There are 2n cards labeled 1, 2, ..., 2n respectively, and the cards are distributed randomly between two players so that each has n cards. Each player takes turns to place one card, and you win if you put down a card so that the current sum of all the played cards is divisible by 2n+1.
For example, if n=10, and the previously placed cards are 5, 8, 9, 19, then if player A now places 1, he wins since 5+8+9+19+1 = 42 is divisible by 2*10+1=21.
Assuming both players want to win, what strategy should one adopt in order to win? Following the strategy, is there a consistent winner of this game?
I agree with David Shin that the second player will win most of the time.
Player 1 should play card three and win instantly. If Player 1 doesn't have the three, then player two should always play his highest card. This way, the total of the cards on the table is always > 2n+1 when its Player 1's turn. Player two will then win on the last card.
Edited on January 10, 2005, 6:13 pm

Posted by Hugo
on 20050110 18:11:17 