Develop Reference

In javascript, how do I add a random amount to a user's balance while controlling how much gets given total?

I'm trying to make it to where when a user does a certain thing, they get between 2 and 100 units. But for every 1,000 requests I want it to add up to 3,500 units given collectively.
Here's the code I have for adding different amounts randomly to a user:
if (Math.floor(Math.random() * 1000) + 1 === 900) {
//db call adding 100
}
else if (Math.floor(Math.random() * 100) + 1 === 90) {
//db call adding 40
}
else if (Math.floor(Math.random() * 30) + 1 === 20) {
//db call adding 10
}
else if (Math.floor(Math.random() * 5) + 1 === 4) {
//db call adding 5
}
else {
//db call adding 2
}
If my math is correct, this should average around 4,332 units per 1,000 calls. But obviously it would vary and I don't want that. I'd also like it to add random amounts instead, as the units added in my example are arbitrary.
EDIT: Guys, Gildor is right that I simply want to have 3,500 units, and give them away within 1,000 requests. It isn't even entirely necessary that it always reaches that maximum of 3,500 either (I could have specified that). The important thing is that I'm not giving users too much, while creating a chance for them to win a bigger amount.
Here's what I have set up now, and it's working well, and will work even better with some tweaking:
Outside of call:
var remaining = 150;
var count = 0;
Inside of call:
count += 1;
if (count === 100) {
remaining = 150;
count = 0;
}
if (Math.floor(Math.random() * 30) + 1 === 20) {
var addAmount = Math.floor(Math.random() * 85) + 15;
if (addAmount <= remaining) {
remaining -= addAmount;
//db call adding addAmount + 2
}
else {
//db call adding 2
}
}
else if (Math.floor(Math.random() * 5) + 1 === 4) {
var addAmount1 = Math.floor(Math.random() * 10) + 1;
if (addAmount1 <= remaining) {
remaining -= addAmount1;
//db call adding addAmount1 + 2
}
else {
//db call adding 2
}
}
else {
//db call adding 2
}
I guess I should have clarified, I want a "random" number with a high likelihood of being small. That's kind of part of the gimmick, where you have low probability of getting a larger amount.

