不同偏置硬币：出现奇数个正面概率 biased coins 专题 Probability / 概率 难度 L4 来源 QuantQuestion 题目详情 You have nnn biased coins with the kth coin having probability 1/(2k+1)1 / (2k + 1)1/(2k+1) of coming up heads. What is the probability of getting an odd number of heads in total? 解析 设第 kkk 枚硬币正面概率为 pk=12k+1p_k=\frac{1}{2k+1}pk=2k+11。令 pnp_npn 表示抛完前 nnn 枚后，正面总数为奇数的概率。 则 pn=pn−1(1−pn(coin))+(1−pn−1)pn(coin)=pn−1(1−12n+1)+(1−pn−1)12n+1.p_n=p_{n-1}(1-p_n^{(coin)})+(1-p_{n-1})p_n^{(coin)} =p_{n-1}\left(1-\frac{1}{2n+1}\right)+(1-p_{n-1})\frac{1}{2n+1}.pn=pn−1(1−pn(coin))+(1−pn−1)pn(coin)=pn−1(1−2n+11)+(1−pn−1)2n+11. 化简得 (2n+1)pn=(2n−1)pn−1+1.(2n+1)p_n=(2n-1)p_{n-1}+1.(2n+1)pn=(2n−1)pn−1+1. 令 an=(2n+1)pna_n=(2n+1)p_nan=(2n+1)pn，并取 p0=0p_0=0p0=0（没有硬币时正面数为 0 是偶数），则 an=an−1+1,a0=0⇒an=n.a_n=a_{n-1}+1,\quad a_0=0\Rightarrow a_n=n.an=an−1+1,a0=0⇒an=n. 因此 pn=n2n+1.\boxed{p_n=\frac{n}{2n+1}}.pn=2n+1n.