Develop Reference

Get a random number focused on center

Is it possible to get a random number between 1-100 and keep the results mainly within the 40-60 range? I mean, it will go out of that range rarely, but I want it to be mainly within that range... Is it possible with JavaScript/jQuery?
Right now I'm just using the basic Math.random() * 100 + 1.

The simplest way would be to generate two random numbers from 0-50 and add them together.
This gives a distribution biased towards 50, in the same way rolling two dice biases towards 7.
In fact, by using a larger number of "dice" (as #Falco suggests), you can make a closer approximation to a bell-curve:
function weightedRandom(max, numDice) {
let num = 0;
for (let i = 0; i < numDice; i++) {
num += Math.random() * (max/numDice);
}
return num;
}
JSFiddle: http://jsfiddle.net/797qhcza/1/

You have some good answers here that give specific solutions; let me describe for you the general solution. The problem is:
I have a source of more-or-less uniformly distributed random numbers between 0 and 1.
I wish to produce a sequence of random numbers that follow a different distribution.
The general solution to this problem is to work out the quantile function of your desired distribution, and then apply the quantile function to the output of your uniform source.
The quantile function is the inverse of the integral of your desired distribution function. The distribution function is the function where the area under a portion of the curve is equal to the probability that the randomly-chosen item will be in that portion.
I give an example of how to do so here:
http://ericlippert.com/2012/02/21/generating-random-non-uniform-data/
The code in there is in C#, but the principles apply to any language; it should be straightforward to adapt the solution to JavaScript.

Taking arrays of numbers, etc. isn't efficient. You should take a mapping which takes a random number between 0 to 100 and maps to the distribution you need. So in your case, you could take f(x)=-(1/25)x2+4x to get a distribution with the most values in the middle of your range.

I might do something like setup a "chance" for the number to be allowed to go "out of bounds". In this example, a 20% chance the number will be 1-100, otherwise, 40-60:
$(function () {
$('button').click(function () {
var outOfBoundsChance = .2;
var num = 0;
if (Math.random() <= outOfBoundsChance) {
num = getRandomInt(1, 100);
} else {
num = getRandomInt(40, 60);
}
$('#out').text(num);
});
function getRandomInt(min, max) {
return Math.floor(Math.random() * (max - min + 1)) + min;
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<button>Generate</button>
<div id="out"></div>
fiddle: http://jsfiddle.net/kbv39s9w/

I needed to solve this problem a few years ago and my solution was easier than any of the other answers.
I generated 3 randoms between the bounds and averaged them. This pulls the result towards the centre but leaves it completely possible to reach the extremities.

It looks stupid but you can use rand twice:
var choice = Math.random() * 3;
var result;
if (choice < 2){
result = Math.random() * 20 + 40; //you have 2/3 chance to go there
}
else {
result = Math.random() * 100 + 1;
}

Sure it is possible. Make a random 1-100. If the number is <30 then generate number in range 1-100 if not generate in range 40-60.

There is a lot of different ways to generate such random numbers. One way to do it is to compute the sum of multiple uniformly random numbers. How many random numbers you sum and what their range is will determine how the final distribution will look.
The more numbers you sum up, the more it will be biased towards the center. Using the sum of 1 random number was already proposed in your question, but as you notice is not biased towards the center of the range. Other answers have propose using the sum of 2 random numbers or the sum of 3 random numbers.
You can get even more bias towards the center of the range by taking the sum of more random numbers. At the extreme you could take the sum of 99 random numbers which each were either 0 or 1. That would be a binomial distribution. (Binomial distributions can in some sense be seen as the discrete version of normal distributions). This can still in theory cover the full range, but it has so much bias towards the center that you should never expect to see it reach the endpoints.
This approach means you can tweak just how much bias you want.

What about using something like this:
var loops = 10;
var tries = 10;
var div = $("#results").html(random());
function random() {
var values = "";
for(var i=0; i < loops; i++) {
var numTries = tries;
do {
var num = Math.floor((Math.random() * 100) + 1);
numTries--;
}
while((num < 40 || num >60) && numTries > 1)
values += num + "<br/>";
}
return values;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>
The way I've coded it allows you to set a couple of variables:
loops = number of results
tries = number of times the function will try to get a number between 40-60 before it stops running through the while loop
Added bonus: It uses do while!!! Awesomeness at its best