As I've commented, 1,000 random numbers between 2 and 100 that add up to 3,500 is an average number of 3.5 which is not consistent with random choices between 2 and 100. You'd have to have nearly all 2 and 3 values in order to achieve that and, in fact couldn't have more than a couple large numbers. Nothing even close to random. So, for this to even be remotely random and feasible, you'd have to pick a total much large than 3,500. A random total of 1,000 numbers between 2 and 100 would be more like 51,000.
Furthermore, you can't dynamically generate each number in a truly random fashion and guarantee a particular total. The main way to guarantee that outcome is to pre-allocate random numbers that add up to the total that are known to achieve that and then random select each number from the pre-allocated scheme, then remove that from the choice for future selections.
You could also try to keep a running total and bias your randomness if you get skewed away form your total, but doing it that way, the last set of numbers may have to be not even close to random in order to hit your total consistently.
A scheme that could work if you reset the total to support what it should be for actual randomness (e.g. to 51,000) would be to preallocated an array of 500 random numbers between 2 and 100 and then add another 500 numbers that are the complements of those. This guarantees the 51 avg number. You can then select each number randomly from the pre-allocated array and then remove it form the array so it won't be selected again. I can add code to do this in a second.
function RandResults(low, high, qty) {
var results = new Array(qty);
var limit = qty/2;
var avg = (low + high) / 2;
for (var i = 0; i < limit; i++) {
results[i] = Math.floor((Math.random() * (high - low)) + low);
//
results[qty - i - 1] = (2 * avg) - results[i];
}
this.results = results;
}
RandResults.prototype.getRand = function() {
if (!this.results.length) {
throw new Error("getRand() called, but results are empty");
}
var randIndex = Math.floor(Math.random() * this.results.length);
var value = this.results[randIndex];
this.results.splice(randIndex, 1);
return value;
}
RandResults.prototype.getRemaining = function() {
return this.results.length;
}
var randObj = new RandResults(2, 100, 1000);
// get next single random value
if (randObj.getRemaining()) {
var randomValue = randObj.getRand();
}
Working demo for a truly random selection of numbers that add up to 51,000 (which is what 1,000 random values between 2 and 100 should add up to): http://jsfiddle.net/jfriend00/wga26n7p/
If what you want is the following: 1,000 numbers that add up to 3,500 and are selected from between the range 2 to 100 (inclusive) where most numbers will be 2 or 3, but occasionally something could be up to 100, then that's a different problem. I wouldn't really use the word random to describe it because it's a highly biased selection.
Here's a way to do that. It generates 1,000 random numbers between 2 and 100, keeping track of the total. Then, afterwards it corrects the random numbers to hit the right total by randomly selected values and decrementing them until the total is down to 3,500. You can see it work here: http://jsfiddle.net/jfriend00/m4ouonj4/
The main part of the code is this:
function RandResults(low, high, qty, total) {
var results = new Array(qty);
var runningTotal = 0, correction, index, trial;
for (var i = 0; i < qty; i++) {
runningTotal += results[i] = Math.floor((Math.random() * (high - low)) + low);
}
// now, correct to hit the total
if (runningTotal > total) {
correction = -1;
} else if (runningTotal < total) {
correction = 1;
}
// loop until we've hit the total
// randomly select a value to apply the correction to
while (runningTotal !== total) {
index = Math.floor(Math.random() * qty);
trial = results[index] + correction;
if (trial >= low && trial <= high) {
results[index] = trial;
runningTotal += correction;
}
}
this.results = results;
}
This meets an objective of a biased total of 3,500 and all numbers between 2 and 100, though the probability of a 2 in this scheme is very high and the probably of a 100 in this scheme is almost non-existent.
And, here's a weighted random generator that adds up to a precise total. This uses a cubic weighting scheme to favor the lower numbers (the probably of a number goes down with the cube of the number) and then after the random numbers are generated, a correction algorithm applies random corrections to the numbers to make the total come out exactly as specified. The code for a working demo is here: http://jsfiddle.net/jfriend00/g6mds8rr/
function RandResults(low, high, numPicks, total) {
var avg = total / numPicks;
var i, j;
// calculate probabilities for each value
// by trial and error, we found that a cubic weighting
// gives an approximately correct sub-total that can then
// be corrected to the exact total
var numBuckets = high - low + 1;
var item;
var probabilities = [];
for (i = 0; i < numBuckets; i++) {
item = low + i;
probabilities[i] = avg / (item * item * item);
}
// now using those probabilities, create a steps array
var sum = 0;
var steps = probabilities.map(function(item) {
sum += item;
return sum;
});
// now generate a random number and find what
// index it belongs to in the steps array
// and use that as our pick
var runningTotal = 0, rand;
var picks = [], pick, stepsLen = steps.length;
for (i = 0; i < numPicks; i++) {
rand = Math.random() * sum;
for (j = 0; j < stepsLen; j++) {
if (steps[j] >= rand) {
pick = j + low;
picks.push(pick);
runningTotal += pick;
break;
}
}
}
var correction;
// now run our correction algorithm to hit the total exactly
if (runningTotal > total) {
correction = -1;
} else if (runningTotal < total) {
correction = 1;
}
// loop until we've hit the total
// randomly select a value to apply the correction to
while (runningTotal !== total) {
index = Math.floor(Math.random() * numPicks);
trial = picks[index] + correction;
if (trial >= low && trial <= high) {
picks[index] = trial;
runningTotal += correction;
}
}
this.results = picks;
}
RandResults.prototype.getRand = function() {
if (!this.results.length) {
throw new Error("getRand() called, but results are empty");
}
return this.results.pop();
}
RandResults.prototype.getAllRand = function() {
if (!this.results.length) {
throw new Error("getAllRand() called, but results are empty");
}
var r = this.results;
this.results = [];
return r;
}
RandResults.prototype.getRemaining = function() {
return this.results.length;
}

As some comments pointed out... the numbers in the question does not quite make sense, but conceptually there are two approaches: calculate dynamically just in time or ahead of time.
To calculate just in time:
You can maintain a remaining variable which tracks how many of 3500 left. Each time when you randomly give some units, subtract the number from remaining until it goes to 0.
In addition, to make sure each time at least 2 units are given, you can start with remaining = 1500 and give random + 2 units each time.
To prevent cases that after 1000 gives there are still balances left, you may need to add some logic to give units more aggressively towards the last few times. However it will result in not-so-random results.
To calculate ahead of time:
Generate a random list with 1000 values in [2, 100] and sums up to 3500. Then shuffle the list. Each time you want to give some units, pick the next item in the array. After 1000 gives, generate another list in the same way. This way you get much better randomized results.
Be aware that both approaches requires some kind of shared state that needs to be handled carefully in a multi-threaded environment.
Hope the ideas help.

