Java "fake" math.random()

Question

I need a function which generates me random numbers but where some numbers can appear more often then others.

Lets say i want random numbers from 1 to 10 where 2 and 3 have twice the chance to appear than the rest.

How can i do this in java? Haven`t found a function for this.

Edit: Code i have so far (Which was a compromise solution, i know it`s not what i want)

int min;
int max;
double min1;
double max1;
for(int j=0;j<Ammount_numbers;j++) {
            if(min1==0) {
                  min1= min + (Math.random() * (max+1-min));
                  max1= min + (Math.random() * (max+1-min));
            }

            double randomnumber = min1 + (Math.random() * ((max1-min1) + 1));
}

Min and Max are the borders (1 to 10 in the example) And with min1 and max1 i generated new, smaller borders so that the randomnumber is clustered in a random range.. And as i said: i know that this is not what i wanted, it was just a compromise solution so i could work with it..

Where is the code which you tried so far so that we could check that for bugs and problems? — Tahir Hussain Mir, Aug 19 '17 at 10:24
fill an array and randomize the index number, adding more 2's and 3's. — Hatted Rooster, Aug 19 '17 at 10:25
@RickAstley you are making him dependable upon others to think. Best thing is encourage him to think himself. Isn't it? — Tahir Hussain Mir, Aug 19 '17 at 10:26
Possible duplicate of [Weighted randomness in Java](https://stackoverflow.com/questions/6737283/weighted-randomness-in-java) — jonrsharpe, Aug 19 '17 at 10:27
@TahirHussainMir: you first comment is irrelevant. The OP is clearly asking how-to. — Yves Daoust, Aug 19 '17 at 10:27
@YvesDaoust I respect your opposition. But SO mentions opinion based should not be answered. Am i right? if not, please enlighten me. — Tahir Hussain Mir, Aug 19 '17 at 10:30
@TahirHussainMir: the question is certainly not opinion-based. There are well-established techniques which are objectively appropriate. If you want to close the question, the "too broad" tag would be more appropriate (but undeserved IMO) — Yves Daoust, Aug 19 '17 at 10:36
i edited what i had so far. i don't need a real "how to". giving me an idea on how to do it is enough cause at the moment i don`t have an idea — Grogak, Aug 19 '17 at 10:37
@Vanna: I know this very well and I have know SO's policy for years, this is not the point. Tahir did not explain this. This question can be answered without writing code. — Yves Daoust, Aug 19 '17 at 10:37

score 0 · Answer 1 · answered Aug 19 '17 at 10:32

0

To implement a generator of a discrete random variable with prescribed probabilities, a well-known technique is to fill an array with the cumulated probabilities. The values will fit in range [0,1].

Then perform uniform drawings and identify the subinterval where the value falls, by dichotomic search (linear can do for a small array).

The subinterval index will follow the desired distribution.

answered Aug 19 '17 at 10:32

Yves Daoust

53,540
8
41
94

Yes. It was a stupid down vote. Still don't know why someone downvoted this. Atleast, he/she must have commented. – Tahir Hussain Mir Aug 19 '17 at 10:42
Okay the first part is clear to me. But what do you mean with "uniform drawings..."? – Grogak Aug 19 '17 at 10:46
@Grogak: using the uniform random number generator. http://keithschwarz.com/darts-dice-coins/ – Yves Daoust Aug 19 '17 at 10:47

score 0 · Answer 2 · answered Aug 19 '17 at 11:07

0

You can do like this Now there is a 50 % chance of getting a 2 or 3

a=(Math.random() * 7) + 4;
b=(Math.random() * 2 ) + 2; // math.random() within a range.
x=Math.random()* 1;
if(x==0){
// print a;
}
else if(x==1){
// print b;
}

answered Aug 19 '17 at 11:07

Varen

1
3

i have just used the concept of Math.random() within the range. this is an excellent article on it https://stackoverflow.com/questions/7961788/math-random-explained – Varen Aug 19 '17 at 11:10

score 0 · Answer 3 · answered Aug 19 '17 at 14:00

Set up the numbers you want at the correct frequencies in an array and pick a random entry from the array.

Random rand = new Random();

// ...

int specialRandom() {
  int[] selections = {1, 2, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10};
  return selections[rand.nextInt(selections.length)];
}

For more complex selection, use the desired frequency, pick a random frequency and return the matching number.

score -1 · Answer 4 · answered Aug 19 '17 at 10:40

-1

You cannot do that with Math.random() itsself, but you could write your own random() method, which uses Math.random() and returns your own random numbers according to your own scale.

answered Aug 19 '17 at 10:40

Dorian Gray

2,781
1
8
25

Java "fake" math.random()

4 Answers4