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math.random always give 0 result

I am using Ubuntu 14.04.3 LTS and I am studying Java from the book. I tried to follow one example on the book with Ubuntu Terminal and I'm using Sublime Text. Here is the code
import java.util.Scanner;
public class RepeatAdditionQuiz{
public static void main(String[] args){
int number1 = (int)(Math.random()%10);
int number2 = (int)(Math.random()%10);
Scanner input = new Scanner(System.in);
System.out.print(
"What is "+number1+" + "+number2+"?");
int answer = input.nextInt();
while(number1+number2 != answer){
System.out.print("Wrong answer. Try again. What is "
+number1+" + "+number2+"? ");
answer = input.nextInt();
}
System.out.println("You got it!");
}
}
But the problem is, when I compiled it and executed it. It gives me result
what is 0 + 0?_
every time. It suppose to give me random number, and yes, it can be 0. But I tried to run it more than 10 times, it keeps giving me 0 + 0, when it's suppose to random from 0-9.
The result of 0 + 0 is fine when I typed 0 as a result, it gets me out of the loop
Did I miss something to make the math library works? How can I fix the randomize issue?

Math.random() returns a double value between 0 (inclusive) and 1 (exclusive). It does not return an integer value. Therefore, when you take the number produced by Math.random() modulo 10, it returns the same double value. The final cast to int makes that value always 0.
Run the following code to see for yourself:
double random = Math.random();
System.out.println(random); // for example 0.5486395326203879
System.out.println(random % 10); // still 0.5486395326203879
System.out.println((int) (random % 10)); // outputs 0
What you really want is to use a Random object and use Random.nextInt(bound). To have a random integer between 0 and 9, you can use:
Random random = new Random();
int value = random.nextInt(10);

Math.random();
Returns a double value with a positive sign, greater than or equal to 0.0 and less than 1.0.
This is the problem. You need to then multiply with 10, so you get a number between 0 and 10 and after that you can cast to int.

public static double random()
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.
(source)
It returns a double that satisfies 0.0 <= double < 1.0, therefore taking the modulo will always result in zero. I recommend the java.util.Random class to achieve what you need. Specifically Random.nextInt(int).

The Math.random() method returns a random double that is from 0 (inclusive) to 1 (exclusive). Performing % 10 doesn't affect this value, but casting it to an int truncates any decimal portion, always yielding 0.
If you want a random number from 0-9, you can multiply Math.random() by 10, instead of taking the remainder when divided by 10.
Alternatively, you can create a java.util.Random object and call nextInt(10).

Read the doc :
Returns a double value with a positive sign, greater than or equal to
0.0 and less than 1.0
Therefore, you will always get less than 1, and the cast to int will round it down to 0. Use, for example, a Random object which has a nextInt(int max) method :
Random rd = new Random();
int number1 = rd.nextInt(10);
int number2 = rd.nextInt(10);

The Math.random() method returns a Double value between 0 and 1, so you will never get a number greater or equal than 1, you will ever get a value that could be 0, but never 1. And as you are taking the residual from this value over 10, you will ever get a 0 as result.
Math.random() % 10 will always be 0, because the random method gives you 0 <= value < 1, and when the % 10 operation takes place, you will get 0.
Check here for more details (oracle documentation)

Math.random() returns a number greater or equal than 0 and less than 1.
What you are looking for is either
int number1 = (int)(Math.random()*10);
int number2 = (int)(Math.random()*10);
or
Random rand = new Random();
int number1 = rand.nextInt(10);
int number2 = rand.nextInt(10);
Also, to get random number from given range, use this for Math.random()
int number 3 = min + (int)(Math.random() * ((max - min) + 1))
and for random.nextInt()
int number 4 = random.nextInt(max - min) + min;

Math.random gives a double value between 0 (incl.)-1 (excl.)!

