Generating random whole numbers in JavaScript in a specific range

Question

How can I generate random whole numbers between two specified variables in JavaScript, e.g. x = 4 and y = 8 would output any of 4, 5, 6, 7, 8?

Answer 1

There are some examples on the Mozilla Developer Network page:

/**
 * Returns a random number between min (inclusive) and max (exclusive)
 */
function getRandomArbitrary(min, max) {
    return Math.random() * (max - min) + min;
}

/**
 * Returns a random integer between min (inclusive) and max (inclusive).
 * The value is no lower than min (or the next integer greater than min
 * if min isn't an integer) and no greater than max (or the next integer
 * lower than max if max isn't an integer).
 * Using Math.round() will give you a non-uniform distribution!
 */
function getRandomInt(min, max) {
    min = Math.ceil(min);
    max = Math.floor(max);
    return Math.floor(Math.random() * (max - min + 1)) + min;
}

---

Here's the logic behind it. It's a simple rule of three:

Math.random() returns a Number between 0 (inclusive) and 1 (exclusive). So we have an interval like this:

[0 .................................... 1)

Now, we'd like a number between min (inclusive) and max (exclusive):

[0 .................................... 1)
[min .................................. max)

We can use the Math.random to get the correspondent in the [min, max) interval. But, first we should factor a little bit the problem by subtracting min from the second interval:

[0 .................................... 1)
[min - min ............................ max - min)

This gives:

[0 .................................... 1)
[0 .................................... max - min)

We may now apply Math.random and then calculate the correspondent. Let's choose a random number:

                Math.random()
                    |
[0 .................................... 1)
[0 .................................... max - min)
                    |
                    x (what we need)

So, in order to find x, we would do:

x = Math.random() * (max - min);

Don't forget to add min back, so that we get a number in the [min, max) interval:

x = Math.random() * (max - min) + min;

That was the first function from MDN. The second one, returns an integer between min and max, both inclusive.

Now for getting integers, you could use round, ceil or floor.

You could use Math.round(Math.random() * (max - min)) + min, this however gives a non-even distribution. Both, min and max only have approximately half the chance to roll:

min...min+0.5...min+1...min+1.5   ...    max-0.5....max
└───┬───┘└────────┬───────┘└───── ... ─────┘└───┬──┘   ← Math.round()
   min          min+1                          max

With max excluded from the interval, it has an even less chance to roll than min.

With Math.floor(Math.random() * (max - min +1)) + min you have a perfectly even distribution.

 min...  min+1...    ...      max-1... max....   (max+1 is excluded from interval)
└───┬───┘└───┬───┘└─── ... ┘└───┬───┘└───┬───┘   ← Math.floor()
   min     min+1               max-1    max

You can't use ceil() and -1 in that equation because max now had a slightly less chance to roll, but you can roll the (unwanted) min-1 result too.

Answer 2

var randomnumber = Math.floor(Math.random() * (maximum - minimum + 1)) + minimum;

Answer 3

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.

Answer 4

Use:

function getRandomizer(bottom, top) {
    return function() {
        return Math.floor( Math.random() * ( 1 + top - bottom ) ) + bottom;
    }
}

Usage:

var rollDie = getRandomizer( 1, 6 );

var results = ""
for ( var i = 0; i<1000; i++ ) {
    results += rollDie() + " ";    // Make a string filled with 1000 random numbers in the range 1-6.
}

Breakdown:

We are returning a function (borrowing from functional programming) that when called, will return a random integer between the the values bottom and top, inclusive. We say 'inclusive' because we want to include both bottom and top in the range of numbers that can be returned. This way, getRandomizer( 1, 6 ) will return either 1, 2, 3, 4, 5, or 6.

('bottom' is the lower number, and 'top' is the greater number)

Math.random() * ( 1 + top - bottom )

Math.random() returns a random double between 0 and 1, and if we multiply it by one plus the difference between top and bottom, we'll get a double somewhere between 0 and 1+b-a.

