掷骰直到出现 6

设 $E$ 表示“当前累计和为偶数时，最终为偶数”的概率，$O$ 表示“当前累计和为奇数时，最终为偶数”的概率。

从偶数状态出发：掷到 6（概率 $1/6$）直接结束且仍为偶数；掷到 1,3,5 会转到奇数状态（共 3 个点数），掷到 2,4 会留在偶数状态（共 2 个点数）。

因此

$$E=\frac{1}{6}\cdot 1+\frac{2}{6}E+\frac{3}{6}O.$$

从奇数状态出发：掷到 6 结束且为奇数（偶数事件失败）；掷到 1,3,5 会转到偶数状态，掷到 2,4 会留在奇数状态：

$$O=\frac{3}{6}E+\frac{2}{6}O.$$

解得 $O=\frac{3}{4}E$，代回得 $E=\frac{4}{7}$。

初始累计为 0（偶数），所以所求概率为 $\boxed{\frac{4}{7}}$。

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英文解析

Let $E$ represent the probability that the current cumulative sum is even, and $O$ represent the probability that the current cumulative sum is even.

From the even state: throwing to 6 (probability $1/6$) directly ends and is still even; throwing to 1,3,5 will go to the odd state (3 points total), and throwing to 2,4 will remain in the even state (2 points total).

Therefore,

$$E=\frac{1}{6}\cdot 1+\frac{2}{6}E+\frac{3}{6}O.$$

From the odd state: throw to the end of 6 and it is odd (even event failure); throw to 1, 3, 5 will go to the even state, and throw to 2, 4 will stay in the odd state:

$$O=\frac{3}{6}E+\frac{2}{6}O.$$

I got $O=\frac{3}{4}E$, and I got $E=\frac{4}{7}$back.

The initial cumulative is 0 (even), so the probability required is $\boxed{\frac{4}{7}}$.