连续两个 6

设 $E$ 为从开始到首次出现连续两个 6 所需掷骰次数的期望。平均先要掷 6 次才会第一次掷出 6。此后下一次再掷：以 $\frac{1}{6}$ 的概率直接再出一个 6，于是只再多掷 1 次；以 $\frac{5}{6}$ 的概率没有出 6，相当于白白多掷了 1 次后重新开始。因此

$$E = 6 + \frac{1}{6} \times 1 + \frac{5}{6}(E+1)$$

解得 $\boxed{E=42}$。

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Original Explanation

We can solve this using a recurrence relation on, $E$, the expected number of rolls. When we start rolling, we expect, on average 6 rolls until a 6 shows up. Once that happens, there is a 1/6 chance that we will roll once more, and a 5/6 chance that we will be, effectively, starting all over again, and so have as many additional expected rolls as when we started. As a result, we can say:

$$E = 6 + \frac{1}{6} \times 1 + \frac{5}{6}(E+1)$$

Solving this, we find that $\boxed{E=42}$