You can write a function that maps random values between [0, 1) to [1, 100] according to weight. Consider this example:
Here, the value 0.95 maps to value between [61, 100].
In fact we have .05 / .1 = 0.5, which, when mapped to [61, 100], yields 81.
Here is the function:
/*
* Function that returns a function that maps random number to value according to map of probability
*/
function createDistributionFunction(data) {
// cache data + some pre-calculations
var cache = [];
var i;
for (i = 0; i < data.length; i++) {
cache[i] = {};
cache[i].valueMin = data[i].values[0];
cache[i].valueMax = data[i].values[1];
cache[i].rangeMin = i === 0 ? 0 : cache[i - 1].rangeMax;
cache[i].rangeMax = cache[i].rangeMin + data[i].weight;
}
return function(random) {
var value;
for (i = 0; i < cache.length; i++) {
// this maps random number to the bracket and the value inside that bracket
if (cache[i].rangeMin <= random && random < cache[i].rangeMax) {
value = (random - cache[i].rangeMin) / (cache[i].rangeMax - cache[i].rangeMin);
value *= cache[i].valueMax - cache[i].valueMin + 1;
value += cache[i].valueMin;
return Math.floor(value);
}
}
};
}
/*
* Example usage
*/
var distributionFunction = createDistributionFunction([
{ weight: 0.1, values: [1, 40] },
{ weight: 0.8, values: [41, 60] },
{ weight: 0.1, values: [61, 100] }
]);
/*
* Test the example and draw results using Google charts API
*/
function testAndDrawResult() {
var counts = [];
var i;
var value;
// run the function in a loop and count the number of occurrences of each value
for (i = 0; i < 10000; i++) {
value = distributionFunction(Math.random());
counts[value] = (counts[value] || 0) + 1;
}
// convert results to datatable and display
var data = new google.visualization.DataTable();
data.addColumn("number", "Value");
data.addColumn("number", "Count");
for (value = 0; value < counts.length; value++) {
if (counts[value] !== undefined) {
data.addRow([value, counts[value]]);
}
}
var chart = new google.visualization.ColumnChart(document.getElementById("chart"));
chart.draw(data);
}
google.load("visualization", "1", { packages: ["corechart"] });
google.setOnLoadCallback(testAndDrawResult);
<script src="https://www.google.com/jsapi"></script>
<div id="chart"></div>

Here's a weighted solution at 3/4 40-60 and 1/4 outside that range.
function weighted() {
var w = 4;
// number 1 to w
var r = Math.floor(Math.random() * w) + 1;
if (r === 1) { // 1/w goes to outside 40-60
var n = Math.floor(Math.random() * 80) + 1;
if (n >= 40 && n <= 60) n += 40;
return n
}
// w-1/w goes to 40-60 range.
return Math.floor(Math.random() * 21) + 40;
}
function test() {
var counts = [];
for (var i = 0; i < 2000; i++) {
var n = weighted();
if (!counts[n]) counts[n] = 0;
counts[n] ++;
}
var output = document.getElementById('output');
var o = "";
for (var i = 1; i <= 100; i++) {
o += i + " - " + (counts[i] | 0) + "\n";
}
output.innerHTML = o;
}
test();
<pre id="output"></pre>

Ok, so I decided to add another answer because I felt like my last answer, as well as most answers here, use some sort of half-statistical way of obtaining a bell-curve type result return. The code I provide below works the same way as when you roll a dice. Therefore, it is hardest to get 1 or 99, but easiest to get 50.
var loops = 10; //Number of numbers generated
var min = 1,
max = 50;
var div = $("#results").html(random());
function random() {
var values = "";
for (var i = 0; i < loops; i++) {
var one = generate();
var two = generate();
var ans = one + two - 1;
var num = values += ans + "<br/>";
}
return values;
}
function generate() {
return Math.floor((Math.random() * (max - min + 1)) + min);
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>

I'd recommend using the beta distribution to generate a number between 0-1, then scale it up. It's quite flexible and can create many different shapes of distributions.
Here's a quick and dirty sampler:
rbeta = function(alpha, beta) {
var a = 0
for(var i = 0; i < alpha; i++)
a -= Math.log(Math.random())
var b = 0
for(var i = 0; i < beta; i++)
b -= Math.log(Math.random())
return Math.ceil(100 * a / (a+b))
}

var randNum;
// generate random number from 1-5
var freq = Math.floor(Math.random() * (6 - 1) + 1);
// focus on 40-60 if the number is odd (1,3, or 5)
// this should happen %60 of the time
if (freq % 2){
randNum = Math.floor(Math.random() * (60 - 40) + 40);
}
else {
randNum = Math.floor(Math.random() * (100 - 1) + 1);
}

The best solution targeting this very problem is the one proposed by BlueRaja - Danny Pflughoeft but I think a somewhat faster and more general solution is also worth mentioning.
When I have to generate random numbers (strings, coordinate pairs, etc.) satisfying the two requirements of
The result set is quite small. (not larger than 16K numbers)
The result set is discreet. (like integer numbers only)
I usually start by creating an array of numbers (strings, coordinate pairs, etc.) fulfilling the requirement (In your case: an array of numbers containing the more probable ones multiple times.), then choose a random item of that array. This way, you only have to call the expensive random function once per item.

Distribution
5% for [ 0,39]
90% for [40,59]
5% for [60,99]
Solution
var f = Math.random();
if (f < 0.05) return random(0,39);
else if (f < 0.95) return random(40,59);
else return random(60,99);
Generic Solution
random_choose([series(0,39),series(40,59),series(60,99)],[0.05,0.90,0.05]);
function random_choose (collections,probabilities)
{
var acc = 0.00;
var r1 = Math.random();
var r2 = Math.random();
for (var i = 0; i < probabilities.length; i++)
{
acc += probabilities[i];
if (r1 < acc)
return collections[i][Math.floor(r2*collections[i].length)];
}
return (-1);
}
function series(min,max)
{
var i = min; var s = [];
while (s[s.length-1] < max) s[s.length]=i++;
return s;
}

You can use a helper random number to whether generate random numbers in 40-60 or 1-100:
// 90% of random numbers should be between 40 to 60.
var weight_percentage = 90;
var focuse_on_center = ( (Math.random() * 100) < weight_percentage );
if(focuse_on_center)
{
// generate a random number within the 40-60 range.
alert (40 + Math.random() * 20 + 1);
}
else
{
// generate a random number within the 1-100 range.
alert (Math.random() * 100 + 1);
}

