Question

我正在尝试制作一个程序，使用随机数字计算骰子卷，并返回不同数量的卷和试验发生的每个总和的概率。然而，该代码仅导致零显示的概率。这似乎是在嵌套for循环中建立匹配变量的问题。想知道我做错了什么。是否根据计数器建立匹配变量？

import java.util.Random;
import java.util.Scanner;

public class DiceProbability
{
public static void main(String[] args)
{
    Scanner in = new Scanner(System.in);
    Random randNumList = new Random();
    System.out.println("How many sides do the dice have: ");
    int diceSides = in.nextInt();
    System.out.println("How many times will the dice be rolled: ");
    int diceRolls = in.nextInt();
    int highestSum = diceSides * 2;
    int diceRoll1 = 0;
    int diceRoll2 = 0;
    int match = 0;
    int totalOfDiceRolls = 0;
    int counter = 0;
    int counter2 = 0;
    System.out.println("Rolls total   " + "Probability");
    for(counter=2;counter<=highestSum;counter ++)
  {
     for(counter2=1;counter2<=diceRolls;counter2 ++)
     {
     diceRoll1 = (randNumList.nextInt(11)+1);
     diceRoll2 = (randNumList.nextInt(11)+1);
     int totalOfRolls = diceRoll1 + diceRoll2;
     if(totalOfDiceRolls != counter)
        {
            match = match + 0;
        }
        else
        {
            match ++;
        }   
     }
     System.out.println(match);
     double probabilityOfSum = (match * 100 / diceRolls);
     System.out.println(counter + ":          " + probabilityOfSum);
     counter2 = 1;
    }
   } 
  }

Answer 1

如果我理解正确的话，你试图计算两个骰子总和的相对频率。如果是这种情况，请编辑您的问题，因为您的询问并不明确，特别是您总是有两个骰子。

如果你有双面骰子，你的数学概率如下：

P(of having 2) = 1/4
P(of having 3) = 2/4
P(of having 4) = 1/4

所有其他人都等于0。

以下是实现此目的的代码。您需要将频率保存在数组中并正确输出除以实验数。

注意：如果你运行一个小于1000的数字，这在数学上是不准确的，所以在这种情况下用户输入实验的数量并不合理......（在我看来）

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    System.out.println("How many sides do one dice have: ");
    int diceSides = in.nextInt();
    int[] results = new int[diceSides * 2 + 1];

        Random rnd = new Random();

    for (int i = 0; i < 10000; i++) {
        int resultdice1 =  rnd.nextInt(diceSides) + 1;
        int resultdice2 =  rnd.nextInt(diceSides) + 1;
        int sum = resultdice1 + resultdice2;

        results[sum] = results[sum] + 1;
    }

    for (int i = 0; i < results.length; i++) {

        System.out.println("Probability to have sum of " + i + " is : " + (double) results[i] / 10000);

    }

}

输出：

有骰子概率计划的问题

1 个答案: