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If you are trying to lose the game as quickly as possible then you can do it in 1 turn and 2 rolls of the dice. This is using the UK version as I'm not sure about the US one. Roll 1: Roll 2 (Because of the doubles): The chance of this is:


27

The answer:


27

Okay. THAT. WAS. INCREDIBLE! It took me two solid hours of work to solve and even longer to write and draw it all up here - hopefully it'll be worth it! To start us off, here is the final maze layout and routes: In the following explanation all colours have been abbreviated to their initials as follows: G=Green, O=Orange, P=Purple, Y=Yellow. An ...


25

So building on kristinalustig's answer they are indeed playing The cards are assigned as follows Alice's first turn Bob's first turn In particular Alice's next move produces Bob's next move produces "you could have got yourself 7 points more if you’d played the five of clubs first." Alice's last move is Alice's score For those unfamiliar with the ...


23

It can be done in jumps. Here is the solution: If you start with the hole elsewhere, you need more moves to reach an end position. Here are all the optimal end positions for each case, but without the move sequence leading up to them.


22

The "best" move for this puzzle is the one most likely to hit the ship in one shot, as the puzzle doesn't allow for further follow up shots. There are 23 possible positions for that ship, and 68 spaces where a shot could be placed. Due to the size of the ship and previously placed shots, we can find both where shots cannot possibly hit the ship, and ...


22

The game is and the result is


20

The word was... And the reason you couldn't play the more common variant was...


20

It is currently: because All that means it is now the


19

First rook is placed at any of Second rook is placed The third rook is placed And then the fourth rook A crappy example game below:


18

D5 covers 4 locations D8 covers 3 F2 covers 2 H2 covers 1 G6 covers 0 E5 is the best; it covers 5 locations.


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15

For the updated version, I'll assume that after the placement of the pieces, the queen moves first. (Otherwise the knight could arrange things so that the queen is captured on the first move.) In this situation, the queen will still always win. Wherever the two pieces start, the queen can get to a space adjacent to the knight in no more than 3 moves. First,...


15

Here is one way you could begin to prove that the board can be cleared for all values of $m$ and $n$. Proof by induction. [incomplete] Case 1: $k = 1$ Here we take k = 1 to mean the smallest possible board with the given conditions $(m \ge n \ge 2)$ which would be a $m = n = 2$ or $2m \times 2n$ = $4\times4$ board. Theorem: For an Othello board of size $...


15

On my puzzle website there is an old puzzle called Hoo-Doo which is essentially this same puzzle. Some other board sizes:


15

I spent quite a while working on this yesterday and today. I only have a partial answer. I started with the assumption that I chose this assumption because: Given this assumption, I tried: However: I might proceed with:


14

The queen can always win. If the queen goes first, obviously it will win (just take the knight). If the knight goes first, there are only 2 places to go (mirrored, so effectively the same): - - - - - + . . . . . | . . K . . | . . . . . | . . . . . | All the queen has to do is move: - - - - - + . . . . . | . . K . . | . . Q . . | . . . . . | All the ...


13

Yes, I found this video on YouTube that has this "perfect game". Terrible music by the way https://www.youtube.com/watch?v=prWG1OFgVqg


13

Using a computer to attempt to beat my original solution I found some Here is one: I reduced the search space by not allowing stones to moves to (although they can jump via) the spots marked with an * below: * * * * · · · · * * * · · · · · * * · · · · · · * · · · · * * * · · · · * * * * 1 2 3 * * * * * 4 5 6 * * * * * 7 8 9 * * * * * I then This ...


12

This kind of puzzle is commonly referred to as "lights out", after a board game by that name. I googled that phrase and came upon a solver at http://www.ueda.info.waseda.ac.jp/~n-kato/lightsout/. So I can't take credit for the answer but it's pretty simple:


12

I initially gave a slightly rambling stream-of-consciousness answer, which I have preserved below in case anyone prefers that. Here is a slightly slicker one. (It is the same argument, just possibly clearer.) It remains to show I have the suspicion that there is a way to express this idea that packs everything into one sentence, but I haven't quite found ...


12

OK, I barely made it with 10 moves, but it seems pretty hard to prove this is optimal (and I suspect the answer may be lower). The last move you make is Rh8, which will be the mate.


12

I got a quick answer but not sure how good it is.


12

This solution is no longer valid. The original posting of the problem had a typo (and lacked the restriction of "no auctions"); with the typo, at least one auction was required to solve, and this answer provided such a solution. For the corrected problem, @dcfyj's post provides the correct answer. It is Player 1: Battleship's turn Here's how the game plays ...


12

Partial answer - some progress made, not sure how it lines up with instructions The given grid can be divided into These turn out to be So... what next?


11

This is known as the "18-Queens Problem". I found this well-known solution by Friedrich Burchard & Friedrich Hariuc (1976) in 96 Half-moves. I can't make a claim to its optimality, but by looking at it and seeing that no better can be found, I'd say it may well be optimal. 1.e4 f5 2.e5 Nf6 3.exf6 e5 4.g4 e4 5.Ne2 e3 6.Ng3 e2 7.h4 f4 8.h5 fxg3 9.h6 g5 ...


11

The upper bound limit is likely to be quite a bit lower than 60. For the purposes of this I will assume that you are WHITE. In order for you to have a valid move in Othello, you need a line formed by the empty space and at least one of each colour. The largest number of valid moves made possible by the presence of a single WHITE counter is 8 - one for each ...


11

The trick is to place them Full list:


11

The other guys already got the answer. (Maybe not with completely airtight arguments, but nevertheless.) The explanations would be a lot easier to follow if they had pictures, so here's one: Legend: Red square: after turn 1, you are in one of these. Green square: after turn 2, you are in one of these, or in some red square (except square 2). Red circle ...


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