If you can use the gaussian function, use it. This function returns normal number with average 0 and sigma 1.
95% of this number are within average +/- 2*sigma. Your average = 50, and sigma = 5 so
randomNumber = 50 + 5*gaussian()

The best way to do that is generating a random number that is distributed equally in a certain set of numbers, and then apply a projection function to the set between 0 and a 100 where the projection is more likely to hit the numbers you want.
Typically the mathematical way of achieving this is plotting a probability function of the numbers you want. We could use the bell curve, but let's for the sake of easier calculation just work with a flipped parabola.
Let's make a parabola such that its roots are at 0 and 100 without skewing it. We get the following equation:
f(x) = -(x-0)(x-100) = -x * (x-100) = -x^2 + 100x
Now, all the area under the curve between 0 and 100 is representative of our first set where we want the numbers generated. There, the generation is completely random. So, all we need to do is find the bounds of our first set.
The lower bound is, of course, 0. The upper bound is the integral of our function at 100, which is
F(x) = -x^3/3 + 50x^2
F(100) = 500,000/3 = 166,666.66666 (let's just use 166,666, because rounding up would make the target out of bounds)
So we know that we need to generate a number somewhere between 0 and 166,666. Then, we simply need to take that number and project it to our second set, which is between 0 and 100.
We know that the random number we generated is some integral of our parabola with an input x between 0 and 100. That means that we simply have to assume that the random number is the result of F(x), and solve for x.
In this case, F(x) is a cubic equation, and in the form F(x) = ax^3 + bx^2 + cx + d = 0, the following statements are true:
a = -1/3
b = 50
c = 0
d = -1 * (your random number)
Solving this for x yields you the actual random number your are looking for, which is guaranteed to be in the [0, 100] range and a much higher likelihood to be close to the center than the edges.

This answer is really good. But I would like to post implementation instructions (I'm not into JavaScript, so I hope you will understand) for different situation.
Assume you have ranges and weights for every range:
ranges - [1, 20], [21, 40], [41, 60], [61, 100]
weights - {1, 2, 100, 5}
Initial Static Information, could be cached:
Sum of all weights (108 in sample)
Range selection boundaries. It basically this formula: Boundary[n] = Boundary[n - 1] + weigh[n - 1] and Boundary[0] = 0. Sample has Boundary = {0, 1, 3, 103, 108}
Number generation:
Generate random number N from range [0, Sum of all weights).
for (i = 0; i < size(Boundary) && N > Boundary[i + 1]; ++i)
Take ith range and generate random number in that range.
Additional note for performance optimizations. Ranges don't have to be ordered neither ascending nor descending order, so for faster range look-up range that has highest weight should go first and one with lowest weight should go last.

Seeking a statistical javascript function to return p-value from a z-score

I need to convert z-scores to percentile. I found reference to a function in the jStat library that I could use (jstat.ztest), but the jStat documentation seems to be ahead of the available library because there is no such function in the currently available version of the library.
I think there is a more recent version of the library on GitHub, which may include the ztest function, but I am a linux novice and could not figure out how to build the library from the instructions. I spent most of a day learning about git bash and cygwin trying to build the library; I finally decided I'd be better off asking here.
So, could anyone point me toward a javascript function that would do what I need?
Alternatively, could anyone point me toward a built version of the jStat library with ztest function included?

I found this in a forum online and it works like a charm.
function GetZPercent(z)
{
//z == number of standard deviations from the mean
//if z is greater than 6.5 standard deviations from the mean
//the number of significant digits will be outside of a reasonable
//range
if ( z < -6.5)
return 0.0;
if( z > 6.5)
return 1.0;
var factK = 1;
var sum = 0;
var term = 1;
var k = 0;
var loopStop = Math.exp(-23);
while(Math.abs(term) > loopStop)
{
term = .3989422804 * Math.pow(-1,k) * Math.pow(z,k) / (2 * k + 1) / Math.pow(2,k) * Math.pow(z,k+1) / factK;
sum += term;
k++;
factK *= k;
}
sum += 0.5;
return sum;
}
And I don't need to include a large library just for the one function.

Just editing the code from Paul's answer for a two-sided t-test
function GetZPercent(z)
{
//z == number of standard deviations from the mean
//if z is greater than 6.5 standard deviations from the mean
//the number of significant digits will be outside of a reasonable
//range
if ( z < -6.5)
return 0.0;
if( z > 6.5)
return 1.0;
if (z > 0) { z = -z;}
var factK = 1;
var sum = 0;
var term = 1;
var k = 0;
var loopStop = Math.exp(-23);
while(Math.abs(term) > loopStop)
{
term = .3989422804 * Math.pow(-1,k) * Math.pow(z,k) / (2 * k + 1) / Math.pow(2,k) * Math.pow(z,k+1) / factK;
sum += term;
k++;
factK *= k;
}
sum += 0.5;
return (2*sum);
}

This seems like such a simple ask but I had a hard time tracking down a library that does this instead of copying some random code snippet. Best I can tell this will calculate z-score from a percentage using the simple-statistics library.
I took their documentation about cumulativestdnormalprobability and backed into the following algorithm. Feels like there should be an easier way but who knows.
https://simplestatistics.org/docs/#cumulativestdnormalprobability
const z_score = inverseErrorFunction((percentile_value - 0.5) / 0.5) * Math.sqrt(2);

