java giving 50% chance

[wassssup]

i know math.random can randomly generate numbers, but i want to know is there a way to let the system decides which variable to choose from a and b?
i want to set both can randomly be selected (50 - 50) chance?

[r035198x]

i know math.random can randomly generate numbers, but i want to know is there a way to let the system decides which variable to choose from a and b?
i want to set both can randomly be selected (50 - 50) chance?

Math.random generates numbers between 0.0 and 1.0
Half of that is 0.5.
Ideally then, numbers in [0.0, 0.5) have a 50% chance of getting generated.

[wassssup]

Math.random generates numbers between 0.0 and 1.0
Half of that is 0.5.
Ideally then, numbers in [0.0, 0.5) have a 50% chance of getting generated.

yeah...but is there other way then math.random? just curious..

[r035198x]

yeah...but is there other way then math.random? just curious..

java.util.Random ?

[BigDaddyLH]

java.util.Random?

Indeed. Method Math.random just uses java.util.Random anyway, and the class has a bunch of useful methods. Method Math.random is just leftover from the early API. When you think you need a source of randomness you should check out class Random first.

[drsmooth]

if u use the java.util.Random, you could use

Random r = new Random(); 
int choice = r.nextInt(2);
if(choice==0)
 //choice 1
else
 //choice b

thats what i would recommend, the nextInt(2); would generate either a 1 or a 0, at what i would assume is a 50-50 chance.

[BigDaddyLH]

if u use the java.util.Random, you could use

Random r = new Random(); 
int choice = r.nextInt(2);
if(choice==0)
 //choice 1
else
 //choice b

thats what i would recommend, the nextInt(2); would generate either a 1 or a 0, at what i would assume is a 50-50 chance.

And if it's a turn-left/turn-right you'd be wanting, there's a method for that in Random. No need to generate 0/1 and then turn it into true/false:

if (rnd.nextBoolean()) {
    ...
} else {
    ...
}