一枚公平硬币 + 其余偏硬币：正面数为偶数的概率

设偏硬币（$n-1$ 枚）出现偶数个正面的概率为 $p$，则出现奇数个正面的概率为 $1-p$。

若公平硬币为 H（概率 $1/2$），要总数为偶数则偏硬币需为奇数；若公平硬币为 T（概率 $1/2$），偏硬币需为偶数。

因此

$$P(\text{偶数})=\frac12(1-p)+\frac12 p=\frac12.$$

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Original Explanation

To obtain an even amount of heads, we have two options. If the fair coin shows $H$, then we must show an odd amount of heads on the other $n-1$ coins to obtain an even total. Alternatively, if the fair coin shows $T$, then we must show an even amount of heads on the other $n-1$ coins to obtain. If $p$ is the probability of an even amount of heads on the other $n-1$ non-fair coins, then by Law of Total Probability, our probability of interest is $\frac{1}{2} \cdot p + \frac{1}{2} \cdot (1-p) = \frac{1}{2}$. This is because the complement of an even amount of heads is an odd amount, and these probabilities are weighted equally in the computation, so they cancel out.