连续两次正面的期望抛掷次数

设状态 A 表示“当前没有连续正面的前缀（上一轮不是 H）”，状态 B 表示“上一轮是 H”。令 $E_A,E_B$ 分别为从状态 A/B 到达 HH 的期望抛掷次数。

从状态 A：下一次抛掷后

• 以 $1/2$ 概率得到 T，仍在 A；
• 以 $1/2$ 概率得到 H，进入 B。

因此

$$E_A = 1 + \frac{1}{2}E_A + \frac{1}{2}E_B.$$

从状态 B：下一次抛掷后

• 以 $1/2$ 概率得到 H，结束；
• 以 $1/2$ 概率得到 T，回到 A。

因此

$$E_B = 1 + \frac{1}{2}\cdot 0 + \frac{1}{2}E_A.$$

联立解得 $E_A=6$，所以答案是 $\boxed{6}$。

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英文解析

Set state A to "there is no current run of consecutive heads (the previous round was not H)", and state B to "The previous round was H". Make $E_A,E_B$ the expected number of tosses from status A/B to HH, respectively.

From State A: After the next toss

• T with a probability of $1/2$, still in A;
• Get H with a probability of $1/2$ and enter B.

Therefore,

$$E_A = 1 + \frac{1}{2}E_A + \frac{1}{2}E_B.$$

From State B: After the next toss

• H with a probability of $1/2$, ending;
• Get T with a probability of $1/2$ and go back to A.

Therefore,

$$E_B = 1 + \frac{1}{2}\cdot 0 + \frac{1}{2}E_A.$$

The answer is $E_A=6$, so the answer is $\boxed{6}$.