium Skeins ($0.05) 3x{B}

Opponents discard down to 2 and -12 life (4). Draw 12.

Delirium Skeins ($0.05) 3x{B}

Opponents discard down to 0 and -8 life (2 triggers remain on stack). Game Over

15 cards (AVG 5.3 life/card)
32 mana (AVG 2.5 life/mana)
80 life loss total (16 on the stack not included.)
$7.40 (AVG 10.8 life/$)

Damned Spirits - Spirits to grave, and turbo mana. Cheapest combo so far.

Swamp ($0.05) {B}
Dark Ritual ($1) 3x{B}
Dark Ritual ($1) 5x{B}
Priest of Gix ($0.25) 5x{B}
Culling the Weak on Priest of Gix ($0.50) 8x{B}
Iname, Death Aspect for 20 common Spirits ($0.20) 2x{B}
Songs of the Damned ($0.10) B + 21 = 22x{B}
Exsanguinate for 20. ($0.05)

Stats

8 cards (AVG 10 life/card) I suppose you could "cheat" to increase the AVG by filling the deck with 52 cheap spirits. This would allow you to Exsanguinate for  52 (AVG 26 life/card). The highest life per card average, but not realistic.
35 mana (AVG 2.5 life/mana)
80 life lost
$4.65 (AVG 17.20 life/$)

Storm Kobolds - Storm for infinite damage (could Brain Freeze for mill and then Burning Inquiry to draw opponents out).

Swamp ($0.05) {B}
Dark Ritual ($1) 3x{B}
Priest of Gix ($0.25) 3x{B}
Burnt Offering on Priest of Gix ($0.10) 2x{B} 3x{R}
Cloudstone Curio ($2.50) 2x{R}
2xCrimson Kobolds ($1 each) For infinite bounce/Storm
Grapeshot ($0.05)

Statistics

8 cards (AVG 10 life/card)
10 mana (AVG 8 life/mana)
80 total life loss (infinite technically)
$5.70 (AVG 14.04 life/$)


text:
Does the Preacher affect the Booty/Loot on the ship or in a den?
In Libertalia, the character card Preacher says:

Discard all booty tiles but one.

Do you discard booty tiles from the ship, from your den, from every den, or some other variation?
An official FAQ, or designer response is what I am looking for.

It is unfortunate that the French rules are no clearer nor the French Priest card, "défaussez toutes vos parts de butin sauf une." Neither

The rules don't specify, but my interpretation agrees with this player aid in that it's meant to be your booty tiles.
EDIT: A BGG post from Stefan Brunell the American publisher here confirms this.

[Q2:] The Preacher has a Dusk effect of "Remove all but one booty token." Is this referring to that day's booty, or your own inventory? Do the removed tokens go back to the bag or do they end up in your inventory?
Second: Own inventory, to the bag.

Publisher is definitive enough for my tastes.


text:
Risk (revised edition 2008): Is it better to roll 1 or 2 Defense dice with the Bonus Defense die Major Reward?
The Risk revised edition rules have a Major Reward that grants the defender a bonus defense die. This die will replace the lowest defender's die rolled (unless the bonus die is the lowest).
What are the expected troop losses for the Attackers and Defenders when the Attacker attacks with 3 troops, and you choose to roll a single Defense die and your bonus Defense die versus when you roll two Defense dice and your bonus Defense die?
I have worked out the math for the single Defense die plus bonus Defense die. This is the easiest calculation, because we only care about the highest die rolled for the attacker (out of 216 possibilites), and the highest die rolled for the defender (out of 36 possibilities).
            |Defender's high die and odds (1-6) |
+-----------+-----+-----+-----+-----+-----+-----+-----+
|           | 1/36| 3/36| 5/36| 7/36| 9/36|11/36|36/36|
+Attacker's +-----+-----+-----+-----+-----+-----++----+----+--------------+
|   High Die| (1) | (2) | (3) | (4) | (5) | (6) ||Wins|Loss|              |
+-------+---+-----+-----+-----+-----+-----+-----++----+----+--------------+
|  1/216|(1)|   1 |   3 |   5 |   7 |   9 |  11 ||  36|   0|   36 of 7776 |
+-------+---+-----+-----+-----+-----+-----+-----++----+----+--------------+
|  7/216|(2)|   7#|  21 |  35 |  49 |  63 |  77 || 245|   7|  252 of 7776 |
+-------+---+-----+-----+-----+-----+-----+-----++----+----+--------------+
| 19/216|(3)|  19#|  57#|  95 | 133 | 171 | 209 || 608|  76|  684 of 7776 |
+-------+---+-----+-----+-----+-----+-----+-----++----+----+--------------+
| 37/216|(4)|  37#| 111#| 185#| 259 | 333 | 407 || 999| 333| 1332 of 7776 |
+-------+---+-----+-----+-----+-----+-----+-----++----+----+--------------+
| 61/216|(5)|  61#| 183#| 305#| 427#| 549 | 671 ||1220| 976| 2196 of 7776 |
+-------+---+-----+-----+-----+-----+-----+-----++----+----+--------------+
| 91/216|(6)|  91#| 273#| 455#| 637#| 819#|1001 ||1001|2275| 3276 of 7776 |
+-------+---+-----+-----+-----+-----+-----+-----++----+----+--------------+
|216/216|                                       ||4109|3667| 7776 of 7776 |
+-------+                                       ++----+----+--------------+
Defender Wins 4109/7776 = 0.52842, expected losses per attack 0.47158
Attacker Wins 3667/7776 = 0.47158, expected losses per attack 0.52842
Ratio of Defender Losses/Attacker Losses = 0.892%

In the 3 Attacker dice versus the 2 Defender dice plus 1 bonus dice, both players throw away their lowest die. So, all that is needed to calculate the losses is to calculate the chances of each two highest die rolls from three dice (out of 216), and then figure out the losses (although, the results are a little more complicated because you have a Win/Loss/Tie (each lose 1).
Defender Wins 0.47614, and ties 0.29308, expected losses 0.75463 troops
Attacker Wins 0.23077, and ties 0.29308, expected losses 1.24537 troops
Ratio of Defender Losses/Attacker Losses = 0.606
I will fill in the table later, but can anyone check my results?