Math.floor( Math.random() * ( 1 + top - bottom ) )

Math.floor rounds the number down to the nearest integer. So we now have all the integers between 0 and top-bottom. The 1 looks confusing, but it needs to be there because we are always rounding down, so the top number will never actually be reached without it. The random decimal we generate needs to be in the range 0 to (1+top-bottom) so we can round down and get an integer in the range 0 to top-bottom:

Math.floor( Math.random() * ( 1 + top - bottom ) ) + bottom

The code in the previous example gave us an integer in the range 0 and top-bottom, so all we need to do now is add bottom to that result to get an integer in the range bottom and top inclusive. :D

---

NOTE: If you pass in a non-integer value or the greater number first you'll get undesirable behavior, but unless anyone requests it I am not going to delve into the argument checking code as it’s rather far from the intent of the original question.

Answer 5

All these solutions are using way too much firepower. You only need to call one function: Math.random();

Math.random() * max | 0;

This returns a random integer between 0 (inclusive) and max (non-inclusive).

Answer 6

Return a random number between 1 and 10:

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

Return a random number between 1 and 100:

Math.floor((Math.random()*100) + 1)

Answer 7

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

Alternative if you are using Underscore.js you can use

_.random(min, max)

Answer 8

If you need a variable between 0 and max, you can use:

Math.floor(Math.random() *  max);

Answer 9

The other answers don't account for the perfectly reasonable parameters of 0 and 1. Instead you should use the round instead of ceil or floor:

function randomNumber(minimum, maximum){
    return Math.round( Math.random() * (maximum - minimum) + minimum);
}

console.log(randomNumber(0,1));  # 0 1 1 0 1 0
console.log(randomNumber(5,6));  # 5 6 6 5 5 6
console.log(randomNumber(3,-1)); # 1 3 1 -1 -1 -1

Answer 10

Cryptographically strong

To get a cryptographically strong random integer number in the range [x,y], try:

let cs = (x,y) => x + (y - x + 1)*crypto.getRandomValues(new Uint32Array(1))[0]/2**32 | 0

console.log(cs(4, 8))

Answer 11

Here's what I use to generate random numbers.

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

Math.random() returns a number between 0 (inclusive) and 1 (exclusive). We multiply this number by the range (max-min). This results in a number between 0 (inclusive), and the range.

For example, take random(2,5). We multiply the random number 0≤x<1 by the range (5-2=3), so we now have a number, x where 0≤x<3.

In order to force the function to treat both the max and min as inclusive, we add 1 to our range calculation: Math.random()*(max-min+1). Now, we multiply the random number by the (5-2+1=4), resulting in an number, x, such that 0≤x<4. If we floor this calculation, we get an integer: 0≤x≤3, with an equal likelihood of each result (1/4).

Finally, we need to convert this into an integer between the requested values. Since we already have an integer between 0 and the (max-min), we can simply map the value into the correct range by adding the minimum value. In our example, we add 2 our integer between 0 and 3, resulting in an integer between 2 and 5.

Answer 12

Use this function to get random numbers in a given range:

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

Answer 13

Here is the Microsoft .NET Implementation of the Random class in JavaScript—

var Random = (function () {
    function Random(Seed) {
        if (!Seed) {
            Seed = this.milliseconds();
        }
        this.SeedArray = [];
        for (var i = 0; i < 56; i++)
            this.SeedArray.push(0);
        var num = (Seed == -2147483648) ? 2147483647 : Math.abs(Seed);
        var num2 = 161803398 - num;
        this.SeedArray[55] = num2;
        var num3 = 1;
        for (var i_1 = 1; i_1 < 55; i_1++) {
            var num4 = 21 * i_1 % 55;
            this.SeedArray[num4] = num3;
            num3 = num2 - num3;
            if (num3 < 0) {
                num3 += 2147483647;
            }
            num2 = this.SeedArray[num4];
        }
        for (var j = 1; j < 5; j++) {
            for (var k = 1; k < 56; k++) {
                this.SeedArray[k] -= this.SeedArray[1 + (k + 30) % 55];
                if (this.SeedArray[k] < 0) {
                    this.SeedArray[k] += 2147483647;
                }
            }
        }
        this.inext = 0;
        this.inextp = 21;
        Seed = 1;
    }

