抛硬币：A 有 n+1 枚，B 有 n 枚

答案恒为 $\frac{1}{2}$。

把 A 的最后一枚硬币单独拿出来。记 $E_1,E_2,E_3$ 为 A 前 $n$ 枚与 B 的 $n$ 枚的比较结果：大于/等于/小于。

由对称性 $P(E_1)=P(E_3)$。

• 若 $E_1$ 发生，A 无论最后一枚是什么都赢；
• 若 $E_3$ 发生，A 无法反超；
• 若 $E_2$ 发生，A 需要最后一枚为正面，概率 $1/2$。

因此胜率为 $P(E_1)+\frac12P(E_2)=\frac12$。

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Original Explanation

Take out A’s last coin and compare the number of heads in A’s first $n$ coins to the number of heads in B’s $n$ coins:

• Let $E_1$ be the event that A’s first $n$ coins have more heads than B’s $n$ coins.
• Let $E_2$ be the event that A’s first $n$ coins have the same number of heads as B’s $n$ coins.
• Let $E_3$ be the event that A’s first $n$ coins have fewer heads than B’s $n$ coins.

By symmetry,

$$P(E_1) = P(E_3) = x,\quad P(E_2) = y,\quad 2x + y = 1.$$

• If $E_1$ occurs, then when we add A’s last coin, A definitely has more heads.
• If $E_2$ occurs, whether A ends up with more heads depends on A’s last coin:
  • If it is heads, A has more total heads.
  • If it is tails, the totals are equal.
  • Each of these happens with probability 0.5.
• If $E_3$ occurs, A cannot surpass B.

Hence, the probability that A gets more heads than B is

$$x + 0.5\,y = 0.5.$$