I have a loot box that I want to fill with a random item. But I want each item to have a different chance of being picked. For example:

  • 5% chance of 10 gold
  • 20% chance of sword
  • 45% chance of shield
  • 20% chance of armor
  • 10% chance of potion

How can I make it so that I select exactly one of the items above, where those percentages are the respective chances of getting the loot?


The Soft-coded Probabilities Solution

The hardcoded probability solution has the disadvantage that you need to set the probabilities in your code. You can't determine them at runtime. It is also hard to maintain.

Here is a dynamic version of the same algorithm.

  1. Create an array of pairs of actual items and weight of each item
  2. When you add an item, the weight of the item needs to be its own weight plus the sum of the weights of all items already in the array. So you should track the sum separately. Especially because you will need it for the next step.
  3. To retrieve an object, generate a random number between 0 and the sum of the weights of all items
  4. iterate the array from start to finish until you found an entry with a weight larger or equal than the random number

Here is a sample implementation in Java in form of a template class which you can instantiate for any object your game uses. You can then add objects with the method .addEntry(object, relativeWeight) and pick one of the entries you added previously with .get()

import java.util.ArrayList;
import java.util.List;
import java.util.Random;

public class WeightedRandomBag<T extends Object> {

    private class Entry {
        double accumulatedWeight;
        T object;

    private List<Entry> entries = new ArrayList<>();
    private double accumulatedWeight;
    private Random rand = new Random();

    public void addEntry(T object, double weight) {
        accumulatedWeight += weight;
        Entry e = new Entry();
        e.object = object;
        e.accumulatedWeight = accumulatedWeight;

    public T getRandom() {
        double r = rand.nextDouble() * accumulatedWeight;

        for (Entry entry: entries) {
            if (entry.accumulatedWeight >= r) {
                return entry.object;
        return null; //should only happen when there are no entries


WeightedRandomBag<String> itemDrops = new WeightedRandomBag<>();

// Setup - a real game would read this information from a configuration file or database
itemDrops.addEntry("10 Gold",  5.0);
itemDrops.addEntry("Sword",   20.0);
itemDrops.addEntry("Shield",  45.0);
itemDrops.addEntry("Armor",   20.0);
itemDrops.addEntry("Potion",  10.0);

// drawing random entries from it
for (int i = 0; i < 20; i++) {

Here is the same class implemented in C# for your Unity, XNA or MonoGame project:

using System;
using System.Collections.Generic;

class WeightedRandomBag<T>  {

    private struct Entry {
        public double accumulatedWeight;
        public T item;

    private List<Entry> entries = new List<Entry>();
    private double accumulatedWeight;
    private Random rand = new Random();

    public void AddEntry(T item, double weight) {
        accumulatedWeight += weight;
        entries.Add(new Entry { item = item, accumulatedWeight = accumulatedWeight });

    public T GetRandom() {
        double r = rand.NextDouble() * accumulatedWeight;

        foreach (Entry entry in entries) {
            if (entry.accumulatedWeight >= r) {
                return entry.item;
        return default(T); //should only happen when there are no entries

And here is one in JavaScript:

var WeightedRandomBag = function() {

    var entries = [];
    var accumulatedWeight = 0.0;

    this.addEntry = function(object, weight) {
        accumulatedWeight += weight;
        entries.push( { object: object, accumulatedWeight: accumulatedWeight });

    this.getRandom = function() {
        var r = Math.random() * accumulatedWeight;
        return entries.find(function(entry) {
            return entry.accumulatedWeight >= r;


  • Can handle any weight ratios. You can have items with astronomically small probability in the set if you want to. The weights also don't need to add up to 100.
  • You can read the items and weights at runtime
  • Memory usage proportional to the number of items in the array


