ALICE CAN WIN the game by ordering the 14 cards as shown (A). Suppose Bob chooses the number four. Alice deals four cards, turning up the six of spades (B). She then deals six cards, turning up the two of spades (C). She deals two more cards, turning up the two of hearts (D). Finally, she deals two cards, turning up the ace of spades (E). But can Alice create another winning 14-card order by making insertions into the arrangement shown (F)? Image: SARA CHEN
Alice and Bob play a game using 14 cards--the ace through seven of hearts and the ace through seven of spades. Bob picks a number between one and seven, then tells Alice the number. Alice deals that many cards, finishing the first round. Bob turns over the last card dealt; the number on that card determines how many cards Alice deals in the second round (an ace is considered one). Again, Bob turns over the last card dealt, and Alice then deals that many more cards, and so on. Alice wins if the last card dealt in some round is the very last of the 14 cards and if it is the ace of spades. Bob wins otherwise.
First, consider a warm-up problem: How can Alice arrange the cards in advance so that no matter which number Bob chooses at the beginning, the game will end with the ace of spades as the last card turned up? One possible answer is shown in illustrations A through E below.
This article was originally published with the title Card Counters.