As already correctly stated by Shane, the equation is an implementation of the Taylor Expansion of the normal cdf. The sum value iterates above and below the "real" value with increasing precision. If the value is close to 1 or 0 there is a very low, but existing, probability that sum will be >1 or <0, because of the (relatively) early break by loopstop.
The deviation is further strengthened by rounding 1/Math.sqrt(2*Math.Pi) to 0.3989422804 and the precision issues of javascript float numbers. Additionally, the provided solution will not work for z-scores >7 or <-7
I updated the code to be more accurate using the decimal.js npm library and to directly return the p-value:
function GetpValueFromZ(_z, type = "twosided")
{
if(_z < -14)
{
_z = -14
}
else if(_z > 14)
{
_z = 14
}
Decimal.set({precision: 100});
let z = new Decimal(_z);
var sum = new Decimal(0);
var term = new Decimal(1);
var k = new Decimal(0);
var loopstop = new Decimal("10E-50");
var minusone = new Decimal(-1);
var two = new Decimal(2);
let pi = new Decimal("3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647")
while(term.abs().greaterThan(loopstop))
{
term = new Decimal(1)
for (let i = 1; i <= k; i++) {
term = term.times(z).times(z.dividedBy(two.times(i)))
}
term = term.times(minusone.toPower(k)).dividedBy(k.times(2).plus(1))
sum = sum.plus(term);
k = k.plus(1);
}
sum = sum.times(z).dividedBy(two.times(pi).sqrt()).plus(0.5);
if(sum.lessThan(0))
sum = sum.abs();
else if(sum.greaterThan(1))
sum = two.minus(sum);
switch (type) {
case "left":
return parseFloat(sum.toExponential(40));
case "right":
return parseFloat((new Decimal(1).minus(sum)).toExponential(40));
case "twosided":
return sum.lessThan(0.5)? parseFloat(sum.times(two).toExponential(40)) : parseFloat((new Decimal(1).minus(sum).times(two)).toExponential(40))
}
}
By increasing the Decimal.js precision value and decreasing the loopstop value you can get accurate p-values for very small (or very high) z-scores for the cost of calculation time.

Rounding up to the nearest 0.05 in JavaScript

Question
Does anyone know of a way to round a float to the nearest 0.05 in JavaScript?
Example
BEFORE | AFTER
2.51 | 2.55
2.50 | 2.50
2.56 | 2.60
Current Code
var _ceil = Math.ceil;
Math.ceil = function(number, decimals){
if (arguments.length == 1)
return _ceil(number);
multiplier = Math.pow(10, decimals);
return _ceil(number * multiplier) / multiplier;
}
Then elsewhere...
return (Math.ceil((amount - 0.05), 1) + 0.05).toFixed(2);
Which is resulting in...
BEFORE | AFTER
2.51 | 2.55
2.50 | 2.55
2.56 | 2.65

Multiply by 20, then divide by 20:
(Math.ceil(number*20)/20).toFixed(2)

Rob's answer with my addition:
(Math.ceil(number*20 - 0.5)/20).toFixed(2)
Otherwise it always rounds up to the nearest 0.05.
** UPDATE **
Sorry has been pointed out this is not what the orig poster wanted.

I would go for the standard of actually dividing by the number you're factoring it to, and rounding that and multiplying it back again after. That seems to be a proper working method which you can use with any number and maintain the mental image of what you are trying to achieve.
var val = 26.14,
factor = 0.05;
val = Math.round(val / factor) * factor;
This will work for tens, hundreds or any number. If you are specifically rounding to the higher number then use Math.ceil instead of Math.round.
Another method specifically for rounding just to 1 or more decimal places (rather than half a place) is the following:
Number(Number(1.5454545).toFixed(1));
It creates a fixed number string and then turns it into a real Number.

I would write a function that does it for you by
move the decimal over two places (multiply by 100)
then mod (%) that inflatedNumber by 5 and get the remainder
subtract the remainder from 5 so that you know what the 'gap'(ceilGap) is between your number and the next closest .05
finally, divide your inflatedNumber by 100 so that it goes back to your original float, and voila, your num will be rounded up to the nearest .05.
function calcNearestPointZeroFive(num){
var inflatedNumber = num*100,
remainder = inflatedNumber % 5;
ceilGap = 5 - remainder
return (inflatedNumber + ceilGap)/100
}
If you want to leave numbers like 5.50 untouched you can always add this checker:
if (remainder===0){
return num
} else {
var ceilGap = 5 - remainder
return (inflatedNumber + ceilGap)/100
}

You need to put -1 to round half down and after that multiply by -1 like the example down bellow.
<script type="text/javascript">
function roundNumber(number, precision, isDown) {
var factor = Math.pow(10, precision);
var tempNumber = number * factor;
var roundedTempNumber = 0;
if (isDown) {
tempNumber = -tempNumber;
roundedTempNumber = Math.round(tempNumber) * -1;
} else {
roundedTempNumber = Math.round(tempNumber);
}
return roundedTempNumber / factor;
}
</script>
<div class="col-sm-12">
<p>Round number 1.25 down: <script>document.write(roundNumber(1.25, 1, true));</script>
</p>
<p>Round number 1.25 up: <script>document.write(roundNumber(1.25, 1, false));</script></p>
</div>

I ended up using this function in my project, successfully:
roundToNearestFiveCents( number: any ) {
return parseFloat((Math.round(number / 0.05) * 0.05).toFixed(2));
}
Might be of use to someone wanting to simply round to the nearest 5 cents on their monetary results, keeps the result a number, so if you perform addition on it further it won't result in string concatenation; also doesn't unnecessarily round up as a few of the other answers pointed out. Also limits it to two decimals, which is customary with finance.