I think your calculations look roughly correct (my calculations disagree at 4 decimal places, but I doubt you're using that many in an actual game!)
Here's my calculated probabilities for a single defence die (plus bonus die):
Defender wins 4109/7776: 0.528421
Attacker wins 3667/7776: 0.471579

Here's my calculated probabilites for two defence dice (plus bonus die):
Defender wins 22218/46656: 0.476209
Attacker wins 10767/46656: 0.230774
Ties 13671/46656: 0.293017

Since the sample space is only a few thousand dice rolls I wrote a python script to loop through all of them.  Here's the code for the first calculation:
from itertools import *
a_score = 0
d_score = 0
for a1,a2,a3,d1,db in product(range(6), repeat=5):
    if max(d1,db) < max(a1,a2,a3):
        a_score += 1
    else:
        d_score += 1
print 'Defender wins %d/%d: %f'%(d_score, 6**5, float(d_score)/6**5)
print 'Attacker wins %d/%d: %f'%(a_score, 6**5, float(a_score)/6**5)

Here's the code for the second:
from itertools import *
a_score = 0
d_score = 0
tie_score = 0
for a1,a2,a3,d1,d2,db in product(range(6), repeat=6):
    a_dice = [a1,a2,a3]
    d_dice = [d1,d2,db]
    a_dice.sort(reverse=True)
    d_dice.sort(reverse=True)
    if d_dice[0] < a_dice[0] and d_dice[1] < a_dice[1]:
        a_score += 1
    elif d_dice[0] >= a_dice[0] and d_dice[1] >= a_dice[1]:
        d_score += 1
    else:
        tie_score += 1
print 'Defender wins %d/%d: %f'%(d_score, 6**6, float(d_score)/6**6)
print 'Attacker wins %d/%d: %f'%(a_score, 6**6, float(a_score)/6**6)
print 'Ties %d/%d: %f'%(tie_score, 6**6, float(tie_score)/6**6)


text:
Can Zombies wound when a Hero rolls doubles?
I was playing Last Night on Earth the other day with some friends and came across a question that I don't believe is clear in the rules.
When a Hero fights a zombie (or the other way around), the hero can only be wounded if the zombie's fight die is tied or higher than the highest hero's fight dice.  What if the hero rolls doubles though?  For example, the hero could roll two 1's, but the zombie rolls a 6.  In that situation, does the hero get wounded?  Or is the zombie considered killed before the hero can be wounded?

The Hero gets wounded. In your example, the Zombie rolled a '6', and Zombies win on ties. The Hero must have a higher die for the Zombie to not win. If the Hero has a higher die, but no doubles then the Zombie is Fended Off. The Zombie is only wounded if the hero has a higher roll and has rolled doubles on any of their Fight Dice. From the rule book (page 15).

Resolving the Fight - A Fight is only resolved after all players have decided not to use any more cards or abilities. Once both players have chosen their highest Fight Dice roll and are done using cards and abilities, the fight is resolved as follows:
If the Zombie wins (has a higher die or is tied for highest), the Hero takes one Wound.
If the Hero has a higher roll, the Zombie is Fended Off. The fight ends and both the Hero and Zombie remain in the space, unaffected.
If the Hero has a higher roll and has rolled doubles on ANY of the Fight Dice (two 5's for instance), the Zombie takes one Wound (removing it from the board).


text:
Would a turn order opposite the disprove order affect the strategy or gameplay in Clue?
The official Clue rules call for a clockwise turn order, and a clockwise order of disproving suggestions. Would changing the order of the turn, or the order of disproving a suggestion so one is the opposite/reverse of the other have any meaning effect on strategy or gameplay?

Miss Scarlet— the player with the red token— always plays first. Play then proceeds, in
  turn, to the first player’s left.
Proving a Suggestion True or False - As soon as you make a Suggestion, your opponents, in turn, try to prove it false. The first to try is the player to your immediate left. This player looks at his or her cards to see if one of the three cards you just named is there. If the player does have one of the cards named, he or she must show it to you and no one else. If the player has more than one of the cards named, he or she selects just one to show you.
If that opponent has none of the cards that you named, then the chance to prove your
  Suggestion false passes, in turn, to the next player on the left.

Shouldn't one of the words "left" in the yellow section above be "right"?  Or is this a quote from the original rules?

@tttppp, No, this is a quote from the rules. My question is in regard to how the game might change if the rules were modified so that the direction of either turn order/disprove order is reversed (opposite) the direction of the other.

The player who reveals his card always learns a little less than everyone else. This is because everyone else learns that he has one of the suggested cards, but the revealer already knew that.
In the normal rules, the next player to act is also more frequently the player who reveals a card. If you reversed the suggestion order, this disadvantage would go away. This probably only has a minor impact, but you'd have to test to be sure.

One variant I've played is to make suggestions to a specific opponent.  That opponent is then first to either show you a card or say they have no matching cards, and it then proceeds clockwise from there, until either one of your opponents shows you a card or noone does.
This allows a bit more strategy in that you can tailor questions for specific opponents.

I like this variant, I think I might try it.

It makes a subtle difference, in that, to prove you wrong, they have to reveal a card in your proposed solution.
If you (Ay) say Col M in the Kitchen with the Pipe, and the next player (Bee) has Col M, the one after (See) has Kitchen and Pipe, you don't find out that See has kitchen or pipe; you only know that Bee has the Col M card.
To find out if Kitchen and Pipe are out there, you need to use a guy that YOU don't have, and Bee doesn't have, so that See has to reveal either kitchen OR pipe.

Also note, in the above example, See knows which card was revealed to Ay, because he has the other two. He can now mark it off. But going the other way, Bee doesn't know which card was revealed to Ay. The reverse can happen in other ways, too.

