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

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?

Sources

• 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Puzzle Games: $207$. -- The Twenty-Two Game
• 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Game Puzzles: $476$. The $22$ Game