I'm trying to devise a (good) way to choose a random number from a range of possible numbers where each number in the range is given a weight. To put it simply: given the range of numbers (0,1,2) choose a number where 0 has an 80% probability of being selected, 1 has a 10% chance and 2 has a 10% chance.

It's been about 8 years since my college stats class, so you can imagine the proper formula for this escapes me at the moment.

Here's the 'cheap and dirty' method that I came up with. This solution uses ColdFusion. Yours may use whatever language you'd like. I'm a programmer, I think I can handle porting it. Ultimately my solution needs to be in Groovy - I wrote this one in ColdFusion because it's easy to quickly write/test in CF.

```
public function weightedRandom( Struct options ) {
var tempArr = [];
for( var o in arguments.options )
{
var weight = arguments.options[ o ] * 10;
for ( var i = 1; i<= weight; i++ )
{
arrayAppend( tempArr, o );
}
}
return tempArr[ randRange( 1, arrayLen( tempArr ) ) ];
}
// test it
opts = { 0=.8, 1=.1, 2=.1 };
for( x = 1; x<=10; x++ )
{
writeDump( weightedRandom( opts ) );
}
```

I'm looking for better solutions, please suggest improvements or alternatives.

Rejection sampling (such as in your solution) is the first thing that comes to mind, whereby you build a lookup table with elements populated by their weight distribution, then pick a random location in the table and return it. As an implementation choice, I would make a higher order function which takes a spec and returns a function which returns values based on the distribution in the spec, this way you avoid having to build the table for each call. The downsides are that the algorithmic performance of building the table is linear by the number of items and there could potentially be a lot of memory usage for large specs (or those with members with very small or precise weights, e.g. {0:0.99999, 1:0.00001}). The upside is that picking a value has constant time, which might be desirable if performance is critical. In JavaScript:

```
function weightedRand(spec) {
var i, j, table=[];
for (i in spec) {
// The constant 10 below should be computed based on the
// weights in the spec for a correct and optimal table size.
// E.g. the spec {0:0.999, 1:0.001} will break this impl.
for (j=0; j<spec[i]*10; j++) {
table.push(i);
}
}
return function() {
return table[Math.floor(Math.random() * table.length)];
}
}
var rand012 = weightedRand({0:0.8, 1:0.1, 2:0.1});
rand012(); // random in distribution...
```

Another strategy is to pick a random number in `[0,1)`

and iterate over the weight specification summing the weights, if the random number is less than the sum then return the associated value. Of course, this assumes that the weights sum to one. This solution has no up-front costs but has average algorithmic performance linear by the number of entries in the spec. For example, in JavaScript:

```
function weightedRand2(spec) {
var i, sum=0, r=Math.random();
for (i in spec) {
sum += spec[i];
if (r <= sum) return i;
}
}
weightedRand2({0:0.8, 1:0.1, 2:0.1}); // random in distribution...
```

Generate a random number R between 0 and 1.

If R in [0, 0.1) -> 1

If R in [0.1, 0.2) -> 2

If R in [0.2, 1] -> 3

If you can't directly get a number between 0 and 1, generate a number in a range that will produce as much precision as you want. For example, if you have the weights for

(1, 83.7%) and (2, 16.3%), roll a number from 1 to 1000. 1-837 is a 1. 838-1000 is 2.

Here are 3 solutions in javascript since I'm not sure which language you want it in. Depending on your needs one of the first two might work, but the the third one is probably the easiest to implement with large sets of numbers.

```
function randomSimple(){
return [0,0,0,0,0,0,0,0,1,2][Math.floor(Math.random()*10)];
}
function randomCase(){
var n=Math.floor(Math.random()*100)
switch(n){
case n<80:
return 0;
case n<90:
return 1;
case n<100:
return 2;
}
}
function randomLoop(weight,num){
var n=Math.floor(Math.random()*100),amt=0;
for(var i=0;i<weight.length;i++){
//amt+=weight[i]; *alternative method
//if(n<amt){
if(n<weight[i]){
return num[i];
}
}
}
weight=[80,90,100];
//weight=[80,10,10]; *alternative method
num=[0,1,2]
```

This is more or less a generic-ized version of what @trinithis wrote, in Java: I did it with ints rather than floats to avoid messy rounding errors.

```
static class Weighting {
int value;
int weighting;
public Weighting(int v, int w) {
this.value = v;
this.weighting = w;
}
}
public static int weightedRandom(List<Weighting> weightingOptions) {
//determine sum of all weightings
int total = 0;
for (Weighting w : weightingOptions) {
total += w.weighting;
}
//select a random value between 0 and our total
int random = new Random().nextInt(total);
//loop thru our weightings until we arrive at the correct one
int current = 0;
for (Weighting w : weightingOptions) {
current += w.weighting;
if (random < current)
return w.value;
}
//shouldn't happen.
return -1;
}
public static void main(String[] args) {
List<Weighting> weightings = new ArrayList<Weighting>();
weightings.add(new Weighting(0, 8));
weightings.add(new Weighting(1, 1));
weightings.add(new Weighting(2, 1));
for (int i = 0; i < 100; i++) {
System.out.println(weightedRandom(weightings));
}
}
```

I use the following

```
function weightedRandom(min, max) {
return Math.round(max / (Math.random() * max + min));
}
```

This is my go-to "weighted" random, where I use an inverse function of "x" (where x is a random between min and max) to generate a weighted result, where the minimum is the most heavy element, and the maximum the lightest (least chances of getting the result)

So basically, using `weightedRandom(1, 5)`

means the chances of getting a 1 are higher than a 2 which are higher than a 3, which are higher than a 4, which are higher than a 5.

