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幸运糖果

Lucky Candy

专题
Probability / 概率
难度
L2

题目详情

你有 50 颗好糖和 50 颗坏糖，要放进两个盒子。

规则：先等概率随机选一个盒子，再从该盒子里等概率随机取出一颗糖。

如何摆放，才能让取到好糖的概率最大？

提示：把所有好糖放一个盒子、所有坏糖放另一个盒子得到的概率是 0.5，但这不是最优。

How do you place 50 good candies and 50 rotten candies in two boxes such that if you choose a box at random and take out a candy at random, it better be good!

We need to maximize the probability of getting a good candy when selecting a random box and a random candy from it.

Hint

Placing all bad in 1 box and all good in another will give the probability of 0.5, which is not the maximum.

解析

最优做法：

  • 盒子 A：放 1 颗好糖。
  • 盒子 B：放剩下的 49 颗好糖 + 50 颗坏糖。

此时取到好糖概率为：

12⋅1+12⋅4999=1−2599⋅12=148198≈74.747%.\frac{1}{2}\cdot 1 + \frac{1}{2}\cdot \frac{49}{99} = 1 - \frac{25}{99}\cdot\frac{1}{2} = \frac{148}{198} \approx 74.747\%.

直觉：让一半概率的“选中盒子 A”直接确保成功，同时另一半概率中尽量不把好糖比例降得太厉害。

Original Explanation

Place 1 good candy in one box and all the remaining (49 good and 50 rotten candies) in the second box.

Solution

Place 1 good candy in one box and all remaining (49 good and 50 rotten candies) in the second box.

This will give a probability of (1/2)∗1+(1/2)∗(49/99)=74.74ˉ%(1/2) * 1 + (1/2) * (49/99) = 74.\bar{74}\%