If you go with first to reveal, you might find out about the pipe or kitchen AND Col M.
Note that, if the first to disprove is the one who played last, you have reason to suspect that they don't have their called suggestion; it changes slightly the strategy in what you pick, because you know that they will get to examine you based upon what they have just tried to solve with.
If you go with simultaneous reveal, you guarantee learning more than what you're supposed to.
If you just change the turn order the other direction, you learn either the pipe or the kitchen, but not the Col.

I am not sure what you are getting at (2nd person who misunderstood me). I am not talking about simultaneous reveals. The difference between a turn order and disprove order being opposite each other, is that your suggestion/suspicion is disproved (if at all), by a guy to your right. When the nest players turn (guy to left) makes a suggestion, the first person who disproved him (if at all) is you. If you use your own cards in hand during suspicions, the next player in TO doesn't see what the player to your right shows you, they see your card. Does this have an impact on how to suggest?

I can't imagine changing turn order would change much. When I play, we usually roll to see who goes first and pick you pick a character on your first turn. It doesn't seem to affect things.
Changing the order of proving a suggestion True or False, could change things if the order was dynamic. I think any fixed order would be equivalent, but if the order were randomly determined after a suggestion is made, that would need to be accounted for in the suggestion. Sometimes I'll be really only testing one person on one card, so I'd have to be more careful if I didn't know the order. This would probably also slow down gameplay and be a bit confusing.
I suppose you could have the suggester (or another player) determine the order of proving accusations false on each turn. This would change the game, at least if I was playing and following the strategy outlined in this answer, because then (following that example) Player B could skip straight to Player A and capitalize more efficiently on what she learned from Player B's turn. Thus I think it would weaken that strategy.

I don't mean changing it each turn, I mean having the turn order be the reverse of the disproving order. In your answer to When to Use Cards From Your Hand, you state that you ususual use 1-2 cards, because it allows for a greater number of players to not disprove your Suggestion. Imagine though, if TO was clockwise, but disproving was counter-clockwise. If your Suggestion went all the way around, the next player in turn order wants to Check to see if you just discovered a card in the envelope, since you are immediately RIGHT, they know the cards in the envelope, or at least 1 in your hand

If they make the same accusation, they know 1 in your hand. Nothing more. They still need to check out one other card (and assume you didn't do the weird all-three-from-your-hand). If they make a different accusation, then they should change it only by adding a card from their hand (or one they know to be in the envelope) otherwise they risk you showing a different card, in which case they still need to check those previous 3 cards.


text:
How likely is it to win a game of Barbarian Prince?
Barbarian Prince is a brutal paragraph based game. Your character must collect 500 gold and return to his homeland to win the game. Unfortunately, there are unfriendly guards, wolves, mobs of peasants, deadly ruins, and worse trying to end your life.

What are your approximate chances of winning a game within 70 days?
What is the best strategy to use when attempting to win?


text:
Does mixing Dominion expansions lessen the power of card combinations in random kingdom setups?
Although they can be mixed together, Dominion sets seem to have many cards that were designed to combo with other cards in the same set. Scout works better with the cards that are both victory cards and actions (Great Hall) or coins (Harem), rather than cards that are just Victory cards, such as those found in the basic set. Seaside's Treasure Map works really well with Warehouse's deck-digging ability. Cornucopia benefits from the Tournament Prizes that help diversify your deck.
Does mixing 3+ sets of Dominion together and selecting 10 random kingdom cards lead to games that are less focused on card interactions? In other words, does individual card quality become more of a driving factor?

Added: A fundamental property of Dominion is card interaction, so to claim (not that anyone has) that a certain expansion doesn't increase the amount of positive interaction between cards/sets would be quite a criticism of said expansion. It's possible, but I haven't seen it yet. (Of course, Alchemy does create a sort of artificial exterior game with the potions, and some dislike it for this reason. But I digress).
So what I should have said from the start is that every set is designed to make games more interesting, especially when combined with other sets. Also, the main way that I have developed strategy over the years is to look at a (random) set of 10 kingdom cards, and start looking at the interactions. As you'll see below, you can use (the cards that you add to your hand with) Scout for two major purposes: trashing, and discarding. So when Scout is in play, I quickly (?) decide if either of these is a viable option given the other cards out there, and then do the same for all the other cards and their main uses.
Consider the combo:
Golem (Tactician, then Library/Minion/etc)
There are other examples, but they aren't fresh in my mind (I quit playing after 2000 something games on isotropic, and 1000 on BSW...).

Added:
While I stand by my answer, it seems that the best answer to this question would involve, at the very minimum, an analysis of how every card interacts with [all the other] cards from other expansions. At worst, we'd have to consider all 10^14 possible sets of kingdom cards!
Here is my analysis of Scout:
Scout: Like the Apothecary, Scout can be used to add 3-4 cards to your hand (which may not have that great of a direct value - e.g. coppers/estates). Then you can combo (for different reasons, effects on your deck) this with:

Ambassador.
Warehouse
Vault (+ Tunnel = BOOM)
Apprentice
Transmute
Forge
Hamlet
Horse Traders
Young Witch
Hunting Party
Crossroads (!! Forgot about this one...)

Furthermore, with Hoard, the strategy of using Scout to get cards in your hand for the purpose of trashing opens up more possibilities. Scout is a good counter to Rabble. Scout might be the only +Action in the set, so then your only option to net an action would be to combine this with Golem, KC, or Throne Room. This is, of course, the last thing that crosses my mind when analyzing a set of 10 cards in-game.

And then there's Scheme, which in itself opens up combos with most Action cards!
Analysis of Black Market:
Besides gaining a card that no other player can have, Black Market opens up a unique set of possibilities due to its extra buy phase. So during the execution of Black Market, you can [put into] play all of the treasures in your hand, possibly leaving you with 1-2 Action cards. In any case, you still have the +$ to spend from those treasures! If the other cards are, say, Library or Minion, then you can draw a bunch more cards and have a longer turn. If you are left with a Tactician, then you can use that to start your next turn with 10 cards, and you still get to spend the money that was played during the Black Market!

