Given a maximum integer value M, I want to generate N integer values that are distributed with a lower frequency when approaching M (preferably M should still have a non-zero probability). I don't really care much about the probability function, let's assume (half of) a normal distribution.

How would I do that if I don't want to keep a history?

for i := 1 to N do GetNextTestValue(M)

What could GetNextTestValue look like?
FWIW I'm doing this in Delphi

  • You may find adug.org.au/MathsCorner/MathsCornerNDist.htm useful.
    – user40980
    Jun 4, 2013 at 12:31
  • 2
    If I understand what you are looking for correctly, you can achieve this by generating two random numbers of the desired range and taking the lowest of the two. Jun 4, 2013 at 17:17
  • Or take random(M/2)+random(M/2). The underlying reason is the same: to achieve the maximum result, the two numbers generated must both be maximal, and the chance of that is quadratically smaller.
    – MSalters
    Jun 6, 2013 at 14:24
  • @MSalters note that adding two random numbers produces a triangle-shaped distribution, with the highest point at M / 2. That does meet the criterion of tapering off near M, but it also produces fewer less than M / 2 going to 0. But that may be OK.
    – sea-rob
    May 20, 2014 at 17:43

3 Answers 3


Building off of GrandmasterB's comment about pick two and select the lowest, I wrote a quick perl script to plot the random numbers and see the distributions - Pick the lowest of two provides one distribution, but others provide other distributions...

First, the code to see how it works and reproduce the results if you so desire.


use strict;
my $count = shift @ARGV;
my $nth = shift @ARGV;

my @nums = ();

for(my $i = 0; $i < 10_000; $i++) {
    my @rnd;
    for(my $j = 0; $j < $count; $j++) {
        push @rnd, int(rand(100));
    @rnd = sort { $a <=> $b } @rnd;

for(my $i = 0; $i < 100; $i++) {
    print "$i\t" . (defined $nums[$i] ? $nums[$i] : 0) . "\n";

The pick min(rand(100),rand(100)) produces a straight line distribution sloping down.

If you min(rand(100),rand(100), rand(100)), you can see a bit of a curve to it.

Exploring this just a bit further... Picking the middle number of 7 produces something that looks more like a nice bell curve.

While the 2nd number of 7 produces a skewed one.

So, from this and the application of some stats to decide what shape you want it (or throwing out samples that don't meet the desired range to get the desired shape and skew)

  • Thats neat. I've used the technique a couple times (including using the lowest of N to adjust the distribution), but never actually plotted it out to see what it looks like. Thanks for posting this! Jun 6, 2013 at 4:38
  • Very nice. I've marked this as the correct solution instead of my own because it's much simpler to implement.
    – Jan Doggen
    Jun 6, 2013 at 6:19
  • BTW can any of you point to the theory behind this - why does it work this way?
    – Jan Doggen
    Jun 6, 2013 at 6:25
  • 1
    @Jan: Simplest to see for the min(rand1, rand2) case: the chance of rand2 being smaller than rand1 is rand1/Max, i.e. linear in rand1. For the result to be 0, either rand1 or rand2 must be 0. For the result to be 1, either rand1 or rand2 must be 1, and the other cannot be 0 - so p(1)<p(0).
    – MSalters
    Jun 6, 2013 at 14:35
  • 1
    you could ask on math.stackexchange.com, but if I remember correctly, the min of N randoms follows a power distribution. There must be a distribution for the "nth of N" distributions, but there are so many distributions that are variations of a bell curve.
    – sea-rob
    May 20, 2014 at 17:29

I found a solution.

The Controul blog by Hristo Dachev describes several methods, of which I have implemented the Marsaglia polar method.
Delphi code for a quick-and-dirty implementation with a standard normal distribution (multiplied to count integer values) is below.

The blog references the Wikipedia article which has pseudo code for any mean and standard deviation.