Javascript shift() always returning 0

I have a deck of cards object that essentially just fills a deck array with 0-51 (52 cards) and then shuffles the cards. I have a cards.deal() method that returns a card. When I use cards.pop() I always get a different number, which is what is expected; however, when I used cards.shift() it always returns 0. I'm content with using pop but would just like to understand for learning purposes why I'm always getting a 0 when I use shift. I've changed the number of shuffles to huge numbers like 10,000 just to make sure it's shuffling enough to actually swap the first element in the array. Could it be possible that the Math.ceil method is not returning a 0, even after 10,000 random numbers generated between 0 and 51? What is going on here?
function Deck(){
var cards = [];
this.shuffle = function(){
var tempCardHolder = 0;
var randomNumber = 0;
var randomNumber2 = 0;
// fill the card array with 52 cards
for (x = 0; x <= 51; x++){
cards[x] = x;
}
// now let's shuffle them up by swapping two cards randomly 50 times
for (y = 0; y < 100; y++){
// make sure we get two random numbers that are different
do {
randomNumber = Math.ceil(Math.random() * 51);
randomNumber2 = Math.ceil(Math.random() * 51);
} while (randomNumber == randomNumber2);
tempCardHolder = cards[randomNumber];
cards[randomNumber] = cards[randomNumber2];
cards[randomNumber2] = tempCardHolder;
}
}
this.deal = function(){
return cards.shift();
}
this.getNumCardsLeft = function(){
return cards.length;
}
}

Use Math.floor to round your numbers. For inclusive ranges, make sure you multiply one larger than your desired maximum value
Here is an example of a random integer generator for inclusive end points
function random_int(max, min) {
min = min || 0;
return min + Math.floor(Math.random() * (max - min + 1));
}

javascript - generate a new random number

I have a variable that has a number between 1-3.
I need to randomly generate a new number between 1-3 but it must not be the same as the last one.
It happens in a loop hundreds of times.
What is the most efficient way of doing this?

May the powers of modular arithmetic help you!!
This function does what you want using the modulo operator:
/**
* generate(1) will produce 2 or 3 with probablity .5
* generate(2) will produce 1 or 3 with probablity .5
* ... you get the idea.
*/
function generate(nb) {
rnd = Math.round(Math.random())
return 1 + (nb + rnd) % 3
}
if you want to avoid a function call, you can inline the code.

Here is a jsFiddle that solves your problem : http://jsfiddle.net/AsMWG/
I've created an array containing 1,2,3 and first I select any number and swap it with the last element. Then I only pick elements from position 0 and 1, and swap them with last element.

var x = 1; // or 2 or 3
// this generates a new x out of [1,2,3] which is != x
x = (Math.floor(2*Math.random())+x) % 3 + 1;

You can randomly generate numbers with the random number generator built in to javascript. You need to use Math.random().
If you're push()-ing into an array, you can always check if the previously inserted one is the same number, thus you regenerate the number. Here is an example:
var randomArr = [];
var count = 100;
var max = 3;
var min = 1;
while (randomArr.length < count) {
var r = Math.floor(Math.random() * (max - min) + min);
if (randomArr.length == 0) {
// start condition
randomArr.push(r);
} else if (randomArr[randomArr.length-1] !== r) {
// if the previous value is not the same
// then push that value into the array
randomArr.push(r);
}
}

As Widor commented generating such a number is equivalent to generating a number with probability 0.5. So you can try something like this (not tested):
var x; /* your starting number: 1,2 or 3 */
var y = Math.round(Math.random()); /* generates 0 or 1 */
var i = 0;
var res = i+1;
while (i < y) {
res = i+1;
i++;
if (i+1 == x) i++;
}

The code is tested and it does for what you are after.
var RandomNumber = {
lastSelected: 0,
generate: function() {
var random = Math.floor(Math.random()*3)+1;
if(random == this.lastSelected) {
generateNumber();
}
else {
this.lastSelected = random;
return random;
}
}
}
RandomNumber.generate();

Generating unique random numbers (integers) between 0 and 'x'

I need to generate a set of unique (no duplicate) integers, and between 0 and a given number.
That is:
var limit = 10;
var amount = 3;
How can I use Javascript to generate 3 unique numbers between 1 and 10?

