Question

We know that the classic range random function is like this:

public static final int random(final int min, final int max) {
    Random rand = new Random();
    return min + rand.nextInt(max - min + 1);  // +1 for including the max
}

I want to create algorithm function for generating number randomly at range between 1..10, but with uneven possibilities like:
1) 1,2,3 -> 3/6 (1/2)
2) 4,5,6,7 -> 1/6
3) 8,9,10 -> 2/6 (1/3)

Above means the function has 1/2 chance to return number between 1 and 3, 1/6 chance to return number between 4 and 7, and 1/3 chance to return number between 8 and 10.

Anyone know the algorithm?

UPDATE:
Actually the range between 1..10 is just served as an example. The function that I want to create would apply for any range of numbers, such as: 1..10000, but the rule is still same: 3/6 for top range (30% portion), 1/6 for middle range (next 40% portion), and 2/6 for bottom range (last 30% portion).

Answer 1

Use the algorithm:

int temp = random(0,5);
if (temp <= 2) {
  return random(1,3);
} else if (temp <= 3) {
 return random(4,7);
} else  {
 return random(8,10);
}

This should do the trick.

EDIT: As requested in your comment:

int first_lo = 1, first_hi = 3000; // 1/2 chance to choose a number in [first_lo, first_hi]
int second_lo = 3001, second_hi = 7000; // 1/6 chance to choose a number in [second_lo, second_hi] 
int third_lo = 7001, third_hi = 10000;// 1/3 chance to choose a number in [third_lo, third_hi] 
int second
int temp = random(0,5);
if (temp <= 2) {
  return random(first_lo,first_hi);
} else if (temp <= 3) {
 return random(second_lo,second_hi);
} else  {
 return random(third_lo,third_hi);
}