    Random.prototype.milliseconds = function () {
        var str = new Date().valueOf().toString();
        return parseInt(str.substr(str.length - 6));
    };

    Random.prototype.InternalSample = function () {
        var num = this.inext;
        var num2 = this.inextp;
        if (++num >= 56) {
            num = 1;
        }
        if (++num2 >= 56) {
            num2 = 1;
        }
        var num3 = this.SeedArray[num] - this.SeedArray[num2];
        if (num3 == 2147483647) {
            num3--;
        }
        if (num3 < 0) {
            num3 += 2147483647;
        }
        this.SeedArray[num] = num3;
        this.inext = num;
        this.inextp = num2;
        return num3;
    };

    Random.prototype.Sample = function () {
        return this.InternalSample() * 4.6566128752457969E-10;
    };

    Random.prototype.GetSampleForLargeRange = function () {
        var num = this.InternalSample();
        var flag = this.InternalSample() % 2 == 0;
        if (flag) {
            num = -num;
        }
        var num2 = num;
        num2 += 2147483646.0;
        return num2 / 4294967293.0;
    };

    Random.prototype.Next = function (minValue, maxValue) {
        if (!minValue && !maxValue)
            return this.InternalSample();
        var num = maxValue - minValue;
        if (num <= 2147483647) {
            return parseInt((this.Sample() * num + minValue).toFixed(0));
        }
        return this.GetSampleForLargeRange() * num + minValue;
    };

    Random.prototype.NextDouble = function () {
        return this.Sample();
    };

    Random.prototype.NextBytes = function (buffer) {
        for (var i = 0; i < buffer.length; i++) {
            buffer[i] = this.InternalSample() % 256;
        }
    };
    return Random;
}());

Use:

var r = new Random();
var nextInt = r.Next(1, 100); // Returns an integer between range
var nextDbl = r.NextDouble(); // Returns a random decimal

Answer 14

I wanted to explain using an example:

Function to generate random whole numbers in JavaScript within a range of 5 to 25

General Overview:

(i) First convert it to the range - starting from 0.

(ii) Then convert it to your desired range ( which then will be very easy to complete).

So basically, if you want to generate random whole numbers from 5 to 25 then:

First step: Converting it to range - starting from 0

Subtract "lower/minimum number" from both "max" and "min". i.e

(5-5) - (25-5)

So the range will be:

0-20 ...right?

Step two

Now if you want both numbers inclusive in range - i.e "both 0 and 20", the equation will be:

Mathematical equation: Math.floor((Math.random() * 21))

General equation: Math.floor((Math.random() * (max-min +1)))

Now if we add subtracted/minimum number (i.e., 5) to the range - then automatically we can get range from 0 to 20 => 5 to 25

Step three

Now add the difference you subtracted in equation (i.e., 5) and add "Math.floor" to the whole equation:

Mathematical equation: Math.floor((Math.random() * 21) + 5)

General equation: Math.floor((Math.random() * (max-min +1)) + min)

So finally the function will be:

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

Answer 15

After generating a random number using a computer program, it is still considered as a random number if the picked number is a part or the full one of the initial one. But if it was changed, then mathematicians do not accept it as a random number and they can call it a biased number.

But if you are developing a program for a simple task, this will not be a case to consider. But if you are developing a program to generate a random number for a valuable stuff such as lottery program, or gambling game, then your program will be rejected by the management if you are not consider about the above case.

So for those kind of people, here is my suggestion:

Generate a random number using Math.random() (say this n):

Now for [0,10) ==>  n*10 (i.e. one digit) and for[10,100) ==> n*100 (i.e., two digits) and so on. Here square bracket indicates that the boundary is inclusive and a round bracket indicates the boundary is exclusive.

