Generate A Weighted Random Number

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.

---

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 |
==================

---

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]

---

8 years late but here's my solution in 4 lines.

Prepare an array of probability mass function such that

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

var pmf = [0.8, 0.1, 0.1]

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 that are more than or equal to this number.

3c) Return its length.

var r = Math.random()
cdf.filter(el => r >= el).length

---

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));
    }
}

---

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 = N@weights/Total@weights // 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 <= Length@weights, i++,
    If[random >= min && random < max,
        Print["Chosen index number: " <> ToString@i]
    ];
    min += weights[[i]];
    If[i == Length@weights,
        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;
}

---

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; }

---

Thanks all, this was a helpful thread. I encapsulated it into a convenience function (Typescript). Tests below (sinon, jest). Could definitely be a bit tighter, but hopefully it's readable.

export type WeightedOptions = {
    [option: string]: number;
};

// Pass in an object like { a: 10, b: 4, c: 400 } and it'll return either "a", "b", or "c", factoring in their respective
// weight. So in this example, "c" is likely to be returned 400 times out of 414
export const getRandomWeightedValue = (options: WeightedOptions) => {
    const keys = Object.keys(options);
    const totalSum = keys.reduce((acc, item) => acc + options[item], 0);

    let runningTotal = 0;
    const cumulativeValues = keys.map((key) => {
        const relativeValue = options[key]/totalSum;
        const cv = {
            key,
            value: relativeValue + runningTotal
        };
        runningTotal += relativeValue;
        return cv;
    });

    const r = Math.random();
    return cumulativeValues.find(({ key, value }) => r <= value)!.key;
};

Tests:

describe('getRandomWeightedValue', () => {
    // Out of 1, the relative and cumulative values for these are:
    //      a: 0.1666   -> 0.16666
    //      b: 0.3333   -> 0.5
    //      c: 0.5      -> 1
    const values = { a: 10, b: 20, c: 30 };

    it('returns appropriate values for particular random value', () => {
        // any random number under 0.166666 should return "a"
        const stub1 = sinon.stub(Math, 'random').returns(0);
        const result1 = randomUtils.getRandomWeightedValue(values);
        expect(result1).toEqual('a');
        stub1.restore();

        const stub2 = sinon.stub(Math, 'random').returns(0.1666);
        const result2 = randomUtils.getRandomWeightedValue(values);
        expect(result2).toEqual('a');
        stub2.restore();

        // any random number between 0.166666 and 0.5 should return "b"
        const stub3 = sinon.stub(Math, 'random').returns(0.17);
        const result3 = randomUtils.getRandomWeightedValue(values);
        expect(result3).toEqual('b');
        stub3.restore();

        const stub4 = sinon.stub(Math, 'random').returns(0.3333);
        const result4 = randomUtils.getRandomWeightedValue(values);
        expect(result4).toEqual('b');
        stub4.restore();

        const stub5 = sinon.stub(Math, 'random').returns(0.5);
        const result5 = randomUtils.getRandomWeightedValue(values);
        expect(result5).toEqual('b');
        stub5.restore();

        // any random number above 0.5 should return "c"
        const stub6 = sinon.stub(Math, 'random').returns(0.500001);
        const result6 = randomUtils.getRandomWeightedValue(values);
        expect(result6).toEqual('c');
        stub6.restore();

        const stub7 = sinon.stub(Math, 'random').returns(1);
        const result7 = randomUtils.getRandomWeightedValue(values);
        expect(result7).toEqual('c');
        stub7.restore();
    });
});

---

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

---

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)]

---

Shortest solution in modern JavaScript

Note: all weights need to be integers

function weightedRandom(items){

  let table = Object.entries(items)
    .flatMap(([item, weight]) => Array(item).fill(weight))

  return table[Math.floor(Math.random() * table.length)]
}

const key = weightedRandom({
  "key1": 1,
  "key2": 4,
  "key3": 8
}) // returns e.g. "key1"

---