[javascript] Generate random number between two numbers in JavaScript

Is there a way to generate a random number in a specified range (e.g. from 1 to 6: 1, 2, 3, 4, 5, or 6) in JavaScript?

Try using:

function random(min, max) {_x000D_
   return Math.round((Math.random() *( Math.abs(max - min))) + min);_x000D_
}_x000D_
console.log(random(1, 6));

I was searching random number generator written in TypeScript and I have written this after reading all of the answers, hope It would work for TypeScript coders.

    Rand(min: number, max: number): number {
        return (Math.random() * (max - min + 1) | 0) + min;
    }   

Math is not my strong point, but I've been working on a project where I needed to generate a lot of random numbers between both positive and negative.

function randomBetween(min, max) {
    if (min < 0) {
        return min + Math.random() * (Math.abs(min)+max);
    }else {
        return min + Math.random() * max;
    }
}

E.g

randomBetween(-10,15)//or..
randomBetween(10,20)//or...
randomBetween(-200,-100)

Of course, you can also add some validation to make sure you don't do this with anything other than numbers. Also make sure that min is always less than or equal to max.

This works for me and produces values like Python's random.randint standard library function:

function randint(min, max) {
   return Math.round((Math.random() * Math.abs(max - min)) + min);
}

console.log("Random integer: " + randint(-5, 5));

Other solutions:

• (Math.random() * 6 | 0) + 1
• ~~(Math.random() * 6) + 1

Try online

ES6 / Arrow functions version based on Francis' code (i.e. the top answer):

const randomIntFromInterval = (min, max) => Math.floor(Math.random() * (max - min + 1) + min);

var x = 6; // can be any number
var rand = Math.floor(Math.random()*x) + 1;

for big number.

var min_num = 900;
var max_num = 1000;
while(true){
    
    let num_random = Math.random()* max_num;
    console.log('input : '+num_random);
    if(num_random >= min_num){
        console.log(Math.floor(num_random));
       break; 
    } else {
        console.log(':::'+Math.floor(num_random));
    }
}

Or, in Underscore

_.random(min, max)

This function can generate a random integer number between (and including) min and max numbers:

function randomNumber(min, max) {
  if (min > max) {
    let temp = max;
    max = min;
    min = temp;
  }

  if (min <= 0) {
    return Math.floor(Math.random() * (max + Math.abs(min) + 1)) + min;
  } else {
    return Math.floor(Math.random() * (max - min + 1)) + min;
  }
}

Example:

randomNumber(-2, 3); // can be -2, -1, 0, 1, 2 and 3
randomNumber(-5, -2); // can be -5, -4, -3 and -2
randomNumber(0, 4); // can be 0, 1, 2, 3 and 4
randomNumber(4, 0); // can be 0, 1, 2, 3 and 4

to return 1-6 like a dice basically,

return Math.round(Math.random() * 5 + 1);

I wrote more flexible function which can give you random number but not only integer.

function rand(min,max,interval)
{
    if (typeof(interval)==='undefined') interval = 1;
    var r = Math.floor(Math.random()*(max-min+interval)/interval);
    return r*interval+min;
}

var a = rand(0,10); //can be 0, 1, 2 (...) 9, 10
var b = rand(4,6,0.1); //can be 4.0, 4.1, 4.2 (...) 5.9, 6.0

Fixed version.

Math.random() is fast and suitable for many purposes, but it's not appropriate if you need cryptographically-secure values (it's not secure), or if you need integers from a completely uniform unbiased distribution (the multiplication approach used in others answers produces certain values slightly more often than others).

In such cases, we can use crypto.getRandomValues() to generate secure integers, and reject any generated values that we can't map uniformly into the target range. This will be slower, but it shouldn't be significant unless you're generating extremely large numbers of values.

To clarify the biased distribution concern, consider the case where we want to generate a value between 1 and 5, but we have a random number generator that produces values between 1 and 16 (a 4-bit value). We want to have the same number of generated values mapping to each output value, but 16 does not evenly divide by 5: it leaves a remainder of 1. So we need to reject 1 of the possible generated values, and only continue when we get one of the 15 lesser values that can be uniformly mapped into our target range. Our behaviour could look like this pseudocode:

Generate a 4-bit integer in the range 1-16.
If we generated  1,  6, or 11 then output 1.
If we generated  2,  7, or 12 then output 2.
If we generated  3,  8, or 13 then output 3.
If we generated  4,  9, or 14 then output 4.
If we generated  5, 10, or 15 then output 5.
If we generated 16 then reject it and try again.

