CODEWITHSUNDEEP
java
David Edgar
David Edgar

Reputation: 505

Generate random number in interval with given probability

I want to generate a random number with a given probability but I'm not sure how to:

I have a two dimensional array: int[ ][ ] aryNumbers = new int[4][]

and for each number(4) I want to generate a int result in a inteval of [1...9] with a prob of 0.5 .Otherwise a number in a interval of [10...99].

Note: I know how to generate a number, but choosing between intervals confused me.

Edit: 

 public int numberAttribution(){
    Random rand = new Random();
    double dbNum = rand.nextDouble();
    int intNum;
    int min1 = 1, max1 = 9, range1 = max1 - min1 + 1;
    int min2 = 10, max2 = 99, range2 =  max2 - min2 + 1;

    if(dbNum < 0.5){
        intNum = rand.nextInt(range1) + min1;
    }else{
        intNum = rand.nextInt(range2) + min2;
    } 
       System.out.print(intNum);
    return intNum;
}

Upvotes: 1

Views: 357

Answers (2)

Andreas
Andreas

Reputation: 5103

I assume this method will be called millions of times a second. In that case it might be wise to only generate one random each time:

  private static int generateRandom() {
    double rand = Math.random();
    if (rand < 0.5) {
      rand *= 18;
      rand += 1;
    } else {
      rand -= 0.5;
      rand *= 180;
      rand += 10;
    }
    return (int) rand;
  }

Upvotes: 1

se1by
se1by

Reputation: 108

So if I got you right, you want a 50% chance to generate a number between 1 and 9, otherwise generate a number between 10 and 99?

In that case you could do it like this:

int randomNumber = 0;
if (Math.random() > 0.5) {
    randomNumber = 1 + (int) (Math.random() * 9);
} else {
    randomNumber = 10 + (int) (Math.random() * 90);
}

Runnable version

Upvotes: 1

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