My solution and test:
let round = function(number, precision = 2, rounding = 0.05) {
let multiply = 1 / rounding;
return parseFloat((Math.round(number * multiply) / multiply)).toFixed(precision);
};
https://jsfiddle.net/maciejSzewczyk/7r1tvhdk/40/

Even though the OP is not explicit about banker rounding, rounding up to the nearest $0.05 (5 cents) should be compatible with banker rounding. What suggested by Arth is more accurate than the accepted answer by Rob W.
(Math.ceil(number*20 - 0.5)/20).toFixed(2)
With banker rounding, you need a basic banker rounding function as suggested at Gaussian/banker's rounding in JavaScript, and I rewrite in TypeScript:
static bankerRound(num: number, decimalPlaces?: number) {
const d = decimalPlaces || 0;
const m = Math.pow(10, d);
const n = +(d ? num * m : num).toFixed(8);
const i = Math.floor(n), f = n - i;
const e = 1e-8;
const r = (f > 0.5 - e && f < 0.5 + e) ?
((i % 2 === 0) ? i : i + 1) : Math.round(n);
return d ? r / m : r;
}
static roundTo5cents(num: number) {
const r = bankerRound(Math.ceil(num * 20 - 0.5) / 20, 2);
return r;
}
The correctness of this algorithm could be verified through MBS Online, e.g. http://www9.health.gov.au/mbs/ready_reckoner.cfm?item_num=60

Different probability for ranges of random numbers

I'm looking for the best way of implementing random number generator, that will allow me to have control over probability from what range the generated number will be returned. To visualize what I'm trying to achieve I have a picture :
So to summarize :
Let's say that my range is 400. At the beginning I'd like to have 5% probability of getting number 0-20. But at some moment in time I'd like to have this probability increased up to 50%. Hope you get the idea.

Hmm, working on your original I had a pretty simple algorithm to generate ranges in an array in the appropriate proportion, then randomly select a range and generate a random number within that range. No doubt it can be optimised if necessary, but it works for me.
It looks like a lot of code, but 3/4 of it is comments, test data and function, the actual randomRange function is only 17 lines of code.
<script type="text/javascript">
function randomRange(dataArray) {
// Helper function
function getRandomInRange(s, f) {
return (Math.random() * (f-s+1) | 0) + s
}
// Generate new data array based on probability
var i, j = dataArray.length;
var oArray = [];
var o;
while (j--) {
o = dataArray[j];
// Make sure probability is an integer
for (i=0, iLen=o.probability|0; i<iLen; i++) {
oArray.push([o.rangeStart, o.rangeEnd]);
}
}
// Randomly select a range from new data array and
// generate a random number in that range
var oEnd = oArray.length;
var range = oArray[getRandomInRange(0, oArray.length - 1)];
return getRandomInRange(range[0], range[1]);
}
// Test data set. Probability just has to be
// representative, so 50/50 === 1/1
var dataArray = [
{
rangeStart: 0,
rangeEnd : 20,
probability: 1
},
{
rangeStart: 21,
rangeEnd : 400,
probability: 1
}
];
// Test function to show range and number is randomly
// selected for given probability
function testIt() {
var el0 = document.getElementById('div0');
var el1 = document.getElementById('div1');
function run() {
var n = randomRange(dataArray);
if (n <= 20) {
el0.innerHTML += '*';
} else {
el1.innerHTML += '*';
}
}
setInterval(run, 500);
}
</script>
<button onclick="testIt();">Generate random number</button>
<div>Numbers 0 - 20</div>
<div id="div0"></div>
<div>Numbers 21 - 400</div>
<div id="div1"></div>

It sounds to me like what you're looking for is a way to generate numbers on a normal (or Gaussian) distribution (take a look at the Wikipedia page if you don't know what that means).
The Box-Muller transformation can be used to generate pairs of normally distributed numbers.
Here is a c++ implementation of the polar form of the Box-Muller transformation that shouldn't be hard to translate to javascript.
// Return a real number from a normal (Gaussian) distribution with given
// mean and standard deviation by polar form of Box-Muller transformation
double x, y, r;
do
{
x = 2.0 * rand() - 1.0;
y = 2.0 * rand() - 1.0;
r = x * x + y * y;
}
while ( r >= 1.0 || r == 0.0 );
double s = sqrt( -2.0 * log(r) / r );
return mean + x * s * stddev;
Where mean is the mean of the normal distribution and stddev is the Standard Deviation of the distribution. This code is from a MersesenneTwister C++ class that I've been using recently that you can find on Rick Wagner's page. You can find some more useful information about the Box-Muller transformation on this page.

Seedable JavaScript random number generator

The JavaScript Math.random() function returns a random value between 0 and 1, automatically seeded based on the current time (similar to Java I believe). However, I don't think there's any way to set you own seed for it.
How can I make a random number generator that I can provide my own seed value for, so that I can have it produce a repeatable sequence of (pseudo)random numbers?