Problems Generating A Math.random Number, Either 0 or 1

I want a random number, either 0 or 1 and then that will be returned to main() as in my code below.
import java.util.Scanner;
public class Exercise8Lab7 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int numFlips = 0;
int heads = 0;
int tails = 0;
String answer;
System.out.print("Please Enter The Number Of Coin Tosses You Want: ");
numFlips = input.nextInt();
for(int x = 1;x <= numFlips; x++){
if(coinToss() == 1){
answer = "Tails";
tails++;
}
else{
answer = "Heads";
heads++;
}
System.out.print("\nCoin Toss " + x + ": " + answer);
}
System.out.println("\n\n====== Overall Results ======" +
"\nPercentage Of Heads: " + (heads/numFlips)*100 + "\nPercentage Of Tails: " + (tails/numFlips)*100);
}
public static int coinToss(){
double rAsFloat = 1 * (2 + Math.random( ) );
int r = (int)rAsFloat;
return r;
}
}
Many solutions had been suggested to use the util.Random option which I have done and works perfectly but I want to sort out why I can't get this to work. Obviously I want the number to be an int myself so I convert it to an int after the random number has been generated. But no matter what I add or multiply the Math.random() by, it will always all either be Heads or all either be Tails. Never mixed.

Try this) It will generate number 0 or 1
Math.round( Math.random() ) ;

You could use boolean values of 0 or 1 based on value of Math.random() as a double between 0.0 and 1.0 and make the random generator much simpler. And you can get rid completely of the coinToss() method.
if(Math.random() < 0.5) {
answer = "Tails";
tails++;
}
Remove the coin toss method and replace the first conditional with the code above.
Math.random(); by itself will return a value between 0.0 and less than 1.0. If the value is in the lower half, [0.0, 0.5), then it has the same probability of being in the upper half, [0.5, 1.0). Therefore you can set any value in the lower half as true and upper as false.

Wierd that no one is using a modulo division for the random number.
This is the simplest implementation you can get:
Random rand = new Random();
int randomValue = rand.nextInt() % 2;

Math.round(Math.random()) will return either 0.0 and 1.0. Since both these values are well within the limits of int range they can be casted to int.
public static int coinToss(){
return (int)Math.round(Math.random());
}

(int)(Math.random()*2) also works fine in this case

its not working because of the integer math you are using, the call to 2+ Math.Random is pretty much always giving you a answer between 0.0 and 1.0.
so assuming that you recieve 0.25 as your result your maths is as follows
double d = 1* (2 + 0.25); // (result = 2
Then you are checking to see if your result == 1 ( which it never will. )
A better result would be to declare java.util.Random as a class variable and call random.nextBoolean() and simply perform your heads/tails calculation on that.
If you were to continue to use Math.random() and lets say
return Math.random() < 0.5
Your results would be ever so slightly skewed due to the fact that Math.random() cannot return 1.0, due to the fact that the java API specification states:
"Returns a double value with a positive sign, greater than or equal to 0.0 and less than 1.0."

Math.random() returns a random float in the range [0.0,1.0)--that means the result can be anything from 0 up to but not including 1.0.
Your code
double rAsFloat = 1 * (2 + Math.random( ) );
will take this number in the [0.0,1.0) range; adding 2 to it gives you a number in the [2.0,3.0) range; multiplying it by 1 does nothing useful; then, when you truncate it to an integer, the result is always 2.
To get integers from this kind of random function, you need to figure out how many different integers you could return, then multiply your random number by that. If you want a "0 or 1" answer, your range is 2 different integers, so multiply Math.random() by 2:
double rAsFloat = 2 * Math.random();
This gives you a random number in the range [0.0,2.0), which can then be 0 or 1 when you truncate to an integer with (int). If, instead, you wanted something that returns 1 or 2, for example, you'd just add 1 to it:
double rAsFloat = 1 + 2 * Math.random();
I think you've already figured out that the Random class gives you what you want a lot more easily. I've decided to explain all this anyway, because someday you might work on a legacy system in some old language where you really do need to work with a [0.0,1.0) random value. (OK, maybe that's not too likely any more, but who knows.)

The problem can be translated to boolean generation as follow :
public static byte get0Or1 {
Random random = new Random();
boolean res= random.nextBoolean();
if(res)return 1;
else return 0;
}

Here it the easiest way I found without using java.util.Random.
Blockquote
Scanner input = new Scanner (System.in);
System.out.println("Please enter 0 for heads or 1 for tails");
int integer = input.nextInt();
input.close();
int random = (int) (Math.random() + 0.5);
if (random == integer) {
System.out.println("correct");
}
else {
System.out.println("incorrect");
}
System.out.println(random);
This will take a random double from (0 to .99) and add .5 to make it (.5 to 1.49). It will also cast it to an int, which will make it (0 to 1). The last line is for testing.

for(int i=0;i<100;i++){
System.out.println(((int)(i*Math.random())%2));
}
use mod it will help you!

