6 次抛硬币恰好出现一段 3 连正面

符合条件的序列形状可分为 4 类：HHH???、?HHH??、??HHH?、???HHH，并要求相邻处不能扩展出更长连串。

计数结果为 12 个序列，因此概率为

$$\frac{12}{2^6}=\frac{12}{64}=\frac{3}{16}.$$

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Original Explanation

We have 4 forms that the sequence can be in, which are $HHH---$, $-HHH--$, $--HHH-$, and $---HHH$. Note that the first and second are respectively the reflection of the fourth and third sequences, so we can just do the first two sequences and then multiply by $2$. For the first sequence, the 4th flip must be tails, but the other two can be either parity, so this yields $4$ possibilities. For the second outcome, the first and second dashes must be $T$, but the last can be either, so this yields $2$ outcomes. Therefore, we have a total of $2(4 + 2) = 12$ sample points that have a single run of three heads, so the probability is $\frac{12}{64} = \frac{3}{16}$.