Well, obviously King's Court will make combos where there might not have been any before.

but are you considering the number of games you have to play that don't have any interesting combos, because the distribution of the cards in your mixed set all have cards that care about different things. The beauty of Scout being played in an Intrigue only game is that there is that if Scout is in the game, there is about a 50% chance that Great Hall or Nobles or Haram is in the game too. When playing with so many different sets, how many random setups do you have to play before you see an interesting interaction? Edited the question to indicate that I was interested in Random kingdoms.

good point about the total kingdom set combinations C(127,10) = 127! / 10! (127 - 10)! =
209,123,798,385,425. It would require marking each card with a 127 bit flag that indicates that this card combos with all the cards with "on" bits. Then there are the actual comparisons (127+1)/(127/2)= 8,128 which need to be done by a human. Quite a difficult problem. I suppose a couple of random cards could be chosen, and analysis performed just on those cards. I wonder how large of a sample, is necessary to justify an accurate answer.

And here I was calculating (125, 10)! :)

You had asked about Donald X.'s response to your Scout analysis; while I can't speak for him, I can give you some thoughts. You're right that Scout is useful for clearing out your deck of otherwise un-useful cards, and so discard-for-benefit cards like Warehouse, Cellar, or Vault become more powerful. But there's lots of limitations to that, most notably that a deck with lots of green in it is less likely to allow you to draw the cards you need for the combo at the same time as Scout.

Thanks for the thoughts, @philosophyguy. Let's not forget that your need not draw any of the cards mentioned in my analysis of Scout, as they all work (to varying degrees)...

It's worth mentioning that you don't ever expect every single pair of cards to have an interesting reaction (with everyone super, no one will be!), and also that "interesting" is difficult to define. All you really need is a few pairs out of the 45 to be interesting. (Also note that with just base+intrigue, and especially with just base, there aren't an overwhelming number of "interesting" interactions.)

"Scout can be used to add 3-4 cards to your hand (which may not have that great of a direct value - e.g. coppers/estates)." If you're using Scout to put Coppers in your hand, you're cheating.

@Pow: Thanks, but I'm sure the meaning is clear from the context! I'll work on my parenthetical phrasing in the future. With 3000+ games logged online, I'd like to think that I haven't been "cheating"...

In general, adding more expansions will allow more interesting combinations to emerge. Scout is one of the notable exceptions; in fact, Donald X. (the creator of the game) mentioned that Scout is one of the cards that wasn't as well balanced outside of Intrigue because it was made earlier and he was still improving in his ability to develop cards (source: http://forum.dominionstrategy.com/index.php?topic=1412.msg23271#msg23271).
Obviously there are some combinations within each set: you note Seaside's Treasure Map/Warehouse combo. But similar combinations exist outside of the set: consider Treasure Map (Seaside) and Stables (Hinterlands). The more you play the game, the better you get at recognizing the strategic choices between sets, as well as within sets.

Wow - I was expecting that quote from Donald X to be years old, but it's weeks old! I'd be curious to see his response to my analysis of Scout (which I worked-up before seeing your/this answer!)

@TheChaz: There are two things to consider.  1. The raw number of combos across all kingdom sets.  Of course this goes up as you add more cards and cards from different sets.  2.  The probability that a given card will be in a viable.  This tends to go down as you add cards from different sets, or equivalently, up as you add more cards from its own set.

Scout works really well with Tunnel and Crossroads from Hinterlands. Any particular theme (in this case, Victory cards that do things) will naturally wax and wane over time.

@MarkWithers: I'm not sure why Tunnel would work well with Scout, unless you have some other Action card in your hand that lets you discard the Tunnel.

@Powerlord Well yeah, you have to have another way of discarding the Tunnel within the set of 10 kingdom cards. But Scout does give you +1 action so that you can use said action and is generally good at filling up your hand with victory cards. I made a deck with 2 Scouts, 1 Jack of All Trades, 3 or 4 Crossroads and as many Tunnels as I could buy. I got to draw my whole deck every turn and often ended up with a fistful of Gold for buying provinces :)

Overall, I have to say yes, somewhat. The cards in a given expansion go together thematically and seem to often be selected with certain combos in mind, which of course you will see more often if you play with just that expansion. However, you'll see those same combos again and again if you just play with the same expansion and you will miss out on some seriously good combos that may never occur to you until you lay the cards out... and often you won't really see them until you're opponent is capitalizing them already. I continue to enjoy Dominion because I continue to see new combos in mixing expansions. I don't think anything in just base/Intrigue could surprise me anymore, but just the other day I learned how awesome Bridge is with Native Village. Hinterlands seems to have cards that play especially well with other sets. I would say the only real exception is Alchemy, where buying a potion is a lot more of a liability if there are only one or two cards with potions in their prices.
I would encourage you to mix sets, but I do think you'll have better games if you don't go completely random. It's nice to do a sanity check and make sure you have a decent price distribution, if there's a reaction it's paired with at least one attack, etc. If you go purely random, good games will definitely be fewer and farther between, but small adjustments can go a long way to making good games that you'd never see limiting your sets. (Veto mode on Isotropic is a good example of a small adjustment to random that makes a big difference in playability.)
(Edit) The reason my first answer is "yes" is because of the two condition you put on the situation: "random kingdom cards" and "3+ sets". I say yes because for pretty much every card, if you could list the set of cards that really work well with it by some metric, I'm guessing that cards in that combo set make up a higher proportion of its own expansion than they do overall. And thus by selecting cards from 3 or sets you decrease probability of selecting another card that works well with it.
Honestly, when playing the game physically, not on a computer, I can rarely be bothered to get out more than 3 boxes, though usually 2 or 3 is what I play with, and I think that's just fine.

You say "yes" in the first line, but the rest of your answer says "no [mixing sets does not diminish card combos]"!