  • Requires some more programming to get right
  • In the worst case, you might have to iterate the whole array (O(n) runtime complexity). So when you have a very large set of items and draw very often, it might become slow. A simple optimization is to put the most probable items first so the algorithm terminates early in most cases. A more complex optimization you can do is to exploit the fact that the array is sorted and do a bisection search. This only takes O(log n) time.
  • You need to build the list in memory before you can use it (although you can easily add items at runtime. Removing items could also be added, but that would require to update the accumulated weights of all items which come after the removed entry, which again has O(n) worst case runtime)
  • 2
    \$\begingroup\$ The C# code could be written using LINQ: return entries.FirstOrDefault(e => e.accumulatedWeight >= r). More importantly, there is a slight possibility that due to floating point precision loss this algorithm will return null if the random value gets just a tiny bit greater than the accumulated value. As a precaution, you might add a small value (say, 1.0) to the last element, but then you would have to explicitly state in your code that the list is final. \$\endgroup\$ – IMil Aug 20 '18 at 21:29
  • 1
    \$\begingroup\$ One small variant on this I've used personally, if you want the weight values in runtime to not be changed to the weight-plus-all-previous value, you can subtract the weight of each passed entry from your random value, stopping when the random value is less than the current items weight (or when subtracting the weight makes the value < 0) \$\endgroup\$ – Lunin Aug 21 '18 at 1:36
  • 2
    \$\begingroup\$ @BlueRaja-DannyPflughoeft premature optimization... the question was about selecting an object from an open loot box. Who is going to open 1000 boxes per second? \$\endgroup\$ – IMil Aug 21 '18 at 5:51
  • 4
    \$\begingroup\$ @IMil: No, the question is a general catch-all for selecting random weighted items. For lootboxes specifically, this answer is probably fine because there are a small number of items and the probabilities don't change (though, since those are usually done on a server, 1000/sec is not unrealistic for a popular game). \$\endgroup\$ – BlueRaja - Danny Pflughoeft Aug 21 '18 at 8:59
  • 4
    \$\begingroup\$ @opa then flag to close as a dupe. Is it really wrong to upvote a good answer just because the question has been asked before? \$\endgroup\$ – Baldrickk Aug 21 '18 at 15:03

Note: I created a C# library for this exact problem

The other solutions are fine if you only have a small number of items and your probabilities never change. However, with lots of items or changing probabilities (ex. removing items after selecting them), you'll want something more powerful.

Here are the two most common solutions (both of which are included in the above library)

Walker's Alias Method

A clever solution that's extremely fast (O(1)!) if your probabilities are constant. In essence, the algorithm creates a 2D dartboard ("alias table") out of your probabilities and throws a dart at it.


There are plenty of articles online about how it works if you'd like to learn more.

The only issue is that if your probabilities change, you need to regenerate the alias table, which is slow. Thus, if you need to remove items after they're picked, this is not the solution for you.

Tree-based solution

The other common solution is to make an array where each item stores the sum of its probability and all the items before it. Then just generate a random number from [0,1) and do a binary search for where that number lands in the list.

This solution is very easy to code/understand, but making a selection is slower than Walker's Alias Method, and changing the probabilities is still O(n). We can improve it by turning the array into a binary-search tree, where each node keeps track of the sum-of-probabilities in all the items in its subtree. Then when we generate the number from [0,1), we can just walk down the tree to find the item it represents.

This gives us O(log n) to pick an item and to change the probabilities! This makes NextWithRemoval() extremely fast!

The results

Here are some quick benchmarks from the above library, comparing these two approaches

         WeightedRandomizer Benchmarks                  |    Tree    |    Table
Add()x10000 + NextWithReplacement()x10:                 |    4 ms    |      2 ms
Add()x10000 + NextWithReplacement()x10000:              |    7 ms    |      4 ms
Add()x10000 + NextWithReplacement()x100000:             |   35 ms    |     28 ms
( Add() + NextWithReplacement() )x10000 (interleaved)   |    8 ms    |   5403 ms
Add()x10000 + NextWithRemoval()x10000:                  |   10 ms    |   5948 ms

So as you can see, for the special case of static (non-changing) probabilities, Walker's Alias method is about 50-100% faster. But in the more dynamic cases, the tree is several orders of magnitude faster!

  • \$\begingroup\$ The tree-based solution also gives us a decent run-time (nlog(n)) when sorting items by weight. \$\endgroup\$ – Nathan Merrill Aug 21 '18 at 8:01
  • 2
    \$\begingroup\$ I'm skeptical of your results, but this is the correct answer. Not sure why this isn't the top answer, considering this is actually the canonical way to handle this problem. \$\endgroup\$ – hythis Aug 21 '18 at 13:47
  • \$\begingroup\$ Which file contains the tree based solution? Second, your benchmark table: is Walker's Alias the "table" column? \$\endgroup\$ – Yakk Aug 21 '18 at 15:31
  • 1
    \$\begingroup\$ @Yakk: The code for the tree-based solution is here. It's built upon an open-source implementation of an AA-tree. And 'yes' to your second question. \$\endgroup\$ – BlueRaja - Danny Pflughoeft Aug 21 '18 at 16:13
  • 1
    \$\begingroup\$ The Walker part is pretty just link-only. \$\endgroup\$ – Acccumulation Aug 22 '18 at 20:22

The Wheel of Fortune solution

You can use this method when the probabilities in your item pool have a rather large common denominator and you need to draw from it very often.

