连续 3 个正面的期望抛掷次数

答案为 14。

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Original Explanation

Let $x$ be the expected number of coin tosses it takes to observe three consecutive heads. On the first toss, there is a $\frac{1}{2}$ probability that you observe a tail and essentially start over with 0 consecutive heads, except that your expected number of coin tosses is $x+1$ since you wasted a toss observing a tail. There is a $\frac{1}{4}$ probability of receiving HT and essentially starting over with 0 consecutive heads, except that your expected number of coin tosses is $x+2$ since you wasted two tosses observing HT. There is a $\frac{1}{8}$ probability of receiving HHT and essentially starting over with 0 consecutive heads, except that your expected number of coin tosses is $x+3$ since you wasted three tosses observing tails HHT. Finally, there is a $\frac{1}{8}$ probability of receiving HHH and requiring three tosses. Thus, we can write this as:

$$x = \frac{1}{2}(x+1) + \frac{1}{4}(x+2) + \frac{1}{8}(x+3) + \frac{1}{8} \times 3 \newline x=14$$