@TheChaz Exactly! I say "Yes, somewhat," to the question "[in purely random kingdom sets from 3+ expansions] does individual card quality become more of a driving factor?" because I think any given card is more likely to combo with cards in its own set. But then I go on to say why I think that--though this sounds like a bad thing--I don't think it is because it adds richness to the game. Then I talk about how adjusting the randomness can more than make up for any perceived downsides.


text:
What is your average VP score rolling 2 versus 4 dice after an initial 3322xx roll?
King of Tokyo has a die rolling mechanic similar to Yahtzee, where you get three rolls and may choose to hold or reroll any die you previously held. At the end of third roll, you must keep the resulting die faces. In King of Tokyo, there are three number faces, '1', '2', and '3' and three non-number faces. You only score Victory Points (VP) if you get a 3-of-a-Kind or more, scoring the die face and no points for non-number die faces or numbers with pairs or less. You initiallyscore the die face of the 3-of-a-Kind, and 1 additional point for each matching die face. For example:
-3 '1's score = 1 VP
-4 '1's score = 2 VP
-4 '2's score = 3 VP
-6 '3's score = 6 VP
-3 '3's and 3 '2's score = 5 VP
You have an initial roll of 3322xx, where 'x' isn't a number. Assume that for your third roll, you always choose to roll dice that will result in the greatest average increase in VP.
What is your expected average VP after your third roll if you choose to roll 2 dice versus 4 dice for your second roll? (i.e. Is it better to roll '22xx' or 'xx' if you want to score the most points on average?)

If you're going to define the strategy of "rerolling 2 vs rerolling 4" you must also define the strategy for the second reroll. I instead think this should be rephrased "What strategy yields the highest expected value for 3322xx?"

@corsiKa, I already did, "what ever results in the greatest VP on average." For example, if your 2nd roll is 33322x, you would decide to roll 22x. Keeping the 33322 and rolling x results in 3+ (1/3)VP. Keeping 333 and rolling 22x results in 3+ (0.5)VP.

I've posted a formal analysis, with code, that says you should get more than 2.9 points for an initial roll of 3,3,2,2,x,x.

And by 2.9 I really mean 3.1.

Your overall expected value is what you might expect: 3.1121.
In other words, you should almost always get your 3-set of 3s, or a 4-set of 2s.
I threw together a simulator to help with this. Now, because I am a lamer, I used Java. It requires Java 7 because it uses <> notation. If you're using an older version, be warned. Here's the simulator, first off. Please feel free to pick it apart for functionality, but as far as style goes I know there's a million improvements to be made. This is a throw-away script if there ever one. There may be copy/paste typos and what not, but the code is actually fairly readable. I will add comments and edit as necessary, although I will group edits together to delay community-wikization.
import java.util.*;
import java.text.*;
class Tokyo {
    // shortcut for iterating over the faces of a die roll
    static final int[] face = {0,0,0,1,2,3};

    public static void main(String...args) {
        // the initial roll - there's no control over this
        final Roll init = new Roll(3,3,2,2,0,0);

        // strat1: reroll all non 3s
        // this gets all possible results and their counts from
        // using strat 1 as our guide. see rollStrat1 on the Roll class
        Map<Roll,Integer> strat1roll1 = init.rollStrat1();
        report(strat1roll1);
        // now, for each of those rolls, we need to simulate using the same
        // strategy - the addTo makes sure we give proper credit to rolls that
        // come up more often
        Map<Roll,Integer> strat1roll2 = new HashMap<>();
        for(Roll roll : strat1roll1.keySet()) { // for each roll from roll 1
            Map<Roll,Integer> curr = roll.rollStrat1(); // reroll according to the strat
            addTo(strat1roll2, curr); // and store them all in the master map
        }
        // now count the results
        report(strat1roll2);

        // strat2: reroll all non doubles - comments ommited because it's the same as above
        Map<Roll,Integer> strat2roll1 = init.rollStrat2();
        report(strat2roll1);

        Map<Roll,Integer> strat2roll2 = new HashMap<>();
        for(Roll roll : strat2roll1.keySet()) {
            Map<Roll,Integer> curr = roll.rollStrat2();
            addTo(strat2roll2, curr);
        }

        report(strat2roll2);

        // strat3: mix the two
        Map<Roll,Integer> strat3roll2 = new HashMap<>();
        for(Roll roll : strat2roll1.keySet()) {
            Map<Roll,Integer> curr = roll.rollStrat3();
            addTo(strat3roll2, curr);
        }
        report(strat3roll2);

    }

    static DecimalFormat df = new DecimalFormat("###.00000");
    static void report(Map<Roll,Integer> map) {
        int sum = 0;
        int count = 0;
        for(Roll roll : map.keySet()) {
            sum += roll.score() * map.get(roll);
            count += map.get(roll);
        }
        double exval = ((double)sum) / (double)count;
        System.out.println("total: " + sum);
        System.out.println("count: " + count);
        System.out.println("exval: " + df.format(exval));
    }

    // just merge the slave map into the master map
    static void addTo(Map<Roll,Integer> master, Map<Roll,Integer> slave) {
        for(Roll roll : slave.keySet()) {
            Integer count = master.get(roll);
            if(count == null) count = 0;
            master.put(roll, count + slave.get(roll));
        }
    }

    // reroll anything that isn't a 0 in the mask
    // if it is a non-0, that is the number that will show up in the roll
    // so if you want to re-roll a 2, in your strat method, turn that 2 into a
    // 0 so it gets re-rolled
    static Map<Roll,Integer> roll(Roll mask) {
        Map<Roll, Integer> rolls = new HashMap<>();
        // iterate over every possibly die roll
        for(int a = 0; a < 6; a++) {
        for(int b = 0; b < 6; b++) {
        for(int c = 0; c < 6; c++) {
        for(int d = 0; d < 6; d++) {
        for(int e = 0; e < 6; e++) {
        for(int f = 0; f < 6; f++) {

            int ra = mask.a == 0 ? face[a] : mask.a;
            int rb = mask.b == 0 ? face[b] : mask.b;
            int rc = mask.c == 0 ? face[c] : mask.c;
            int rd = mask.d == 0 ? face[d] : mask.d;
            int re = mask.e == 0 ? face[e] : mask.e;
            int rf = mask.f == 0 ? face[f] : mask.f;