One more variant
rand.nextInt(2);
As it described in docs it will return random int value between 0 (inclusive) and the specified value (exclusive)

Random number in Java where one number has higher odds

I understand that in Java I can generate a random number with the following code:
Random rand=new Random()
int x=rand.nextInt(1);
I am interested generating either the number zero or one. But I want that number one has 90% higher probability of of being generated than zero.
How can I achieve that?
thanks
EDIT:
Thanks everyone. It's working.

Generate a random number from 0 to 9. If the number is 0, you return zero. If the number is 1-9, you return one.

Heres a pretty compact way to express it
Random rand=new Random();
int x = ((rand.nextInt(10) == 0)) ? 0 : 1;

This would do it:
int result;
if (Math.random() < 0.9) {
result = 1;
}
else {
result = 0;
}
Or more concise:
int result = (Math.random() < 0.9) ? 1 : 0;

read nextInt(int) manual which says:
Returns a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value (exclusive), drawn from this random number generator's sequence. The general contract of nextInt is that one int value in the specified range is pseudorandomly generated and returned. All n possible int values are produced with (approximately) equal probability. The method nextInt(int n) is implemented by class Random as if by:
change your code to
Random rand=new Random();
int x=rand.nextInt(10);
return (x == 0) ? 0 : 1;
then run it again

You can write it like this.
int x = (int) (Math.random() / 0.9); // 90% chance of 0
or
int x = (int) (Math.random() + 0.9); // 90% chance of 1

How to get a random between 1 - 100 from randDouble in Java?

Okay, I'm still fairly new to Java. We've been given an assisgnment to create a game where you have to guess a random integer that the computer had generated. The problem is that our lecturer is insisting that we use:
double randNumber = Math.random();
And then translate that into an random integer that accepts 1 - 100 inclusive. I'm a bit at a loss. What I have so far is this:
//Create random number 0 - 99
double randNumber = Math.random();
d = randNumber * 100;
//Type cast double to int
int randomInt = (int)d;
However, the random the lingering problem of the random double is that 0 is a possibility while 100 is not. I want to alter that so that 0 is not a possible answer and 100 is. Help?

or
Random r = new Random();
int randomInt = r.nextInt(100) + 1;

You're almost there. Just add 1 to the result:
int randomInt = (int)d + 1;
This will "shift" your range to 1 - 100 instead of 0 - 99.

The ThreadLocalRandom class provides the int nextInt(int origin, int bound) method to get a random integer in a range:
// Returns a random int between 1 (inclusive) & 101 (exclusive)
int randomInt = ThreadLocalRandom.current().nextInt(1, 101)
ThreadLocalRandom is one of several ways to generate random numbers in Java, including the older Math.random() method and java.util.Random class. The advantage of ThreadLocalRandom is that it is specifically designed be used within a single thread, avoiding the additional thread synchronization costs imposed by the other implementations. Therefore, it is usually the best built-in random implementation to use outside of a security-sensitive context.
When applicable, use of ThreadLocalRandom rather than shared Random objects in concurrent programs will typically encounter much less overhead and contention.

Here is a clean and working way to do it, with range checks! Enjoy.
public double randDouble(double bound1, double bound2) {
//make sure bound2> bound1
double min = Math.min(bound1, bound2);
double max = Math.max(bound1, bound2);
//math.random gives random number from 0 to 1
return min + (Math.random() * (max - min));
}
//Later just call:
randDouble(1,100)
//example result:
//56.736451234

I will write
int number = 1 + (int) (Math.random() * 100);

double random = Math.random();
double x = random*100;
int y = (int)x + 1; //Add 1 to change the range to 1 - 100 instead of 0 - 99
System.out.println("Random Number :");
System.out.println(y);

How to generate a random BigInteger value in Java?

I need to generate arbitrarily large random integers in the range 0 (inclusive) to n (exclusive). My initial thought was to call nextDouble and multiply by n, but once n gets to be larger than 253, the results would no longer be uniformly distributed.
BigInteger has the following constructor available:
public BigInteger(int numBits, Random rnd)
Constructs a randomly generated BigInteger, uniformly distributed over the range 0 to (2numBits - 1), inclusive.
How can this be used to get a random value in the range 0 - n, where n is not a power of 2?

