先抛硬币再掷骰子

设 $N$ 为首次出现正面所在的抛掷次数。

由于硬币公平， $$\mathbb{P}(N=n)=\left(\frac{1}{2}\right)^n,\quad n=1,2,\dots$$

在给定 $N=n$ 的条件下，公平六面骰掷 $n$ 次都没有出现 1 的概率为 $$\left(\frac{5}{6}\right)^n$$ 所以至少出现一个 1 的概率是 $$1-\left(\frac{5}{6}\right)^n$$

对 $n$ 求全概率： $$\mathbb{P}(\text{至少出现一个 }1) = \sum_{n=1}^{\infty} \left[1-\left(\frac{5}{6}\right)^n\right]\left(\frac{1}{2}\right)^n$$

化简得 $$1-\sum_{n=1}^{\infty}\left(\frac{5}{12}\right)^n$$

这是等比级数， $$\sum_{n=1}^{\infty}\left(\frac{5}{12}\right)^n = \frac{\frac{5}{12}}{1-\frac{5}{12}} = \frac{5}{7}$$

因此最终答案为 $$\boxed{\mathbb{P}(\text{至少出现一个 }1)=1-\frac{5}{7}=\frac{2}{7}}$$

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Original Explanation

Let N be the flip on which the first heads appears.

Step 1: Distribution of N. Since the coin is fair, $$\mathbb{P}(N = n) = \left(\frac{1}{2}\right)^n, \quad n = 1,2,\dots$$

Step 2: Conditional probability given $N = n$. If we roll a fair six-sided die n times, the probability of observing no 1's is $$\left(\frac{5}{6}\right)^n$$ Therefore, the probability of observing at least one 1 is $$1 - \left(\frac{5}{6}\right)^n$$

Step 3: Unconditional probability. $$\mathbb{P}(\text{at least one 1}) = \sum_{n=1}^{\infty} \left[1 - \left(\frac{5}{6}\right)^n \right] \left(\frac{1}{2}\right)^n$$

$$= 1 - \sum_{n=1}^{\infty} \left(\frac{5}{6} \cdot \frac{1}{2}\right)^n = 1 - \sum_{n=1}^{\infty} \left(\frac{5}{12}\right)^n$$

The remaining sum is geometric: $$\sum_{n=1}^{\infty} \left(\frac{5}{12}\right)^n = \frac{\frac{5}{12}}{1 - \frac{5}{12}} = \frac{5}{7}$$

Final answer: $$\boxed{\mathbb{P}(\text{at least one 1}) = 1 - \frac{5}{7} = \frac{2}{7}}$$