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16

In Solving Quantum Tic-Tac-Toe, Ishizeki and Matsuura use a computer to search the game space to find solutions, but don't specifically speak to strategy. However, there are a few strategies we can glean from their results: Go for a half-point victory, not a full point victory. In their search they found the first player, X, cannot guarantee a win by being ...


13

Note: This answer below is using 1st Edition scoring rules (Cities only scored once). Although parts of the analysis are valuable to calculating the highest score possible (36 Scoring opportunities, try to score with Meeples that have a high point value per the 36 turns, the highest score of 278 is not correct. I will have to do further analysis to determine ...


10

One of the clearest examples of game theory in Dominion shows up quite frequently in 2 player endgames, and it's called the Penultimate Province Rule (PPR). Basically, you should avoid buying the second to last Province if buying the remaining Province would allow your opponent to win. Imagine a game where my opponent and I each have 3 Provinces, with 2 ...


10

Because if you're only ever going to call "cheat" when I play my last card, I'll cheat like crazy on every turn other than my last card and do everything in my power to make sure that I'm not cheating on my last card.


9

You're correct in that it's usually better to let other people call cheaters out, since there's a personal risk but no personal reward. However, imagine the case where it's about to be my turn, there's a large stack on the table, and I have no option but to cheat. Here, it might be best for me to call "cheat" on the person who just played in order to clear ...


9

As Hackworth notes, testing all possible layouts is obviously infeasible. However, it might be possible to get a decent upper bound on the points. I think we can safely assume that meeples will not be a scarce resource when playing in this way (since we can choose to complete the map in any order, it should not be difficult to ensure we always have meeples ...


8

Is there an optimal way of placing the tiles in order to maximize one player's score? Yes With a finite amount of tiles, a finite number of legal placements for any tile at any point in the game, and a finite amount of meeple plays after placing a tile, it should be obvious that there are a finite number of possible games. Every game has a definite result ...


7

You can call "cheat" even when you aren't 100% positive someone is lying to keep the fear of being called out real. Like @PhilipKendall said, if you are only going to call "cheat" when you can prove it or when you are at the last opportunity to keep someone from winning, there is never any fear of being called as long as you make relatively conservative ...


6

I suspect that you're right that in principle and possibly even in practice an optimal AI could be built for this game to determine who can win from the starting position, though it would likely take more computing power than you're likely to have readily available. The best approach for brute-force solving a game like this is something close to what ...


5

The maximum theoretical score is 338. I previously answered the question for 1st Edition rules (cities only scored once by farmers), so here is my second stab at this question. First, it should be rather clear to everyone that since there are only 72 tiles and one is the starting tile, that the maximum number of scoring opportunities is 36 if you go first. ...


4

The maximum possible score is 342. 2 Farms with 16 cities = 96 5 Farms with 2 cities = 30 7 Roads with a total of 14 end pieces and 32 non end pieces = 46 6 Cloisters = 54 16 Cities worth a total of 116 Total = 342, 36 meeples placed. The answer user1873 gave is quite good, but overlooks that 16 cities are possible. It was also extremely non trivial to ...


3

Some cards are very dependent on the actions your opponents, and some are not. e.g. buy Tribute if you see your opponent to the left buy lots of different cards, don't buy it if they are not. However, in most cases, if your strategy can be influenced by opponents actions, then it is best to proactively consider those potential influences even before they ...


3

I don't know if you'd consider these "rules issues" or not; they're certainly not as bad as an instant on the field, but the rules as currently worded don't allow for subtypes to be associated with supertypes. If you make Tribal a supertype and start messing around with an object's types,things don't work out the same way; see the following examples. Maybe ...


3

This is a generalized version of @Guvante's answer. O (player 2) always wins in 4. The diagonals are not necessary to achieve a win. Examples are included in bold, but this strategy works for all choices made by X. X places a piece at (a,b): (1,1) O chooses a row that is not a, and places two pieces in that row, but not in column b: (2,2) and (2,5) Now, ...


2

The winning strategy for such a small Hex board is shown in this basic strategy guide. Like tic-tac-toe, on a 4x4 board white will always win by opening on the main diagonal, because for every counter that black can make, there is another way for white to force the win. Once white can form a "two-bridge" by placing the second piece in a non-adjacent space ...


2

My co-worker found what I believe is a winning strategy for the first player, but now I find earlier evidence by others as well. This is for the original version of the game, where you can send your opponent to an already won field and he has to place his mark there. It seems that the question is still open for the updated version where he can then choose ...


2

From a game theory perspective, you'd have to analyze this in more concrete scenarios. Yes, if someone says "one ace" on the first turn where everyone else has the same card holding and nobody happens to have three or four aces, it's a pretty silly risk to call them on it - which is why nobody does call them on it in that scenario. (Although, if you have ...


1

The answer to this question will be very heavily dependent on your group, and their groupthink. In my experience, often bidding 3 on the blade would not result in getting it (I would probably be 2nd or 3rd, with someoen else going all in on the sword), while a bid of 2 on the Raven would probably be enough to get me at least one star (with someone else ...


1

This is actually a variation on the Game Theory problem The Volunteer's Dilemma. Essentially, there is a cost to you calling cheat (The chance of being wrong and taking cards), and a benefit to all for you doing it (Stopping an opponent). There are a lot of different possible equilibria, but if every player acts the same, then the only equilibrium (state ...


1

The biggest piece of strategy is knowing how to move a pile of cards from one stack to the next. The basic algorithm is a recursive algorithm. To move n cards from stack a to stack b, you need to: move the n-1 cards from stack a to stack c move the nth card to stack b move the n-1 cards from stack c to stack b. To move a stack of three ...


1

I have flip-flopped a few times, but think that Player 2 has an unbeatable strategy. The trick is that Player 1 needs an X in each row and each column while Player 2 needs a "box". Four O's that form a rectangle with no X's that share a row or column with said rectangle. First two O's sharing a row with self and separated by 1 square. Never put below or ...


1

Sort of. A computer could brute force the optimal move each round based upon the decisions made by the opponent. You could even construct a matrix(huge) of all possible moves for a game. But there is no single path of one players choices that would guarantee a win. Your opponent can make choices that counter your choices effectively and give them the ...



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