返回题库

四对二

Deuces of Four

专题
Discrete Math / 离散数学
难度
L3

题目详情

Raj 和 Rachel 正在玩一场特殊的网球比赛，规则如下：

  1. 计分从 0 开始，当有人赢得一分时，他们的分数就会增加 1（因此在第一分之后，分数要么是 1-0，要么是 0-1）。
  2. 如果一方比另一方先得 4 分，则一方获胜，unlessunless 比分是 3-3（平手）。如果是平局，玩家必须赢两分才能赢得比赛。

有多少种方法可以达到4-4的分数？

Raj and Rachel are playing a special game of tennis with the following rules:

  1. Scoring starts at 0, when someone wins a point, their score goes up by 1 (thus after the first point, the score is either 1-0 or 0-1).
  2. If a player gets 4 points before the other player, they win, unlessunless the score was 3-3 (deuce). If it was a deuce, a player has to win by two points to win the game.

How many ways are there to get to a score of 4-4?

解析

这是一个有趣的问题，要求我们考虑各种游戏的可能性。嗯，实际上，真的只有 1。比赛比分达到 4-4 的唯一方法是在某个时刻比赛打成 3-3 平。让我们从逻辑上思考这个事实。比如说，如果 Raj 在 Rachel 达到 3 之前达到 4 分（比如 4-2 的比分），那么比赛就会结束，Rachel 永远没有机会达到 4 分。 4-1 和 4-0 的比分也是如此，当然，如果雷切尔领先，情况也是如此。

由此我们得出结论：两名球员必须各得 3 分。让我们计算一下可能发生这种情况的有效方式的数量。总共得分为 6 分。我们可以将它们视为插槽。在这 6 个中，其中 3 个将属于 Rachel 或 Raj。为了获得将 6 个插槽中的 3 个分配给 Rachel 或 Raj 的方式总数，我们可以使用 (63)6 \choose 3。我们正在选择将六分中的哪三分分配给任一球员。

因此，有 (63)6 \choose 3 的方法可以达到 3-3 的分数。但是4-4怎么样？从这里开始，我敢说，问题是微不足道的。从 3-3 我们可以通过两种方式达到 4-4：或者 Rachel 得分然后 Raj 得分，或者 Raj 得分然后 Rachel 得分。这意味着从3-3达到4-4有两种方法。

综合这些结果，我们得到了达到 4-4 分的总共方法： (63)∗2\boxed{{6 \choose 3} * 2}

Original Explanation

This is an interesting question which asks us to consider various game possibilities. Well, actually, really only 1. The only way for the game to reach a score of 4-4 is if at some point the game was tied up at 3-3. Let's think about this fact logically. If, say, Raj reached 4 before Rachel reached 3 (like a score of 4-2), the game would end and Rachel would never have a chance to reach 4 points. The same is true with scores of 4-1 and 4-0, and of course, is true if Rachel is the one ahead.

This leads us to the conclusion that both players must reach 3 points each. Let's count the number of valid ways this could happen. In total, there are 6 points scored. We can think of them as slots. Of these 6, 3 of them are going to either Rachel or Raj. To get the total number of ways we could assign 3 of the 6 slots to either Rachel or Raj, we can use (63)6 \choose 3. We are choosing which three of the six points to assign to either player.

So, there are (63)6 \choose 3 ways to reach a score of 3-3. But how about 4-4? From here, the question is, dare I say, trivial. From 3-3 we can reach 4-4 two ways: either Rachel scores then Raj scores, or Raj scores then Rachel scores. This means there are two ways to reach 4-4 from 3-3.

Combing these results gives us a total number of ways to reach a score of 4-4 as:

(63)∗2\boxed{{6 \choose 3} * 2}