One option is http://davidbau.com/seedrandom which is a seedable RC4-based Math.random() drop-in replacement with nice properties.

If you don't need the seeding capability just use Math.random() and build helper functions around it (eg. randRange(start, end)).
I'm not sure what RNG you're using, but it's best to know and document it so you're aware of its characteristics and limitations.
Like Starkii said, Mersenne Twister is a good PRNG, but it isn't easy to implement. If you want to do it yourself try implementing a LCG - it's very easy, has decent randomness qualities (not as good as Mersenne Twister), and you can use some of the popular constants.
EDIT: consider the great options at this answer for short seedable RNG implementations, including an LCG option.
function RNG(seed) {
// LCG using GCC's constants
this.m = 0x80000000; // 2**31;
this.a = 1103515245;
this.c = 12345;
this.state = seed ? seed : Math.floor(Math.random() * (this.m - 1));
}
RNG.prototype.nextInt = function() {
this.state = (this.a * this.state + this.c) % this.m;
return this.state;
}
RNG.prototype.nextFloat = function() {
// returns in range [0,1]
return this.nextInt() / (this.m - 1);
}
RNG.prototype.nextRange = function(start, end) {
// returns in range [start, end): including start, excluding end
// can't modulu nextInt because of weak randomness in lower bits
var rangeSize = end - start;
var randomUnder1 = this.nextInt() / this.m;
return start + Math.floor(randomUnder1 * rangeSize);
}
RNG.prototype.choice = function(array) {
return array[this.nextRange(0, array.length)];
}
var rng = new RNG(20);
for (var i = 0; i < 10; i++)
console.log(rng.nextRange(10, 50));
var digits = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'];
for (var i = 0; i < 10; i++)
console.log(rng.choice(digits));

If you want to be able to specify the seed, you just need to replace the calls to getSeconds() and getMinutes(). You could pass in an int and use half of it mod 60 for the seconds value and the other half modulo 60 to give you the other part.
That being said, this method looks like garbage. Doing proper random number generation is very hard. The obvious problem with this is that the random number seed is based on seconds and minutes. To guess the seed and recreate your stream of random numbers only requires trying 3600 different second and minute combinations. It also means that there are only 3600 different possible seeds. This is correctable, but I'd be suspicious of this RNG from the start.
If you want to use a better RNG, try the Mersenne Twister. It is a well tested and fairly robust RNG with a huge orbit and excellent performance.
EDIT: I really should be correct and refer to this as a Pseudo Random Number Generator or PRNG.
"Anyone who uses arithmetic methods to produce random numbers is in a state of sin."
--- John von Neumann

I use a JavaScript port of the Mersenne Twister:
https://gist.github.com/300494
It allows you to set the seed manually. Also, as mentioned in other answers, the Mersenne Twister is a really good PRNG.

The code you listed kind of looks like a Lehmer RNG. If this is the case, then 2147483647 is the largest 32-bit signed integer, 2147483647 is the largest 32-bit prime, and 48271 is a full-period multiplier that is used to generate the numbers.
If this is true, you could modify RandomNumberGenerator to take in an extra parameter seed, and then set this.seed to seed; but you'd have to be careful to make sure the seed would result in a good distribution of random numbers (Lehmer can be weird like that) -- but most seeds will be fine.

The following is a PRNG that may be fed a custom seed. Calling SeedRandom will return a random generator function. SeedRandom can be called with no arguments in order to seed the returned random function with the current time, or it can be called with either 1 or 2 non-negative inters as arguments in order to seed it with those integers. Due to float point accuracy seeding with only 1 value will only allow the generator to be initiated to one of 2^53 different states.
The returned random generator function takes 1 integer argument named limit, the limit must be in the range 1 to 4294965886, the function will return a number in the range 0 to limit-1.
function SeedRandom(state1,state2){
var mod1=4294967087
var mul1=65539
var mod2=4294965887
var mul2=65537
if(typeof state1!="number"){
state1=+new Date()
}
if(typeof state2!="number"){
state2=state1
}
state1=state1%(mod1-1)+1
state2=state2%(mod2-1)+1
function random(limit){
state1=(state1*mul1)%mod1
state2=(state2*mul2)%mod2
if(state1<limit && state2<limit && state1<mod1%limit && state2<mod2%limit){
return random(limit)
}
return (state1+state2)%limit
}
return random
}
Example use:
var generator1=SeedRandom() //Seed with current time
var randomVariable=generator1(7) //Generate one of the numbers [0,1,2,3,4,5,6]
var generator2=SeedRandom(42) //Seed with a specific seed
var fixedVariable=generator2(7) //First value of this generator will always be
//1 because of the specific seed.
This generator exhibit the following properties:
It has approximately 2^64 different possible inner states.
It has a period of approximately 2^63, plenty more than anyone will ever realistically need in a JavaScript program.
Due to the mod values being primes there is no simple pattern in the output, no matter the chosen limit. This is unlike some simpler PRNGs that exhibit some quite systematic patterns.
It discards some results in order to get a perfect distribution no matter the limit.
It is relatively slow, runs around 10 000 000 times per second on my machine.
Bonus: typescript version