The following code uses similar logic, but generates a 32-bit integer instead, because that's the largest common integer size that can be represented by JavaScript's standard number type. (This could be modified to use BigInts if you need a larger range.) Regardless of the chosen range, the fraction of generated values that are rejected will always be less than 0.5, so the expected number of rejections will always be less than 1.0 and usually close to 0.0; you don't need to worry about it looping forever.

const randomInteger = (min, max) => {_x000D_
  const range = max - min;_x000D_
  const maxGeneratedValue = 0xFFFFFFFF;_x000D_
  const possibleResultValues = range + 1;_x000D_
  const possibleGeneratedValues = maxGeneratedValue + 1;_x000D_
  const remainder = possibleGeneratedValues % possibleResultValues;_x000D_
  const maxUnbiased = maxGeneratedValue - remainder;_x000D_
_x000D_
  if (!Number.isInteger(min) || !Number.isInteger(max) ||_x000D_
       max > Number.MAX_SAFE_INTEGER || min < Number.MIN_SAFE_INTEGER) {_x000D_
    throw new Error('Arguments must be safe integers.');_x000D_
  } else if (range > maxGeneratedValue) {_x000D_
    throw new Error(`Range of ${range} (from ${min} to ${max}) > ${maxGeneratedValue}.`);_x000D_
  } else if (max < min) {_x000D_
    throw new Error(`max (${max}) must be >= min (${min}).`);_x000D_
  } else if (min === max) {_x000D_
    return min;_x000D_
  } _x000D_
_x000D_
  let generated;_x000D_
  do {_x000D_
    generated = crypto.getRandomValues(new Uint32Array(1))[0];_x000D_
  } while (generated > maxUnbiased);_x000D_
_x000D_
  return min + (generated % possibleResultValues);_x000D_
};_x000D_
_x000D_
console.log(randomInteger(-8, 8));          // -2_x000D_
console.log(randomInteger(0, 0));           // 0_x000D_
console.log(randomInteger(0, 0xFFFFFFFF));  // 944450079_x000D_
console.log(randomInteger(-1, 0xFFFFFFFF));_x000D_
// Error: Range of 4294967296 covering -1 to 4294967295 is > 4294967295._x000D_
console.log(new Array(12).fill().map(n => randomInteger(8, 12)));_x000D_
// [11, 8, 8, 11, 10, 8, 8, 12, 12, 12, 9, 9]

This is about nine years late, but randojs.com makes this a simple one-liner:

rando(1, 6)

You just need to add this to the head of your html document, and you can do pretty much whatever you want with randomness easily. Random values from arrays, random jquery elements, random properties from objects, and even preventing repetitions if needed.

<script src="https://randojs.com/1.0.0.js"></script>

Adding float with fixed precision version based on the int version in @Francisc's answer:

function randomFloatFromInterval (min, max, fractionDigits) {
  const fractionMultiplier = Math.pow(10, fractionDigits)
  return Math.round(
    (Math.random() * (max - min) + min) * fractionMultiplier,
  ) / fractionMultiplier
}

so:

randomFloatFromInterval(1,3,4) // => 2.2679, 1.509, 1.8863, 2.9741, ...

and for int answer

randomFloatFromInterval(1,3,0) // => 1, 2, 3

function randomIntFromInterval(min, max) { // min and max included 
  return Math.floor(Math.random() * (max - min + 1) + min);
}

What it does "extra" is it allows random intervals that do not start with 1. So you can get a random number from 10 to 15 for example. Flexibility.

This should work:

const getRandomNum = (min, max) => Math.floor(Math.random() * (max - min + 1)) + min

Inspite of many answers and almost same result. I would like to add my answer and explain its working. Because it is important to understand its working rather than copy pasting one line code. Generating random numbers is nothing but simple maths.

CODE:

function getR(lower, upper) {

  var percent = (Math.random() * 100);
  // this will return number between 0-99 because Math.random returns decimal number from 0-0.9929292 something like that
  //now you have a percentage, use it find out the number between your INTERVAL :upper-lower 
  var num = ((percent * (upper - lower) / 100));
  //num will now have a number that falls in your INTERVAL simple maths
  num += lower;
  //add lower to make it fall in your INTERVAL
  //but num is still in decimal
  //use Math.floor>downward to its nearest integer you won't get upper value ever
  //use Math.ceil>upward to its nearest integer upper value is possible
  //Math.round>to its nearest integer 2.4>2 2.5>3   both lower and upper value possible
  console.log(Math.floor(num), Math.ceil(num), Math.round(num));
}

The top rated solution is not mathematically correct as same as comments under it -> Math.floor(Math.random() * 6) + 1.

Task: generate random number between 1 and 6.

Math.random() returns floating point number between 0 and 1 (like 0.344717274374 or 0.99341293123 for example), which we will use as a percentage, so Math.floor(Math.random() * 6) + 1 returns some percentage of 6 (max: 5, min: 0) and adds 1. The author got lucky that lower bound was 1., because percentage floor will "maximumly" return 5 which is less than 6 by 1, and that 1 will be added by lower bound 1.