            Roll roll = new Roll(ra, rb, rc, rd, re, rf);
            roll = roll.sort();
            if(rolls.containsKey(roll)) {
                rolls.put(roll, rolls.get(roll) + 1);
            } else {
                rolls.put(roll, 1);
            }
        }}}}}} // my code-review spidy sense is very pissed at me for this
        return rolls;
    }

    // unused, YAGNI I know... but this does illustrate all possible rolls
    static Map<Roll,Integer> possibleRolls() {
        return roll(new Roll(0,0,0,0,0,0));
    }

    // this represnts a roll, but it also represents a mask - lame, I know
    static class Roll {
        final int a,b,c,d,e,f; // the dice you have
        Roll(int a, int b, int c, int d, int e, int f) {
            this.a=a; this.b=b; this.c=c;
            this.d=d; this.e=e; this.f=f;
        }
        // welcome to Java - assumes they are properly sorted - a bad practice that
        // should be fixed if this grew beyond its current scope
        public boolean equals(Object o) {
            Roll ros = (Roll)o;
            Roll sor = this;
            return ros.a == sor.a && ros.b == sor.b && ros.c == sor.c
                && ros.d == sor.d && ros.e == sor.e && ros.f == sor.f;
        }
        // order doesn't matter for the result, but this is a perfect hash, that's cool
        public int hashCode() {
            return (a+1) * (b+1) * (c+1) * (d+1) * (e+1) * (f+1);
        }

        Roll sort() { // delegate sorting to Arrays =)
            int[] p = {a,b,c,d,e,f};
            Arrays.sort(p);
            return new Roll(p[0],p[1],p[2],p[3],p[4],p[5]);
        }

        int[] counts() { // convenience method
            int[] counts = new int[4];
            counts[a]++; counts[b]++; counts[c]++;
            counts[d]++; counts[e]++; counts[f]++;
            return counts;
        }

        int score() {
            int[] counts = counts();

            // trust the math =), or don't and work it out for yourself =)
            return (counts[1] > 2 ? counts[1] - 2 : 0)
                 + (counts[2] > 2 ? counts[2] - 1 : 0)
                 + (counts[3] > 2 ? counts[3] : 0);
        }

        Map<Roll,Integer> rollStrat1() {
            return keepAll(3);
        }

        Map<Roll,Integer> keepAll(int k) {
        // reroll all non ks
            int ra = a == k ? k : 0; // the mask will have ks or 0s
            int rb = b == k ? k : 0;
            int rc = c == k ? k : 0;
            int rd = d == k ? k : 0;
            int re = e == k ? k : 0;
            int rf = f == k ? k : 0;
            return roll(new Roll(ra,rb,rc,rd,re,rf));
        }

        Map<Roll,Integer> rollStrat2() {
            // reroll anything that doesn't have 2x of something, except 0 - of course reroll 0s
            int[] counts = counts();
            int ra = counts[a] > 1 && a > 0 ? a : 0;
            int rb = counts[b] > 1 && b > 0 ? b : 0;
            int rc = counts[c] > 1 && c > 0 ? c : 0;
            int rd = counts[d] > 1 && d > 0 ? d : 0;
            int re = counts[e] > 1 && e > 0 ? e : 0;
            int rf = counts[f] > 1 && f > 0 ? f : 0;
            return roll(new Roll(ra,rb,rc,rd,re,rf));
        }

        Map<Roll,Integer> rollStrat3() {
            int[] counts = counts();
            if(counts[3] > 2 && counts[2] > 2) {
                Map<Roll,Integer> map = new HashMap<>();
                map.put(this,score());
                return map;
            }
            if(counts[3] > 2) {
                return keepAll(3);
            }
            if(counts[2] > 2) {
                return keepAll(2);
            }
            return rollStrat1();
        }
    }

}

What the results are when it is run is (these values have changed - see edit history).
c:\files\j>java Tokyo
total: 81756
count: 46656
exval: 1.75231
total: 5081940
count: 1632960
exval: 3.11210
total: 73872
count: 46656
exval: 1.58333
total: 1283040
count: 466560
exval: 2.75000
total: 1165885
count: 419909
exval: 2.77652

What does this tell us? Well, first it validates the system by coming up with values nearly identical to those calculated by other methods for the first roll. That's excellent news.
Now, let's look at the actual results. The number of different values explodes for strategy 1, mostly because it is rerolling 4 different dice on the first pass, and probably an average of 3.5 or so for the second pass. The count seems so much higher because of the uniqueness, but I adjusted for this 'problem' in the code.
The second strategy starts off less, and does stand for improvement. For example, if you don't get your 2s the second time, would it be better to dump all the 2s and go for 3s? This is where I implement strat 3.
Rather than start off with the results from strat 1, which are only focusing on the 3s, I instead use the first roll of strat 2 to focus on acquiring doubles. Then, I shift focus: if I have both sets, haha, we're done here... thanks... If I have a set of 3s or a set of 2s, I keep it and ditch the other set: after all, it's silly to try to aim for that 1/6 chance to pick up the other set - much better odds in ditching those dice and put them into something that's already bearing fruit. Lastly, if I don't have a set, I default to just rolling all non 3s, which is the same as the original 'strat 1' and is shown to have a higher single-roll result than strat 2. And that yields an excellent result but still not better.
Note the 5 reports:
strat 1 roll 1
strat 1 roll 2

strat 2 roll 1
strat 2 roll 2

strat 3 roll 2 (its roll 1 is strat 2's roll 1)

So the end result is that in this scenario, thus far the strategies indicate you should simply keep your 3s and go for it.

