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

const rndInt = randomIntFromInterval(1, 6)
console.log(rndInt)

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.

Important

If you wanted to get a random integer between 1 (and only 1) and 6, you would calculate:

    const rndInt = Math.floor(Math.random() * 6) + 1
    console.log(rndInt)

Where:

• 1 is the start number
• 6 is the number of possible results (1 + start (6) - end (1))

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.

TL;DR

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

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

EXPLANATION BELOW

integer - A number which is not a fraction; a whole number

We need to get a random number , say X between min and max. X, min and max are all integers

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

For, any given X, the above equation satisfies only when r has range of [0,1] For any other values of r the above equation is unsatisfied.

Learn more about ranges [x,y] or (x,y) here

Our next step is to find a function which always results in a value which has a range of [0,1]

Now, the range of r i.e [0,1] is very similar to Math.random() function in Javascript. 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)

Random Function using Math.random() 0 <= r < 1

Notice that in Math.random() left bound is inclusive and the right bound is exclusive. This means min + (max - min) * r will evaluate to having a range from [min, max)

To include our right bound i.e [min,max] we increase the right bound by 1 and floor the result.

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

To get the random number

generateRandomInteger(-20, 20);

Other solutions:

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

Try online

Or, in Underscore

_.random(min, max)

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

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

Get a random integer between 0 and 400

let rand = Math.round(Math.random() * 400)

document.write(rand)

Get a random integer between 200 and 1500

let range = {min: 200, max: 1500}
let delta = range.max - range.min

const rand = Math.round(range.min + Math.random() * delta)

document.write(rand)

Using functions

function randBetween(min, max){
  let delta = max - min
  return Math.round(min + Math.random() * delta)
}

document.write(randBetween(10, 15));

// JavaScript ES6 arrow function

const randBetween = (min, max) => {
  let delta = max - min
  return Math.round(min + Math.random() * delta)
}

document.write(randBetween(10, 20))

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.

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);

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) => {
  const range = max - min;
  const maxGeneratedValue = 0xFFFFFFFF;
  const possibleResultValues = range + 1;
  const possibleGeneratedValues = maxGeneratedValue + 1;
  const remainder = possibleGeneratedValues % possibleResultValues;
  const maxUnbiased = maxGeneratedValue - remainder;

  if (!Number.isInteger(min) || !Number.isInteger(max) ||
       max > Number.MAX_SAFE_INTEGER || min < Number.MIN_SAFE_INTEGER) {
    throw new Error('Arguments must be safe integers.');
  } else if (range > maxGeneratedValue) {
    throw new Error(`Range of ${range} (from ${min} to ${max}) > ${maxGeneratedValue}.`);
  } else if (max < min) {
    throw new Error(`max (${max}) must be >= min (${min}).`);
  } else if (min === max) {
    return min;
  } 

  let generated;
  do {
    generated = crypto.getRandomValues(new Uint32Array(1))[0];
  } while (generated > maxUnbiased);

  return min + (generated % possibleResultValues);
};

console.log(randomInteger(-8, 8));          // -2
console.log(randomInteger(0, 0));           // 0
console.log(randomInteger(0, 0xFFFFFFFF));  // 944450079
console.log(randomInteger(-1, 0xFFFFFFFF));
// Error: Range of 4294967296 covering -1 to 4294967295 is > 4294967295.
console.log(new Array(12).fill().map(n => randomInteger(8, 12)));
// [11, 8, 8, 11, 10, 8, 8, 12, 12, 12, 9, 9]

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).

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

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;
    }   

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));
}

to return 1-6 like a dice basically,

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

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

Crypto-strong

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

console.log( rand(1,6) );

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

Using random function, which can be reused.

function randomNum(min, max) {
  return Math.floor(Math.random() * (max - min + 1)) + min;
}
randomNum(1, 6);

This should work:

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

If the starting number is 1, as in your example (1-6), you can use Math.ceil() method instead of Math.floor().

Math.ceil(Math.random() * 6)

instead of

Math.floor(Math.random() * 6) + 1

Let's not forget other useful Math methods.

Short Answer: It's achievable using a simple array.

you can alternate within array elements.

This solution works even if your values are not consecutive. Values don't even have to be a number.

let array = [1, 2, 3, 4, 5, 6];
const randomValue = array[Math.floor(Math.random() * array.length)];

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>

Try using:

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

This simple function is handy and works in ANY cases (fully tested). Also, the distribution of the results has been fully tested and is 100% correct.

function randomInteger(pMin = 1, pMax = 1_000_000_000)
//Author: Axel Gauffre. 
//Here: https://stackoverflow.com/a/74636954/5171000
//Inspired by: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/random#getting_a_random_number_between_two_values
//
//This function RETURNS A RANDOM INTEGER between pMin (INCLUDED) and pMax (INCLUDED).
//  - pMin and pMax should be integers.
//  - HOWEVER, if pMin and/or pMax are FLOATS, they will be ROUNDED to the NEAREST integer.
//  - NEGATIVE values ARE supported.
//  - The ORDER of the 2 arguments has NO consequence: If pMin > pMax, then pMin and pMax will simply be SWAPPED.
//  - If pMin is omitted, it will DEFAULT TO 1.
//  - If pMax is omitted, it will DEFAULT TO 1 BILLION.
//
//This function works in ANY cases (fully tested).
//Also, the distribution of the results has been fully tested and is 100% correct.
{
    pMin = Math.round(pMin);
    pMax = Math.round(pMax);
    if (pMax < pMin) { let t = pMin; pMin = pMax; pMax = t;}
    return Math.floor(Math.random() * (pMax+1 - pMin) + pMin);
}

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;
}

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));