Dice (III)（概率，期望）

Posted 2022-11-05 如风如影�

Dice (III)

题目链接: link.

原题描述
Given a dice with n sides, you have to find the expected number of times you have to throw that dice to see all its faces at least once. Assume that the dice is fair, that means when you throw the dice, the probability of occurring any face is equal.

For example, for a fair two sided coin, the result is 3. Because when you first throw the coin, you will definitely see a new face. If you throw the coin again, the chance of getting the opposite side is 0.5, and the chance of getting the same side is 0.5. So, the result is

1 + (1 + 0.5 * (1 + 0.5 * …))

= 2 + 0.5 + 0.52 + 0.53 + …

= 2 + 1 = 3

输入
5
1
2
3
6
100

输出
Case 1: 1
Case 2: 3
Case 3: 5.5
Case 4: 14.7
Case 5: 518.7377517640

题意：有一个n面的骰子，每次投掷每个面都是等概率出现，求所有面都出现的期望投掷次数。

题目转化为假设现在已经有m个面投掷出来，要投出m+1个面的期望投掷次数。概率即为(n-m)/n，根据性质概率倒数即为期望n/(n-m)。n（1/1+1/2+1/3+1/4+……+1/n）。

概率期望的性质证明

#include <bits/stdc++.h>
using namespace std;
int main()

    int t;
    cin>>t;
    for(int i=1;i<=t;i++)
    
           double sum=0;
           int n;
           cin>>n;
           for(int j=1;j<=n;j++)
           
               sum=sum+n*1.0/j;
           
            printf("Case %d: %.9lf\\n",i,sum);    
    
    return 0;