掷N个骰子时获得所有可能值的可能性

给定一个表示骰子数量的整数N ，任务是找到将N个骰子放在一起可获得的每个可能值的概率。

例子：

Input: N = 1 
Output: 
1: 0.17 
2: 0.17 
3: 0.17 
4: 0.17 
5: 0.17 
6: 0.17 
Explanation: On throwing a dice, the probability of all values from [1, 6] to appear at the top is 1/6 = 0.17

Input: N = 2 
Output: 
2: 0.028 
3: 0.056 
4: 0.083 
5: 0.11 
6: 0.14 
7: 0.17 
8: 0.14 
9: 0.11 
10: 0.083 
11: 0.056 
12: 0.028 
Explanation: The possible values of the sum of the two numbers that appear at the top on throwing two dices together ranges between [2, 12]. 

方法：想法是使用动态编程和DP表存储每个可能值的概率。

• 存储掷1个骰子时可能出现的所有6个数字的概率。
• 现在，对于N = 2，[2，12]之间所有可能的和的概率等于两个数字相加后得出的概率之和。例如，

Probability of 4 on throwing 2 dices = (Probability of 1 ) * ( Probability of 3) + (Probability of 2) * ( Probability of 2) + (Probability of 3 ) * ( Probability of 1) 
 

• 因此，对于N个骰子，

Probability of Sum S = (Probability of 1) * (Probability of S – 1 using N -1 dices) + (Probability of 2) * (Probability of S – 2 using N-1 dices) + ….. + (Probability of 6) * (Probability of S – 6 using N -1 dices) 
 

• 因此，为了解决该问题，我们需要使用以下关系，使用自顶向下方法将dp [] []表从2填充到N：

dp[i][x] = dp[1][y] + dp[i-1][z] where x = y + z and i denotes the number of dices 
 

• 显示为N存储的所有概率作为答案。

下面是上述方法的实现：

C++

// C++ Program to calculate
// the probabilty of
// all the possible values
// that can be obtained
// throwing N dices
 
#include 
using namespace std;
 
void dicesSum(int n)
{
    // Store the probablities
    vector > dp(n + 1);
    // Precompute the probabilities
    // for values possible using 1 dice
    dp[1] = { { 1, 1 / 6.0 },
              { 2, 1 / 6.0 },
              { 3, 1 / 6.0 },
              { 4, 1 / 6.0 },
              { 5, 1 / 6.0 },
              { 6, 1 / 6.0 } };
 
    // Compute the probabilies
    // for all values from 2 to N
    for (int i = 2; i <= n; i++) {
        for (auto a1 : dp[i - 1]) {
            for (auto a2 : dp[1]) {
                dp[i][a1.first + a2.first]
                    += a1.second * a2.second;
            }
        }
    }
    // Print the result
    for (auto a : dp[n]) {
        cout << a.first << " "
             << setprecision(2)
             << a.second
             << endl;
    }
}
 
// Driver code
int main()
{
    int n = 2;
    dicesSum(n);
 
    return 0;
}

Java

// Java program to calculate
// the probabilty of all the
// possible values that can
// be obtained throwing N dices
import java.io.*;
import java.util.*;
 
class GFG{
 
static void dicesSum(int n)
{
     
    // Store the probablities
    double[][] dp = new double[n + 1][6 * n + 1];
 
    // Precompute the probabilities
    // for values possible using 1 dice
    for(int i = 1; i <= 6; i++)
        dp[1][i] = 1 / 6.0;
 
    // Compute the probabilies
    // for all values from 2 to N
    for(int i = 2; i <= n; i++)
        for(int j = i - 1; j <= 6 * (i - 1); j++)
            for(int k = 1; k <= 6; k++)
            {
                dp[i][j + k] += (dp[i - 1][j] *
                                 dp[1][k]);
            }
 
    // Print the result
    for(int i = n; i <= 6 * n; i++)
    {
        System.out.println(i + " " +
                           Math.round(dp[n][i] * 1000.0) /
                                                 1000.0);
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 2;
     
    dicesSum(n);
}
}
 
// This code is contributed by jithin

Python3

# Python3 program to calculate
# the probabilty of all the
# possible values that can
# be obtained throwing N dices
def diceSum(n):
     
    # Inititalize a 2d array upto
    # (n*total sum possible) sum
    # with value 0
    dp = [[ 0 for j in range(n * 6)]
              for i in range(n + 1)]
                   
    # Store the probability in a
    # single throw for 1,2,3,4,5,6
    for i in range(6):
        dp[1][i] = 1 / 6
         
    # Compute the probabilies
    # for all values from 2 to N
    for i in range(2, n + 1):
        for j in range(len(dp[i - 1])):
            for k in range(6):
                     
                if (dp[i - 1][j] != 0 and
                    dp[i - 1][k] != 0):
                    dp[i][j + k] += (dp[i - 1][j] *
                                     dp[1][k])
     
    # Print the result
    for i in range(len(dp[n]) - n + 1):
        print("%d %0.3f" % (i + n, dp[n][i]))
 
# Driver code
n = 2
 
# Call the function
diceSum(n)
 
# This code is contributed by dipesh99kumar

C#

// C# program to calculate
// the probabilty of all the
// possible values that can
// be obtained throwing N dices
using System;
class GFG {
     
    static void dicesSum(int n)
    {
          
        // Store the probablities
        double[,] dp = new double[n + 1,6 * n + 1];
      
        // Precompute the probabilities
        // for values possible using 1 dice
        for(int i = 1; i <= 6; i++)
            dp[1,i] = 1 / 6.0;
      
        // Compute the probabilies
        // for all values from 2 to N
        for(int i = 2; i <= n; i++)
            for(int j = i - 1; j <= 6 * (i - 1); j++)
                for(int k = 1; k <= 6; k++)
                {
                    dp[i,j + k] += (dp[i - 1,j] *
                                     dp[1,k]);
                }
      
        // Print the result
        for(int i = n; i <= 6 * n; i++)
        {
            Console.WriteLine(i + " " +
                               Math.Round(dp[n,i] * 1000.0) /
                                                     1000.0);
        }
    }
 
  static void Main() {
    int n = 2;
  
    dicesSum(n);
  }
}
 
// This code is contributed by divyesh072019

输出：

2 0.028
3 0.056
4 0.083
5 0.11
6 0.14
7 0.17
8 0.14
9 0.11
10 0.083
11 0.056
12 0.028

时间复杂度： O(N ² )
辅助空间： O(N ² )