硬币配对 III

令 $k$ 为当前反面数（$k\in\{0,1,2,3,4\}$），记 $f(k)$ 为在“还剩 $k$ 个反面”时最终正面数的期望。

边界：$f(0)=4$，$f(1)=3$。

当 $k\ge 2$ 时，重掷两枚反面：

• 以概率 $1/4$ 变成 HH，反面数减 2；
• 以概率 $1/2$ 变成 HT/TH，反面数减 1；
• 以概率 $1/4$ 仍是 TT，反面数不变。

因此

$$f(k)=\frac14 f(k-2)+\frac12 f(k-1)+\frac14 f(k) \Rightarrow f(k)=\frac13 f(k-2)+\frac23 f(k-1).$$

由此可算得 $f(4)=\frac{88}{27}$。由于初始 4 枚硬币的反面数分布对称，最终总体期望也为

$$\boxed{\frac{88}{27}}\approx 3.2593.$$

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Original Explanation

Let $T$ be the total number of heads and $X$ be the total number of heads that appear on the first flipping of the coins. Then $$\mathbb{E}[T] = \mathbb{E}[\mathbb{E}[T \mid X]] = \sum_{k=0}^{4}\mathbb{E}[T \mid X =k]\mathbb{P}[X = k]$$ by the Law of Total Expectation. $X \sim \text{Binom}\left(4,\dfrac{1}{2}\right)$ because $X$ counts the number of heads appearing in 4 flips of a fair coin.

If $X = 4$, then obviously we don't flip any coins again, so $\mathbb{E}[T \mid X = 4] = 4$. Similarly, if $X = 3$, then we aren't able to flip just one tails, so $\mathbb{E}[T \mid X = 3] = 3$.

If $X = 2$, then we are going to continually flip the two tail coins again until we don't obtain $TT$. Conditioned on the fact that we don't obtain $TT$, 2 of the 3 equally-likely outcomes result in us having 3 heads, while 1 of the 3 results in having 4 heads, so $\mathbb{E}[T \mid X = 2] = \dfrac{2}{3} \cdot 3 + \dfrac{1}{3} \cdot 4 = \dfrac{10}{3}$.

Iterating this logic, if $X = 1$, then we flip two tails until they don't appear $TT$. With probability $\dfrac{2}{3}$ we end up with 2 heads, but we can iterate the process again from there on the remaining two tails and reach an expected $\dfrac{10}{3}$ heads. With probability $\dfrac{1}{3}$, we reach 3 heads and we're done. Therefore, $\mathbb{E}[T \mid X = 1] = \dfrac{2}{3} \cdot \dfrac{10}{3} + \dfrac{1}{3} \cdot 3 = \dfrac{29}{9}$.

Lastly, if $X = 0$, with probability $\dfrac{2}{3}$, we end up with 1 head, so we can iterate the process again to have an expected number of heads of $\dfrac{29}{9}$. With probability $\dfrac{1}{3}$, we end up with 2 heads, and we iterate the process again to get an expected number of heads of $\dfrac{10}{3}$. Therefore,

$$\mathbb{E}[T \mid X = 0] = \dfrac{2}{3} \cdot \dfrac{29}{9} + \dfrac{1}{3} \cdot \dfrac{10}{3} = \dfrac{88}{27}$$

Plugging the values and the PMF of $X$ into our expression from the beginning, we get that

$$\mathbb{E}[T] = \dfrac{1}{16} \cdot 4 + \dfrac{4}{16} \cdot 3 + \dfrac{6}{16} \cdot \dfrac{10}{3} + \dfrac{4}{16} \cdot \dfrac{29}{9} + \dfrac{1}{16} \cdot \dfrac{88}{27} = \dfrac{88}{27}$$