Get a random number focused on center

Question

Is it possible to get a random number between 1-100 and keep the results mainly within the 40-60 range? I mean, it will go out of that range rarely, but I want it to be main

Answer 1

Taking arrays of numbers, etc. isn't efficient. You should take a mapping which takes a random number between 0 to 100 and maps to the distribution you need. So in your case, you could take f(x)=-(1/25)x2+4x to get a distribution with the most values in the middle of your range.

Answer 2

I needed to solve this problem a few years ago and my solution was easier than any of the other answers.

I generated 3 randoms between the bounds and averaged them. This pulls the result towards the centre but leaves it completely possible to reach the extremities.

Answer 3

This answer is really good. But I would like to post implementation instructions (I'm not into JavaScript, so I hope you will understand) for different situation.

---

Assume you have ranges and weights for every range:

ranges - [1, 20], [21, 40], [41, 60], [61, 100]
weights - {1, 2, 100, 5}

---

Initial Static Information, could be cached:

1. Sum of all weights (108 in sample)
2. Range selection boundaries. It basically this formula: Boundary[n] = Boundary[n - 1] + weigh[n - 1] and Boundary[0] = 0. Sample has Boundary = {0, 1, 3, 103, 108}

Number generation:

1. Generate random number N from range [0, Sum of all weights).
2. for (i = 0; i < size(Boundary) && N > Boundary[i + 1]; ++i)
3. Take ith range and generate random number in that range.

---

Additional note for performance optimizations. Ranges don't have to be ordered neither ascending nor descending order, so for faster range look-up range that has highest weight should go first and one with lowest weight should go last.

Answer 4

I might do something like setup a "chance" for the number to be allowed to go "out of bounds". In this example, a 20% chance the number will be 1-100, otherwise, 40-60:

$(function () {
    $('button').click(function () {
        var outOfBoundsChance = .2;
        var num = 0;
        if (Math.random() <= outOfBoundsChance) {
            num = getRandomInt(1, 100);
        } else {
            num = getRandomInt(40, 60);
        }
        $('#out').text(num);
    });
    
    function getRandomInt(min, max) {
        return Math.floor(Math.random() * (max - min + 1)) + min;
    }
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

<button>Generate</button>
<div id="out"></div>

fiddle: http://jsfiddle.net/kbv39s9w/

Answer 5

Ok, so I decided to add another answer because I felt like my last answer, as well as most answers here, use some sort of half-statistical way of obtaining a bell-curve type result return. The code I provide below works the same way as when you roll a dice. Therefore, it is hardest to get 1 or 99, but easiest to get 50.

var loops = 10; //Number of numbers generated
var min = 1,
    max = 50;
var div = $("#results").html(random());

function random() {
    var values = "";
    for (var i = 0; i < loops; i++) {
        var one = generate();
        var two = generate();
        var ans = one + two - 1;
        var num = values += ans + "<br/>";
    }
    return values;
}

function generate() {
    return Math.floor((Math.random() * (max - min + 1)) + min);
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>

Answer 6

If you can use the gaussian function, use it. This function returns normal number with average 0 and sigma 1.

95% of this number are within average +/- 2*sigma. Your average = 50, and sigma = 5 so

randomNumber = 50 + 5*gaussian()