足球比赛

Soccer Match

专题: Probability / 概率
难度: L2
来源: OpenQuant

题目详情

$A$ 和 $B$ 两支球队在两场独立比赛中踢了两次足球。 $A$ 队在第 1 场比赛中的平均射门准确度高于 $B$ 队在第 1 场比赛中的平均射门准确度。

$A$ 队在第 2 场比赛中的平均准确度也高于 $B$ 队在第 2 场比赛中的平均准确度。

根据这些数据，我们是否可以得出结论， $A$ 队在两场比赛中的平均准确率大于 $B$ 队的平均准确率？如果是，请解释原因。如果答案是否定的，请提供反例。

Two teams, $A$ and $B$ , played soccer twice in two independent matches. The average accuracy of shots taken by team $A$ in match 1 was greater than the average accuracy of shots taken by team $B$ in match 1.

The average accuracy of team $A$ in match 2 was also greater than the average accuracy of team $B$ in match 2.

Based on this data, can we conclude that the average accuracy of team $A$ is greater than the average accuracy of team $B$ in the two matches together? If yes, explain why. If the answer is no, provide a counter-example.

解析

我们不能得出这样的结论： $A$ 队两场比赛的平均准确度比 $B$ 队更高。其原因是由于一种称为“辛普森悖论”的现象，即趋势出现在多组数据中，但当将各组数据组合在一起时，趋势就会消失或逆转。

让我们考虑以下示例。在第 1 场比赛中， $A$ 队的得分为 1 / 100，准确率为 1%， $B$ 队的得分为 0 / 1，准确率为 0%。在第 2 场比赛中， $A$ 队的得分为 1 / 1，准确度为 100%， $B$ 队的得分为 99 / 100，准确度为 99%。

在每场个人比赛中， $A$ 队伍的准确率高于 $B$ 队伍，但 $A$ 队伍的总体准确率为 $(1 + 1) / (100 + 1) = 2 / 101 =$ ~1.98%， $B$ 队伍的总体准确率为 $(0 + 99) / (1 + 100) = 99 / 101 =$ ~98%。

Original Explanation

We can not conclude that the average accuracy of the two matches together is greater for team $A$ than for team $B$ . The reason for this is due to phenomenon known as Simpson's paradox in which a trend appears in several groups of data but disappears or reverses when the groups are combined.

Let's consider the following example. In match 1, team $A$ goes 1 / 100 giving them an accuracy of 1% and team $B$ goes 0 / 1, giving them an accuracy of 0%. In match 2, team $A$ goes 1 / 1, giving them an accuracy of 100%, and team $B$ goes 99 / 100 giving them an accuracy of 99%.

In each individual game, team $A$ has a better accuracy than team $B$ , but overall team $A$ has an accuracy of $(1 + 1) / (100 + 1) = 2 / 101 =$ ~1.98% and team $B$ has an overall accuracy of $(0 + 99) / (1 + 100) = 99 / 101 =$ ~98%.