Then remove the rest after the decimal point. (i.e., get the floor) - using Math.floor(). This can be done.

If you know how to read the random number table to pick a random number, you know the above process (multiplying by 1, 10, 100 and so on) does not violate the one that I was mentioned at the beginning (because it changes only the place of the decimal point).

Study the following example and develop it to your needs.

If you need a sample [0,9] then the floor of n10 is your answer and if you need [0,99] then the floor of n100 is your answer and so on.

Now let’s enter into your role:

You've asked for numbers in a specific range. (In this case you are biased among that range. By taking a number from [1,6] by roll a die, then you are biased into [1,6], but still it is a random number if and only if the die is unbiased.)

So consider your range ==> [78, 247] number of elements of the range = 247 - 78 + 1 = 170; (since both the boundaries are inclusive).

/* Method 1: */
    var i = 78, j = 247, k = 170, a = [], b = [], c, d, e, f, l = 0;
    for(; i <= j; i++){ a.push(i); }
    while(l < 170){
        c = Math.random()*100; c = Math.floor(c);
        d = Math.random()*100; d = Math.floor(d);
        b.push(a[c]); e = c + d;
        if((b.length != k) && (e < k)){  b.push(a[e]); }
        l = b.length;
    }
    console.log('Method 1:');
    console.log(b);

/* Method 2: */

    var a, b, c, d = [], l = 0;
    while(l < 170){
        a = Math.random()*100; a = Math.floor(a);
        b = Math.random()*100; b = Math.floor(b);
        c = a + b;
        if(c <= 247 || c >= 78){ d.push(c); }else{ d.push(a); }
        l = d.length;
    }
    console.log('Method 2:');
    console.log(d);

Note: In method one, first I created an array which contains numbers that you need and then randomly put them into another array.

In method two, generate numbers randomly and check those are in the range that you need. Then put it into an array. Here I generated two random numbers and used the total of them to maximize the speed of the program by minimizing the failure rate that obtaining a useful number. However, adding generated numbers will also give some biasedness. So I would recommend my first method to generate random numbers within a specific range.

In both methods, your console will show the result (press F12 in Chrome to open the console).

Answer 16

For a random integer with a range, try:

function random(minimum, maximum) {
  var bool = true;

  while (bool) {
    var number = (Math.floor(Math.random() * maximum + 1) + minimum);
    if (number > 20) {
      bool = true;
    } else {
      bool = false;
    }
  }

  return number;
}

Answer 17

function getRandomInt(lower, upper)
{
    //to create an even sample distribution
    return Math.floor(lower + (Math.random() * (upper - lower + 1)));

    //to produce an uneven sample distribution
    //return Math.round(lower + (Math.random() * (upper - lower)));

    //to exclude the max value from the possible values
    //return Math.floor(lower + (Math.random() * (upper - lower)));
}

To test this function, and variations of this function, save the below HTML/JavaScript to a file and open with a browser. The code will produce a graph showing the distribution of one million function calls. The code will also record the edge cases, so if the the function produces a value greater than the max, or less than the min, you.will.know.about.it.

<html>
    <head>
        <script type="text/javascript">
        function getRandomInt(lower, upper)
        {
            //to create an even sample distribution
            return Math.floor(lower + (Math.random() * (upper - lower + 1)));

            //to produce an uneven sample distribution
            //return Math.round(lower + (Math.random() * (upper - lower)));

            //to exclude the max value from the possible values
            //return Math.floor(lower + (Math.random() * (upper - lower)));
        }

        var min = -5;
        var max = 5;

        var array = new Array();

        for(var i = 0; i <= (max - min) + 2; i++) {
          array.push(0);
        }

        for(var i = 0; i < 1000000; i++) {
            var random = getRandomInt(min, max);
            array[random - min + 1]++;
        }

        var maxSample = 0;
        for(var i = 0; i < max - min; i++) {
            maxSample = Math.max(maxSample, array[i]);
        }

