Random encounter not so random

Hello i am having some problems generating random numbers with C# Now i have this function.

public Color getRandomColor()
{
    Color1 = new Random().Next(new Random().Next(0, 100), new Random().Next(200, 255));
    Color2 = new Random().Next(new Random().Next(0, 100), new Random().Next(200, 255));
    Color3 = new Random().Next(new Random().Next(0, 100), new Random().Next(200, 255));
    Color color = Color.FromArgb(Color1, Color2, Color3);
    Console.WriteLine("R: " + Color1 + " G: " + Color2 + " B: " + Color3 + " = " + color.Name);
    return color;
}

Now you might notice that there are ALOT of new Random() there, that is because i want to weed out the probability that it could be a same instance error.

I now run this function 8 times, a couple of times. Now here are the out puts.

R: 65 G: 65 B: 65 = ff414141
R: 242 G: 242 B: 242 = fff2f2f2
R: 205 G: 205 B: 205 = ffcdcdcd
R: 40 G: 40 B: 40 = ff282828
R: 249 G: 249 B: 249 = fff9f9f9
R: 249 G: 249 B: 249 = fff9f9f9
R: 94 G: 94 B: 94 = ff5e5e5e
R: 186 G: 186 B: 186 = ffbababa

R: 142 G: 142 B: 142 = ff8e8e8e
R: 190 G: 190 B: 190 = ffbebebe
R: 19 G: 19 B: 19 = ff131313
R: 119 G: 119 B: 119 = ff777777
R: 119 G: 119 B: 119 = ff777777
R: 75 G: 75 B: 75 = ff4b4b4b
R: 169 G: 169 B: 169 = ffa9a9a9
R: 127 G: 127 B: 127 = ff7f7f7f

R: 73 G: 73 B: 73 = ff494949
R: 27 G: 27 B: 27 = ff1b1b1b
R: 125 G: 125 B: 125 = ff7d7d7d
R: 212 G: 212 B: 212 = ffd4d4d4
R: 174 G: 174 B: 174 = ffaeaeae
R: 0 G: 0 B: 0 = ff000000
R: 0 G: 0 B: 0 = ff000000
R: 220 G: 220 B: 220 = ffdcdcdc

As you can see this is not so random again, but why dose this happens? and how can i counter it?

You're creating a new Random for each single value you need.

Try creating a unique Random object and calling the .Next() function multiple times.

public Color getRandomColor()
{
    Random rand = new Random();

    Color1 = rand.Next(rand.Next(0, 100), rand.Next(200, 255));
    Color2 = rand.Next(rand.Next(0, 100), rand.Next(200, 255));
    Color3 = rand.Next(rand.Next(0, 100), rand.Next(200, 255));
    Color color = Color.FromArgb(Color1, Color2, Color3);
    Console.WriteLine("R: " + Color1 + " G: " + Color2 + " B: " + Color3 + " = " + color.Name);
    return color;
}

Taken from MSDN documentation on Random object :

By default, the parameterless constructor of the Random class uses the system clock to generate its seed value, while its parameterized constructor can take an Int32 value based on the number of ticks in the current time. However, because the clock has finite resolution, using the parameterless constructor to create different Random objects in close succession creates random number generators that produce identical sequences of random numbers

Each new Random() is seeded with the current time.

If you create several Random instances in quick succession, and your machine runs it fast enough that they are close enough together to get the same seed, they will return the same sequence of values!

Use a single Random, calling .Next() to get each value.

Now you might notice that there are ALOT of new Random() there, that is because i want to weed out the probability that it could be a same instance error.

This is a fallacy; a stream of numbers from a single Random is evenly distributed - effort has gone into making it random.

You are creating many Random objects sequentially, which causes them to be seeded with the same/almost the same timestamp. Thus generating random numbers but equal among them.

Try creating a single instance of Random and using that for all your Random number needs.

Random works better when using one instance of it.

Try this:

public Color getRandomColor()
{
    var random = new Random();
    Color1 = random.Next(random.Next(0, 100), random.Next(200, 255));
    Color2 = random.Next(random.Next(0, 100), random.Next(200, 255));
    Color3 = random.Next(random.Next(0, 100), random.Next(200, 255));
    Color color = Color.FromArgb(Color1, Color2, Color3);
    Console.WriteLine("R: " + Color1 + " G: " + Color2 + " B: " + Color3 + " = " + color.Name);
    return color;
}