If you program in Typescript, I adapted the Mersenne Twister implementation that was brought in Christoph Henkelmann's answer to this thread as a typescript class:
/**
* copied almost directly from Mersenne Twister implementation found in https://gist.github.com/banksean/300494
* all rights reserved to him.
*/
export class Random {
static N = 624;
static M = 397;
static MATRIX_A = 0x9908b0df;
/* constant vector a */
static UPPER_MASK = 0x80000000;
/* most significant w-r bits */
static LOWER_MASK = 0x7fffffff;
/* least significant r bits */
mt = new Array(Random.N);
/* the array for the state vector */
mti = Random.N + 1;
/* mti==N+1 means mt[N] is not initialized */
constructor(seed:number = null) {
if (seed == null) {
seed = new Date().getTime();
}
this.init_genrand(seed);
}
private init_genrand(s:number) {
this.mt[0] = s >>> 0;
for (this.mti = 1; this.mti < Random.N; this.mti++) {
var s = this.mt[this.mti - 1] ^ (this.mt[this.mti - 1] >>> 30);
this.mt[this.mti] = (((((s & 0xffff0000) >>> 16) * 1812433253) << 16) + (s & 0x0000ffff) * 1812433253)
+ this.mti;
/* See Knuth TAOCP Vol2. 3rd Ed. P.106 for multiplier. */
/* In the previous versions, MSBs of the seed affect */
/* only MSBs of the array mt[]. */
/* 2002/01/09 modified by Makoto Matsumoto */
this.mt[this.mti] >>>= 0;
/* for >32 bit machines */
}
}
/**
* generates a random number on [0,0xffffffff]-interval
* #private
*/
private _nextInt32():number {
var y:number;
var mag01 = new Array(0x0, Random.MATRIX_A);
/* mag01[x] = x * MATRIX_A for x=0,1 */
if (this.mti >= Random.N) { /* generate N words at one time */
var kk:number;
if (this.mti == Random.N + 1) /* if init_genrand() has not been called, */
this.init_genrand(5489);
/* a default initial seed is used */
for (kk = 0; kk < Random.N - Random.M; kk++) {
y = (this.mt[kk] & Random.UPPER_MASK) | (this.mt[kk + 1] & Random.LOWER_MASK);
this.mt[kk] = this.mt[kk + Random.M] ^ (y >>> 1) ^ mag01[y & 0x1];
}
for (; kk < Random.N - 1; kk++) {
y = (this.mt[kk] & Random.UPPER_MASK) | (this.mt[kk + 1] & Random.LOWER_MASK);
this.mt[kk] = this.mt[kk + (Random.M - Random.N)] ^ (y >>> 1) ^ mag01[y & 0x1];
}
y = (this.mt[Random.N - 1] & Random.UPPER_MASK) | (this.mt[0] & Random.LOWER_MASK);
this.mt[Random.N - 1] = this.mt[Random.M - 1] ^ (y >>> 1) ^ mag01[y & 0x1];
this.mti = 0;
}
y = this.mt[this.mti++];
/* Tempering */
y ^= (y >>> 11);
y ^= (y << 7) & 0x9d2c5680;
y ^= (y << 15) & 0xefc60000;
y ^= (y >>> 18);
return y >>> 0;
}
/**
* generates an int32 pseudo random number
* #param range: an optional [from, to] range, if not specified the result will be in range [0,0xffffffff]
* #return {number}
*/
nextInt32(range:[number, number] = null):number {
var result = this._nextInt32();
if (range == null) {
return result;
}
return (result % (range[1] - range[0])) + range[0];
}
/**
* generates a random number on [0,0x7fffffff]-interval
*/
nextInt31():number {
return (this._nextInt32() >>> 1);
}
/**
* generates a random number on [0,1]-real-interval
*/
nextNumber():number {
return this._nextInt32() * (1.0 / 4294967295.0);
}
/**
* generates a random number on [0,1) with 53-bit resolution
*/
nextNumber53():number {
var a = this._nextInt32() >>> 5, b = this._nextInt32() >>> 6;
return (a * 67108864.0 + b) * (1.0 / 9007199254740992.0);
}
}
you can than use it as follows:
var random = new Random(132);
random.nextInt32(); //return a pseudo random int32 number
random.nextInt32([10,20]); //return a pseudo random int in range [10,20]
random.nextNumber(); //return a a pseudo random number in range [0,1]
check the source for more methods.

Here is quite an effective but simple javascript PRNG function that I like to use:
// The seed is the base number that the function works off
// The modulo is the highest number that the function can return
function PRNG(seed, modulo) {
str = `${(2**31-1&Math.imul(48271,seed))/2**31}`
.split('')
.slice(-10)
.join('') % modulo
return str
}
I hope this is what you're looking for.