Create an array of options. But put each element into it multiple times, with the number of duplicates of each element proportional to its chance of appearing. For the example above, all elements have probabilities which are multipliers of 5%, so you can create an array of 20 elements like this:

10 gold

Then simply pick a random element of that list by generating one random integer between 0 and the length of the array - 1.


  • You need to build the array the first time you want to generate an item.
  • When one of your elements is supposed to have a very low probability, you end up with a really large array, which can require a lot of memory.


  • When you already have the array and want to draw from it multiple times, then it is very fast. Just one random integer and one array access.
  • 3
    \$\begingroup\$ As a hybrid solution to avoid the second disadvantage, you can designate the last slot as "other," and handle it via other means, such as Philipp's array approach. Thus you might fill that last slot with an array offering a 99.9% chance of a potion, and just a 0.1% chance of an Epic Scepter of the Apocalypse. Such a two tiered approach leverages the advantages of both approaches. \$\endgroup\$ – Cort Ammon Aug 20 '18 at 21:04
  • 1
    \$\begingroup\$ I use somewhat a variation of this in my own project. What I do is calculate each item & weight, and store those in an array, [('gold', 1),('sword',4),...], sum up all of the weights, and then roll a random number from 0 to the sum, then iterate the array and calculate where the random number lands (ie a reduce). Works fine for arrays that are updated often, and no major memory hog. \$\endgroup\$ – user39686 Aug 21 '18 at 0:57
  • 1
    \$\begingroup\$ @Thebluefish That solution is described in my other answer "The Soft-coded Probabilities Solution" \$\endgroup\$ – Philipp Aug 21 '18 at 12:06

The Hard-coded Probabilities Solution

The most simple way find a random item from a weighted collection is to traverse down a chain of if-else statements, where each if-else increases in probably, as the previous one does not hit.

int rand = random(100); //Random number between 1 and 100 (inclusive)
if(rand <= 5) //5% chance
    print("You found 10 gold!");
else if(rand <= 25) //20% chance
    print("You found a sword!");
else if(rand <= 70) //45% chance
    print("You found a shield!");
else if(rand <= 90) //20% chance
    print("You found armor!");
else //10% chance
    print("You found a potion!");

The reason the conditionals are equal to its chance plus all of the previous conditionals chances is because the previous conditionals have already eliminated the possibility of it being those items. So for the shield's conditional else if(rand <= 70), 70 is equal to the 45% chance of the shield, plus the 5% chance of the gold and 20% chance of the sword.


  • Easy to program, because it requires no data structures.


  • Hard to maintain, because you need to maintain your drop-rates in your code. You can't determine them at runtime. So if you want something more future proof, you should check the other answers.
  • 3
    \$\begingroup\$ This would be really annoying to maintain. E.g. if you wish to remove gold, and make potion takes its spot, you need to adjust the probabilities of all items between them. \$\endgroup\$ – Alexander Aug 20 '18 at 19:39
  • 1
    \$\begingroup\$ To avoid the issue that @Alexander mentions, you can instead subtract the current rate at each step, instead of adding it to each condition. \$\endgroup\$ – Alex Jones Aug 21 '18 at 4:53

In C# you could use a Linq scan to run your accumulator to check against a random number in the range 0 to 100.0f and .First() to get. So like one line of code.

So something like:

var item = a.Select(x =>
    sum += x.prob;
    if (rand < sum)
        return x.item;
        return null;

sum is a zero initialized integer and a is a list of prob/item structs/tuples/instances. rand is a previously generated random number in the range.

This simply accumulates the sum over the list of ranges until it exceeds the previously selected random number, and returns either the item or null, where null would be returned if the random number range (e.g. 100) is less than the total weighting range by mistake, and the random number selected is outside the total weighting range.

However, you will notice that weights in OP closely match a normal distribution (Bell Curve). I think in general you will not want specific ranges, you will tend to want a distribution that tapers off either around a bell curve or just on a decreasing exponential curve (for example). In this case you could just use a mathematical formula to generate an index into an array of items, sorted in order of preferred probability. A good example is CDF in normal distribution

Also an example here.