        //create a bar graph to show the sample distribution
        var maxHeight = 500;
        for(var i = 0; i <= (max - min) + 2; i++) {
            var sampleHeight = (array[i]/maxSample) * maxHeight;

            document.write('<span style="display:inline-block;color:'+(sampleHeight == 0 ? 'black' : 'white')+';background-color:black;height:'+sampleHeight+'px">&nbsp;[' + (i + min - 1) + ']:&nbsp;'+array[i]+'</span>&nbsp;&nbsp;');
        }
        document.write('<hr/>');
        </script>
    </head>
    <body>

    </body>
</html>

Answer 18

Here is a function that generates a random number between min and max, both inclusive.

const randomInt = (max, min) => Math.round(Math.random() * (max - min)) + min;

Answer 19

To get a random number say between 1 and 6, first do:

0.5 + (Math.random() * ((6 - 1) + 1))

This multiplies a random number by 6 and then adds 0.5 to it. Next round the number to a positive integer by doing:

Math.round(0.5 + (Math.random() * ((6 - 1) + 1))

This round the number to the nearest whole number.

Or to make it more understandable do this:

var value = 0.5 + (Math.random() * ((6 - 1) + 1))
var roll = Math.round(value);
return roll;

In general, the code for doing this using variables is:

var value = (Min - 0.5) + (Math.random() * ((Max - Min) + 1))
var roll = Math.round(value);
return roll;

The reason for taking away 0.5 from the minimum value is because using the minimum value alone would allow you to get an integer that was one more than your maximum value. By taking away 0.5 from the minimum value you are essentially preventing the maximum value from being rounded up.

Answer 20

Using the following code, you can generate an array of random numbers, without repeating, in a given range.

function genRandomNumber(how_many_numbers, min, max) {

    // Parameters
    //
    //   how_many_numbers: How many numbers you want to
    //                     generate. For example, it is 5.
    //
    //   min (inclusive):  Minimum/low value of a range. It
    //                     must be any positive integer, but
    //                     less than max. I.e., 4.
    //
    //   max (inclusive):  Maximum value of a range. it must
    //                     be any positive integer. I.e., 50
    //
    //   Return type: array

    var random_number = [];
    for (var i = 0; i < how_many_numbers; i++) {
        var gen_num = parseInt((Math.random() * (max-min+1)) + min);
        do {
            var is_exist = random_number.indexOf(gen_num);
            if (is_exist >= 0) {
                gen_num = parseInt((Math.random() * (max-min+1)) + min);
            }
            else {
                random_number.push(gen_num);
                is_exist = -2;
            }
        }
        while (is_exist > -1);
    }
    document.getElementById('box').innerHTML = random_number;
}

Answer 21

Random whole number between lowest and highest:

function randomRange(low, high) {
  var range = (high-low);
  var random = Math.floor(Math.random()*range);
  if (random === 0) {
    random += 1;
  }
  return low + random;
}

It is not the most elegant solution, but something quick.

Answer 22

I found this simple method on W3Schools:

Math.floor((Math.random() * max) + min);

Answer 23

Here is an example of a JavaScript function that can generate a random number of any specified length without using Math.random():

function genRandom(length)
{
  const t1 = new Date().getMilliseconds();
  var min = "1", max = "9";
  var result;
  var numLength = length;
  if (numLength != 0)
  {
     for (var i = 1; i < numLength; i++)
     {
        min = min.toString() + "0";
        max = max.toString() + "9";
     }
  }
  else
  {
     min = 0;
     max = 0;
     return;
  }

  for (var i = min; i <= max; i++)
  {
       // Empty Loop
  }

  const t2 = new Date().getMilliseconds();
  console.log(t2);
  result = ((max - min)*t1)/t2;
  console.log(result);
  return result;
}

Answer 24

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]

Answer 25

Use:

<!DOCTYPE html>
<html>
    <head>
        <meta charset="utf-8" />
    </head>

    <body>
        <script>
            /*
                Assuming that window.crypto.getRandomValues
                is available, the real range would be from
                0 to 1,998 instead of 0 to 2,000.