Thank you, #aaaaaaaaaaaa (Accepted Answer)
I really needed a good non-library solution (easier to embed)
so... i made this class to store the seed and allow a Unity-esque "Next" ... but kept the initial Integer based results
class randS {
constructor(seed=null) {
if(seed!=null) {
this.seed = seed;
} else {
this.seed = Date.now()%4645455524863;
}
this.next = this.SeedRandom(this.seed);
this.last = 0;
}
Init(seed=this.seed) {
if (seed = this.seed) {
this.next = this.SeedRandom(this.seed);
} else {
this.seed=seed;
this.next = this.SeedRandom(this.seed);
}
}
SeedRandom(state1,state2){
var mod1=4294967087;
var mod2=4294965887;
var mul1=65539;
var mul2=65537;
if(typeof state1!="number"){
state1=+new Date();
}
if(typeof state2!="number"){
state2=state1;
}
state1=state1%(mod1-1)+1;
state2=state2%(mod2-1)+1;
function random(limit){
state1=(state1*mul1)%mod1;
state2=(state2*mul2)%mod2;
if(state1<limit && state2<limit && state1<mod1%limit && state2<mod2%limit){
this.last = random;
return random(limit);
}
this.last = (state1+state2)%limit;
return (state1+state2)%limit;
}
this.last = random;
return random;
}
}
And then checked it with these... seems to work well with random (but queryable) seed value (a la Minecraft) and even stored the last value returned (if needed)
var rng = new randS(9005646549);
console.log(rng.next(20)+' '+rng.next(20)+' '+rng.next(20)+' '+rng.next(20)+' '+rng.next(20)+' '+rng.next(20)+' '+rng.next(20));
console.log(rng.next(20) + ' ' + rng.next(20) + ' ' + rng.last);
which should output (for everybody)
6 7 8 14 1 12 6
9 1 1
EDIT: I made the init() work if you ever needed to reseed, or were testing values (this was necessary in my context as well)

Note: This code was originally included in the question above. In the interests of keeping the question short and focused, I've moved it to this Community Wiki answer.
I found this code kicking around and it appears to work fine for getting a random number and then using the seed afterward but I'm not quite sure how the logic works (e.g. where the 2345678901, 48271 & 2147483647 numbers came from).
function nextRandomNumber(){
var hi = this.seed / this.Q;
var lo = this.seed % this.Q;
var test = this.A * lo - this.R * hi;
if(test > 0){
this.seed = test;
} else {
this.seed = test + this.M;
}
return (this.seed * this.oneOverM);
}
function RandomNumberGenerator(){
var d = new Date();
this.seed = 2345678901 + (d.getSeconds() * 0xFFFFFF) + (d.getMinutes() * 0xFFFF);
this.A = 48271;
this.M = 2147483647;
this.Q = this.M / this.A;
this.R = this.M % this.A;
this.oneOverM = 1.0 / this.M;
this.next = nextRandomNumber;
return this;
}
function createRandomNumber(Min, Max){
var rand = new RandomNumberGenerator();
return Math.round((Max-Min) * rand.next() + Min);
}
//Thus I can now do:
var letters = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'];
var numbers = ['1','2','3','4','5','6','7','8','9','10'];
var colors = ['red','orange','yellow','green','blue','indigo','violet'];
var first = letters[createRandomNumber(0, letters.length)];
var second = numbers[createRandomNumber(0, numbers.length)];
var third = colors[createRandomNumber(0, colors.length)];
alert("Today's show was brought to you by the letter: " + first + ", the number " + second + ", and the color " + third + "!");
/*
If I could pass my own seed into the createRandomNumber(min, max, seed);
function then I could reproduce a random output later if desired.
*/

OK, here's the solution I settled on.
First you create a seed value using the "newseed()" function. Then you pass the seed value to the "srandom()" function. Lastly, the "srandom()" function returns a pseudo random value between 0 and 1.
The crucial bit is that the seed value is stored inside an array. If it were simply an integer or float, the value would get overwritten each time the function were called, since the values of integers, floats, strings and so forth are stored directly in the stack versus just the pointers as in the case of arrays and other objects. Thus, it's possible for the value of the seed to remain persistent.
Finally, it is possible to define the "srandom()" function such that it is a method of the "Math" object, but I'll leave that up to you to figure out. ;)
Good luck!
JavaScript:
// Global variables used for the seeded random functions, below.
var seedobja = 1103515245
var seedobjc = 12345
var seedobjm = 4294967295 //0x100000000
// Creates a new seed for seeded functions such as srandom().
function newseed(seednum)
{
return [seednum]
}
// Works like Math.random(), except you provide your own seed as the first argument.
function srandom(seedobj)
{
seedobj[0] = (seedobj[0] * seedobja + seedobjc) % seedobjm
return seedobj[0] / (seedobjm - 1)
}
// Store some test values in variables.
var my_seed_value = newseed(230951)
var my_random_value_1 = srandom(my_seed_value)
var my_random_value_2 = srandom(my_seed_value)
var my_random_value_3 = srandom(my_seed_value)
// Print the values to console. Replace "WScript.Echo()" with "alert()" if inside a Web browser.
WScript.Echo(my_random_value_1)
WScript.Echo(my_random_value_2)
WScript.Echo(my_random_value_3)
Lua 4 (my personal target environment):
-- Global variables used for the seeded random functions, below.
seedobja = 1103515.245
seedobjc = 12345
seedobjm = 4294967.295 --0x100000000
-- Creates a new seed for seeded functions such as srandom().
function newseed(seednum)
return {seednum}
end
-- Works like random(), except you provide your own seed as the first argument.
function srandom(seedobj)
seedobj[1] = mod(seedobj[1] * seedobja + seedobjc, seedobjm)
return seedobj[1] / (seedobjm - 1)
end
-- Store some test values in variables.
my_seed_value = newseed(230951)
my_random_value_1 = srandom(my_seed_value)
my_random_value_2 = srandom(my_seed_value)
my_random_value_3 = srandom(my_seed_value)
-- Print the values to console.
print(my_random_value_1)
print(my_random_value_2)
print(my_random_value_3)