骰子点数分布概率问题

今天学习算法的时候碰见一个骰子点数分布概率的问题：

Dice simulation.

The following code computes the exact probability distribu-tion for the sum of two dice:

    int SIDES = 6;
    double[] dist = new double[2*SIDES+1];
    for (int i = 1; i <= SIDES; i++)

       for (int j = 1; j <= SIDES; j++)
          dist[i+j] += 1.0;

    for (int k = 2; k <= 2*SIDES; k++)
       dist[k] /= 36.0;

The value dist[i] is the probability that the dice sum to k. Run experiments to vali-date this calculation simulating N dice throws, keeping track of the frequencies of oc-currence of each value when you compute the sum of two random integers between 1and 6. How large does N have to be before your empirical results match the exact resultsto three decimal places? 

其实这道题要做出来的关键就是求出投了n个骰子之后所有点数和出现的分布：

先贴代码： 

/**

 *  run experiments to validate this calculation simulating N dice throws,

 *  keeping track of the frequencies of occurrence of each value when you compute the sum of two random integers between 1 and 6. 

 *  How large does N have to be before your empirical results match the exact results to three decimal places?

 * @author xiehang

 */

public class Ex1_1_35 {

public static void main(String[] args) {

int side = 6;

final int n = 3; //number of dice

double total =  Math.pow(6, n);

int[] a1 = new int [6*n+1]; // a1 is for calculate

int[] a2 = new int [6*n+1]; // a2 is for store the result 

int result = 0;

//for the first dice

for(int i= 1; i <=side; i++ ){

a1[i]++;

}

a2 = Util.arrayCopy(a1);

System.out.println("When we have 1 dice");

Util.printArray(a1);

//for the dice after 1

for(int i = 2; i <= n; i++ ){

// each dice will scan from number 1 to the biggest number

for(int k = 0; k < 6 * i +1; k++){

/**

* 原理：

* 先定义第一个骰子的所有可能： 1-6各出现一次

* 再接下去的骰子进入循环：

* 已知下一次点数k出现的和等于上一次的点数为k-1,k-2,k-3,k-4.k-5.k-6出现的和

* e.g. 点数7出现的是点数1，2，3，4，5，6出现的和

* 以这个原理循环下去

*/

for(int j = 1; j <= 6;j++){

if(k-j >=0){ // k-j 相当于上面出现的k-1,k-2

result += a2[k-j];

}

/**

* 解释为何要用a1,a2两个array操作：

* 如果只用一个a1操作的话，

* 前面的数变了之后会影响之后的数

* 所以要用两个array操作

* 

* 为何不能用a1[k] += a2[k-j]：

* a1[k] 之前是有一个值的

* 所以在加上去也会让结果不正确

* 所以引入一个新的数储存了以后再变0不断转存就可以了

*/

}

a1[k] = result;

result = 0;

}

a2 = Util.arrayCopy(a1);

System.out.println("When we have " + i + " dice: ");

Util.printArray(a2);

}

System.out.println("End");

Util.printArray(a2);

System.out.println("Here is the rate: ");

for(int i = 1; i < 6*n+1; i++ ){

System.out.println("The rate for number "+ i +" appears is: " + String.format("%.2f", a2[i]/total*100) + "%");

}

}

}