Use a loop:
BigInteger randomNumber;
do {
randomNumber = new BigInteger(upperLimit.bitLength(), randomSource);
} while (randomNumber.compareTo(upperLimit) >= 0);
on average, this will require less than two iterations, and the selection will be uniform.
Edit: If your RNG is expensive, you can limit the number of iterations the following way:
int nlen = upperLimit.bitLength();
BigInteger nm1 = upperLimit.subtract(BigInteger.ONE);
BigInteger randomNumber, temp;
do {
temp = new BigInteger(nlen + 100, randomSource);
randomNumber = temp.mod(upperLimit);
} while (s.subtract(randomNumber).add(nm1).bitLength() >= nlen + 100);
// result is in 'randomNumber'
With this version, it is highly improbable that the loop is taken more than once (less than one chance in 2^100, i.e. much less than the probability that the host machine spontaneously catches fire in the next following second). On the other hand, the mod() operation is computationally expensive, so this version is probably slower than the previous, unless the randomSource instance is exceptionally slow.

The following method uses the BigInteger(int numBits, Random rnd) constructor and rejects the result if it's bigger than the specified n.
public BigInteger nextRandomBigInteger(BigInteger n) {
Random rand = new Random();
BigInteger result = new BigInteger(n.bitLength(), rand);
while( result.compareTo(n) >= 0 ) {
result = new BigInteger(n.bitLength(), rand);
}
return result;
}
The drawback to this is that the constructor is called an unspecified number of times, but in the worst case (n is just slightly greater than a power of 2) the expected number of calls to the constructor should be only about 2 times.

The simplest approach (by quite a long way) would be to use the specified constructor to generate a random number with the right number of bits (floor(log2 n) + 1), and then throw it away if it's greater than n. In the worst possible case (e.g. a number in the range [0, 2n + 1) you'll throw away just under half the values you create, on average.

Why not constructing a random BigInteger, then building a BigDecimal from it ?
There is a constructor in BigDecimal : public BigDecimal(BigInteger unscaledVal, int scale) that seems relevant here, no ? Give it a random BigInteger and a random scale int, and you'll have a random BigDecimal. No ?

Here is how I do it in a class called Generic_BigInteger available via:
Andy Turner's Generic Source Code Web Page
/**
* There are methods to get large random numbers. Indeed, there is a
* constructor for BigDecimal that allows for this, but only for uniform
* distributions over a binary power range.
* #param a_Random
* #param upperLimit
* #return a random integer as a BigInteger between 0 and upperLimit
* inclusive
*/
public static BigInteger getRandom(
Generic_Number a_Generic_Number,
BigInteger upperLimit) {
// Special cases
if (upperLimit.compareTo(BigInteger.ZERO) == 0) {
return BigInteger.ZERO;
}
String upperLimit_String = upperLimit.toString();
int upperLimitStringLength = upperLimit_String.length();
Random[] random = a_Generic_Number.get_RandomArrayMinLength(
upperLimitStringLength);
if (upperLimit.compareTo(BigInteger.ONE) == 0) {
if (random[0].nextBoolean()) {
return BigInteger.ONE;
} else {
return BigInteger.ZERO;
}
}
int startIndex = 0;
int endIndex = 1;
String result_String = "";
int digit;
int upperLimitDigit;
int i;
// Take care not to assign any digit that will result in a number larger
// upperLimit
for (i = 0; i < upperLimitStringLength; i ++){
upperLimitDigit = new Integer(
upperLimit_String.substring(startIndex,endIndex));
startIndex ++;
endIndex ++;
digit = random[i].nextInt(upperLimitDigit + 1);
if (digit != upperLimitDigit){
break;
}
result_String += digit;
}
// Once something smaller than upperLimit guaranteed, assign any digit
// between zero and nine inclusive
for (i = i + 1; i < upperLimitStringLength; i ++) {
digit = random[i].nextInt(10);
result_String += digit;
}
// Tidy values starting with zero(s)
while (result_String.startsWith("0")) {
if (result_String.length() > 1) {
result_String = result_String.substring(1);
} else {
break;
}
}
BigInteger result = new BigInteger(result_String);
return result;
}

