连续正面

让我们从求解 $\mathbb{E}(X_1)$ 开始：

$\mathbb{E}(X_1)$ 是直到我们恰好看到 1 个正面为止的预期翻转次数。我们可以利用总概率定律将期望值的计算分成两部分的加权平均值：一个部分对应于第一次出现正面时的翻转次数，另一个对应于出现背面时的翻转次数。得出以下总和：

$\newline$ $\mathbb{E}(X_1) = \mathbb{P}(Tail)*[1+\mathbb{E}(X_1)] + \mathbb{P}(Head)*1$ $\newline$

正如你从上面的总和中看到的，如果我们得到一个反面，我们本质上是“重新开始”，并且我们需要对单次反面翻转 (1) 和获得 1 个正面的预期翻转次数求和，即 $\mathbb{E}(X_1)$。另一方面，如果我们得到正面，我们只需返回值 1，因为我们需要翻转 1 次才能看到正面。这就形成了一种递归关系，我们可以代入各个概率的值，求解$\mathbb{E}(X_1)$我们发现它等于2。

$\newline$ 现在我们来求解 $\mathbb{E}(X_2)$： 通过与上述相同的逻辑，我们可以将期望计算为加权平均值，得出以下总和：

$\newline$ $\mathbb{E}(X_2) = \mathbb{P}(Tail)*(1+\mathbb{E}(X_2)) + \mathbb{P}(Head)*[\mathbb{P}(Tail)*(2+\mathbb{E}(X_2)) + \mathbb{P}(Head)*2]$ $\newline$

在第一个表达式中，我们具有相同的逻辑，如果我们得到尾部，我们会通过额外的滚动重新启动该过程。然而，在第二个求和中，我们有嵌套递归的情况，我们得到了正面。如果我们得到正面，我们可以得到反面并重新开始计算，但需要额外掷两次。或者，我们可以获得第二个连续的正面并完成该过程，值为 2。代入相应的概率并求解，得出 $\mathbb{E}(X_2)$ 值为 6。

$\newline$ 让我们从求解 $\mathbb{E}(X_3)$ 开始： 求解此期望会产生嵌套表达式：

$\newline$ $\mathbb{E}(X_3) = \mathbb{P}(Tail)*(1+\mathbb{E}(X_3)) + \mathbb{P}(Head)*[\mathbb{P}(Tail)*(2+\mathbb{E}(X_3)) + \mathbb{P}(Head)*[\mathbb{P}(Tail)*(3+\mathbb{E}(X_3) + \mathbb{P}(Head)*3]]$ $\newline$

代入相应的概率并求解 $\mathbb{E}(X_3)$，结果为 14。

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Original Explanation

Let's start by solving for $\mathbb{E}(X_1)$:

$\mathbb{E}(X_1)$ is the expected number of flips until we see exactly 1 head. We can use the Law of Total Probability to split the calculation of expectation into a weighted average of two parts: one which corresponds to the number of flips if we first get a head, and the other the number of flips which correspond to a tail. This yields the following sum:

$\newline$ $\mathbb{E}(X_1) = \mathbb{P}(Tail)*[1+\mathbb{E}(X_1)] + \mathbb{P}(Head)*1$ $\newline$

As you can see from the sum above, if we get a tails we essentially 'restart' and, we're left summing the single tails flip (1) and the expected number of flips to get 1 head, $\mathbb{E}(X_1)$. On the other hand, if we get a heads, we would simply return the value 1 because it took 1 flips for us to witness the head. This forms a kind of recursive relationship, we can substitute in the values for respective probabilities, and solving for $\mathbb{E}(X_1)$ we find it equals 2.

$\newline$ Let's now solve for $\mathbb{E}(X_2)$: By the same logic as above, we can compute the expectation as a weighted average, yielding the following sum:

$\newline$ $\mathbb{E}(X_2) = \mathbb{P}(Tail)*(1+\mathbb{E}(X_2)) + \mathbb{P}(Head)*[\mathbb{P}(Tail)*(2+\mathbb{E}(X_2)) + \mathbb{P}(Head)*2]$ $\newline$

In the first expression, we have the same logic where we restart the process with one additional roll if we get a tails. In the second sum, however, we have the nested recursive case where we get a heads. If we get a heads, we can either get a tails and restart with calculation, but with 2 additional rolls. Or, we can get a second consecutive heads and complete the process, with a value of 2. Plugging in the respective probabilities and solving yields an $\mathbb{E}(X_2)$ value of 6.

$\newline$ Let's start by solving for $\mathbb{E}(X_3)$: Solving for this expectation yields the nested expression:

$\newline$ $\mathbb{E}(X_3) = \mathbb{P}(Tail)*(1+\mathbb{E}(X_3)) + \mathbb{P}(Head)*[\mathbb{P}(Tail)*(2+\mathbb{E}(X_3)) + \mathbb{P}(Head)*[\mathbb{P}(Tail)*(3+\mathbb{E}(X_3) + \mathbb{P}(Head)*3]]$ $\newline$

Plugging in the respective probabilities and solving for $\mathbb{E}(X_3)$ yields 14.