[Prob] (Coupon collector)

Suppose there are n types of toys, which you are collecting one by one, with the goal of getting a complete set. When collecting toys, the toy types are random (as is sometimes the case, for example, with toys included in cereal boxes or included with kids' meals from a fast food restaurant).

Assume that each time you collect a toy, it is equally likely to be any of the n types. What is the expected number of toys needed until you have a complete set?

Solution: Let N be the number of toys needed; we want to find E(N). Our strategy will be to break up N into a sum of simpler r.v.s so that we can apply linearity. So write

N = N1 + N2 + · · · + Nn,

where N1 is the number of toys until the first toy type you haven't seen before (which is always 1, as the first toy is always a new type), N2 is the additional number of toys until the second toy type you haven't seen before, and so forth.

with

相关推荐
缘友一世14 小时前
现代密码学【3】之密码学形式化分析与可证明安全基础
安全·密码学·概率论
byzh_rc15 小时前
[模式识别-从入门到入土] 拓展-EM算法
算法·机器学习·概率论
无水先生2 天前
随机变量在代数运算中的误差传播(2/2)
概率论·统计学
图像生成小菜鸟2 天前
Score Based diffusion model 数学推导
算法·机器学习·概率论
*星星之火*3 天前
【大白话 AI 答疑】 第7篇熵、交叉熵与交叉熵损失的概念梳理及计算示例
人工智能·机器学习·概率论
咚咚王者4 天前
人工智能之数学基础 概率论与统计:第一章 基础概念
人工智能·概率论
_Li.4 天前
机器学习-贝叶斯公式
人工智能·机器学习·概率论
咚咚王者4 天前
人工智能之数学基础 概率论与统计:第二章 核心定理
人工智能·概率论
Y_fulture5 天前
datawhale组队学习:第一章习题
学习·机器学习·概率论
醒过来摸鱼6 天前
空间直线方程
线性代数·概率论