For those who are still asking this question and are looking for a way to generate arbitrarily large random BigIntegers within a positive integer range, this is what I came up with. This random generator works without trying bunch of numbers until one fits the range. Instead it will generate a random number directly that will fit the given range.
private static BigInteger RandomBigInteger(BigInteger rangeStart, BigInteger rangeEnd){
Random rand = new Random();
int scale = rangeEnd.toString().length();
String generated = "";
for(int i = 0; i < rangeEnd.toString().length(); i++){
generated += rand.nextInt(10);
}
BigDecimal inputRangeStart = new BigDecimal("0").setScale(scale, RoundingMode.FLOOR);
BigDecimal inputRangeEnd = new BigDecimal(String.format("%0" + (rangeEnd.toString().length()) + "d", 0).replace('0', '9')).setScale(scale, RoundingMode.FLOOR);
BigDecimal outputRangeStart = new BigDecimal(rangeStart).setScale(scale, RoundingMode.FLOOR);
BigDecimal outputRangeEnd = new BigDecimal(rangeEnd).add(new BigDecimal("1")).setScale(scale, RoundingMode.FLOOR); //Adds one to the output range to correct rounding
//Calculates: (generated - inputRangeStart) / (inputRangeEnd - inputRangeStart) * (outputRangeEnd - outputRangeStart) + outputRangeStart
BigDecimal bd1 = new BigDecimal(new BigInteger(generated)).setScale(scale, RoundingMode.FLOOR).subtract(inputRangeStart);
BigDecimal bd2 = inputRangeEnd.subtract(inputRangeStart);
BigDecimal bd3 = bd1.divide(bd2, RoundingMode.FLOOR);
BigDecimal bd4 = outputRangeEnd.subtract(outputRangeStart);
BigDecimal bd5 = bd3.multiply(bd4);
BigDecimal bd6 = bd5.add(outputRangeStart);
BigInteger returnInteger = bd6.setScale(0, RoundingMode.FLOOR).toBigInteger();
returnInteger = (returnInteger.compareTo(rangeEnd) > 0 ? rangeEnd : returnInteger); //Converts number to the end of output range if it's over it. This is to correct rounding.
return returnInteger;
}
How does it work?
First it generates a String with random numbers with the same length as the maximum range. For example: with given range of 10-1000 it will generate some number between 0000 and 9999 as a String.
Then it creates BigDecimals to represent the maximum possible value (9999 in previous example) and minimum value (0) and converts the range parameter BigIntegers to BigDecimals. Also in this step to the given range maximum value is added 1 in order to correct rounding errors in the next step.
Then using this formula the generated random number is mapped to the given range:
(generated - inputRangeStart) / (inputRangeEnd - inputRangeStart) * (outputRangeEnd - outputRangeStart) + outputRangeStart
After that it will do a last check whether or not the mapped number fits the given range and sets it to the given range maximum if it doesn't. This is done in order to correct rounding errors.

Just use modular reduction
new BigInteger(n.bitLength(), new SecureRandom()).mod(n)

Compile this F# code into a DLL and you can also reference it in your C# / VB.NET programs
type BigIntegerRandom() =
static let internalRandom = new Random()
/// Returns a BigInteger random number of the specified number of bytes.
static member RandomBigInteger(numBytes:int, rand:Random) =
let r = if rand=null then internalRandom else rand
let bytes : byte[] = Array.zeroCreate (numBytes+1)
r.NextBytes(bytes)
bytes.[numBytes] <- 0uy
bigint bytes
/// Returns a BigInteger random number from 0 (inclusive) to max (exclusive).
static member RandomBigInteger(max:bigint, rand:Random) =
let rec getNumBytesInRange num bytes = if max < num then bytes else getNumBytesInRange (num * 256I) bytes+1
let bytesNeeded = getNumBytesInRange 256I 1
BigIntegerRandom.RandomBigInteger(bytesNeeded, rand) % max
/// Returns a BigInteger random number from min (inclusive) to max (exclusive).
static member RandomBigInteger(min:bigint, max:bigint, rand:Random) =
BigIntegerRandom.RandomBigInteger(max - min, rand) + min