Henry Ernest Dudeney/Modern Puzzles/Puzzle Games

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:

$\qquad \begin {array} {|c|c|c|} \hline \text X & \text O & \text O \\ \hline \text X & \text X & \text O \\ \hline \text O & & \text X \\ \hline \end{array}$

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.

$\qquad \begin{matrix} \boxed {A \heartsuit} & \boxed {A \spadesuit} & \boxed {A \diamondsuit} & \boxed {A \clubsuit} \\ \boxed {2 \heartsuit} & \boxed {2 \spadesuit} & \boxed {2 \diamondsuit} & \boxed {2 \clubsuit} \\ \boxed {3 \heartsuit} & \boxed {3 \spadesuit} & \boxed {3 \diamondsuit} & \boxed {3 \clubsuit} \\ \boxed {4 \heartsuit} & \boxed {4 \spadesuit} & \boxed {4 \diamondsuit} & \boxed {4 \clubsuit} \\ \end{matrix}$

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.