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

by : $207$

 * The Twenty-Two Game
 * Here is a variation of our little "Thirty-one Game" (: 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?