Sample distribution generated by the program (10000 points): enter image description here

Delphi XE2 .pas file:

unit uMarsaglia;
// Based on     http://blog.controul.com/2009/04/standard-normal-distribution-in-as3/
// Refers to    http://en.wikipedia.org/wiki/Marsaglia_polar_method


  Winapi.Windows, Winapi.Messages, System.SysUtils, System.Variants, System.Classes, Vcl.Graphics, System.Math,
  Vcl.Controls, Vcl.Forms, Vcl.Dialogs, VCLTee.TeEngine, Vcl.ExtCtrls,
  VCLTee.TeeProcs, VCLTee.Chart, Vcl.StdCtrls, VCLTee.Series;

  TFrmMarsaglia = class(TForm)
    Chart1: TChart;
    Series1: TBarSeries;
    procedure FormShow(Sender: TObject);
    FReady : Boolean;
    FCache : Real;
    function StandardNormal: Real;

  FrmMarsaglia: TFrmMarsaglia;


{$R *.dfm}

function Ln(R: Real): Real;
   Result := Log10(R) / Log10(2.7182818);

procedure TFrmMarsaglia.FormShow(Sender: TObject);
   cTestMax = 1000000;
   cScale = 1000;
   cMaxScale = 5*cScale; // We don't count generated numbers higher than that
   A : Array[1..cMaxScale] of Integer;
   i: integer;
   Nr: Integer;
   rScale: Real;
   for Nr := 1 to cMaxScale do A[Nr] := 0;
   rScale := cScale;
   for Nr := 1 to cTestMax do
      i := Trunc(rScale * StandardNormal);
      if i <= cMaxScale then Inc(A[i]);
   lMax := 0;
   for Nr := 1 to cMaxScale do if A[Nr] > lMax then lMax := A[Nr];
   for Nr := 1 to cMaxScale do Series1.Add(A[Nr]);

function TFrmMarsaglia.StandardNormal: Real;
//  Returns real numbers between N(0,1) that when repeated have a normal distribution
var x,y,w,l: Real;
   if FReady then
   begin                // Return a cached result from a previous call if available.
      FReady := false;
      Result := FCache;

   // Repeat extracting uniform values in the range (-1,1) until 0 < w = x*x + y*y < 1
      x := Random;
      y := Random;
      w := x * x + y * y;
   until (w > 0) and (w < 1);

   l := Ln(w);
   w := sqrt (-2 * l / w);

   FReady := true;
   FCache := x * w;         //  Cache one of the outputs
   Result := y * w;         //  and return the other.


Delphi .frm file using a TChart (TeeChart) showing the results:

object FrmMarsaglia: TFrmMarsaglia
  Left = 0
  Top = 0
  Caption = 'Marsaglia polar method'
  ClientHeight = 626
  ClientWidth = 964
  Color = clBtnFace
  Font.Charset = DEFAULT_CHARSET
  Font.Color = clWindowText
  Font.Height = -11
  Font.Name = 'Tahoma'
  Font.Style = []
  OldCreateOrder = False
  OnShow = FormShow
  PixelsPerInch = 96
  TextHeight = 13
  object Chart1: TChart
    Left = 0
    Top = 0
    Width = 964
    Height = 626
    BackWall.Visible = False
    BottomWall.Visible = False
    Foot.Visible = False
    LeftWall.Visible = False
    Legend.Visible = False
    Title.Text.Strings = (
    Title.Visible = False
    View3D = False
    Align = alClient
    BevelOuter = bvNone
    TabOrder = 0
    ExplicitLeft = 272
    ExplicitWidth = 692
    ColorPaletteIndex = 13
    object Series1: TBarSeries
      Marks.Arrow.Visible = True
      Marks.Callout.Brush.Color = clBlack
      Marks.Callout.Arrow.Visible = True
      Marks.Visible = False
      XValues.Name = 'X'
      XValues.Order = loAscending
      YValues.Name = 'Bar'
      YValues.Order = loNone

Note that 'not keeping a history' is not satisfied: there's one boolean and one real number to remember state. That's good enough for me; I was afraid of having to maintain arrays of already generated numbers.


If rolling two and taking the least is too much ;) you could scale a single uniform random so that the distribution favors small results (pseudocode):

def scaled_random(M):
    r = random()
    return M * r * r
  • random() produces a uniform random number from 0..1

  • When r is small, r * r will be even smaller, so there are lots of ways to get numbers close to zero.

  • As r gets closer to 1, then r * r gets closer to 1, so the return value is closer to M. But very few numbers will get that high.

Note that this is going to be a lot faster than the least of 2 random numbers, because multiplication (r * r) is a lot faster than generating a second random. Random number generation is slow compared to basic operations -- even "fast" random number generators are just "fast" relatively.


Also, some math libraries support a "power" distribution for random numbers -- I believe that would also do the trick.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.