Henry Ernest Dudeney/Modern Puzzles/Puzzle Games
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Henry Ernest Dudeney: Modern Puzzles: Puzzle Games
$202$ - Noughts and Crosses
- Every child knows how to play this ancient game.
- You make a square of nine cells, and each of the two players, playing alternately, puts his mark
- (a nought or a cross, as the case may be) in a cell with the object of getting three in a line.
- Whichever player gets three in a line wins.
- In this game, cross has won:
- $\begin {array} {|c|c|c|} \hline \text X & \text O & \text O \\ \hline \text X & \text X & \text O \\ \hline \text O & & \text X \\ \hline \end{array}$
- I have said in my book, The Canterbury Puzzles,
- that between two players who thoroughly understand the play every game should be drawn,
- for neither party could ever win except through the blundering of his opponent.
- Can you prove this?
- Can you be sure of not losing a game against an expert opponent?
$203$ - The Horse-Shoe Game
- This little game is an interesting companion to our "Noughts and Crosses".
- There are two players.
- One has two white counters, the other two black.
- Playing alternately, each places a counter on a vacant point, where he leaves it.
- When all are played, you slide only, and the player is beaten who is so blocked that he cannot move.
- In the example, Black has just placed his lower counter.
- White now slides his lower one to the centre, and wins.
- Black should have played to the centre himself, and won.
- Now, which player ought to win at this game?
$204$ - Turning the Die
- This is played with a single die.
- The first player calls any number he chooses, from $1$ to $6$, and the second player throws the die at hazard.
- Then they take it in turns to roll over the die in any direction they choose, but never giving it more than a quarter turn.
- The score increases as they proceed, and the player wins who manages to score $25$ or forces his opponent to score beyond $25$.
- I will give an example game.
- Player $A$ calls $6$, and $B$ happens to throw $3$, making the score $9$.
- Now $A$ decides to turn up $1$, scoring $10$;
- $B$ turns up $3$, scoring $13$;
- $A$ turns up $6$, scoring $19$;
- $B$ turns up $3$, scoring $22$;
- $A$ turns up $1$, scoring $23$;
- and $B$ turns up $2$, scoring $25$ and winning.
- What call should $A$ make in order to have the best chance at winning?
- Remember that the numbers on opposite sides of a correct die always sum to $7$, that is, $1 - 6$, $2 - 5$, $3 - 4$.
$205$ - The Three Dice
- Mason and Jackson were playing with three dice.
- The player won whenever the numbers thrown added up to one of the two numbers he selected at the beginning of the game.
- As a matter of fact, Mason selected $7$ and $13$, and one of his winning throws was $6$, $4$, $3$.
- What were his chances of winning a throw?
- And what two other numbers should Jackson have selected for his own throws to make his chances of winning exactly equal?
$206$ - The $37$ Puzzle Game
- Here is a beautiful new puzzle game, absurdly simple to play but quite fascinating.
- To most people it will seem to be practically a game of chance -- equal for both players --
- but there are pretty subtleties in it, and I will show how to win with certainty.
- Place the five dominoes $1$, $2$, $3$, $4$, $5$, on the table.
- There are two players, who play alternately.
- The first player places a coin on any domino, say the $5$, which scores $5$;
- then the second player removes the coin to another domino, say to the $3$,
- and adds that domino, scoring $8$;
- then the first player removes the coin again, say to the $1$, scoring $9$; and so on.
- The player who scores $37$, or forces his opponent to score more than $37$, wins.
- Remember, the coin must be removed to a different domino at each play.
$207$ - The Twenty-Two Game
- Here is a variation of our little "Thirty-one Game" (The Canterbury Puzzles: No. $79$).
- Lay out the $16$ cards as shown.
- $\begin{matrix} \boxed {A \heartsuit} & \boxed {A \spadesuit} & \boxed {A \diamondsuit} & \boxed {A \clubsuit} \\ \boxed {2 \heartsuit} & \boxed {2 \spadesuit} & \boxed {2 \diamondsuit} & \boxed {2 \clubsuit} \\ \boxed {3 \heartsuit} & \boxed {3 \spadesuit} & \boxed {3 \diamondsuit} & \boxed {3 \clubsuit} \\ \boxed {4 \heartsuit} & \boxed {4 \spadesuit} & \boxed {4 \diamondsuit} & \boxed {4 \clubsuit} \\ \end{matrix}$
- Two players alternately turn down a card and add it to the common score,
- and the player who makes the score of $22$, or forces his opponent to go beyond that number, wins.
- For example, $A$ turns down a $4$, $B$ turns down $3$ (counting $7$), $A$ turns down a $4$ (counting $11$),
- $B$ plays a $2$ (counting $13$), $A$ plays $1$ (counting $14$), $B$ plays $3$ ($17$), and whatever $A$ does, $B$ scores the winning $22$ next play.
- Now, which player should always win, and how?
$208$ - The Nine Squares Game
- Make the simple square diagram show and provide a box of matches.
- The side of the large square is three matches in length.
- The game is, playing one match at a time alternately, to enclose more of those small squares than your opponent.
- For every small square that you enclose, you not only score one point, but you play again.
- This illustration shows an illustrative game in progress.
- Twelve matches are placed, my opponent and myself having made six plays each, and, as I had first play, it is now my turn to play a match.
- What is my best line of play in order to win most squares?
- If I play $FG$ my opponent will play $BF$ and score one point.
- Then, as he has the right to play again, he will score another with $EF$ and again with $IJ$, and still again with $IJ$, and still again with $GK$.
- If he now plays $CD$, I have nothing better than $DH$ (scoring one), but, as I have to play again, I am compelled, whatever I do, to give him all the rest.
- So he will win by $8$ to $1$ -- a bad defeat for me.
- Now, what should I have played instead of that disastrous $FG$?
- There is room for a lot of skilful play in the game, and it can never end in a draw.