Calling a Method 1 out of 100 Times

Question

I need a 1 in a 100 chance of some functionality being called. Right now I am using the following code, but it's being called more often than 1% of the time.

int randomItem = (int)(Math.random()*100);
if (randomItem >= 90 && randomItem <= 100)
{
    // Do something ...
}

Answer 1

You're adding the sprite when the random number is between 90 and 100. This makes a 10% chance. Just test if Math.random() is < 0.01.

Answer 2

You have to say something like if (Math.random() * 100 >= 99.0), or just without the scaling, if (Math.random() >= 0.99), or with the integer, if (randomItem == 99).

Math.random() returns a floating point number in the range [0.0, 1.0), so 100 times this gives you a range [0.0, 100.0), and conversion to integer gives an integer in the range [0, 100).

The conversion to integer may not be necessary if you don't need it for anything else.

Answer 3

You should replace

randomItem >= 90 && randomItem <= 100

with

randomItem == 99

Answer 4

Right now you have a 1 int 10 chance. To get a 1 in 100 chance do this:

int chance = (int)(100*Math.random())
if (chance == 1) {
    //Code
}

Answer 5

Math.Random returns a number between 1 and 0. To get a 1% chance you want your code to look somewhat like this:

Math.Random() <= 0.01

Also, the nextDouble method of the Random class is used the same way.

r.nextDouble() < 0.01; //r is a instance of Random

Both of these will give you a 1% chance.