Hello,

I have a bit of programming dilemma which I cannot seem to solve my self.

What if I have an array of numbers and I want to randomly select entry from the array so that the higher the number in the array the more likely it gets picked?

For example array like this:

[1000,800,400,200]

Should have highest change of selecting the highest number (1000) and lowest change for the lowest number.

How to calculate this?

thanks!

Sort the numbers, then define a non-uniform probability distribution. (Search for that and similar terms for examples.)

A memory intensive way to go about it is:

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

#include <cstdlib>

int main(void)
{
    vector<long unsigned int> numbers;
    
    for(size_t i = 0; i < 1000; i++)
        numbers.push_back(1000);
    
    for(size_t i = 0; i < 800; i++)
        numbers.push_back(800);
    
    for(size_t i = 0; i < 400; i++)
        numbers.push_back(400);
    
    for(size_t i = 0; i < 200; i++)
        numbers.push_back(200);
    
    if(numbers.size() > RAND_MAX)
    {
        cout << "Need a better PRNG" << endl;
        return 0;
    }
    
    long unsigned int pseudo_random_number = numbers[rand() % numbers.size()];
    
    cout << pseudo_random_number << endl;
    
    return 1;
}

sjhalayka: You don't need to shuffle the array if you're picking from it randomly.

The main information missing from the question is a rigorous definition of the desired distribution. How much more likely should one number be picked over another, and what decides that?

Usually the simple approach here is to have an explicitly-weighted random selection. The standard algorithm for this is roulette wheel/fitness proportionate selection. (The Java 'rouletteSelect' function on that page is probably the most sensible implementation.)

Thank you Kylotan. I'll change the source code.

@Lactose I googled one article but no help yet from there, will continue looking.

@sjhalayka Thanks that works! I wonder if there is another approach that's not so heavy... 

@Kylotan The higher the number the higher the change of being picked.

It's still not clear - are the numbers you're selecting also the exact proportions that you want them to be picked in? Or are the proportions different to the numbers?

2 minutes ago, Kylotan said:

It's still not clear - are the numbers you're selecting also the exact proportions that you want them to be picked in? Or are the proportions different to the numbers?

proportions are different to the numbers. all I need is select just one number from the array as long as it's random.

Right, so what I'm saying (and what Lactose said) is that you need to specify those proportions somehow. 'Higher' isn't enough information - how much higher, in each case? The roulette wheel selection method I linked to allows you to choose those proportions explicitly.

Ok here's the deal, I have been working on code to do this but it doesn't seem right, I'm unsure if I got the math right.

I made sample program where there's no arrays but I'm trying to calculate the probability for each number. 100 / 100 for example should have 50% probability.

Or in another example max number being 100 the number 10 should have 10% change of being picked.

(max number just means the highest number found in array)

here's the program:

// Example program
#include <iostream>
#include <string>

using namespace std;

int maxNum = 100;

int main()
{
 for(float i = 0; i <= maxNum; i += 10)
 {
     float change = i / ((float)maxNum) / 2.0;
     
     cout << (int)i << " / " << maxNum << " = " << change << " " << ( int(100 * change)) << "% " << endl;
 }
}

Which outputs:

0 / 100 = 0 0% 
10 / 100 = 0.05 5% 
20 / 100 = 0.1 10% 
30 / 100 = 0.15 15% 
40 / 100 = 0.2 20% 
50 / 100 = 0.25 25% 
60 / 100 = 0.3 30% 
70 / 100 = 0.35 35% 
80 / 100 = 0.4 40% 
90 / 100 = 0.45 45% 
100 / 100 = 0.5 50% 

where the final (100 / 100) seems to be correct result but the five first calculations doesn't seem right. 10 / 100 should be 10% but I'm not sure if I got that part right.

any ideas how to make this work?