Use the basic Math methods:
Math.random() returns a random number between 0 and 1 (including 0, excluding 1).
Multiply this number by the highest desired number (e.g. 10)
Round this number downward to its nearest integer
Math.floor(Math.random()*10) + 1
Example:
//Example, including customisable intervals [lower_bound, upper_bound)
var limit = 10,
amount = 3,
lower_bound = 1,
upper_bound = 10,
unique_random_numbers = [];
if (amount > limit) limit = amount; //Infinite loop if you want more unique
//Natural numbers than exist in a
// given range
while (unique_random_numbers.length < limit) {
var random_number = Math.floor(Math.random()*(upper_bound - lower_bound) + lower_bound);
if (unique_random_numbers.indexOf(random_number) == -1) {
// Yay! new random number
unique_random_numbers.push( random_number );
}
}
// unique_random_numbers is an array containing 3 unique numbers in the given range

Math.floor(Math.random() * (limit+1))
Math.random() generates a floating point number between 0 and 1, Math.floor() rounds it down to an integer.
By multiplying it by a number, you effectively make the range 0..number-1. If you wish to generate it in range from num1 to num2, do:
Math.floor(Math.random() * (num2-num1 + 1) + num1)
To generate more numbers, just use a for loop and put results into an array or write them into the document directly.

function generateRange(pCount, pMin, pMax) {
min = pMin < pMax ? pMin : pMax;
max = pMax > pMin ? pMax : pMin;
var resultArr = [], randNumber;
while ( pCount > 0) {
randNumber = Math.round(min + Math.random() * (max - min));
if (resultArr.indexOf(randNumber) == -1) {
resultArr.push(randNumber);
pCount--;
}
}
return resultArr;
}
Depending on range needed the method of returning the integer can be changed to: ceil (a,b], round [a,b], floor [a,b), for (a,b) is matter of adding 1 to min with floor.

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

for(i = 0;i <amount; i++)
{
var randomnumber=Math.floor(Math.random()*limit)+1
document.write(randomnumber)
}

Here’s another algorithm for ensuring the numbers are unique:
generate an array of all the numbers from 0 to x
shuffle the array so the elements are in random order
pick the first n
Compared to the method of generating random numbers until you get a unique one, this method uses more memory, but it has a more stable running time – the results are guaranteed to be found in finite time. This method works better if the upper limit is relatively low or if the amount to take is relatively high.
My answer uses the Lodash library for simplicity, but you could also implement the algorithm described above without that library.
// assuming _ is the Lodash library
// generates `amount` numbers from 0 to `upperLimit` inclusive
function uniqueRandomInts(upperLimit, amount) {
var possibleNumbers = _.range(upperLimit + 1);
var shuffled = _.shuffle(possibleNumbers);
return shuffled.slice(0, amount);
}

Something like this
var limit = 10;
var amount = 3;
var nums = new Array();
for(int i = 0; i < amount; i++)
{
var add = true;
var n = Math.round(Math.random()*limit + 1;
for(int j = 0; j < limit.length; j++)
{
if(nums[j] == n)
{
add = false;
}
}
if(add)
{
nums.push(n)
}
else
{
i--;
}
}

var randomNums = function(amount, limit) {
var result = [],
memo = {};
while(result.length < amount) {
var num = Math.floor((Math.random() * limit) + 1);
if(!memo[num]) { memo[num] = num; result.push(num); };
}
return result; }
This seems to work, and its constant lookup for duplicates.

