java - 三个浮点数的概率

Question

我如何在代码中实现这一点 我正在接受三个单独的浮点数的用户输入，这些浮点数必须加到一个（即 .333333、.333333、.333333）这些数字是数字（-1,0,1）的概率随机挑选的。

if( new Random().nextDouble() <= 0.333334){array[i]=randomNumber(-1,0,1)?

或类似的规定？

Answer 1

三个浮点数加起来正好为 1.0的可能性非常低，因为许多（大多数）实数不能完全表示为浮点数。你能做的最好的就是输入两个数字并计算第三个数字，这样可以保证它们加起来等于 1.0。

Answer 2

public static void main(String[] args) {
    double[] probs = readProbabilities();
    double random = new Random().nextDouble();
    int randomNumber;
    if (random <= probs[0]) {
        randomNumber = -1;
    } else if (random <= (probs[0] + probs[1])) {
        randomNumber = 0;
    } else {
        randomNumber = 1;
    }
    System.out.println("Random Number is " + randomNumber);
}

public static double[] readProbabilities() {
    Scanner sc = new Scanner(System.in);
    double first, second, third;
    System.out.print("Please insert 1st probability: ");
    first = sc.nextDouble();
    while (first < 0.0 || first > 1.0) {
        System.out.print("Must be between 0.0 and 1.0, try again: ");
        first = sc.nextDouble();
    }
    System.out.print("Please insert 2nd probability: ");
    second = sc.nextDouble();
    while (second < 0.0 || (first + second) > 1.0 ) {
        System.out.print("Must be between 0.0 and " + (1.0 - first) + ":");
        second = sc.nextDouble();
    }
    third = 1.0 - (first + second);
    System.out.println("3rd Possibility is " + third);
    return new double[] {first, second, third};
}

问题？