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硬币配对 I

Coin Pair I

专题
Probability / 概率
难度
L2

题目详情

面前有 4 枚公平硬币。你先同时抛掷 4 枚并观察结果。之后你可以反复选择任意两枚硬币一起翻面（一次必须翻两枚，不能只翻 1 枚；翻面会把正面变反面、反面变正面）。你可以重复任意次。

目标：让最后正面数尽量多。若你最优操作，最终正面数的期望是多少？

You have 44 fair coins in front of you. You flip all 44 coins and observe their outcomes. You can then choose to turn over any 22 coins at a time, but you cannot turn just 11 coin without turning another. (A turn moves a coin from heads to tails or tails to heads). You can repeat turning over pairs of coins as many times as you want. The goal of the game is to get the most amount of heads, assuming you play optimally, what is the expected number of heads you will have?

解析

若初始正面数为偶数（0、2、4），可以通过翻转两枚把它变成 4 个正面；若初始为 1 个正面，可翻转两枚反面变成 3 个正面；若初始为 3 个正面则不必操作。

初始正面数为偶数的概率为

(40)+(42)+(44)24=1+6+116=12.\frac{\binom{4}{0}+\binom{4}{2}+\binom{4}{4}}{2^4}=\frac{1+6+1}{16}=\frac{1}{2}.

因此最终以概率 1/21/2 得到 4 个正面，以概率 1/21/2 得到 3 个正面，期望为

12⋅4+12⋅3=72.\frac{1}{2}\cdot 4+\frac{1}{2}\cdot 3=\frac{7}{2}.

Original Explanation

We can start by acknowledging, we can turn over any pair of coins after flipping. From here, we can go over all possible cases for our 4 flips. If we obtain 44 heads from the start, we wouldn't turn over any. If we obtain no heads, we can turn over two pairs of 22 coins to get all heads. Similarly, we can do this if we obtain 22 tails. If we obtain 33 heads, we have no benefit by turning over any coins. If we obtain 11 head, we turn over a pair of tails and end with 33 heads. The probability we end with an even number of heads from the 44 flips is 1+6+116=12\frac{1 + 6 + 1}{16} = \frac{1}{2}, as there are (42)=6\binom{4}{2} = 6 ways to obtain 22 heads and 22 tails.

Therefore, the probability we end with either 33 or 44 heads from the 44 flips is 12\frac{1}{2}. The answer is just 4+32=72\frac{4+3}{2} = \frac{7}{2}.