Does strat 2 have an issue on roll 2 if you rolled two 1s?  So if you end up with 332211 is it just keeping that?  Probably won't make much of a difference, but worth correcting.  Also, for strat 3 if you don't end up with either set, shouldn't you default back to strat 1, which has a better 1 roll expected value?

332211 is as useful as 3322xx - the 1s get rerolled trying to get either a 3 or a 2 in strat 2 (and in strat 3, actually). So nothing to correct there. As far as strat 3 reverting to strat 2 over strat 1 for the failover: in that scenario, you have 2 2s and 2 3s. You are correct, but the engine does not agree with you. I do agree with you, though - therefore it must be the engine that is incorrect. I'm unable to find the bug, but I suspect it has to do with how it counts the total and the score, and how it weights the more-likely scenarios.

I removed the continues and commented that out. My results are still in line with the hand-calculated data for the first run, but the second run now appears more accurate. And it changes the result significantly.

updated to reflect the massive bug that changes the outcome, and increases your odds by .2 :D

Nice.  Rerolling everything but 3s is also better from a non-VP view, as those 2s are completely worthless if you don't match them, but things like energy and hits might be worthwhile.  Unless of course you are trying to avoid hitting.

I don't understand Tokyo, so that doesn't mean anything to me. But, being a gamer, I could see 3 being better than 2. :)

I've added a rundown of every initial roll here: http://pastebin.com/TFGQ7DH6 - what's interesting to note is that strat 2 wins the most, and that strat 3 scores better on (0,0,0,0,0,0) than it does on (0,0,1,1,2,2) - indicating that there is definitely a strat 4 that works better overall. note that in that pastebin, a 4 on the end is the 3rd bit being set - so 7 (being 111 in binary) means they all tie. The ideal strat4 would have bit4 set in all cases.

It also shows that strat 2 gets you 2.48 EV when you have (0,0,0,0,0,0) - so you can expect that or greater from all rolls.

I tried to replicate your strategies to find the discrepancies between our expected scores. I got much lower expected scores when I tried the strategies you've suggested.  The script I wrote to test your strategies is here (together with the output) https://ideone.com/SFJOOQ

Summary
You should keep the two threes and rethrow the rest. Your expected score is 2.81 (versus an expected score of 2.64 for keeping all four dice).
I calculated this using a script on my phone, but the values seemed to disagree with everyone else, so I rewrote it in python.  Here's the python version for anyone to pick apart.
It takes a long time to run (much longer than the optimised version I wrote for my phone), so here is the output.
Method
If you've looked at the output then you'll notice the script's doing a lot more than answering just this question.  Rather it's calculating the best move for any set of dice showing, and with any number of throws left.  It starts from the simple case of "no throws left" and increments the number of throws. This allows it to search all possible sets of kept dice, and compute the highest expected score from each.
Optimal play
Taking the best moves given by the script I was able to find the optimal play for any given game state. After the first roll it's possible to be in one of 84 positions.  Since we're treating claws, hearts and lightning bolts (4s, 5s and 6s) all as X's, you're more likely to have thrown Xs than e.g. 2s. The 84 positions are shown below:

The image has got shrunk a bit small, but if you open it in a new tab or zoom in then it's readable. Each square shows a particular roll you may have got, and highlights the dice you should keep. I've colour-coded the positions by whether you should keep 1s (red), 2s (green) or 3s (blue), some combination (yellow, cyan and magenta), or no dice (grey). The bottom of each square shows your expected score.
Once you have kept some of the dice and rerolled the rest you end up in one of the positions in the figure below. Again, this image shows which dice you should keep and what your expected score is.

After rolling for the third time, you end up in one of the positions in the figure below. This time the diagram shows you the actual score, and the colours correspond to which dice are getting you points.

Example
Here's an example of how to use the images above. Imagine that for our first roll we get three 1s, two 2s and a 3. We look in the top figure and find the bit that looks like this:

This tells us our best bet is to keep the 2s and 3s, and reroll the 1s. Our expected score is 2.15. Rerolling those three dice, we get two 1s and a claw (an X). We look at the second image from above and find the bit that looks like this:

We are told that our best strategy now is to keep the 2s and reroll the rest. Our expected score has dropped to 1.25.  Rerolling we get two more 2s, another 1, and a heart (an X). This puts us in this square from the third diagram above:

We score 3 points for a set of four 2s.
Other results
While playing around with this problem I found several other interesting results.
The average expected score with three rolls left is 2.24. This drops to 1.34 if you haven't rolled any 1s, 2s or 3s in the first throw. It drops to 0.40 if you've still got nothing with one roll left.
The best strategy is usually to keep a single number and rethrow the rest, but occasionally you should keep more than one number (e.g. for (2,2,3,6,6,6) with 2 throws left you should keep (2,2,3)).
If on your first roll you get (1, 1, 1, 1, 3, 3) you should keep the 3s (expected score 2.81), but if it's your second roll you should keep the 1s (expected score 2.33).