These answers either don't give unique values, or are so long (one even adding an external library to do such a simple task).
1. generate a random number.
2. if we have this random already then goto 1, else keep it.
3. if we don't have desired quantity of randoms, then goto 1.
function uniqueRandoms(qty, min, max){
var rnd, arr=[];
do { do { rnd=Math.floor(Math.random()*max)+min }
while(arr.includes(rnd))
arr.push(rnd);
} while(arr.length<qty)
return arr;
}
//generate 5 unique numbers between 1 and 10
console.log( uniqueRandoms(5, 1, 10) );
...and a compressed version of the same function:
function uniqueRandoms(qty,min,max){var a=[];do{do{r=Math.floor(Math.random()*max)+min}while(a.includes(r));a.push(r)}while(a.length<qty);return a}

/**
* Generates an array with numbers between
* min and max randomly positioned.
*/
function genArr(min, max, numOfSwaps){
var size = (max-min) + 1;
numOfSwaps = numOfSwaps || size;
var arr = Array.apply(null, Array(size));
for(var i = 0, j = min; i < size & j <= max; i++, j++) {
arr[i] = j;
}
for(var i = 0; i < numOfSwaps; i++) {
var idx1 = Math.round(Math.random() * (size - 1));
var idx2 = Math.round(Math.random() * (size - 1));
var temp = arr[idx1];
arr[idx1] = arr[idx2];
arr[idx2] = temp;
}
return arr;
}
/* generating the array and using it to get 3 uniques numbers */
var arr = genArr(1, 10);
for(var i = 0; i < 3; i++) {
console.log(arr.pop());
}

I think, this is the most human approach (with using break from while loop), I explained it's mechanism in comments.
function generateRandomUniqueNumbersArray (limit) {
//we need to store these numbers somewhere
const array = new Array();
//how many times we added a valid number (for if statement later)
let counter = 0;
//we will be generating random numbers until we are satisfied
while (true) {
//create that number
const newRandomNumber = Math.floor(Math.random() * limit);
//if we do not have this number in our array, we will add it
if (!array.includes(newRandomNumber)) {
array.push(newRandomNumber);
counter++;
}
//if we have enought of numbers, we do not need to generate them anymore
if (counter >= limit) {
break;
}
}
//now hand over this stuff
return array;
}
You can of course add different limit (your amount) to the last 'if' statement, if you need less numbers, but be sure, that it is less or equal to the limit of numbers itself - otherwise it will be infinite loop.

Just as another possible solution based on ES6 Set ("arr. that can contain unique values only").
Examples of usage:
// Get 4 unique rnd. numbers: from 0 until 4 (inclusive):
getUniqueNumbersInRange(4, 0, 5) //-> [5, 0, 4, 1];
// Get 2 unique rnd. numbers: from -1 until 2 (inclusive):
getUniqueNumbersInRange(2, -1, 2) //-> [1, -1];
// Get 0 unique rnd. numbers (empty result): from -1 until 2 (inclusive):
getUniqueNumbersInRange(0, -1, 2) //-> [];
// Get 7 unique rnd. numbers: from 1 until 7 (inclusive):
getUniqueNumbersInRange(7, 1, 7) //-> [ 3, 1, 6, 2, 7, 5, 4];
The implementation:
function getUniqueNumbersInRange(uniqueNumbersCount, fromInclusive, untilInclusive) {
// 0/3. Check inputs.
if (0 > uniqueNumbersCount) throw new Error('The number of unique numbers cannot be negative.');
if (fromInclusive > untilInclusive) throw new Error('"From" bound "' + fromInclusive
+ '" cannot be greater than "until" bound "' + untilInclusive + '".');
const rangeLength = untilInclusive - fromInclusive + 1;
if (uniqueNumbersCount > rangeLength) throw new Error('The length of the range is ' + rangeLength + '=['
+ fromInclusive + '…' + untilInclusive + '] that is smaller than '
+ uniqueNumbersCount + ' (specified count of result numbers).');
if (uniqueNumbersCount === 0) return [];
// 1/3. Create a new "Set" – object that stores unique values of any type, whether primitive values or object references.
// MDN - https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set
// Support: Google Chrome 38+(2014.10), Firefox 13+, IE 11+
const uniqueDigits = new Set();
// 2/3. Fill with random numbers.
while (uniqueNumbersCount > uniqueDigits.size) {
// Generate and add an random integer in specified range.
const nextRngNmb = Math.floor(Math.random() * rangeLength) + fromInclusive;
uniqueDigits.add(nextRngNmb);
}
// 3/3. Convert "Set" with unique numbers into an array with "Array.from()".
// MDN – https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/from
// Support: Google Chrome 45+ (2015.09+), Firefox 32+, not IE
const resArray = Array.from(uniqueDigits);
return resArray;
}
The benefits of the current implementation:
Have a basic check of input arguments – you will not get an unexpected output when the range is too small, etc.
Support the negative range (not only from 0), e. g. randoms from -1000 to 500, etc.
Expected behavior: the current most popular answer will extend the range (upper bound) on its own if input bounds are too small. An example: get 10000 unique numbers with a specified range from 0 until 10 need to throw an error due to too small range (10-0+1=11 possible unique numbers only). But the current top answer will hiddenly extend the range until 10000.

