偶数次正面

我们将 $E$ 称为 Paul 翻转硬币偶数次的事件，将 $F_n$ 称为 $n$ 次翻转的结果。 $$P(E) = \frac12P(E | F_1 = heads) + \frac12P(E | F_1 = tails)$$ 根据对称性，$P(E | F_1 = heads) = P(E | F_1 = tails)$，由于它们相同，因此也等于 $P(E)$。我们第一次翻转得到什么似乎并不重要。让我们检查一下 $F_2$ $$P(E) = \frac12P(E | F_1 = F_2) + \frac12P(E | F_1 \not= F_2)$$ 在 Paul 的第一次和第二次翻转比赛中，他发现 $2$ 翻转是连续翻转，因此， $$P(E | F_1 = F_2) = 1$$ 如果他的前两次翻转不匹配，他的进步水平与 $F_1$ 相同，但多了一次翻转。因此， $$P(E | F_1 \not= F_2) = 1 - P(E)$$ 我们现在有： $$P(E) = \frac12 + \frac12(1- P(E)) \Longrightarrow P(E) = \boxed{\frac23}$$

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Original Explanation

Let's call $E$ the event Paul flips the coin an even number of times and $F_n$ the outcome of the $n$th flip.

$$P(E) = \frac12P(E | F_1 = heads) + \frac12P(E | F_1 = tails)$$

By symmetry, $P(E | F_1 = heads) = P(E | F_1 = tails)$, and since they are the same, they also equal $P(E)$. It does not seem to matter what we get on the first flip. Let's examine $F_2$

$$P(E) = \frac12P(E | F_1 = F_2) + \frac12P(E | F_1 \not= F_2)$$

In the case Paul's first and second flip match, he found consecutive flips with $2$ flips so,

$$P(E | F_1 = F_2) = 1$$

In the case his first two flips don't match, he is at the same level of progress as $F_1$ but with an additional flip. Therefore,

$$P(E | F_1 \not= F_2) = 1 - P(E)$$

We now have:

$$P(E) = \frac12 + \frac12(1- P(E)) \Longrightarrow P(E) = \boxed{\frac23}$$