If after the second roll, you have 3322xx:
Re-rolling 4 dice
Chance of one 3 being rolled = 4C1 * 1/6 * 5/6 * 5/6 * 5/6 = 500/1296
Chance of two 3s being rolled = 4C2 * 1/6 * 1/6 * 5/6 * 5/6 = 150/1296
Chance of three 3s being rolled = 4C3 * 1/6 * 1/6 * 1/6 * 5/6 = 20/1296
Chance of four 3s being rolled = 4C4 * 1/6 * 1/6 * 1/6 * 1/6 = 1/1296
Chance of three 1s being rolled = 4C1 * 1/6 * 1/6 * 1/6 * 4/6 = 16/1296
Chance of four 1s being rolled = 4C4 * 1/6 * 1/6 * 1/6 * 1/6 = 1/1296
Chance of three 1s and a 3 being rolled = 4C1 * 1/6 * 1/6 * 1/6 * 1/6 = 4/1296
Chance of three 2s being rolled = 4C1 * 1/6 * 1/6 * 1/6 * 4/6 = 16/1296
Chance of four 2s being rolled = 4C4 * 1/6 * 1/6 * 1/6 * 1/6 = 1/1296
Chance of three 2s and a 3 being rolled = 4C1 * 1/6 * 1/6 * 1/6 * 1/6 = 4/1296
Chance of zero 3s being rolled = 1 - the above probabilities = 583/1296
Expected average is the sum of all probabilities times the VPs that outcome yields:
(500 / 1296) * 3 = 1.157
(150 / 1296) * 4 = 0.463
(20 / 1296) * 5 = 0.077
(1 / 1296) * 6 = 0.005
(16 / 1296) * 1 = 0.012
(1 / 1296) * 2 = 0.001
(4 / 1296) * 4 = 0.012
(16 / 1296) * 2 = 0.025
(1 / 1296) * 3 = 0.002
(4 / 1296) * 5 = 0.015
(583 / 1296) * 0 = 0
Expected value for re-rolling 4 dice is 1.157 + 0.463 + 0.077 + 0.005 + 0.012 + 0.001 + 0.012 + 0.025 + 0.002 + 0.015 = 1.769 VPs
Re-rolling 2 dice
Chance of rolling two 3s = 2C2 * 1/6 * 1/6 = 1/36
Chance of rolling one 3 = 2C1 * 1/6 * 4/6 = 8/36
Chance of rolling two 2s = 2C2 * 1/6 * 1/6 = 1/36
Chance of rolling one 2 = 2C1 * 1/6 * 4/6 = 8/36
Chance of rolling one 3 and one 2 = 2C1 * 1/6 * 1/6 = 2/36
Chance of rolling zero 2s or 3s = 1 - above probabilities = 16/36
Expected average is the sum of all probabilities times the VPs that outcome yields:
(1/36) * 4 = .111
(8/36) * 3 = .666
(1/36) * 3 = .083
(8/36) * 2 = .444
(2/36) * 5 = .278
(16/36) * 0 = 0
Expected value for re-rolling 2 dice is .111 + .666 + .083 + .444 + .278 = 1.582VPs
Summary: On your last reroll in this situation, reroll the 2s. Rerolling 2 dice gives an average of 1.582 VPs compared to 1.769 VPs for rerolling 4 dice.
Note: This only works out which is better for the final reroll. It does not calculate what is better if the two rerolls are available.

I think you double-counted the one 3 and one 2 scenarios.  It should be chance of rolling one 3 and one that isn't a 2 or 3 to avoid that, making them 8/36 instead.

This question is specifically about the initial first roll, so only answering for the third roll isn't an answer. Additionally, if you roll 111, or 1111, or 222, etc. When rolling 4 dice you still score points. You need to factor those into your answer.

@user1873 Yes, you are correct about needing to factor those in. I did specify what my answer covers in the hope it's of some help.

Your expected value for 2 dice still seems wrong.  The values for both the 1/36 cases are off.

updated to account for rolling 1111, 1113, 111, 2222, 2223 and 222.
@bwarner updated those values. Good spot.

first let's examine rerolling only 2 dice.
since we can't get VPs from 1s, we want 2s and 3s.
there's 1/6 chance of either, so the options we like are:

1/36 of 2,2, worth 3 points
1/36 of 3,3, worth 4 points
2/36 of 2,3, worth 5 points
8/36 of 2,X, worth 2 points
8/36 of 3,X, worth 3 points

for a total average of 1.5833
rerolling 4 dice is to complicated to do by hand like that, so I wrote a short Python script that returned an average of 1.7523
so, baring any specific evolutions or items, reroll those 2s!
EDIT
assuming we reroll twice, as originally asked
if we know how much every dice pool is worth (on average, with optimal strategy) before final reroll, we can make an educated decision about the first reroll.
it's obvious we want to keep the 3s. we're not sure about the 2s, but definitely keep 3s.
so, we will definitely have 2 3s in the final roll. which options are better to keep then 2 3s?

obviously any other pair or single number is worse then (3,3), as is 3 1s (to keep before final reroll on average). (3,3,2,2,2) is better the either (3,3) or (2,2,2)
(3,3) is on average worth 1.7523
(1,1,1,1) is on average worth 2.3333
(3,3,2,2,2) is on average worth 2.6666
(2,2,2,2) is on average worth 3.3333
(3,3,3) is on average worth 3.5138
(3,3,3,1,1,1) is on average worth 4
(3,3,3,2,2,2) is on average worth 5

knowing that, we can write a script that will test all the possibilities if we keep only 3s or keep both, and use the highest average it qualifies for (if we get 3,3,2,2,2,2 we use 3.3333 as our value, despite our roll matching other, lesser options).
running said script I got a value of 2.6834 for keeping only 3s and 2.62024 for keeping both.
they're about 1 more then the case with 1 reroll (quite reasonable), but the numbers are lower then corsiKa's expected 3.1121, which is unexpected.
I will investigate further, as I love King of Tokyo and puzzles that I can play with and not just brute-force
can provide code used for calculations, if desired

"for a total average of 1.5833," that is only for the 2nd roll. You need to continue the calculation for each of those intermediate steps. (I.e. Starting from (16/36) 3322xx, you get the best odds rolling all 4 dice (AVG 1.7523 supposedly), but that is just one state, you also have 33322x, 333322, 33222x, 332222. For those states you are probably rolling 3/2/3/2 dice respectively on the 3rd roll (haven't worked out the exact math)

true, I just assumed you're on your final roll. I will update accordingly, but since it's better to reroll 2s on final roll (in other words going for 3s), I would be extremely surprised if 2nd roll would be any different.

Why do you say "we can't get VPs from 1s"? 3 1s is worth 1 point, 4 1s is worth 2, and 5 1s is worth 3.

this question wrongly assumed we are rerolling only once. we rolled (3,3,2,2,X,X). if we reroll only Xs ("rerolling only 2 dice"), there will never be more then 2 1s.