Random numbers probability

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Hi
I have a enum which consists of four elements (Up, Down, Left, Right).
I want to select randomly an element.
Th current problem is that most of the applications run gives me Up/Down.
How do I give a 25% probability for each element?

My code is simple:

C#

Random rnd = new Random(DateTime.Now.Millisecond);
int randomGenerated = 0;
for (int index = 0; index < 20; index++)
{
    randomGenerated = rnd.Next(1, 301) / 100;
    Move((Directions)randomGenerated);
}

As you can see I tried to generate between 1 and 301 to give better chances for each element to be selected but it didn't help.

Thanks
Shahar

Posted 9-Jun-12 22:54pm

Shahar Eldad

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2 solutions

Solution 2

I didn't understand your problem...

C#

enum Direction
{
  Up = 0, Down, Left, Right
}

static Direction GetRandomDirection(Random random)
{
    return (Direction)random.Next((int)Direction.Up, (int)Direction.Right + 1);
}

This works fine for me.

As Random is normally distributed, P(Up) = P(Down) = P(Left) = P(Right) = 1/4 = 25%.

Posted 9-Jun-12 23:03pm

Henning Dieterichs

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OriginalGriff · Accepted Answer · 2012-06-09T23:01:00

Don't. Just do this:

C#

Random rnd = new Random();
int randomGenerated = 0;
for (int index = 0; index < 20; index++)
{
    randomGenerated = rnd.Next(4);
    Move((Directions)randomGenerated);
}

Random numbers are just that. By restricting your number space to 0-300 and then dividing by a hundred , you are reducing the chances of getting a random distribution:

0 : 100 chances in 301
1 : 100 chances in 301
2 : 100 chances in 301
3 :   1 chance  in 301

[edit]Typo: "ten" for "a hundred" - OriginalGriff[/edit]