连续 10 个正面的期望抛掷次数

经典结果：首次出现 $n$ 连正面的期望抛掷次数为 $2^{n+1}-2$。

代入 $n=10$ 得

$$2^{11}-2=2046.$$

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Original Explanation

We are going to solve this more generally for $n$ consecutive heads. Let $f_{k}$ be the expected number of flips of the coin needed to obtain $n$ heads when you have $0 \leq k \leq n$ heads consecutively obtained already. Our boundary condition is that $f_n = 0$, as you already have all $n$ heads obtained. We can see that $f_{n-1} = \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot (1 + f_0) = 1 + \frac{1}{2}f_0$, as with probability $1/2$, you reach $n$ heads, and with probability $1/2$, you start over. Similarly, we have that

$$f_{n-2} = 1 + \frac{1}{2}f_{n-1} + \frac{1}{2}f_0 = 1 + \frac{1}{2}\left(1 + \frac{1}{2}f_0\right) + \frac{1}{2}f_0 = \frac{3}{2} + \frac{3}{4}f_0$$

By doing this again, we obtain

$$f_{n-3} = 1 + \frac{1}{2}f_{n-2} + \frac{1}{2}f_0 = 1 + \frac{1}{2}\left(\frac{3}{2} + \frac{3}{4}f_0\right) + \frac{1}{2}f_0 = \frac{7}{4} + \frac{7}{8}f_0$$

A pattern now arises. By recursively applying this process, we can see that

$$f_{n-k} = \left(2 - \frac{1}{2^{k-1}}\right) + \left(1 - \frac{1}{2^k}\right)f_0$$

By plugging in $k = n$, we obtain

$$f_0 = f_0\left(1 - \frac{1}{2^n}\right) + 2 - \frac{1}{2^{n-1}} \iff \frac{1}{2^n}f_0 = 2 - \frac{1}{2^{n-1}} \iff f_0 = 2^{n+1} - 2$$

In particular, for $n = 10$, our answer is $2^{11} - 2 = 2046$.