                See the JavaScript documentation
                for an explanation:

                  https://developer.mozilla.org/en-US/docs/Web/API/RandomSource/getRandomValues
            */
            var array = new Uint8Array(2);
            window.crypto.getRandomValues(array);
            console.log(array[0] + array[1]);
        </script>
    </body>
</html>

Uint8Array creates an array filled with a number up to three digits which would be a maximum of 999. This code is very short.

Answer 26

This is my take on a random number in a range, as in I wanted to get a random number within a range of base to exponent. E.g., base = 10, exponent = 2, gives a random number from 0 to 100, ideally, and so on.

If it helps using it, here it is:

// Get random number within provided base + exponent
// By Goran Biljetina --> 2012

function isEmpty(value) {
    return (typeof value === "undefined" || value === null);
}

var numSeq = new Array();

function add(num, seq) {
    var toAdd = new Object();
    toAdd.num = num;
    toAdd.seq = seq;
    numSeq[numSeq.length] = toAdd;
}

function fillNumSeq (num, seq) {
    var n;
    for(i=0; i<=seq; i++) {
        n = Math.pow(num, i);
        add(n, i);
    }
}

function getRandNum(base, exp) {
    if (isEmpty(base)) {
        console.log("Specify value for base parameter");
    }
    if (isEmpty(exp)) {
        console.log("Specify value for exponent parameter");
    }

    fillNumSeq(base, exp);

    var emax;
    var eseq;
    var nseed;
    var nspan;
    emax = (numSeq.length);
    eseq = Math.floor(Math.random()*emax) + 1;
    nseed = numSeq[eseq].num;
    nspan = Math.floor((Math.random())*(Math.random()*nseed)) + 1;
    return Math.floor(Math.random()*nspan) + 1;
}

console.log(getRandNum(10, 20), numSeq);

//Testing:
//getRandNum(-10, 20);
//console.log(getRandNum(-10, 20), numSeq);
//console.log(numSeq);

Answer 27

Ionuț G. Stan wrote a great answer, but it was a bit too complex for me to grasp. So, I found an even simpler explanation of the same concepts at Math.floor( Math.random () * (max - min + 1)) + min) Explanation by Jason Anello.

Note: The only important thing you should know before reading Jason's explanation is a definition of "truncate". He uses that term when describing Math.floor(). Oxford dictionary defines "truncate" as:

Shorten (something) by cutting off the top or end.

Answer 28

This I guess, is the most simplified of all the contributions.

maxNum = 8,
minNum = 4

console.log(Math.floor(Math.random() * (maxNum - minNum) + minNum))

console.log(Math.floor(Math.random() * (8 - 4) + 4))

This will log random numbers between 4 and 8 into the console, 4 and 8 inclusive.

Answer 29

A function called randUpTo that accepts a number and returns a random whole number between 0 and that number:

var randUpTo = function(num) {
    return Math.floor(Math.random() * (num - 1) + 0);
};

A function called randBetween that accepts two numbers representing a range and returns a random whole number between those two numbers:

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

A function called randFromTill that accepts two numbers representing a range and returns a random number between min (inclusive) and max (exclusive)

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

A function called randFromTo that accepts two numbers representing a range and returns a random integer between min (inclusive) and max (inclusive):

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

Answer 30

You can you this code snippet,

let randomNumber = function(first, second) {
    let number = Math.floor(Math.random()*Math.floor(second));
    while(number < first) {

        number = Math.floor(Math.random()*Math.floor(second));
    }
    return number;
}

Answer 31

My method of generating a random number between 0 and n, where n <= 10 (n excluded):

Math.floor((Math.random() * 10) % n)

Answer 32

I made this function which takes into account options like min, max, exclude (a list of ints to exclude), and seed (in case you want a seeded random generator).

