集齐 p 种优惠券：期望盒数 Cereal coupon 专题 Probability / 概率 难度 L4 来源 QuantQuestion 收藏 标记掌握 个人笔记 题目详情 概率题：集齐 p 种优惠券：期望盒数。 英文原题 Each box of cereal contains a coupon. If there are ppp kinds of coupons, how many boxes of cereal have to be bought on average to obtain at least one coupon of each kind? 解析 设已集齐 k−1k-1k−1 种时，再得到一个新种类的概率为 p−(k−1)p\frac{p-(k-1)}{p}pp−(k−1)，因此等待的新盒数期望为 E[Nk]=1(p−k+1)/p=pp−k+1.\mathbb{E}[N_k]=\frac{1}{(p-k+1)/p}=\frac{p}{p-k+1}.E[Nk]=(p−k+1)/p1=p−k+1p. 总期望 E[N]=∑k=1ppp−k+1=p∑j=1p1j=pHp.\mathbb{E}[N]=\sum_{k=1}^{p}\frac{p}{p-k+1}=p\sum_{j=1}^{p}\frac{1}{j}=\boxed{pH_p}.E[N]=k=1∑pp−k+1p=pj=1∑pj1=pHp. 大 ppp 时近似 plnp+γp+12p\ln p+\gamma p+\tfrac12plnp+γp+21。 英文解析 E[Nk]=1(p−k+1)/p=pp−k+1.\mathbb{E}[N_k]=\frac{1}{(p-k+1)/p}=\frac{p}{p-k+1}.E[Nk]=(p−k+1)/p1=p−k+1p. E[N]=∑k=1ppp−k+1=p∑j=1p1j=pHp.\mathbb{E}[N]=\sum_{k=1}^{p}\frac{p}{p-k+1}=p\sum_{j=1}^{p}\frac{1}{j}=\boxed{pH_p}.E[N]=k=1∑pp−k+1p=pj=1∑pj1=pHp.