Another example is that you could take a random value from 90 degrees to 180 degrees to get the lower right quadrant of a circle, take the x component using cos(r) and use that to index into a prioritized list.

With different formulae you could have a general approach where you just input a prioritized list of any length (e.g. N) and map the outcome of the formula (e.g.: cos(x) is 0 to 1) by multiplication (e.g.: Ncos(x) = 0 to N) to get the index.

  • 3
    \$\begingroup\$ Could you give us this line of code if it's just one line? I'm not as familiar with C# so I don't know what you mean. \$\endgroup\$ – HEGX64 Aug 21 '18 at 1:33
  • \$\begingroup\$ @HEGX64 added but using mobile and editor not working. Can you edit? \$\endgroup\$ – Sentinel Aug 21 '18 at 8:50
  • 4
    \$\begingroup\$ Can you change this answer to explain the concept behind it, rather than a specific imlementation in a specific language? \$\endgroup\$ – Raimund Krämer Aug 21 '18 at 9:39
  • \$\begingroup\$ @RaimundKrämer Erm, done? \$\endgroup\$ – Sentinel Aug 21 '18 at 21:05
  • \$\begingroup\$ Downvote without explanation = useless and antisocial. \$\endgroup\$ – WGroleau Aug 22 '18 at 22:40

Probabilities don’t need to be hard-coded. The items and the thresholds can be together in an array.

for X in items’range loop
  If items (X).threshold < random() then
     Announce (items(X).name)
     Exit loop
  End if
End loop

You do have to accumulate the thresholds still, but you can do it when creating a parameter file instead of coding it.

  • 3
    \$\begingroup\$ Could you elaborate on how to calculate the correct threshold? For example, if you have three items with 33% chance each, how would you build this table? Since a new random() is generated each time, the first would need 0.3333, the second would need 0.5 and the last would need 1.0. Or did I read the algorithm wrong? \$\endgroup\$ – pipe Aug 22 '18 at 7:02
  • \$\begingroup\$ You compute it the way others did in their answers. For equal probabilities of X items, the first threshold is 1/X, the second, 2/X, etc. \$\endgroup\$ – WGroleau Aug 22 '18 at 18:43
  • \$\begingroup\$ Doing that for 3 items in this algorithm would make the thresholds 1/3, 2/3 and 3/3 but the outcome probabilities 1/3, 4/9 and 2/9 for the first, second and third item. Do you really mean to have the call to random() in the loop? \$\endgroup\$ – pipe Aug 23 '18 at 5:54
  • \$\begingroup\$ No, that's definitely a bug. Each check needs the same random number. \$\endgroup\$ – WGroleau Aug 23 '18 at 21:34

I done this function: https://github.com/thewheelmaker/GDscript_Weighted_Random Now! in your case you can use it like this:


It gives just a number between 0 to 4 but you can put it in array where you got the items.


Or in function:


Here is the code. I made it on GDscript, you can, but it can alter other language, also check for logic errors:

func on_normal_case(arrayy,transformm):
    var random_num=0
    var sum=0
    var summatut=0
    #func sumarrays_inarray(array):
    for i in range(arrayy.size()):
#func no_fixu_random_num(here_range,start_from):
#Randomies be pressed down
#first start from zero
    if 0<=random_num and random_num<=arrayy[0]:
        return 0+ transformm
    for i in range(arrayy.size()-1):
        #they must pluss together
        #if array[i]<=random_num and random_num<array[i+1]:
        if summatut<random_num and random_num<=summatut+arrayy[i+1]:
            #return i+1+transform
            return i+1+ transformm


It works like this: on_normal_case([50,50],0) This gives 0 or 1, it has same probability both.

on_normal_case([50,50],1) This gives 1 or 2, it has same probability both.

on_normal_case([20,80],1) This gives 1 or 2, it has bigger change to get two.

on_normal_case([20,80,20,20,30],1) This give random numbers range 1-5 and bigger numbers are more likely than smaller numbers.

on_normal_case([20,80,0,0,20,20,30,0,0,0,0,33],45) This throw dices between numbers 45,46,49,50,51,56 you see when there is zero it never occure.

So it function returns just one random number that depends lenght of that arrayy array and transformm number, and ints in the array are probability weights that a number might occure, where that number is location on the array, pluss transformm number.


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