How to get random numbers to equal a specific number in javascript?

Question

Okay, so my problem is that I have three random number generators and I want the value of all three randomly generated numbers (generated from 1-5) to add up to 9. For example, I randomly generate a 4, a 3 and a 2. And on another click of the button, a 5, a 3 and a 1. I've tried for a while to do this but just can not figure it out.

        function statGen() {
            var x = document.getElementById("number");
            x.innerHTML = Math.floor((Math.random() * 5) + 1);
            var a = document.getElementById("agl");
            a.innerHTML = Math.floor((Math.random() * 4) + 1);
            var l = document.getElementById("lck");
            l.innerHTML = Math.floor((Math.random() * 3) + 1);
        }

 <button id="statbutton" onclick="statGen()">Get Numbers</button>       
        <p id="main">Main:</p>
        <p id="number"></p>
        <br/>
        <p id="Agl">Agl:</p>
        <p id="agl"></p>
        <br/>
        <p id="Lck">Lck:</p>
        <p id="lck"></p>

I have tried to run it through a loop multiple times until it was the number 9, but none of those worked.

Answer 1

All of the answers so far do not have good randomness properties. By choosing the first number, then the second, then the third, you have a large bias towards the extremes.

Let me illustrate with a dice example. Say you want to roll three dice where they add up to 8. If you roll them one at a time, there's a 1/6 chance that the permutation you end up with will be [6, 1, 1]. That's because after you've rolled the 6 (at 1/6 probability), the only acceptable remaining numbers are 1 and 1.

In reality, [6, 1, 1] is not nearly 1/6th of all acceptable rolls. For example, starting with a 3 gives [3, 1, 4], [3, 2, 3], [3, 3, 2], [3, 4, 1]. So having a 3 as the first number should be four times as likely as having a 6. But that's not the case with these other approaches!

So there's a few ways to get "good" randomness. The simplest is actually to just roll all three "dice", and if the results don't meet your constraint, roll them all again.

function roll() {
   return Math.floor(Math.random() * 5 + 1); //1-5 uniformly
}

function chooseNumbers() {
   var x = 0, y = 0, z = 0;

   while (x + y + z !== 9) {
      x = roll();
      y = roll();
      z = roll();
   }
   return [x, y, z];
}

This will be extremely fast for most purposes, but you do end up throwing out a decent amount of rolls. Another option is to enumerate (using code, or manually) all of the permutations you could have, and then choose an index using a uniform random number.

var permutations = [
   [5, 3, 1],
   [5, 2, 2],
   [5, 1, 3],
   [4, 4, 1],
   //...
   [1, 3, 5]
];

function chooseNumbers() {
   return permutations[Math.floor(Math.random() * permutations.length)];
}

Answer 2

last random number r3 depends on r1 and r2 so it's considered to be random.

    function statGen() {
        var x = document.getElementById("number");
        var r1=Math.floor((Math.random() * 9) + 1);
        var r2=Math.floor((Math.random() * (8-r1)) + 1);
        var r3=9-r1-r2;
        x.innerHTML = r1;
        var a = document.getElementById("agl");
        a.innerHTML = r2;
        var l = document.getElementById("lck");
        l.innerHTML = r3;
    }

Answer 3

Something like this? (This is pseudocode):

var first = Math.floor((Math.random() * 5) + 1);

var second = Math.floor((Math.random() * 8-first) + 1);

var third = Math.floor((Math.random() * 9 - (first+second);

If you need 3 numbers guaranteed, the sum of the first two should be 8 or less, unless you accept 0 as the 3rd possible number.

Answer 4

Mark Peters outlines two methods in his answer. The first method is sort of a brute force method: generate groups of three random numbers until you find one whose sum is 9. The second method is to tabulate all 28 possible permutations, and then randomly select one.

The first method might be slower, and the second will consume more memory. As long as you only need to execute this once, and the required sum of the three numbers is small (like 9) it doesn't make much difference which one you use.

However, if the required sum is large, you might need a different approach. Suppose, the required sum was 1 million. The brute force approach could be quite slow. Tabulating all of the permutations might consume too much memory (or disk space, depending on the implementation).

The total number of permutations is the sum of integers 1 through the maximum possible value. That sum can be quickly calculated by using this equation from Mathematics Stack Exchange:

Take the average of the first number and the last number, and multiply by the number of numbers.

So, in this case, the total number of permutations would be almost 1,000,000 x 500,000; about 500,000,000,000 permutations.

So instead of computing all the possible permutations, we can generate a random number n and calculate only the nth permutation given a required sum:

function getPermutation( n, requiredSum) {
  var remainder = n;
  var step = requiredSum - 2;

  var firstNumber = 1;

  while(remainder > step){
    firstNumber++;
    remainder -= step;
    step--;
  }

  var secondNumber = remainder;

  var thirdNumber = requiredSum - firstNumber - secondNumber;

  return [ firstNumber, secondNumber, thirdNumber ];

}

Example here:

$(document).ready( function(){
  $("button").click( function() {
    $("div").text( JSON.stringify( get3Numbers(1000000) ) );
  });
});

function get3Numbers(requiredSum){
  var maximum = requiredSum - 2;

  var permutations = maximum * ( (maximum + 1.0) / 2.0);

  var randomNumber = Math.floor((Math.random() * permutations) + 1);

  var threeNumbers = getPermutation(randomNumber, requiredSum);

  var sumOfNumbers = threeNumbers[0] + threeNumbers[1] + threeNumbers[2];

  return {
    threeNumbers: threeNumbers,
    total:  sumOfNumbers
  };
}

function getPermutation( n, requiredSum) {
  var remainder = n;
  var step = requiredSum - 2;

  var firstNumber = 1;

  while(remainder > step){
    firstNumber++;
    remainder -= step;
    step--;
  }

  var secondNumber = remainder;

  var thirdNumber = requiredSum - firstNumber - secondNumber;

  return [ firstNumber, secondNumber, thirdNumber ];

}

div {
    border: 1px solid silver;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div>&nbsp;</div>
<button>Get three numbers!</button>

Answer 5

Here's one way to do it.

function statGen() {
    var x = document.getElementById("number");
    var firstNumber = Math.floor((Math.random() * 5) + 1);
    x.innerHTML = firstNumber;
    var a = document.getElementById("agl");
    if (firstNumber == 5) {
        var secondNumber = Math.floor((Math.random() * 4) + 1);
    } else {
        var secondNumber = Math.floor((Math.random() * 5) + 1);
    }
    a.innerHTML = secondNumber;
    var l = document.getElementById("lck");
    l.innerHTML = 9-(firstNumber + secondNumber);
 }

 <button id="statbutton" onclick="statGen()">Get Numbers</button>       
        <p id="main">Main:</p>
        <p id="number"></p>
        <br/>
        <p id="Agl">Agl:</p>
        <p id="agl"></p>
        <br/>
        <p id="Lck">Lck:</p>
        <p id="lck"></p>