两骰连掷：先出现两个 7 还是先出现 6 次偶和

把“无关结果”（奇数和且不为 7）忽略，只看“相关掷”：要么掷出 7（概率 $1/6$），要么掷出偶数和（概率 $1/2$）。

在相关掷中，掷出 7 的条件概率为

$$\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{2}}=\frac14.$$

到第 7 次相关掷时，必然已经达到“2 个 7”或“6 次偶数和”之一。我们要的是前 7 次相关掷中至少出现 2 次 7 的概率。

令 $Y\sim\mathrm{Bin}(7,1/4)$，则

$$\mathbb{P}(Y\ge 2)=1-\left(\frac{3}{4}\right)^7-7\cdot\frac14\left(\frac{3}{4}\right)^6 n=\boxed{\frac{4547}{8192}}\approx 0.555.$$

---

Original Explanation

The probability of rolling a sum of seven in a single roll of a pair of fair dice is $\frac{6}{36} = \frac{1}{6}$ , since

$$7 = 1 + 6 = 2 + 5 = 3 + 4 = 4 + 3 = 5 + 2 = 6 + 1.$$

Similarly, the probability of rolling an even sum in a single roll of a pair of fair dice is $\frac{1}{2}$ .

Observe that the rolls of odd sums other than seven can be ignored and will be, hence, declared as irrelevant. In addition, note that by the 7 th relevant roll we must have either 2 sevens or 6 even numbers. Thus, we get 2 sevens before 6 even numbers if and only if we get at least 2 sevens in the first 7 relevant rolls.

The probability of getting a seven in each relevant roll is

$$\frac{\frac{1}{6}}{\frac{1}{6} + \frac{1}{2}} = \frac{1}{4}$$

The probability of getting at most 1 seven in 7 relevant rolls is

$$\left(\begin{array}{c}{{7}}\\ {{0}}\end{array}\right)\cdot\left(\frac{3}{4}\right)^{7}+\left(\begin{array}{c}{{7}}\\ {{1}}\end{array}\right)\cdot\left(\frac{1}{4}\right)^{1}\left(\frac{3}{4}\right)^{6}=\frac{3645}{8192}$$

Therefore, the probability of getting at least 2 sevens in 7 relevant rolls, or, equivalently, the probability of getting 2 sevens before 6 even numbers is

$$1 - \frac{3645}{8192} = \frac{4547}{8192}\approx 0.55$$