get_random_int = function(args={})
{
    let def_args =
    {
        min: 0,
        max: 1,
        exclude: false,
        seed: Math.random
    }

    args = Object.assign(def_args, args)

    let num = Math.floor(args.seed() * (args.max - args.min + 1) + args.min)

    if(args.exclude)
    {
        let diff = args.max - args.min
        let n = num

        for(let i=0; i<diff*2; i++)
        {
            if(args.exclude.includes(n))
            {
                if(n + 1 <= args.max)
                {
                    n += 1
                }

                else
                {
                    n = args.min
                }
            }

            else
            {
                num = n
                break
            }
        }
    }

    return num
}

It can be used like:

let n = get_random_int
(
    {
        min: 0,
        max: some_list.length - 1,
        exclude: [3, 6, 5],
        seed: my_seed_function
    }
)

Or more simply:

let n = get_random_int
(
    {
        min: 0,
        max: some_list.length - 1
    }
)

Then you can do:

let item = some_list[n]

Gist: https://gist.github.com/madprops/757deb000bdec25776d5036dae58ee6e

Answer 33

• random(min, max) generates a random number between min (inclusive) and max (exclusive)

• Math.floor rounds a number down to the nearest integer

    function generateRandomInteger(min, max) { 
        return Math.floor(random(min, max)) 
    }
  

So to generate a random integer between 4 and 8 inclusive, call the above function with the following arguments:

generateRandomInteger(4, 9)

Answer 34

For best performance, you can simply use:

var r = (Math.random() * (maximum - minimum + 1) ) << 0

Answer 35

// Example
function ourRandomRange(ourMin, ourMax) {
    return Math.floor(Math.random() * (ourMax - ourMin + 1)) + ourMin;
}

ourRandomRange(1, 9);

// Only change code below this line.
function randomRange(myMin, myMax) {
    var a = Math.floor(Math.random() * (myMax - myMin + 1)) + myMin;
    return a; // Change this line
}

// Change these values to test your function
var myRandom = randomRange(5, 15);

Answer 36

Using modern JavaScript + Lodash:

const generateRandomNumbers = (max, amount) => {
  const numbers = [...Array(max).keys()];
  const randomNumbers = sampleSize(numbers, amount);

  return randomNumbers.sort((a, b) => a - b);
};

Also, a TypeScript version:

const generateRandomNumbers = (max: number, amount: number) => {
  const numbers = [...Array(max).keys()];
  const randomNumbers: number[] = sampleSize(numbers, amount);

  return randomNumbers.sort((a: number, b: number) => a - b);
};

Answer 37

This implementation works when both inputs are integers.

function randomRange(myMin, myMax) {
  return Math.floor(
    Math.random() * (Math.ceil(myMax) - Math.floor(myMin) + 1) + myMin
  );
}

Answer 38

All of the existing answers generate a random number between some min and a max (or possibly between zero and a max) but I wanted the most random possible whole number.

Since Math.random() generates a 16 digit decimal number, the way to get maximally random positive integers is:

Math.floor(Math.random()*10**16)

Answer 39

Problems with the accepted answer

It's worth noting that the accepted answer does not properly handle cases where min is greater than max. Here's an example of that:

min = Math.ceil(2);
max = Math.floor(1);
for(var i = 0; i < 25; i++) {
  console.log(Math.floor(Math.random() * (max - min + 1)) + min);
}

In addition, it's a bit wordy and unclear to read if you're unfamiliar with this little algorithm.

Why is Randojs a better solution?

Randojs handles cases where min is greater than max automatically (and it's cryptographically secure):

for(var i = 0; i < 25; i++) console.log(rando(2, 1));

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

It also handles negatives, zeros, and everything else you'd expect. If you need to do floats or use other variable types, there are options for that as well, but I won't talk about them here. They're on the site so you can delve deeper there if needed. The final reason is pretty obvious. Stylistically, it's is much cleaner and easier to read.

---

TL;DR. Just give me the solution...

randojs.com makes this and a ton of other common randomness stuff robust, reliable, as simple/readable as this:

console.log(rando(20, 30));

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

Answer 40

Here's something I found on a webpage:

function randomInt(e,t){return Math.floor(Math.random()*(t-e+1)+e)}