领先 2 分

把分差视为随机游走：每局 Alice 赢则 $+1$（概率 $p=0.3$），Alice 输则 $-1$（概率 $q=0.7$）。从 0 出发，吸收边界为 $+2$（Alice 组胜）与 $-2$（Alice 组负）。

将状态平移到 $\{0,1,2,3,4\}$，起点为 2，目标为先到 4（胜）而非 0（负）。设 $r=q/p=\frac{7}{3}$。

带偏赌徒破产的命中概率为

$$\mathbb{P}(\text{先到 4}\mid\text{从 2 出发})=\frac{1-r^2}{1-r^4}.$$

代入得 $r^2=49/9,\ r^4=2401/81$，因此

$$\frac{1-49/9}{1-2401/81}=\frac{-40/9}{-2320/81}=\frac{9}{58}.$$

所以 Alice 赢下整组的概率为 $\boxed{\frac{9}{58}}$。

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英文解析

Consider the score difference as a random walk: Alice wins $+1$ (probability $p=0.3$) and Alice loses $-1$ (probability $q=0.7$). From 0, the absorption boundary is $+2$ (Alice wins) and $-2$ (Alice negative).

Pan to $\{0,1,2,3,4\}$ starting at 2 and aiming for 4 (wins) first instead of 0 (negative). Set $r=q/p=\frac{7}{3}$.

The probability of hitting a bankruptcy with a biased gambler is

$$\mathbb{P}(\text{hit 4 first}\mid\text{starting from 2})=\frac{1-r^2}{1-r^4}.$$

into $r^2=49/9, r^4=2401/81$, so

$$\frac{1-49/9}{1-2401/81}=\frac{-40/9}{-2320/81}=\frac{9}{58}.$$

So the odds of Alice winning the set are $\boxed{\frac{9}{58}}$.