c++ - 计算类似于二项式和的条件概率

我正在考虑一个有任意人数的社会。每个人只有两个选择。他或她保留当前的选择，或者她切换。在我要编写的代码中，用户输入了切换人员的可能性。

为了弄清楚我要做什么，假设用户告诉计算机社会中有3个人，每个人选择切换的概率由(p1，p2，p3)给出。考虑人1。他有切换p1的可能性。用他作为我们计算的基础，给人1作为基础的概率，即社会中没有人选择切换的概率由下式给出:

P_ {1}(0)=(1-p2)*(1-p3)

并且以人1为基数，社会中恰好有一个人选择切换的概率由下式给出:

P_ {1}(1)= p2 *(1-p3)+(1-p2)* p3。

如果不写出总和中的每一项，我将无法弄清楚如何用C++编写此概率函数。我考虑使用二项式系数，但由于用户输入的不同，需要考虑任意多个概率，因此我无法计算出总和的封闭式表达式。

我附上了我所拥有的。概率函数只是我要尝试做的一部分，但它也是最难的部分。我将概率函数命名为probab，而函数中for循环中的内容显然是错误的。

编辑:基本上，我想计算选择一个子集的概率，其中该子集中的每个元素都有不同的被选择概率。

我将不胜感激有关此问题的任何提示。请注意，我是C++的初学者，因此，任何有关提高我的编程技能的技巧也将受到赞赏。

#include <iostream>
#include <vector>

using namespace std;
unsigned int factorial(unsigned int n);
unsigned int binomial(unsigned int bin, unsigned int cho);
double probab(int numOfPeople, vector<double> probs, int p, int num);

int main() {

    char correctness;
    int numOfPeople = 0;
    cout << "Enter the # of people: ";
    cin >> numOfPeople;
    vector<double> probs(numOfPeople); // Create a vector of size numOfPeople;

    for (int i = 1; i < numOfPeople+1; i++) {
        cout << "Enter the probability of person "<< i << " will accept change: ";
        cin >> probs[i-1];
    }
    cout << "You have entered the following probabilities of accepting change: (";
    for (int i = 1; i < numOfPeople+1; i++) {
        cout << probs[i-1];
        if (i == numOfPeople) {
            cout << ")";
        }
        else {
            cout << ",";
        }
    }
    cout << endl;
    cout << "Is this correct? (Enter y for yes, n for no): ";
    cin >> correctness;
    if (correctness == 'n') {
        return 0;
    }

    return 0;
}

unsigned int factorial(unsigned int n){                                     // Factorial function
    unsigned int ret = 1;

    for(unsigned int i = 1; i <= n; ++i) {
        ret *= i;
    }
    return ret;
}

unsigned int binomial(unsigned int totl, unsigned int choose) {             // Binomial function
    unsigned int bin = 0;
    bin = factorial(totl)/(factorial(choose)*factorial(totl-choose));
    return bin;
}

double probab(int numOfPeople, vector<double> probs, int p, int num) {      // Probability function
    double prob = 0;

    for (int i = 1; i < numOfPeople; i++) {
        prob += binomial(numOfPeople, i-1)/probs[p]*probs[i-1];
    }
    return prob;
}

最佳答案

供将来参考，对于任何尝试这样做的人，概率函数将类似于:

double probability (vector<double> &yesprobabilities, unsigned int numOfPeople, unsigned int yesNumber, unsigned int startIndex) {
    double kprobability = 0;
    // Not enough people!
    if (numOfPeople-1 < yesNumber) {
        kprobability = 0;
    }
    // n == k, the only way k people will say yes is if all the remaining people say yes.
    else if (numOfPeople-1 == yesNumber) {
        kprobability = 1;
        for (int i = startIndex; i < numOfPeople-1; ++i) {
            kprobability = kprobability * yesprobabilities[i];
        }
    }
    else if (yesprobabilities[startIndex] == 1) {
        kprobability += probability(yesprobabilities,numOfPeople-1,yesNumber-1,startIndex+1);
    }
    else {
        // The first person says yes, k - 1 of the other persons have to say yes.
        kprobability += yesprobabilities[startIndex] * probability(yesprobabilities,numOfPeople-1,yesNumber-1,startIndex+1);
        // The first person says no, k of the other persons have to say yes.
        kprobability += (1 - yesprobabilities[startIndex]) * probability(yesprobabilities,numOfPeople-1,yesNumber,startIndex+1);
    }
    return probability;
}

这里使用一种称为递归函数的东西。这对我来说是全新的，非常有启发性。我将此归功于Math Stack交流的Calle。我稍微修改了他的版本，在一些帮助下采用了 vector 而不是数组。

关于c++ - 计算类似于二项式和的条件概率，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/22727196/