Might not be useful for your use case but probably useful for people googling this same question.

After a 100 iterations try, it gave me:

```
==================
| Result | Times |
==================
| 1 | 55 |
| 2 | 28 |
| 3 | 8 |
| 4 | 7 |
| 5 | 2 |
==================
```

How about

int [ ] numbers = { 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1 , 2 } ;

then you can randomly select from numbers and 0 will have an 80% chance, 1 10%, and 2 10%

This one is in Mathematica, but it's easy to copy to another language, I use it in my games and it can handle decimal weights:

```
weights = {0.5,1,2}; // The weights
weights = [email protected]/[email protected] // Normalize weights so that the list's sum is always 1.
min = 0; // First min value should be 0
max = weights[[1]]; // First max value should be the first element of the newly created weights list. Note that in Mathematica the first element has index of 1, not 0.
random = RandomReal[]; // Generate a random float from 0 to 1;
For[i = 1, i <= [email protected], i++,
If[random >= min && random < max,
Print["Chosen index number: " <> [email protected]]
];
min += weights[[i]];
If[i == [email protected],
max = 1,
max += weights[[i + 1]]
]
]
```

*(Now I'm talking with a lists first element's index equals 0)* The idea behind this is that having a normalized list *weights* there is a chance of *weights[n]* to return the index *n*, so the distances between the min and max at step *n* should be *weights[n]*. The total distance from the minimum min *(which we put it to be 0)* and the maximum max is the sum of the list *weights*.

The good thing behind this is that you don't append to any array or nest for loops, and that increases heavily the execution time.

Here is the code in C# without needing to normalize the *weights* list and deleting some code:

```
int WeightedRandom(List<float> weights) {
float total = 0f;
foreach (float weight in weights) {
total += weight;
}
float max = weights [0],
random = Random.Range(0f, total);
for (int index = 0; index < weights.Count; index++) {
if (random < max) {
return index;
} else if (index == weights.Count - 1) {
return weights.Count-1;
}
max += weights[index+1];
}
return -1;
}
```

here is the input and ratios : 0 (80%), 1(10%) , 2 (10%)

lets draw them out so its easy to visualize.

```
0 1 2
-------------------------------------________+++++++++
```

lets add up the total weight and call it TR for total ratio. so in this case 100. lets randomly get a number from (0-TR) or (0 to 100 in this case) . 100 being your weights total. Call it RN for random number.

so now we have TR as the total weight and RN as the random number between 0 and TR.

so lets imagine we picked a random # from 0 to 100. Say 21. so thats actually 21%.

**WE MUST CONVERT/MATCH THIS TO OUR INPUT NUMBERS BUT HOW ?**

lets loop over each weight (80, 10, 10) and keep the sum of the weights we already visit. the moment the sum of the weights we are looping over is greater then the random number RN (21 in this case), we stop the loop & return that element position.

```
double sum = 0;
int position = -1;
for(double weight : weight){
position ++;
sum = sum + weight;
if(sum > 21) //(80 > 21) so break on first pass
break;
}
//position will be 0 so we return array[0]--> 0
```

lets say the random number (between 0 and 100) is 83. Lets do it again:

```
double sum = 0;
int position = -1;
for(double weight : weight){
position ++;
sum = sum + weight;
if(sum > 83) //(90 > 83) so break
break;
}
//we did two passes in the loop so position is 1 so we return array[1]---> 1
```

8 years late but here's my solution in 3 lines.

1) Prepare an array of **probability mass function** such that

pmf[array_index] = P(X=array_index):

```
var pmf = [0.8, 0.1, 0.1]
```

2) Prepare an array for the corresponding **cumulative distribution function** such that

cdf[array_index] = F(X=array_index):

```
var cdf = pmf.map((sum => value => sum += value)(0))
// [0.8, 0.9, 1]
```

3a) Generate a random number.

3b) Get an array of elements which are more than or equal to this number.

3c) Return its length.

```
cdf.filter(el => Math.random() >= el).length
```

I suggest to use a continuous check of the probability and the rest of the random number.

This function sets first the return value to the last possible index and iterates until the rest of the random value is smaller than the actual probability.

The probabilities have to sum to one.

```
function getRandomIndexByProbability(probabilities) {
var r = Math.random(),
index = probabilities.length - 1;
probabilities.some(function (probability, i) {
if (r < probability) {
index = i;
return true;
}
r -= probability;
});
return index;
}
var i,
probabilities = [0.8, 0.1, 0.1],
count = probabilities.map(function () { return 0; });
for (i = 0; i < 1e6; i++) {
count[getRandomIndexByProbability(probabilities)]++;
}
console.log(count);
```

`.as-console-wrapper { max-height: 100% !important; top: 0; }`

I have a slotmachine and I used the code below to generate random numbers. In probabilitiesSlotMachine the keys are the output in the slotmachine, and the values represent the weight.

```
const probabilitiesSlotMachine = [{0 : 1000}, {1 : 100}, {2 : 50}, {3 : 30}, {4 : 20}, {5 : 10}, {6 : 5}, {7 : 4}, {8 : 2}, {9 : 1}]
var allSlotMachineResults = []
probabilitiesSlotMachine.forEach(function(obj, index){
for (var key in obj){
for (var loop = 0; loop < obj[key]; loop ++){
allSlotMachineResults.push(key)
}
}
});
```

Now to generate a random output, I use this code:

```
const random = allSlotMachineResults[Math.floor(Math.random() * allSlotMachineResults.length)]
```

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