The problems occurs when lower bound is greater than 1. For instance, Task: generate random between 2 and 6.

(following author's logic) Math.floor(Math.random() * 6) + 2, it is obviously seen that if we get 5 here -> Math.random() * 6 and then add 2, the outcome will be 7 which goes beyond the desired boundary of 6.

Another example, Task: generate random between 10 and 12.

(following author's logic) Math.floor(Math.random() * 12) + 10, (sorry for repeating) it is obvious that we are getting 0%-99% percent of number "12", which will go way beyond desired boundary of 12.

So, the correct logic is to take the difference between lower bound and upper bound add 1, and only then floor it (to substract 1, because Math.random() returns 0 - 0.99, so no way to get full upper bound, thats why we adding 1 to upper bound to get maximumly 99% of (upper bound + 1) and then we floor it to get rid of excess). Once we got the floored percentage of (difference + 1), we can add lower boundary to get the desired randomed number between 2 numbers.

The logic formula for that will be: Math.floor(Math.random() * ((up_boundary - low_boundary) + 1)) + 10.

P.s.: Even comments under the top-rated answer were incorrect, since people forgot to add 1 to the difference, meaning that they will never get the up boundary (yes it might be a case if they dont want to get it at all, but the requirenment was to include the upper boundary).

I discovered a great new way to do this using ES6 default parameters. It is very nifty since it allows either one argument or two arguments. Here it is:

function random(n, b = 0) {
    return Math.random() * (b-n) + n;
}

Example

Return a random number between 1 and 10:

Math.floor((Math.random() * 10) + 1);

The result could be: 3

Try yourself: here

--

or using lodash / undescore:

_.random(min, max)

Docs: - lodash - undescore

Math.random()

Returns an integer random number between min (included) and max (included):

function randomInteger(min, max) {
  return Math.floor(Math.random() * (max - min + 1)) + min;
}

Or any random number between min (included) and max (not included):

function randomNumber(min, max) {
  return Math.random() * (max - min) + min;
}

Useful examples (integers):

// 0 -> 10
Math.floor(Math.random() * 11);

// 1 -> 10
Math.floor(Math.random() * 10) + 1;

// 5 -> 20
Math.floor(Math.random() * 16) + 5;

// -10 -> (-2)
Math.floor(Math.random() * 9) - 10;

** And always nice to be reminded (Mozilla):

Math.random() does not provide cryptographically secure random numbers. Do not use them for anything related to security. Use the Web Crypto API instead, and more precisely the window.crypto.getRandomValues() method.

jsfiddle: https://jsfiddle.net/cyGwf/477/

Random Integer: to get a random integer between min and max, use the following code

function getRandomInteger(min, max) {
  min = Math.ceil(min);
  max = Math.floor(max);
  return Math.floor(Math.random() * (max - min)) + min;
}

Random Floating Point Number: to get a random floating point number between min and max, use the following code

function getRandomFloat(min, max) {
  return Math.random() * (max - min) + min;
}

Reference: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/random

TL;DR

function generateRandomInteger(min, max) {
  return Math.floor(min + Math.random()*(max + 1 - min))
}

To get the random number generateRandomInteger(-20, 20);

EXPLANATION BELOW

We need to get a random integer, say X between min and max.

Right?

i.e min <= X <= max

If we subtract min from the equation, this is equivalent to

0 <= (X - min) <= (max - min)

Now, lets multiply this with a random number r which is

0 <= (X - min) * r <= (max - min) * r

Now, lets add back min to the equation

min <= min + (X - min) * r <= min + (max - min) * r

Now, lets chose a function which results in r such that it satisfies our equation range as [min,max]. This is only possible if 0<= r <=1

OK. Now, the range of r i.e [0,1] is very similar to Math.random() function result. Isn't it?

The Math.random() function returns a floating-point, pseudo-random number in the range [0, 1); that is, from 0 (inclusive) up to but not including 1 (exclusive)

For example,

Case r = 0

min + 0 * (max-min) = min

Case r = 1

min + 1 * (max-min) = max

Random Case using Math.random 0 <= r < 1

min + r * (max-min) = X, where X has range of min <= X < max

The above result X is a random numeric. However due to Math.random() our left bound is inclusive, and the right bound is exclusive. To include our right bound we increase the right bound by 1 and floor the result.

function generateRandomInteger(min, max) {
  return Math.floor(min + Math.random()*(max + 1 - min))
}

To get the random number

generateRandomInteger(-20, 20);

Crypto-strong random integer number in range [a,b] (assumption: a < b )

let rand= (a,b)=> a+(b-a+1)*crypto.getRandomValues(new Uint32Array(1))[0]/2**32|0_x000D_
_x000D_
console.log( rand(1,6) );