抛硬币领先

Lead Count

专题: Probability / 概率
难度: L4
来源: QuantQuestion

个人笔记

题目详情

抛一枚公平硬币 10 次。

定义：在前 $n$ 次抛掷里，若某一面（正面或反面）出现次数严格大于 $n/2$ ，则称该面在第 $n$ 次后“领先”。

问：第一次抛掷得到的那一面，在第 10 次后仍然领先的概率是多少？

英文原题

Suppose you flip a fair coin $10$ times. We say a certain outcome (heads or tails) is leading after $n$ flips if there are strictly larger than $\frac{n}{2}$ flips of that outcome within the first $n$ flips. Find the probability that the outcome of the first flip of the coin is leading after $10$ flips.

解析

若第一次为正面，则“第一次结果领先”要求 10 次中正面数 $\ge 6$ 。

若第一次为反面，则要求正面数 $\le 4$ （等价于反面数 $\ge 6$ ）。

由对称性，这两种情况概率相同，且它们恰好把所有结果二分。

因此所求概率为 $\boxed{\frac{1}{2}}$ 。

英文解析

If it is positive for the first time, the first result lead requires a positive number of $\ge 6$ out of 10 times.

If it is negative for the first time, the positive number $\le 4$ (equivalent to the negative number $\ge 6$ ) is required.

By symmetry, both cases have the same probability, and they just happen to bisect all the results.

Therefore, the probability required is $\boxed{\frac{1}{2}}$ .