java - 名称结果与概率不匹配

Question

我正在尝试编写这个程序，随机选择 3 个名称当选择一个名字时，每次弹出该名字时，该名字再次出现的概率就会降低 10%。

问题是，当我将 for 循环中的所有名称更改为一个名称时，我得到了我将所有内容更改为的名称的 90% 以及其他 2 个名称之一的 10%。

String [] arr = new String[30];

int i; int b; int g;

for (i = 0; i < 11; i++)
{
    arr[i] = "moshe";
}

for (b = 9; b < 20; b++)
{
    arr[b] = "Nir";
}

for (g = 22; g < 29; g++)
{
    arr[g] = "Yoad";
}

double letsdomath = Math.random()*arr.length; // Exp: return the no. 10 / 30 letsdomath = 10

if (letsdomath < 11) // if i get moshe
{
    for (i = 0; i <= letsdomath - 1; i++)
    {
        arr[i] = "Nir"; // Originally it would be Moshe here
        arr[i + 1] = "Nir";
    }
}

if (letsdomath > 11 && letsdomath < 21) // if i get nir
{
    for (b = 0; b <= letsdomath -1; b++)
    {
        arr[b] = "Nir";
        arr[b + 1] = "Nir"; // Originally it would be Yoad here
    }
}

if (letsdomath > 21 && letsdomath < 30) // if i get yoad
{
    for (g = 0; g <= letsdomath -1; g++)
    {
        arr[g] = "Nir"; // Originally it would be Yoad here
        arr[29] = "Nir"; // Originally it would be Moshe here
    }
}
System.out.println(arr[(int) letsdomath]);

预期结果:
让 Nir ​​这个名字每次都弹出

实际结果:
Nir 这个名字大约有 90% 的时间出现，而 Yoad 这个名字大约有 10% 的时间出现。

Answer 1

让我们重新审视您对想要做的事情的描述:为什么不将您所说的每件事都实现为“我们可以做的事情”，然后运行“我们喜欢的尽可能多的迭代”，只需按顺序运行这些迭代？

例如，让我们从基础知识开始:

import java.util.*; 
import java.lang.Math;

public class Test  {
  String[] names;
  int len;
  double[] probabilities, thresholds;

  public static void main(String[] args) {
    new Test();
  }

  public Test() {
    init();
    int steps = 5;
    for (int i=0; i<steps; i++) {
      // do the thing!
    }
  }

  public void init() {
    // set up a probability distribution
    names = new String[]{"name1", "name2", "name3"};
    len = names.length;
    thresholds = new double[len];
    probabilities = new double[len];
    for (int i=0; i<len; i++) {
      probabilities[i] = 1./len;
    }
  }
}

然后，我们确保有一种方法可以查找概率阈值:如果概率为 [0.4, 0.3, 0.3]，那么我们需要阈值 [0, 0.4, 0.7]，以便我们可以轻松地处理以下事实:随机数 >= 0 但 < 0.4 应解析为索引 0，数字 >= 0.4 但 < 0.7 应解析为索引 1，等等:

// turns [0.3, 0.4, 0.3] into [0, 0.3, 0.7]
public void setThresholds() {
  double tally = 0;
  for (int i=0; i<len; i++){
      thresholds[i] = tally;
      tally += probabilities[i];
  }
}

然后，让我们根据我们选择的名称索引定义重新平衡概率的函数:

// turns [0.4, 0.3, 0.3] with pos=0 into [0.36, 0.32, 0.32]
public void updateProbabilities(int namePos) {
  double sprinkle = (probabilities[namePos] * 0.1) / (len - 1.);
  probabilities[namePos] *= 0.9;
  for (int i=0; i<len; i++) {
    if (i == namePos) continue;
    probabilities[i] += sprinkle;
  }
}

正确，完成所有设置后，我们现在可以更新 public Test() 来运行一百万次更新传递，依靠这些函数按我们的预期工作(您应该这样做，当然，请验证):

  public Test() {
    init();

    int steps = 5;

    for (int i=0; i<steps; i++) {
      setThresholds();
      double randomValue = Math.random();

      // find the associated name by finding the index of
      // the threshold that is higher than our random value.
      int namePos = findIndex(randomValue);
      if (namePos == -1) {
        namePos = names.length;
      }
      namePos--;

      updateProbabilities(namePos);

      // String name = names[namePos];
      // System.out.println("step " + i + ": picked " + name + " (index " + namePos + ") based on " + randomValue);
      // System.out.println("new probabilities: " + Arrays.toString(probabilities));
    }

    System.out.println("Final probabilities: " + Arrays.toString(probabilities));
  }

  public int findIndex(double randomValue) {
    for (int i=0; i<len; i++) {
      if (thresholds[i] > randomValue) return i;
    }
    return -1;
  }

将这些中间控制台日志注释掉，因为您不想看到一千个中介......当然，除非您这样做。