随机数概率 [英] Random numbers probability

问题描述

嗨
我有一个枚举，它由四个元素(上，下，左，右)组成.
我想随机选择一个元素.
当前的问题是，大多数运行的应用程序都使我无法正常运行.
如何给每个元素25％的概率?

我的代码很简单:

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:

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);
}

如您所见，我尝试生成1到301之间的值，以便为每个元素的选择提供更好的机会，但这没有帮助.

谢谢
Shahar

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

推荐答案

Don''t.只需执行以下操作即可:

Don''t. Just do this:

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

随机数就是这样.通过将数字空间限制为0-300，然后除以100，可以减少获得随机分布的机会:

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]错字:十"代表一百"-OriginalGriff [/edit]

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

我不明白您的问题...

I didn''t understand your problem...

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

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

这对我来说很好.

由于Random是正态分布，因此P(Up)= P(Down)= P(Left)= P(Right)= 1/4 = 25％.

This works fine for me.

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

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