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Home/ Questions/Q 6567701
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T14:22:06+00:00 2026-05-25T14:22:06+00:00

public class SumOfTwoDice { public static void main(String[] args) { int SIDES = 6;

  • 0
public class SumOfTwoDice { 
    public static void main(String[] args) {
        int SIDES = 6;
        int a = 1 + (int) (Math.random() * SIDES);
        int b = 1 + (int) (Math.random() * SIDES);
        int sum = a + b;
        System.out.println(sum);
    }
}

Here is the above code to find out the sum of two random integers between 1 and 6 or any given number.

The below is my own written code, Is this fine. The way i am achieving the Sum of Two random integers. Is this Correct ???

public class TestSample {
    public static void main(String[] args) { 

        int a = Integer.parseInt(args[0]); // 1
        int b = Integer.parseInt(args[1]); // 6
        double ran = Math.random();
        System.out.println("Random Number" + ran);
        double random;

        if(a < b)
            random = (b-a)*ran + a;
        else
            random = (a-b)*ran + b;

        double sum = random + random;
        System.out.println("Random Number" +(int)sum);
    }
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T14:22:07+00:00Added an answer on May 25, 2026 at 2:22 pm

    You have to use Math.random() again to generate a new random number between 0 and 1. Should be something like this:

     public class TestSample {
            public static void main(String[] args) { 
    
            int a = Integer.parseInt(args[0]); // 1
            int b = Integer.parseInt(args[1]); // 6
            double random1, random2;
    
            if(a < b) {
                random1 = (b-a)*Math.random() + a;
                random2 = (b-a)*Math.random() + a;
            }
            else {
                random1 = (a-b)*Math.random() + b;
                random2 = (a-b)*Math.random() + b;
            }
            double sum = random1 + random2;
            System.out.println("Random Number" +(int)sum);
        }
    }
    
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