在C ++中N次抛出2个骰子得到求和的概率

2024-04-27 12:24:04 364

示例

Input-: sum = 12, N = 1 Output-: Probability = 1/36 Explanation-: if a pair of dice is thrown once then the combinations will be (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6). Out of these combinations we can get sum 12 at pair (6, 6) only therefore probability would be 1/36 Input-: sum = 4 and N = 6 Output-: probability is : 1/2985984

以下程序中使用的方法如下-

输入sum和N的值，N表示掷骰子的次数

为了计算N次投掷2个骰子时发生总和的概率，请应用以下公式：（ 有利/总计）^N

现在计算投掷2个骰子1次的总和发生的概率，让我们得出1

用于计算发生投掷2个骰子的总和N次的概率为-

概率2=（概率1）^N.即，概率1提高到幂N

算法

Start
Step 1-> declare function to calculate the probability
   int probability(int sum, int times)
   Declare and set float res = 0.0 and total = 36.0
   Declare and set long int probab = 0
   Loop For i = 1 and i <= 6 and i++
      Loop For j = 1 and j <= 6 and j++
         IF ((i + j) = sum)
            Set res++
         End
      End
   End
   Declare and set int gcd1 = __gcd((int)res, (int)total)
   Declare and set res = res / (float)gcd1
   Set total = total / (float)gcd1
   Set probab = pow(total, times)
   return probab
Step 2-> In main()   Declare and set int sum = 4 and times = 6
   Call probability(sum, times)  
Stop

示例

#include <bits/stdc++.h>
using namespace std;
//计算N次投掷2次骰子时求和的概率的函数
int probability(int sum, int times) {
   float res = 0.0, total = 36.0;
   long int probab = 0;
   for (int i = 1; i <= 6; i++) {
      for (int j = 1; j <= 6; j++) {
         if ((i + j) == sum)
         res++;
      }
   }
   int gcd1 = __gcd((int)res, (int)total);
   res = res / (float)gcd1;
   total = total / (float)gcd1;
   probab = pow(total, times);
   return probab;
}
int main() {
   int sum = 4, times = 6;
   cout<<"probability is : ";
   cout << "1" << "/" << probability(sum, times);
   return 0;
}

输出结果

probability is : 1/2985984

在C ++中N次抛出2个骰子得到求和的概率

示例

算法

示例

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