多一枚硬币的优势

比较 Emilia 的前 $n$ 枚与 Ron 的 $n$ 枚正面数。设三种事件：

• Emilia 前 $n$ 枚正面更多；
• 正面数相同；
• Emilia 前 $n$ 枚正面更少。

由于对称性，第一与第三种事件概率相等。

若 Emilia 前 $n$ 枚已领先，则额外一枚硬币不影响她仍领先；若落后，也不可能靠一枚硬币反超超过 1 的差距；只有在“打平”时，额外一枚硬币以 $1/2$ 概率（掷出正面）让她获胜。

最终可得 Emilia 获胜概率恒为 $1/2$。

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Original Explanation

We can solve using symmetry by comparing the number of heads in Emilia and Ron's first $n$ coins. Consider the 3 results: $E_1$ is the event that Emilia has more heads than Ron; $E_2$ is the event that Emilia and Ron have the same number of heads; $E_3$ is the event that Emilia has fewer heads than Ron. By symmetry, $P(E_1) = P(E_3) = a$ and $P(E_2) = b$. Since $\sum_{\omega \in \Omega} P(\omega) = 1$, $2a + b = 1$. For $E_1$, Emilia will always have more heads than Ron, regardless of Emilia's final coin result. Furthermore, for $E_3$, Emilia will never have more heads than Ron, regardless of Emilia's final coin result. Therefore, only event $E_2$ depends on Emilia's $n+1$th coin flip, in which Emilia will win half the time (if she gets a heads). Thus, Emilia's total probability of winning is: $a + 0.5b = a + (1-2a) = \frac{1}{2}$.