How to generate random number to express the probability

Question

I don't know how to make it in JAVA. Sorry everybody. My case is with 51% probability, I have to do something. and, with 49% probability, I don't have to do anything. I think I need to generate a random number, which will reference, express the probability.

how can I make it suitable to my case in Java? Thank you in advanced!

Possible duplicate of [Generating random integers in a range with Java](http://stackoverflow.com/questions/363681/generating-random-integers-in-a-range-with-java) — Stig Tore, Nov 17 '15 at 14:28
@JaroslawPawlak - What's wrong with `new Random().nextDouble()`? — TheLostMind, Nov 17 '15 at 14:29
@JaroslawPawlak - Why does he need to do division if he gets a value between 0 and 1? :) — TheLostMind, Nov 17 '15 at 14:31
@VinodMadyalkar Ah, you are right, my mistake. It somehow seems simpler to me when the random can return one of two possible values instead of checking whether the return value is lower than the average, and we when we have infinite number of possibilities. But you are right, `nextDouble()` will work fine here. — Jaroslaw Pawlak, Nov 17 '15 at 14:34
plz read again my case. I edited my question in order to make it obvious. thank u! — My Will, Nov 17 '15 at 14:50

gonzo · Accepted Answer · 2015-11-17T14:35:51.073

3

You can use the Random class. It has methods such as Random.nextInt where you can give it an upper bound and it will give you a random number between 0 (inclusive) and that number (exclusive). There are also other methods like Random.nextBoolean which returns 50% chance of true or false.

edited Nov 17 '15 at 14:35

answered Nov 17 '15 at 14:29

gonzo

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so If the chance is 49% and 51% then how to I make it ? – My Will Nov 17 '15 at 14:47
1

@MyWill I would use `Random.nextDouble` for that. `if (new Random().nextDouble()<=.51){}` – gonzo Nov 17 '15 at 14:51
2

if (r.nextInt(99) < 51) { .. } – Rustam Nov 17 '15 at 14:53
1

Yup you can also go the `nextInt` way like @Rustam has mentioned. – gonzo Nov 17 '15 at 14:53
@MyWill glad I was able to help. Feel free to accept my answer so it can help other SO users. :) – gonzo Nov 19 '15 at 16:41

score 0 · Answer 2 · answered Nov 17 '15 at 14:34

You can use Math.random function alternatively. The tutorial is here

Quoting javadoc.

Returns a double value with a positive sign, greater than or equal to 0.0 and less than 1.0. Returned values are chosen pseudorandomly with (approximately) uniform distribution from that range.

Razib · Answer 3 · 2015-11-17T14:46:07.213

If you want to generate integer then you can use nextInt() method like this -

 Random randomGenerator = new Random();
 for (int i = 1; i <= 10; ++i){
   int randomInt = randomGenerator.nextInt(100);
   System.out.println("Generated : " + randomInt);
 }

If you want double you can use nextDouble() method -

 Random randomGenerator = new Random();
 for (int i = 1; i <= 10; ++i){
   int randomInt = randomGenerator.nextDouble(100);
   System.out.println("Generated : " + randomInt);
 }

And if you want to generate random between a range then you can do -

int shift=0;
int range=6;
Random ran = new Random();
int x = ran.nextInt(range) + shift;

This code will generate random number (int) upto 6 (from 0 to 5). If you want to generate random number shifting the lower limit then you can change the shif value. For example changing the shift to 2 will give you all random number greater than or equal 2.

How to generate random number to express the probability

3 Answers3