Henry Ernest Dudeney/Modern Puzzles/207 - The Twenty-Two Game/Solution

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Modern Puzzles by Henry Ernest Dudeney: $207$

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?


Solution

The first player can always win by playing $1$.


Proof

Apart from the cards running out, the winning series is:

$7, 12, 17, 22$

If you can score $17$ and leave at least one pair adding up to $5$ of both kinds: $1 - 4$ and $3 - 2$, you must win.

If you can score $12$ and leave at least two pairs adding up to $5$ of both kinds, you must win.

If you can score $7$ and leave at least $3$ pairs adding up to $5$ of both kinds, you must win.


Call the first player $A$ and the second player $B$.

So, if the $A$ plays $3$ or $4$, $B$ plays $4$ and $3$, gets $7$, and wins.

Nothing now stops $B$ scoring $12$, then $17$, then $22$.


If $A$ plays $2$, then $B$ replies with either $3$ or $2$.

So, for example: $2 - 3$, $2 - 3$, $2 - 3$, $2 - 3$ making $20$, and there is no $2$ left to make $22$.

$A$ must then play $1$ leaving another $1$ for $B$ to play and win.

Similarly: $2 - 3$, $1 - 3$, 3 - 2$, 3 - 2$ giving $19$ and again $B$ wins, as there is no $3$ left to make $22$.

And so on.


So, $A$ plays $1$.

Examples:

$1 - 1$, $4 - 1$, $4 - 1$, $4$ making $16$ and $A$ wins.
$1 - 3$, $1 - 2$, $4 - 1$, $4 - 1$, $4$ making $21$ and $A$ wins.
$1 - 4$, $2$ making $7$ and $A$ wins.
$1 - 2$, $4$ making $7$ and $A$ wins.

$\blacksquare$


Sources

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