I wrote this C# code a few years back, derived from a Wikipedia-documented algorithm, which I forget now (feel free to comment...). Uniqueness is guaranteed for the lifetime of the HashSet. Obviously, if you will be using a database, you could store the generated numbers there. Randomness was ok for my needs, but probably can be improved using a different RNG. Note: count must be <= max - min (duh!) and you can easily modify to generate ulongs.
private static readonly Random RndGen = new Random();
public static IEnumerable<int> UniqueRandomIntegers(int count, int min, int max)
{
var rv = new HashSet<int>();
for (var i = max - min - count + 1; i <= max - min; i++)
{
var r = (int)(RndGen.NextDouble() * i);
var v = rv.Contains(r) ? i : r;
rv.Add(v);
yield return v;
}
}

Randomized Array, Sliced
Similar to #rory-okane's answer, but without lodash.
Both Time Complexity and Space Complexity = O(n) where n=limit
Has a consistent runtime
Supports a positive or negative range of numbers
Theoretically, this should support a range from 0 to ±2^32 - 1
This limit is due to Javascript arrays only supporting 2^32 - 1 indexes as per the ECMAScript specification
I stopped testing it at 10^8 because my browser got weird around here and strangely only negative numbers to -10^7 - I got an Uncaught RangeError: Invalid array length error (shrug)
Bonus feature: Generate a randomized array of n length 0 to limit if you pass only one argument
let uniqueRandomNumbers = (limit, amount = limit) => {
let array = Array(Math.abs(limit));
for (let i = 0; i < array.length; i++) array[i] = i * Math.sign(limit);
let currentIndex = array.length;
let randomIndex;
while(currentIndex > 0) {
randomIndex = Math.floor(Math.random() * currentIndex--);
[array[currentIndex], array[randomIndex]] = [array[randomIndex], array[currentIndex]];
}
return array.slice(0, Math.abs(amount));
}
console.log(uniqueRandomNumbers(10, 3));
console.log(uniqueRandomNumbers(-10, 3));
//bonus feature:
console.log(uniqueRandomNumbers(10));
Credit:
I personally got here because I was trying to generate random arrays of n length. Other SO questions that helped me arrive at this answer for my own use case are below. Thank you everyone for your contributions, you made my life better today.
Most efficient way to create a zero filled JavaScript array?
How to randomize (shuffle) a JavaScript array?
Also the answer from #ashleedawg is where I started, but when I discovered the infinite loop issues I ended up at the sliced randomized array approach.

const getRandomNo = (min, max) => {
min = Math.ceil(min);
max = Math.floor(max);
return Math.floor(Math.random() * (max - min + 1)) + min;
}
This function returns a random integer between the specified values. The value is no lower than min (or the next integer greater than min if min isn't an integer) and is less than (but not equal to) max.
Example
console.log(`Random no between 0 and 10 ${getRandomNo(0,10)}`)

Here's a simple, one-line solution:
var limit = 10;
var amount = 3;
randoSequence(1, limit).slice(0, amount);
It uses randojs.com to generate a randomly shuffled array of integers from 1 through 10 and then cuts off everything after the third integer. If you want to use this answer, toss this within the head tag of your HTML document:
<script src="https://randojs.com/1.0.0.js"></script>

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.