三个浮点数的概率

Question

How can I implement this into code I am taking a user input of three separate floats that must add to one (ie .333333,.333333,.333333) those numbers are the probability of a number (-1,0,1) being picked at random.

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

Or something along those lines?

Answer 1

The likelihood that three floats will add to exactly 1.0 is very low, because many (most) real numbers cannot be represented exactly as floats. The best you could do is enter two numbers and calculate the third, which would guarantee that they would add up to 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};
}

Questions?