1993 Putnam B2

Consider the following game played with a deck of $2n$ cards numbered

from 1 to $2n$. The deck is randomly shuffled and $n$ cards are dealt to

each of two players. Beginning with $A$, the players take turns

discarding one of their remaining cards and announcing its number. The

game ends as soon as the sum of the numbers on the discarded cards is

divisible by $2n+1$. The last person to discard wins the game. Assuming

optimal strategy by both $A$ and $B$, what is the probability that $A$ wins?

考虑使用一副编号为 $2n$ 的牌玩以下游戏

从 1 到 $2n$。牌组被随机洗牌，并且将 $n$ 牌发给

每两名球员。从$A$开始，玩家轮流

丢弃剩余的一张牌并宣布其号码。的

当弃牌上的数字之和达到时游戏结束

可以被$2n+1$整除。最后丢弃的人赢得了游戏。假设

$A$ 和 $B$ 的最优策略，$A$ 获胜的概率是多少？