9 枚硬币那方正面更多的概率

设 $A\sim\mathrm{Bin}(8,1/2)$。设 B 的前 8 枚正面数为 $X\sim\mathrm{Bin}(8,1/2)$，第 9 枚为 $Y\sim\mathrm{Bern}(1/2)$，且独立，则 $B=X+Y$。

所求

$$P(B>A)=P(X+Y>A)=\tfrac12 P(X\ge A)+\tfrac12 P(X>A).$$

由于 $X$ 与 $A$ 同分布且独立，有

$$P(X>A)=P(A>X)=\frac{1-P(X=A)}{2},\quad P(X\ge A)=P(X>A)+P(X=A).$$

代入得

$$P(B>A)=\tfrac12\bigl(P(X>A)+P(X=A)\bigr)+\tfrac12 P(X>A)=P(X>A)+\tfrac12P(X=A)=\frac12.$$

因此 $\boxed{P(B>A)=\frac12}$。

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Original Explanation

A known symmetry argument: $P(B>A) = \tfrac12$.
One intuitive reasoning: If you imagine B’s extra coin is set aside, compare B’s first 8 coins to A’s 8. Probability that B has strictly more heads among those 8 is 1/2 minus half the tie probability; but the 9th coin can break ties. Detailed analysis shows it indeed is $1/2$.

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